Quasi-coherent sheaves and complexes
Hennion, Benjamin
Definitions of quasi-coherent complex on a derived Artin stack, t-structure, direct and inverse images, base change formula.
 So I'm going to do 3-6 3-things in this talk. I'm going to, well, suppose, define the material, the drag category, or the drag stack, and then we're going to go to the object structure. Okay, because you know, you know what happened in the wrong drag, as a great jump lead. And at the end, I'm going to talk about the next thing formula, which we have on a common scheme, when there are compact and easy to draw. So, first, it's given like X for the scheme, just a scheme. We define the drag category of, well, the question for here and she is last, well, we define the question for here and she is last, and then we say the complex of the question for here and she is last, and that's what we write. Here, we won't do that. Okay, well, we'll do that, but I'm not exactly. The first thing is, well, we first define this on the find and then we do it, obviously. So, we have a factor on this PGA, non-positive degree. Okay. To PGA categories. That's okay. Why is this factor, why is it sent A to the DG categories of A to D1? A to D1. And by that I mean, and down, yeah, here I'm in non-positive degree, here I'm just monitoring. And that's A of the DG. That's A of A. Okay. Basically, one more. Yeah, okay, well, this is what I want. Well, A is for you to do it. Okay, so, in the talk before, in the talk before, well, we saw something like that, here, perfect modules. Okay, and yeah, he said, Fransma could not prove that there, that this was the one where perfect modules were stacked. Okay, so it satisfies the distribution and it's still about this one. Okay, so this one is stacked in the DG category, whatever that means. Okay, so. Now, I think I'm going to write this one. So, I'm going to use, I want to do that, okay, obviously if it has a decent combination, then I will be able to do this. This formally is like, is done using universal property of the category of DRI stack, okay. And what is that? Well, if I have a function from CGA, up into some category C, excuse me, you can use the category with all the limits. Small for. And if this function, let's say, yeah, satisfies the decent condition, so, well, I mean, if you take the dishing of being a stack in DG categories, for example, you pop it, then it extends, here, you have to do me enough, enter. It extends essentially uniquely in a function like that, which is, I don't know, let's write it like that, which preserves coordinates. So, here we are at the same degree, but of course, I have to put up and up, so, however, this can be formulaed in the existence of essentially unity, which is on equivalent to two spaces. So, well, yeah, okay, well, if I want to apply that to this thing, I get something like that. And that, I'm sure, is LQ code, okay, so, which, since the stack, it's throughout the category. So, basically, you just take the huge diagram of all the affines going to your stack at X, and take a, you know, the only way to do it is to call it. So, here, I mean, by F, S, what do you say, the limit of stack A, when you say A, of Rx. We don't have to do the whole thing, anyway, if it's a nice enough stack. Yeah, I mean, there's geometric, you can turn it into a simple shot, but yeah. Okay, so this is into the limit of TQR of the rate. Okay, so, the definition is, I mean, given the tools I have, pretty easy. Now, computing is such a thing, usually. You don't. Well, you try to, so I give two of the examples, but, well, when I do the pipe, it's easy. Okay, so. That's quite a moment. Yeah, I'm going to, this one was hard. Well, I have, okay, I can, if I have X, the drive stack, and G, a group argument on X, I can perform the quotient stack of X by G, and its drive category is, well, what does it look like? This one is the quotient of, okay, I'm going to grab it here. This one is the quotient of something like that, so I have some group of it like that. Okay, let's say, this is the one, the quotient of this thing. And so, here I have the projection, you want here I have the action group. So, if I'm going to take the drive, the drive category, I could achieve all these things. I get the diagram the other way around, and it is pretty obvious that these gears, the Q-core of X, which are G group arguments, okay? So, I mean, this is a definition in fact when the X and G are Y, so when they're nice and nice and easy, I think it is. Now, our number of causes, let's say G is in the time, I'm going to take the structure of it. So, I have the algebraic team, then I have the G and the group scheme. And this thing is, well, what you can see, okay, this is G-provided classic algorithm scheme for this point, so this is like the point is, okay, so this is a chain complex on which G acts, okay? And so, it's actually complexes of representation of G, so this is a stratification theorem, okay? So, this is just a representation of G and you take the drive category by that I mean the DG category of complex little G representation. Yeah. Because if you think of the left hand side, it's, you know, these are complexes with an actual G in the reservoir sense, so like you have a complex and then you put it back on G, then it has an auto-equivalence which is on which it represents the actual G and you put it back on, put it back in two different ways in G cross G, you have homo-tropies between these actions and so on, whereas the right hand side is really complex of representation in the, you know, in this case sense. So, I mean this is not, no, I just wanted to say it's not a good statement. Do you think there have been some other assumptions? Plats or, you know, maps? Plats or an, an evasive? Yeah. Where does the parkas come in? Not sure. I mean, the problem is that it is not a strat and when you put it back on G cross G you might have a T-Rap G cross G, you know, in the picture on the left hand side that you are not going to find on the right hand side. I mean, I don't know. I prefer to be safe. Let's say, right. I mean, it's a problem, sorry, yeah, it's a problem in a totally logical way, which is if you have an action one stack which is not T-Rap, which is positive impactant and mind-dialing, there's an F-Q-Q that stack is the rapid-view of the geese. That's a special case. So one stack, do you mind if I break? I'll change the name to that. I find that all is present in class. Thank you. Thank you to all the experts in the category, the derived category of conferences of quasi-trainers, in a way, to be... And the thing is that there are examples of schemes which are not that might be available for which this is wrong, only schemes. Okay, so... The same thing, okay. This is a specification thing. Yeah, I agree, this is what we wanted to leave. On this side, we have complexes, okay, which are x of the current sheet that's complex. And this one is each level is a percentage human. I see. The same thing as this one, I mean, this one was the complex one, which gx and this one is a change of geosification. Okay. So, of course... So, a huge difference for... When I compare the map from x and y in the derived class, I mean, because my function, the value of the function, I get a map of that. And in fact, to whose categories, because you can ask presentable, you can continue the categories instead of the categories. And I just want to use this polynomial, so I can get to write it, John. Okay, and I'm going to write some Q code here, because you can define the derived category on normal, but in coherent sheets. And those two do nothing inside, if I have lower star and lower star Q code upon all the same. My last statement on the case of the lead, that in some kind of a condition, then it's too... Actually the same. Okay, of course when we just draw the lines, it's basically... Okay, so now we'll talk about the t-structure of... Sorry, I have a question. So you just said sometimes it's the same as the product, I guess, for all x models. Does this mean that it's always, I don't know, fully placed for embedding from LQ code x into the category of all x models? Yeah, yeah, yeah. This is a full-time category of, let's say, L of x, which I would, for the line. Well, it's the way to say it, but it's actually... You wrote LQ code x is the limit of that, you know, of L of x of A of y of x. You replace limit by a lag-limit, meaning you have transition coefficients and not the equivalence anymore. So it's like the sections of the golden inventory of all of that. So there is a limit in the category, there is a lag-limit. And this Lx is the lag-limit thing. And inside you have the objects which all transition coefficients are equivalences. So are we reducing the resistance of these things, at first, for its preserve of all the maps? At this point? Yeah. Well, basically, one of the things is called image. Well, when you do this one... You're doing a call-image of a diagram of... Well, yeah, you surely get maps, but why do you have preserve of all of these? Because it's a limit of maps who preserve all the limits between maps, and they're going to be in the categories. Okay. It's just basically everything that you're doing. Because it works for a lifetime. Yeah, yeah. Okay, so... Now, yeah, key structure on a L, you could... Of s, when s is our... Okay? We'll see why in this assumption. And so, first, let's see what happens if x is defined. We'll do that, of course. So, first, x is defined. Now, a, d, d, d, m equals... And so, the key structure is, to that, how the category... a d, d, m equals a non-positive degree. And, actually, then you have the same... the same categories with a degree over something, whatever, m. And you have something like that, too. Like, in a degree greater or equal to 1, or non-negative degrees or whatever. So, that's your thing. And you do this such like a few actions, you must have something like that. Okay, so whenever you have an object which is a non-positive degree and an object which is a degree greater or equal to 1, then there is no non-triple map between them. Okay, so this has to be the autogruble. And that's why this one, actually, can be defined as the autogruble. The left or the right autogruble is this one. So, how do we say that? How do we define them? Well, obviously, we know what this should be. We know what this should be. And so, in here, fsi of m is 0, i is greater than 0, and the central goes for m, and the central goes for m. So, this defines the G structure on ATGM, the algorithm of the GTA code. So, what's important is that the heart of this structure is actually a heart of this structure. It is just a view of a normal system. Is this supposed to be a normal term on heart, or is this a philosophical statement? It's a technical term. Oh, okay, so, what do you define exactly with this? The heart is a section of this one, and this one. Okay? This one would be fine for any n here, and any n here, I mean, we don't need to be doing one. The theorem is that the heart is a billion. That's not completely obvious. This is the category. Are you going to define what the t-thorcher is? Well, the t-thorcher is, but that is the two things. But in general, we're not... No, no, well, on c, okay, c is some category, and you ask that, you ask... You ask to have c......the working zero. In fact, you have that, you have this thing, or two, you need to... And you ask 20x and c. 20x and c. There is a lot of x. And you ask this... Okay, so if you guys do your final category, that's just the final, just the two which final, carry it all off in these tables, and you can get it together, and you can get it in the final five seconds. So you demand that c is stable? Well, this one, half. Okay, again, we can do it for just finalized categories. As derived categories on derived... As you're right, John, you know, you're more of a category of derived categories in this final agent, so you can do that in finalized categories, and as that, you're just in the final, and if you period a lot in the stable-due category, so this is just five or five-inch segments. You know, it's just a way to formalize the fact that your standard category has a few subcategories which is a real agent, and that any object has a kind of prosnical decomposition, whose piece is on in the house. If you try to formalize this, you'll see that it's a little bit more complex. Okay, so... And the half of the g-squature is always intersection of the c in all these degrees in the degree. Okay. So, this is the final g-squature, and actually when you have such things, then you have a center which associated with its is from Haitian beer, and another center here. And the other from Haitian is from Haitian beer. Now, how do we do that? Well, if you have a map... You have a map A to B, obviously, B, G, A, and A. And you want to know when, I mean, when one does a survey center to preserve this g-squature. So, the answer is when you use that. So, I'll say that. And then... This is a map in here, and you have a training model M, a g-module M, and a list of maps. So, here it can apply space change factor, I think it's the right... And... Okay, because when I say we think together, this one is the fiber of this one, and we can also do that. We can also... When we're in B, here, this one is in B, whereas identity, and here we have the communication. This is a location of the g-squature on B, and we can do that on B, G, 1, B. And so... So, I think these are two sequences, actually are the same when my autism is flat Y. Well, you have to prove that one lose may have maps like that. And it's enough to prove that one lose is an equivalent. And this one is, I'm flat Y is that in the situation that I'm at, that HB is flat. That's when I say that overall M, a g-module, is equal to the normal Y. Another big rule of today's change is this... Today's change is a normal theory. Okay? So, to prove this lemma, I mean, one side is easy, is you take M, which one is it? If it's flat, then it satisfies this condition, because you take... No, it's not very right. If you add that, and you want to show that this basic formula is as five, okay, I have five, I write a long exact sequence of homology of this fiber, and because I have that, I can prove that every time, there's another multiple case. I might write it. So, say I have M1, M0, M2, and M... and this is a fiber, okay? So, that's a Gaussian diagram. I have a long exact sequence of my mat, and the one... and the one... and then I'm going to... plus one, I guess. Okay, so... you write it in case, you write it in case, you take it into base change, and you write the same thing with the fiber of the base change, and you see that the homology groups, because of this relation here, you put the two long exact sequence, one side to the other, and then you have isomorphisms, which implies that actually isomorphisms are the fibers, when you have change base and change base, and then you take the fiber, and I'll... sort of write it. So, there's this side of the lemma, this side of the lemma is the monoclonal, try to prove. So, using that, it's really easy to know this one is an equivalent, okay? And so, this one is due. So, what's important in that? Yeah, so, this means that flat morphism preserves a T-structure, okay? And in fact, it's an equivalent, okay? So morphism is flat, and you have to preserve the T-structure, right? Preserving the T-structure, I mean, exactly that. Okay, so zoos are exactly the same. So, those two sequences are equivalent. Okay. Okay, so, the flat, yeah. Right back down. That means it preserves the structure. Okay, I'm going to... There's one. There's a flat wall to the relative structure, and I can take, now, x be the right stack, and x is the alternate. So, why do we need x to be on? I need x to have a flat address, okay? So I can use... Because the quasi-quadric, the category of quasi-quadric machine on x is just gluing of a AG module for A, cloud, or X. And then, when I have a relation to A to B, I can do it, so... Can I ask you a question? There's proof about that preserving the structure. It seems you only needed some proof that it preserved push-outs, is that right? No, it always preserved push-outs. Right. So, no. But how do you... But isn't that enough to deduce that to not? I need to prove it. So, I think one of the things, the equivalence, and I don't know if my model is not flat, then there should not be... There may not be equivalences. There's no reason for them to be. You're saying that the co-molg would be the slider there by not being... Yeah....or the former there by not being. Yeah, that's what I'm saying. But there's all lemmings, or is it... You're saying you can't do them without a frat ass? Well, there needs to be a few co-molg groups of this thing. And to do so, I need to slam up. And to prove them, since then, I'd actually exactly say that the co-molg group of this one is a co-molg group of this one. Well, a co-molg group of lower than you. And what's important is also that, intentionally by being, you're still got something which is in degree, not in degree. Okay. And this one will make you prove that you're a learned. We did. I don't know. Okay. So, yes, we have a flat ass. And so, well, I'm just saying that we can glue T-structure. I'll just do the structure test. One thing about it, why is it not of a T-structure over X? So, I have... I have three four Xs on number three. And why is that? This is just the limit over the A to X flat. And morphology and screen therapy, flat T, are a more pre-hand. So, I'm going to call it... What? Call it? Call it. No, yes. I mean, X is a little bit... So, this one is going to be a bit funny to you. Yeah. So, 10-4 is the same thing. And so, the art on the T-structure... So, it's a section of this one and all of this one is zero. And so, the tilt of this. And it's actually... so that's quasi-coherent sheets over X. But it's just a definition of A. But this is actually the same thing as a category of quasi-coherent sheets of truncation of X. So, if you take the thought of this T-structure, you forget the geratin structure. So, you're having an enhancement of a stack. And if you take the geratin structure, you don't change the part of your geratin category, but you change the geratin category. Okay, so... I'm going to take the 20 minutes to try and explain the Bayesian formula. So, to try it on Bayesian formula, I think you can play through... and see. So... Um... So, Bayesian formula. Well, if you have a diagram like that, so we take X, Y, Z, and... So, the way that you see that is the hardest, is that you can actually take the limited preserved intersection to categories and how I should figure this out. I don't understand what you said. So, if I were to figure out what the intersection between these two T-structure are, then I would go to column X. Because I take a limit, I'd be just trying to take intersection... Yeah, yeah, yeah. Because the limit preserved on the intersection, that's how I know the two categories. Yeah, that's good. That's what makes you happy. And so, if I take something like that, so, Cartesian diagram in derived stats, we don't make these structure are, but we have, so to say, these, these, these, you. We have two ways of getting from the derived category of the algebraic category of X. So, these two ways are basically... Is this the book? I'll just have some of those cool back-up elements over here. Yeah, yeah, cool. So, what do you do with Z and......the last one. So, one way is to first......move forward and go back, basically. This one and the second one is......to show you through that one. And that's... And you have......for the information that you can use to learn that, which is just given by the adjunction. The adjunction can be......backed as this. So, what... The question is, the basic formula is, when is this transformation in coordinates? Okay? So, we're going to......to prove, to catch a clue. So, let's call the transformation alpha. And so, F, M, S, Y, Z......and......and......and the complex. Rides...... etmeüne......men. Alumnus and men. Okay, so is it the same statement as the one you derived from? Okay. You can make a very good base change. There are all skips and one with a very good base change. Oh yeah, yeah, here we don't need, that's the important part, we don't need actually flat, which we need in non-derived case. Okay, we don't need actually flat and why that? Because our category is stable, so the base change always presents the products we need. Okay, so I mean are you going to define the stable case and the term in that case? You use the same statement. Well, the difference is that the base change always preserves cofibers, okay, so they're joined. So because it's stable it also preserves spiders. That's all. I mean I think maybe the more important point is that the fiber product is derived. Yeah, yeah, yeah. So I mean if F is flat then we could just naively compute the fiber product and that would be the correct thing. If F is not flat then is that different from the classical fiber product? That's what's making this perfect. So if it would be real different, this is not what we're trying to do in this situation. So you guys have any point to say about this? Okay, so you can't say that you have to use a product that you can't use. Yeah, I'm going to give that to you. Okay. So we'll try to prove that when X and Y are fine. And then we'll see how, we'll see if not creative. But we'll test how to, to, to, to generalize from X and Y are fine too. And then you, doesn't come back crazy crazy crazy. You're right. So first X and Y are fine. You're right. So this one is F. F is FK. So because Z is fairly compact, okay, this means that I have to write down this diagram again. So Z is, uh, as the compact and well separated. So in fact, in fact, and well separated. So in the right category of the right stack, the problem is the final problem is the fine scheme. The right scheme. And what should I do with that? Well, so we have to prove that. Um, we cannot go to Z. Um, since that, this is true, okay. So that's alpha and equivalence. It's first stable by an angle in it, and then it's the same as the fine. So first case for one one, let's say, one whole point is Z. So Z is a spec C. Okay, so C B is just a spec of tensor to drive tensor five. And the right is pure. Usually fine. Okay, so Y is, um, first Y is just a, um, a spec of a particular module M, uh, C D G module. Then this thing is just a definition and, um, I don't know if you like that. So the problem with Y is that the problem with Y is that it's not a good thing. So, um, what should I say? I don't know. And then the other thing is, yeah, the other thing is this one. So it's a H A of D, which is A of D. So this is the definition of the homology. What we need to, this is just a forgetful function from D G module to D G module. So this is just M C, there's an A D G module. Yeah, yeah, yeah, yeah. Okay. Um, so, uh, yeah, this thing is the same thing. It's a homology. Well, let's say it's just, I first restrict you to the same, okay, so I can take M and I transfer it with C and then I forget that it is a B tensor C module but I just say it to the B module. Okay, so, I, yeah, it's also a B, that's why that's the problem. Yeah, yeah, yeah, that's the problem. So, I'm going to take the same thing, okay, and then I'm going to transfer it to the C module. No, no, please leave it over. B is C, okay. What is the number? What is the C? What is the C? You can't say the C. The C is over C. C is over C. What's that? Mine is M, that's M, and those two are equivalent. Okay? Yeah, okay. So, it's time to pull it back in. So, now I've proved that this kind of isomorphism is stable when I'm moving a finite number of Zs, okay? And the important thing is that the only thing you can do is understand things. This is just......the number of Zs. And so, the thing is, you can understand what this map is. Find here, and then pull this diagram. And so, this......what is that? Well, if I just E, in the diagram, the integral of the graph, the graph can go quite pretty cheap of Z. Then I have projections, only need to find, for this finite number of finds, which Z is the probability. So, I have projections in some... let's say Z-alpha, even more, but Z is the probability of......the spec of Z-alpha. I'm going to start to find it. Okay, so I have these projections, and then I can go and forget about Z-alpha in knowledge. I can go to the knowledge picture, so I get into this thing, and so for any alphas, works well in harmony. And this one, this one are compatible, and so in here, I have a diagram indexed by this same alphas, okay? So, I take the limits of this diagram, and that's my P start, and I take the limit of P. So, I use one alphas, but let's say Z is quite alphas, and I take the limit of P. So, now, by predicting this, and I want to use some data, this one will just be the limit of everything, because, in fact, again, the right-hand side product deserves the finite limits. So... Sorry, Professor, so the alphas are the morphine from... Yeah, yeah, so this is from the spec of... Sorry. Yeah. My alphas are morphine from spec of the alphas, please. Um... Yeah. When I take it, so my finite limit is here, and so the finite limit will just go out of the tensor, drag the subproduct, so it will be here, I mean, okay? So, just will still be an increment if Z is such a finite form. Okay, you just write and write that the tensor product, the right-hand product, can use the finite limit, and you get much more. Okay? I think that the whole step is to say that the full value of Z is the limit of C. Yeah, yeah. That's why it makes it more... Yeah, yeah. It's a category of zero-ass tags to the top of the... the top of us, so... um, the problem is how you do those all, so if Z is the finite form of C alpha, then ZB is the finite form of C alpha's tensor, drag tensor, we go right. Okay. In the... in the spite of the... in spite of the... Yeah, yeah, yeah. Yeah. Um, yeah, so I'm going to write that a unit of Z alpha and this one is set for the coordinate of the aspect of C alpha and so it's anything. Okay. So, you're doing that, doing that, and you're doing this ability to get that when the... you can't lose quality. Now, the question I wanted to... try and look at that is how to... to get the basis to do finite. I mean, here, you said, okay, when you had a diagram like that, and... and you also said, okay, when X, Y, and Z are the finite, this works. Now, when X and Y are the finite, and Z are the finite, then this works. So, check basis formula first. Now, seeing that X could be a finite is... it's easy because... you write that, I mean, we go into the right category of X, so here we can just test the equivalence on every... on every aspect A or X. Okay? So, seeing that X can be a finite, you ought to see that Y can be a finite. So, how... this is the part that we need... we need to see that... so, first, we write every of non-quadrant equations about every sheet of the scene, which I did also find that... I'm just naming them... write a p star, just to remember, this is not possible to compute. And... so, we said earlier that this one does not compute. Actually, in our case, it does when we do an R-close compact, it was separated, then this can be used, and that's the reason why... what we have to do, actually, is like testing whether or not, if I take E, P, U, I go there, I do that, and I have to test whether or not it's quite coherent or not. But this can be done in... on any... a finite or a Y, and then, in something like that, that future of state, you think, and then, L, R, it's there somewhere. Yeah, so if I want to... this is Y. Okay. So this is one I know it can use. So to test at this one, can use... I just have to tell you that this one can use for any state A over Y. Okay. So that's how I get to Y as a fine. Okay, and then, X as a fine, and Z as a fine. We go back and get... and I think I'm... okay. Here. APPLAUSE Questions or comments? Can we observe where the state is last step? It is... So the base change formula always works in the non-cosqlian formula, just by definition. If you write down what's the first formula of non-cosqlian formula, it doesn't say base change just by definition. So if you know that the first formula preserves cosqlian G, you actually prove the base change formula. And now what does it mean for the first formula to preserve cosqlian G? A cosqlian G is just something that flips on a fine and it has to be stable by base change. So it's actually step 1.2. It tells you that the first formula will preserve cosqlian G. Okay. I have a question. So as you said, there's a space for the polar bear from the quasi-perherent modules and some stack x into just usual OX modules. So now I said I have an OX module. How do I actually check the real amount of this quasi-perherent? Well, that's kind of what you just said. You could try to... I mean, you have the challenge like that. Okay. And you should be like... On the fine side of the whole section. Okay. So... What's really... What you need to do is just... Okay, if I take my sheet here, E, and I want to know if it's quite coherent, I take every... I can do it. We don't need Z. We can restrict to Z at 5, okay? Because it's quite coherent. If I'm only here, there's a push forward pullback. It's fine. Oh, really? Yeah. It's quite coherent. Yeah. So let's say this is Z. E is quasi-perherent. Even on a......radiified map, it's quite coherent. Ah, so that's very nice. Yeah, I should say that. That's the definition. By definition, OZ module is a sheet of module on the big side of the fine scheme to resume. It is defined on the fine scheme. Checking this quasi-perherent is just saying that the value over B is just the value over A and Z. Vibranatum. So it's just a definition. And so this gives a much more complete definition than what we said. Yeah, I mean, what's a G? For any of the points, spec A to Z, I have the day module, A of N. Any time I have a map from spec B to spec A, I have a map from E of A and Z of B to E of B. That's it. And, you know, combatable with this composition is Vism. And the sheet is quasi-perherent, if it's Vism, I'm sorry, what's the next? Okay, nice. Great. So I have a question. What if we, if we start using in DG categories, I mean, we can't believe that this conceptual saving that we can save all DG categories or a common stable alternative work category, and I'm wrong, but is it just a historical reason? I mean, the leaves, the leaves are DG categories for this? Why do you use GK as a word? Because you spoke about GK as a word. Yeah, I didn't do this. That's a historical reason, I mean. One and a half years. It's just a model. Now, the thing is, it's special. I mean, you shouldn't care if you use what was wrong, there's no reason except that, you know, it takes half a case to define what's a DG category, and it takes hundreds of pages to define what's a linear stable DG category, so that's the only reason. There's no reason to decide that. But this conceptual saving and characteristic heroes have proven that there's a head-eye on it. Well, it seems that the alternative is not used in the paper, but it's true already. The other doesn't know why characters exist in this paper. We were discussing with Marco before, but he doesn't know why. I guess, Christine, there's a need for a DG. If he doesn't see it. If he asks, I'll tell you because he doesn't know why. I don't know why.
 So I'm going to do 3-6 3-things in this talk.  I'm going to, well, suppose, define the material, the drag category, or the drag stack,  and then we're going to go to the object structure.  Okay, because you know, you know what happened in the wrong drag, as a great jump lead.  And at the end, I'm going to talk about the next thing formula, which we have on a common scheme,  when there are compact and easy to draw.  So, first, it's given like X for the scheme, just a scheme.  We define the drag category of, well, the question for here and she is last,  well, we define the question for here and she is last, and then we say the complex of the question for here and she is last,  and that's what we write.  Here, we won't do that.  Okay, well, we'll do that, but I'm not exactly.  The first thing is, well, we first define this on the find and then we do it, obviously.  So, we have a factor on this PGA, non-positive degree.  Okay.  To PGA categories.  That's okay.  Why is this factor, why is it sent A to the DG categories of A to D1?  A to D1.  And by that I mean, and down, yeah, here I'm in non-positive degree, here I'm just monitoring.  And that's A of the DG.  That's A of A.  Okay.  Basically, one more.  Yeah, okay, well, this is what I want.  Well, A is for you to do it.  Okay, so, in the talk before, in the talk before,  well, we saw something like that, here, perfect modules.  Okay, and yeah, he said, Fransma could not prove that there, that this was the one where perfect modules were stacked.  Okay, so it satisfies the distribution and it's still about this one.  Okay, so this one is stacked in the DG category, whatever that means.  Okay, so.  Now, I think I'm going to write this one.  So, I'm going to use, I want to do that, okay, obviously if it has a decent combination, then I will be able to do this.  This formally is like, is done using universal property of the category of DRI stack, okay.  And what is that?  Well, if I have a function from CGA, up into some category C, excuse me, you can use the category with all the limits.  Small for.  And if this function, let's say, yeah, satisfies the decent condition, so, well, I mean,  if you take the dishing of being a stack in DG categories, for example, you pop it, then it extends, here, you have to do me enough, enter.  It extends essentially uniquely in a function like that, which is, I don't know, let's write it like that, which preserves coordinates.  So, here we are at the same degree, but of course, I have to put up and up, so, however,  this can be formulaed in the existence of essentially unity, which is on equivalent to two spaces.  So, well, yeah, okay, well, if I want to apply that to this thing, I get something like that.  And that, I'm sure, is LQ code, okay, so, which, since the stack, it's throughout the category.  So, basically, you just take the huge diagram of all the affines going to your stack at X, and take a, you know, the only way to do it is to call it.  So, here, I mean, by F, S, what do you say, the limit of stack A, when you say A, of Rx.  We don't have to do the whole thing, anyway, if it's a nice enough stack.  Yeah, I mean, there's geometric, you can turn it into a simple shot, but yeah.  Okay, so this is into the limit of TQR of the rate.  Okay, so, the definition is, I mean, given the tools I have, pretty easy.  Now, computing is such a thing, usually.  You don't.  Well, you try to, so I give two of the examples, but, well, when I do the pipe, it's easy.  Okay, so.  That's quite a moment.  Yeah, I'm going to, this one was hard.  Well, I have, okay, I can, if I have X, the drive stack, and G, a group argument on X,  I can perform the quotient stack of X by G, and its drive category is,  well, what does it look like?  This one is the quotient of, okay, I'm going to grab it here.  This one is the quotient of something like that, so I have some group of it like that.  Okay, let's say, this is the one, the quotient of this thing.  And so, here I have the projection, you want here I have the action group.  So, if I'm going to take the drive, the drive category, I could achieve all these things.  I get the diagram the other way around, and it is pretty obvious that these gears,  the Q-core of X, which are G group arguments, okay?  So, I mean, this is a definition in fact when the X and G are Y, so when they're nice and  nice and easy, I think it is. Now, our number of causes, let's say G is in the time,  I'm going to take the structure of it.  So, I have the algebraic team, then I have the G and the group scheme.  And this thing is, well, what you can see, okay, this is G-provided classic algorithm  scheme for this point, so this is like the point is, okay, so this is a chain complex  on which G acts, okay? And so, it's actually complexes of representation of G,  so this is a stratification theorem, okay? So, this is just a representation of G  and you take the drive category by that I mean the DG category of complex little  G representation.  Yeah.  Because if you think of the left hand side, it's, you know, these are complexes with  an actual G in the reservoir sense, so like you have a complex and then you put it back  on G, then it has an auto-equivalence which is on which it represents the actual G  and you put it back on, put it back in two different ways in G cross G, you have  homo-tropies between these actions and so on, whereas the right hand side is really  complex of representation in the, you know, in this case sense.  So, I mean this is not, no, I just wanted to say it's not a good statement.  Do you think there have been some other assumptions?  Plats or, you know, maps? Plats or an, an evasive?  Yeah.  Where does the parkas come in?  Not sure.  I mean, the problem is that it is not a strat and when you put it back on G cross G  you might have a T-Rap G cross G, you know, in the picture on the left hand side  that you are not going to find on the right hand side. I mean, I don't know.  I prefer to be safe. Let's say, right.  I mean, it's a problem, sorry, yeah, it's a problem in a totally logical way,  which is if you have an action one stack which is not T-Rap, which is positive  impactant and mind-dialing, there's an F-Q-Q that stack is the rapid-view  of the geese. That's a special case.  So one stack, do you mind if I break? I'll change the name to that.  I find that all is present in class.  Thank you.  Thank you to all the experts in the category, the derived category  of conferences  of quasi-trainers, in a way, to be...  And the thing is that there are examples of schemes which are not  that might be available for which this is wrong, only schemes.  Okay, so...  The same thing, okay.  This is a specification thing.  Yeah, I agree, this is what we wanted to leave.  On this side, we have complexes, okay,  which are x of the current sheet that's complex.  And this one is each level is a percentage human.  I see.  The same thing as this one, I mean, this one was the complex one,  which gx and this one is a change of geosification.  Okay.  So, of course...  So, a huge difference for...  When I compare the map from x and y in the derived class,  I mean, because my function, the value of the function,  I get a map of that.  And in fact, to whose categories, because you can ask presentable,  you can continue the categories instead of the categories.  And I just want to use this polynomial, so I can get to write it, John.  Okay, and I'm going to write some Q code here,  because you can define the derived category on normal, but in coherent sheets.  And those two do nothing inside,  if I have lower star and lower star Q code upon all the same.  My last statement on the case of the lead,  that in some kind of a condition, then it's too...  Actually the same.  Okay, of course when we just draw the lines, it's basically...  Okay, so now we'll talk about the t-structure of...  Sorry, I have a question.  So you just said sometimes it's the same as the product, I guess, for all x models.  Does this mean that it's always, I don't know,  fully placed for embedding from LQ code x into the category of all x models?  Yeah, yeah, yeah.  This is a full-time category of, let's say, L of x, which I would, for the line.  Well, it's the way to say it, but it's actually...  You wrote LQ code x is the limit of that, you know, of L of x of A of y of x.  You replace limit by a lag-limit, meaning you have transition coefficients  and not the equivalence anymore.  So it's like the sections of the golden inventory of all of that.  So there is a limit in the category, there is a lag-limit.  And this Lx is the lag-limit thing.  And inside you have the objects which all transition coefficients are equivalences.  So are we reducing the resistance of these things, at first, for its preserve of all the maps?  At this point?  Yeah.  Well, basically, one of the things is called image.  Well, when you do this one...  You're doing a call-image of a diagram of...  Well, yeah, you surely get maps, but why do you have preserve of all of these?  Because it's a limit of maps who preserve all the limits between maps,  and they're going to be in the categories.  Okay.  It's just basically everything that you're doing.  Because it works for a lifetime.  Yeah, yeah.  Okay, so...  Now, yeah, key structure on a L, you could...  Of s, when s is our...  Okay?  We'll see why in this assumption.  And so, first, let's see what happens if x is defined.  We'll do that, of course.  So, first, x is defined.  Now, a, d, d, d, m equals...  And so, the key structure is, to that, how the category...  a d, d, m equals a non-positive degree.  And, actually, then you have the same...  the same categories with a degree over something, whatever, m.  And you have something like that, too.  Like, in a degree greater or equal to 1,  or non-negative degrees or whatever.  So, that's your thing.  And you do this such like a few actions,  you must have something like that.  Okay, so whenever you have an object which is a non-positive degree  and an object which is a degree greater or equal to 1,  then there is no non-triple map between them.  Okay, so this has to be the autogruble.  And that's why this one, actually, can be defined as the autogruble.  The left or the right autogruble is this one.  So, how do we say that?  How do we define them?  Well, obviously, we know what this should be.  We know what this should be.  And so, in here, fsi of m is 0,  i is greater than 0,  and the central goes for m,  and the central goes for m.  So, this defines the G structure on ATGM,  the algorithm of the GTA code.  So, what's important is that the heart of this structure  is actually a heart of this structure.  It is just a view of a normal system.  Is this supposed to be a normal term on heart,  or is this a philosophical statement?  It's a technical term.  Oh, okay, so, what do you define exactly with this?  The heart is a section of this one, and this one.  Okay?  This one would be fine for any n here,  and any n here, I mean, we don't need to be doing one.  The theorem is that the heart is a billion.  That's not completely obvious.  This is the category.  Are you going to define what the t-thorcher is?  Well, the t-thorcher is,  but that is the two things.  But in general, we're not...  No, no, well, on c, okay,  c is some category, and you ask that, you ask...  You ask to have c... ...the working zero.  In fact, you have that, you have this thing,  or two, you need to...  And you ask 20x and c.  20x and c.  There is a lot of x.  And you ask this...  Okay, so if you guys do your final category,  that's just the final, just the two which final,  carry it all off in these tables,  and you can get it together,  and you can get it in the final five seconds.  So you demand that c is stable?  Well, this one, half.  Okay, again, we can do it for just finalized categories.  As derived categories on derived...  As you're right, John, you know,  you're more of a category of derived categories  in this final agent, so you can do that  in finalized categories, and as that,  you're just in the final,  and if you period a lot in the stable-due category,  so this is just five or five-inch segments.  You know, it's just a way to formalize  the fact that your standard category  has a few subcategories which is a real agent,  and that any object has a kind of  prosnical decomposition, whose piece is on in the house.  If you try to formalize this,  you'll see that it's a little bit more complex.  Okay, so...  And the half of the g-squature is always intersection of  the c in all these degrees in the degree.  Okay.  So, this is the final g-squature,  and actually when you have such things,  then you have a center which associated with its  is from Haitian beer, and another center here.  And the other from Haitian is from Haitian beer.  Now, how do we do that?  Well, if you have a map...  You have a map A to B,  obviously, B, G, A, and A.  And you want to know when, I mean,  when one does a survey center to preserve this g-squature.  So, the answer is when you use that.  So, I'll say that.  And then...  This is a map in here,  and you have a training model M,  a g-module M,  and a list of maps.  So, here it can apply space change factor,  I think it's the right...  And...  Okay, because when I say we think together,  this one is the fiber of this one,  and we can also do that.  We can also...  When we're in B, here,  this one is in B,  whereas identity,  and here we have the communication.  This is a location of the g-squature on B,  and we can do that on B, G, 1, B.  And so...  So, I think these are two sequences,  actually are the same when my autism is flat Y.  Well, you have to prove that one lose may have maps like that.  And it's enough to prove that one lose is an equivalent.  And this one is,  I'm flat Y is that in the situation that I'm at,  that HB is flat.  That's when I say that overall M,  a g-module,  is equal to the normal Y.  Another big rule of today's change  is this...  Today's change  is a normal theory.  Okay?  So, to prove this lemma,  I mean, one side is easy,  is you take M,  which one is it?  If it's flat,  then it satisfies this condition,  because you take...  No, it's not very right.  If you add that,  and you want to show that this basic formula  is as five,  okay, I have five,  I write a long exact sequence  of homology of this fiber,  and because I have that,  I can prove that every time,  there's another multiple case.  I might write it.  So, say I have M1, M0, M2, and M...  and this is a fiber, okay?  So, that's a Gaussian diagram.  I have a long exact sequence  of my mat,  and the one...  and the one...  and then I'm going to...  plus one, I guess.  Okay, so...  you write it in case,  you write it in case,  you take it into base change,  and you write the same thing with  the fiber of the base change,  and you see that the homology groups,  because of this relation here,  you put the two long exact sequence,  one side to the other,  and then you have isomorphisms,  which implies that actually isomorphisms are the fibers,  when you have change base and change base,  and then you take the fiber,  and I'll...  sort of write it.  So, there's this side of the lemma,  this side of the lemma is the monoclonal,  try to prove.  So, using that,  it's really easy to know this one is an equivalent, okay?  And so, this one is due.  So, what's important in that?  Yeah, so, this means that flat morphism  preserves a T-structure, okay?  And in fact, it's an equivalent, okay?  So morphism is flat,  and you have to preserve the T-structure, right?  Preserving the T-structure, I mean, exactly that.  Okay, so zoos are exactly the same.  So, those two sequences are equivalent.  Okay.  Okay, so, the flat, yeah.  Right back down.  That means it preserves the structure.  Okay, I'm going to...  There's one.  There's a flat wall to the relative structure, and I can take, now,  x be the right stack, and x is the alternate.  So, why do we need x to be on? I need x to have a flat address, okay?  So I can use...  Because the quasi-quadric, the category of quasi-quadric machine on x  is just gluing of a AG module for A, cloud, or X.  And then, when I have a relation to A to B, I can do it, so...  Can I ask you a question?  There's proof about that preserving the structure.  It seems you only needed some proof that it preserved push-outs, is that right?  No, it always preserved push-outs.  Right.  So, no.  But how do you...  But isn't that enough to deduce that to not?  I need to prove it.  So, I think one of the things, the equivalence, and I don't know if my model is not flat,  then there should not be...  There may not be equivalences.  There's no reason for them to be.  You're saying that the co-molg would be the slider there by not being...  Yeah. ...or the former there by not being.  Yeah, that's what I'm saying.  But there's all lemmings, or is it...  You're saying you can't do them without a frat ass?  Well, there needs to be a few co-molg groups of this thing.  And to do so, I need to slam up.  And to prove them, since then, I'd actually exactly say that the co-molg group of this one  is a co-molg group of this one.  Well, a co-molg group of lower than you.  And what's important is also that, intentionally by being,  you're still got something which is in degree, not in degree.  Okay.  And this one will make you prove that you're a learned.  We did.  I don't know.  Okay.  So, yes, we have a flat ass.  And so, well, I'm just saying that we can glue T-structure.  I'll just do the structure test.  One thing about it, why is it not of a T-structure over X?  So, I have...  I have three four Xs on number three.  And why is that?  This is just the limit over the A to X flat.  And morphology and screen therapy, flat T, are a more pre-hand.  So, I'm going to call it...  What?  Call it?  Call it.  No, yes.  I mean, X is a little bit...  So, this one is going to be a bit funny to you.  Yeah.  So, 10-4 is the same thing.  And so, the art on the T-structure...  So, it's a section of this one and all of this one is zero.  And so, the tilt of this.  And it's actually... so that's quasi-coherent sheets over X.  But it's just a definition of A.  But this is actually the same thing as a category of quasi-coherent sheets  of truncation of X.  So, if you take the thought of this T-structure,  you forget the geratin structure.  So, you're having an enhancement of a stack.  And if you take the geratin structure,  you don't change the part of your geratin category,  but you change the geratin category.  Okay, so...  I'm going to take the 20 minutes to try and explain the Bayesian formula.  So, to try it on Bayesian formula,  I think you can play through...  and see.  So...  Um...  So, Bayesian formula.  Well, if you have a diagram like that,  so we take X, Y, Z, and...  So, the way that you see that is the hardest,  is that you can actually take the limited preserved intersection to categories  and how I should figure this out.  I don't understand what you said.  So, if I were to figure out what the intersection between these two T-structure are,  then I would go to column X.  Because I take a limit, I'd be just trying to take intersection...  Yeah, yeah, yeah.  Because the limit preserved on the intersection,  that's how I know the two categories.  Yeah, that's good.  That's what makes you happy.  And so, if I take something like that,  so, Cartesian diagram in derived stats,  we don't make these structure are,  but we have, so to say,  these, these, these, you.  We have two ways of getting from the derived category of the algebraic category of X.  So, these two ways are basically...  Is this the book?  I'll just have some of those cool back-up elements over here.  Yeah, yeah, cool.  So, what do you do with Z and... ...the last one.  So, one way is to first... ...move forward and go back, basically.  This one and the second one is... ...to show you through that one.  And that's...  And you have... ...for the information that you can use to learn that,  which is just given by the adjunction.  The adjunction can be... ...backed as this.  So, what...  The question is, the basic formula is,  when is this transformation in coordinates?  Okay?  So, we're going to... ...to prove, to catch a clue.  So, let's call the transformation alpha.  And so, F, M, S, Y, Z... ...and... ...and... ...and the complex.  Rides... ... etmeüne... ...men.  Alumnus and men.  Okay, so is it the same statement as the one you derived from?  Okay.  You can make a very good base change.  There are all skips and one with a very good base change.  Oh yeah, yeah, here we don't need, that's the important part, we don't need actually flat, which we need in non-derived case.  Okay, we don't need actually flat and why that?  Because our category is stable, so the base change always presents the products we need.  Okay, so I mean are you going to define the stable case and the term in that case?  You use the same statement.  Well, the difference is that the base change always preserves cofibers, okay, so they're joined.  So because it's stable it also preserves spiders.  That's all.  I mean I think maybe the more important point is that the fiber product is derived.  Yeah, yeah, yeah.  So I mean if F is flat then we could just naively compute the fiber product and that would be the correct thing.  If F is not flat then is that different from the classical fiber product?  That's what's making this perfect.  So if it would be real different, this is not what we're trying to do in this situation.  So you guys have any point to say about this?  Okay, so you can't say that you have to use a product that you can't use.  Yeah, I'm going to give that to you.  Okay.  So we'll try to prove that when X and Y are fine.  And then we'll see how, we'll see if not creative.  But we'll test how to, to, to, to generalize from X and Y are fine too.  And then you, doesn't come back crazy crazy crazy.  You're right.  So first X and Y are fine.  You're right.  So this one is F.  F is FK.  So because Z is fairly compact, okay,  this means that I have to write down this diagram again.  So Z is, uh, as the compact and well separated.  So in fact, in fact, and well separated.  So in the right category of the right stack,  the problem is the final problem is the fine scheme.  The right scheme.  And what should I do with that?  Well, so we have to prove that.  Um, we cannot go to Z.  Um, since that, this is true, okay.  So that's alpha and equivalence.  It's first stable by an angle in it,  and then it's the same as the fine.  So first case for one one, let's say,  one whole point is Z.  So Z is a spec C.  Okay, so C B is just a spec of tensor to drive tensor five.  And the right is pure.  Usually fine.  Okay, so Y is, um,  first Y is just a, um,  a spec of a particular module M, uh, C D G module.  Then this thing is just a definition  and, um, I don't know if you like that.  So the problem with Y is that the problem with Y is  that it's not a good thing.  So, um, what should I say?  I don't know.  And then the other thing is,  yeah, the other thing is this one.  So it's a H A of D, which is A of D.  So this is the definition of the homology.  What we need to, this is just a forgetful function  from D G module to D G module.  So this is just M C, there's an A D G module.  Yeah, yeah, yeah, yeah.  Okay.  Um, so, uh, yeah, this thing is the same thing.  It's a homology.  Well, let's say it's just,  I first restrict you to the same, okay,  so I can take M and I transfer it with C  and then I forget that it is a B tensor C module  but I just say it to the B module.  Okay, so, I, yeah, it's also a B, that's why that's the problem.  Yeah, yeah, yeah, that's the problem.  So, I'm going to take the same thing, okay,  and then I'm going to transfer it to the C module.  No, no, please leave it over.  B is C, okay.  What is the number?  What is the C?  What is the C?  You can't say the C.  The C is over C.  C is over C.  What's that?  Mine is M, that's M,  and those two are equivalent.  Okay?  Yeah, okay.  So, it's time to pull it back in.  So, now I've proved that this kind of isomorphism is stable  when I'm moving a finite number of Zs, okay?  And the important thing is that  the only thing you can do is understand things.  This is just... ...the number of Zs.  And so, the thing is,  you can understand what this map is.  Find here,  and then pull this diagram.  And so, this... ...what is that?  Well, if I just E,  in the diagram, the integral of the graph,  the graph can go quite pretty cheap of Z.  Then I have projections,  only need to find,  for this finite number of finds,  which Z is the probability.  So, I have projections in some...  let's say Z-alpha,  even more,  but Z is the probability of... ...the spec of Z-alpha.  I'm going to start to find it.  Okay, so I have these projections,  and then I can go and forget  about Z-alpha in knowledge.  I can go to the knowledge picture,  so I get into this thing,  and so for any alphas,  works well in harmony.  And this one,  this one are compatible,  and so in here, I have a diagram  indexed by this same alphas, okay?  So, I take the limits of this diagram,  and that's my P start,  and I take the limit of P.  So, I use one alphas,  but let's say Z is quite alphas,  and I take the limit of P.  So, now, by predicting this,  and I want to use some data,  this one will just be the limit of everything,  because, in fact, again,  the right-hand side product  deserves the finite limits.  So...  Sorry, Professor,  so the alphas are the morphine from...  Yeah, yeah, so this is from the spec of...  Sorry.  Yeah.  My alphas are morphine from spec of the alphas, please.  Um...  Yeah.  When I take it,  so my finite limit is here,  and so the finite limit will just go out  of the tensor,  drag the subproduct,  so it will be here, I mean, okay?  So,  just will still be an increment if Z is such a finite form.  Okay, you just write and write that  the tensor product, the right-hand product,  can use the finite limit,  and you get much more.  Okay?  I think that the whole step is to say  that the full value of Z is the limit of C.  Yeah, yeah.  That's why it makes it more...  Yeah, yeah.  It's a category of zero-ass tags to the top of the...  the top of us, so...  um,  the problem is how you do those all,  so if Z is the finite form of C alpha,  then ZB is the finite form of C alpha's tensor,  drag tensor, we go right.  Okay.  In the... in the spite of the...  in spite of the...  Yeah, yeah, yeah.  Yeah.  Um, yeah, so I'm going to write that  a unit of Z alpha  and this one is set for the coordinate  of the aspect of C alpha  and so it's anything.  Okay.  So, you're doing that, doing that,  and you're doing this ability to get that  when the... you can't lose quality.  Now, the question I wanted to...  try and look at that is how to...  to get the basis to do finite.  I mean, here, you said, okay,  when you had a diagram like that,  and...  and you also said, okay, when X, Y, and Z are the finite,  this works.  Now, when X and Y are the finite,  and Z are the finite,  then this works.  So, check basis formula first.  Now,  seeing that X could be a finite is...  it's easy because...  you write that, I mean, we go into the right category of X,  so here we can just test the equivalence  on every... on every aspect A or X.  Okay?  So, seeing that X can be a finite,  you ought to see that Y can be a finite.  So, how...  this is the part that we need...  we need to see that...  so, first, we write every of non-quadrant equations  about every sheet of the scene,  which I did also find that...  I'm just naming them...  write a p star,  just to remember,  this is not possible to compute.  And...  so, we said earlier that this one does not compute.  Actually, in our case,  it does when we do an R-close compact,  it was separated, then this can be used,  and that's the reason why...  what we have to do, actually,  is like testing whether or not,  if I take E, P, U,  I go there, I do that,  and I have to test whether or not it's quite coherent or not.  But this can be done in...  on any... a finite or a Y,  and then, in something like that,  that future of state,  you think,  and then, L, R,  it's there somewhere.  Yeah, so if I want to...  this is Y.  Okay.  So this is one I know it can use.  So to test at this one, can use...  I just have to tell you that this one can use for any state A over Y.  Okay.  So that's how I get to Y as a fine.  Okay, and then,  X as a fine, and Z as a fine.  We go back and get...  and I think I'm...  okay.  Here.  APPLAUSE  Questions or comments?  Can we observe where the state is last step?  It is...  So the base change formula always works in the non-cosqlian formula,  just by definition.  If you write down what's the first formula  of non-cosqlian formula,  it doesn't say base change just by definition.  So if you know that the first formula  preserves cosqlian G,  you actually prove the base change formula.  And now what does it mean for the first formula  to preserve cosqlian G?  A cosqlian G is just something that flips on a fine  and it has to be stable by base change.  So it's actually step 1.2.  It tells you that the first formula  will preserve cosqlian G.  Okay.  I have a question.  So as you said, there's a space for the polar bear  from the quasi-perherent modules  and some stack x into just usual OX modules.  So now I said I have an OX module.  How do I actually check the real amount of this quasi-perherent?  Well, that's kind of what you just said.  You could try to...  I mean, you have the challenge like that.  Okay.  And you should be like...  On the fine side of the whole section.  Okay.  So...  What's really...  What you need to do is just...  Okay, if I take my sheet here,  E, and I want to know if it's quite coherent,  I take every...  I can do it.  We don't need Z.  We can restrict to Z at 5, okay?  Because it's quite coherent.  If I'm only here, there's a push forward pullback.  It's fine.  Oh, really?  Yeah.  It's quite coherent.  Yeah.  So let's say this is Z.  E is quasi-perherent.  Even on a... ...radiified map,  it's quite coherent.  Ah, so that's very nice.  Yeah, I should say that.  That's the definition.  By definition,  OZ module is a sheet of module  on the big side of the fine scheme to resume.  It is defined on the fine scheme.  Checking this quasi-perherent is just saying that  the value over B is just the value over A and Z.  Vibranatum.  So it's just a definition.  And so this gives a much more complete definition than what we said.  Yeah, I mean, what's a G?  For any of the points,  spec A to Z,  I have the day module, A of N.  Any time I have a map from spec B to spec A,  I have a map from E of A and Z of B to E of B.  That's it.  And, you know, combatable with this composition is Vism.  And the sheet is quasi-perherent,  if it's Vism, I'm sorry, what's the next?  Okay, nice.  Great.  So I have a question.  What if we,  if we start using  in DG categories,  I mean,  we can't believe that  this conceptual saving that  we can save all  DG categories or  a common stable alternative work category,  and I'm wrong, but is it just a historical reason?  I mean, the leaves,  the leaves are DG categories for this?  Why do you use GK as a word?  Because you spoke about GK as a word.  Yeah, I didn't do this.  That's a historical reason, I mean.  One and a half years.  It's just a model.  Now, the thing is, it's special.  I mean, you shouldn't care  if you use what was wrong, there's no reason  except that, you know,  it takes half a case to define  what's a DG category,  and it takes hundreds of pages to define  what's a linear stable DG category,  so that's the only reason.  There's no reason to decide that.  But this conceptual saving  and characteristic heroes have proven  that there's a head-eye on it.  Well, it seems that the alternative  is not used in the paper,  but it's true already.  The other doesn't know  why characters exist in this paper.  We were discussing with Marco before,  but he doesn't know why.  I guess, Christine,  there's a need for a DG.  If he doesn't see it.  If he asks, I'll tell you because he doesn't know why.  I don't know why.
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So I'm going to do 3-6 3-things in this talk. I'm going to, well, suppose, define the material, the drag category, or the drag stack, and then we're going to go to the object structure. Okay, because you know, you know what happened in the wrong drag, as a great jump lead. And at the end, I'm going to talk about the next thing formula, which we have on a common scheme, when there are compact and easy to draw. So, first, it's given like X for the scheme, just a scheme. We define the drag category of, well, the question for here and she is last, well, we define the question for here and she is last, and then we say the complex of the question for here and she is last, and that's what we write. Here, we won't do that. Okay, well, we'll do that, but I'm not exactly. The first thing is, well, we first define this on the find and then we do it, obviously. So, we have a factor on this PGA, non-positive degree. Okay. To PGA categories. That's okay. Why is this factor, why is it sent A to the DG categories of A to D1? A to D1. And by that I mean, and down, yeah, here I'm in non-positive degree, here I'm just monitoring. And that's A of the DG. That's A of A. Okay. Basically, one more. Yeah, okay, well, this is what I want. Well, A is for you to do it. Okay, so, in the talk before, in the talk before, well, we saw something like that, here, perfect modules. Okay, and yeah, he said, Fransma could not prove that there, that this was the one where perfect modules were stacked. Okay, so it satisfies the distribution and it's still about this one. Okay, so this one is stacked in the DG category, whatever that means. Okay, so. Now, I think I'm going to write this one. So, I'm going to use, I want to do that, okay, obviously if it has a decent combination, then I will be able to do this. This formally is like, is done using universal property of the category of DRI stack, okay. And what is that? Well, if I have a function from CGA, up into some category C, excuse me, you can use the category with all the limits. Small for. And if this function, let's say, yeah, satisfies the decent condition, so, well, I mean, if you take the dishing of being a stack in DG categories, for example, you pop it, then it extends, here, you have to do me enough, enter. It extends essentially uniquely in a function like that, which is, I don't know, let's write it like that, which preserves coordinates. So, here we are at the same degree, but of course, I have to put up and up, so, however, this can be formulaed in the existence of essentially unity, which is on equivalent to two spaces. So, well, yeah, okay, well, if I want to apply that to this thing, I get something like that. And that, I'm sure, is LQ code, okay, so, which, since the stack, it's throughout the category. So, basically, you just take the huge diagram of all the affines going to your stack at X, and take a, you know, the only way to do it is to call it. So, here, I mean, by F, S, what do you say, the limit of stack A, when you say A, of Rx. We don't have to do the whole thing, anyway, if it's a nice enough stack. Yeah, I mean, there's geometric, you can turn it into a simple shot, but yeah. Okay, so this is into the limit of TQR of the rate. Okay, so, the definition is, I mean, given the tools I have, pretty easy. Now, computing is such a thing, usually. You don't. Well, you try to, so I give two of the examples, but, well, when I do the pipe, it's easy. Okay, so. That's quite a moment. Yeah, I'm going to, this one was hard. Well, I have, okay, I can, if I have X, the drive stack, and G, a group argument on X, I can perform the quotient stack of X by G, and its drive category is, well, what does it look like? This one is the quotient of, okay, I'm going to grab it here. This one is the quotient of something like that, so I have some group of it like that. Okay, let's say, this is the one, the quotient of this thing. And so, here I have the projection, you want here I have the action group. So, if I'm going to take the drive, the drive category, I could achieve all these things. I get the diagram the other way around, and it is pretty obvious that these gears, the Q-core of X, which are G group arguments, okay? So, I mean, this is a definition in fact when the X and G are Y, so when they're nice and nice and easy, I think it is. Now, our number of causes, let's say G is in the time, I'm going to take the structure of it. So, I have the algebraic team, then I have the G and the group scheme. And this thing is, well, what you can see, okay, this is G-provided classic algorithm scheme for this point, so this is like the point is, okay, so this is a chain complex on which G acts, okay? And so, it's actually complexes of representation of G, so this is a stratification theorem, okay? So, this is just a representation of G and you take the drive category by that I mean the DG category of complex little G representation. Yeah. Because if you think of the left hand side, it's, you know, these are complexes with an actual G in the reservoir sense, so like you have a complex and then you put it back on G, then it has an auto-equivalence which is on which it represents the actual G and you put it back on, put it back in two different ways in G cross G, you have homo-tropies between these actions and so on, whereas the right hand side is really complex of representation in the, you know, in this case sense. So, I mean this is not, no, I just wanted to say it's not a good statement. Do you think there have been some other assumptions? Plats or, you know, maps? Plats or an, an evasive? Yeah. Where does the parkas come in? Not sure. I mean, the problem is that it is not a strat and when you put it back on G cross G you might have a T-Rap G cross G, you know, in the picture on the left hand side that you are not going to find on the right hand side. I mean, I don't know. I prefer to be safe. Let's say, right. I mean, it's a problem, sorry, yeah, it's a problem in a totally logical way, which is if you have an action one stack which is not T-Rap, which is positive impactant and mind-dialing, there's an F-Q-Q that stack is the rapid-view of the geese. That's a special case. So one stack, do you mind if I break? I'll change the name to that. I find that all is present in class. Thank you. Thank you to all the experts in the category, the derived category of conferences of quasi-trainers, in a way, to be... And the thing is that there are examples of schemes which are not that might be available for which this is wrong, only schemes. Okay, so... The same thing, okay. This is a specification thing. Yeah, I agree, this is what we wanted to leave. On this side, we have complexes, okay, which are x of the current sheet that's complex. And this one is each level is a percentage human. I see. The same thing as this one, I mean, this one was the complex one, which gx and this one is a change of geosification. Okay. So, of course... So, a huge difference for... When I compare the map from x and y in the derived class, I mean, because my function, the value of the function, I get a map of that. And in fact, to whose categories, because you can ask presentable, you can continue the categories instead of the categories. And I just want to use this polynomial, so I can get to write it, John. Okay, and I'm going to write some Q code here, because you can define the derived category on normal, but in coherent sheets. And those two do nothing inside, if I have lower star and lower star Q code upon all the same. My last statement on the case of the lead, that in some kind of a condition, then it's too... Actually the same. Okay, of course when we just draw the lines, it's basically... Okay, so now we'll talk about the t-structure of... Sorry, I have a question. So you just said sometimes it's the same as the product, I guess, for all x models. Does this mean that it's always, I don't know, fully placed for embedding from LQ code x into the category of all x models? Yeah, yeah, yeah. This is a full-time category of, let's say, L of x, which I would, for the line. Well, it's the way to say it, but it's actually... You wrote LQ code x is the limit of that, you know, of L of x of A of y of x. You replace limit by a lag-limit, meaning you have transition coefficients and not the equivalence anymore. So it's like the sections of the golden inventory of all of that. So there is a limit in the category, there is a lag-limit. And this Lx is the lag-limit thing. And inside you have the objects which all transition coefficients are equivalences. So are we reducing the resistance of these things, at first, for its preserve of all the maps? At this point? Yeah. Well, basically, one of the things is called image. Well, when you do this one... You're doing a call-image of a diagram of... Well, yeah, you surely get maps, but why do you have preserve of all of these? Because it's a limit of maps who preserve all the limits between maps, and they're going to be in the categories. Okay. It's just basically everything that you're doing. Because it works for a lifetime. Yeah, yeah. Okay, so... Now, yeah, key structure on a L, you could... Of s, when s is our... Okay? We'll see why in this assumption. And so, first, let's see what happens if x is defined. We'll do that, of course. So, first, x is defined. Now, a, d, d, d, m equals... And so, the key structure is, to that, how the category... a d, d, m equals a non-positive degree. And, actually, then you have the same... the same categories with a degree over something, whatever, m. And you have something like that, too. Like, in a degree greater or equal to 1, or non-negative degrees or whatever. So, that's your thing. And you do this such like a few actions, you must have something like that. Okay, so whenever you have an object which is a non-positive degree and an object which is a degree greater or equal to 1, then there is no non-triple map between them. Okay, so this has to be the autogruble. And that's why this one, actually, can be defined as the autogruble. The left or the right autogruble is this one. So, how do we say that? How do we define them? Well, obviously, we know what this should be. We know what this should be. And so, in here, fsi of m is 0, i is greater than 0, and the central goes for m, and the central goes for m. So, this defines the G structure on ATGM, the algorithm of the GTA code. So, what's important is that the heart of this structure is actually a heart of this structure. It is just a view of a normal system. Is this supposed to be a normal term on heart, or is this a philosophical statement? It's a technical term. Oh, okay, so, what do you define exactly with this? The heart is a section of this one, and this one. Okay? This one would be fine for any n here, and any n here, I mean, we don't need to be doing one. The theorem is that the heart is a billion. That's not completely obvious. This is the category. Are you going to define what the t-thorcher is? Well, the t-thorcher is, but that is the two things. But in general, we're not... No, no, well, on c, okay, c is some category, and you ask that, you ask... You ask to have c......the working zero. In fact, you have that, you have this thing, or two, you need to... And you ask 20x and c. 20x and c. There is a lot of x. And you ask this... Okay, so if you guys do your final category, that's just the final, just the two which final, carry it all off in these tables, and you can get it together, and you can get it in the final five seconds. So you demand that c is stable? Well, this one, half. Okay, again, we can do it for just finalized categories. As derived categories on derived... As you're right, John, you know, you're more of a category of derived categories in this final agent, so you can do that in finalized categories, and as that, you're just in the final, and if you period a lot in the stable-due category, so this is just five or five-inch segments. You know, it's just a way to formalize the fact that your standard category has a few subcategories which is a real agent, and that any object has a kind of prosnical decomposition, whose piece is on in the house. If you try to formalize this, you'll see that it's a little bit more complex. Okay, so... And the half of the g-squature is always intersection of the c in all these degrees in the degree. Okay. So, this is the final g-squature, and actually when you have such things, then you have a center which associated with its is from Haitian beer, and another center here. And the other from Haitian is from Haitian beer. Now, how do we do that? Well, if you have a map... You have a map A to B, obviously, B, G, A, and A. And you want to know when, I mean, when one does a survey center to preserve this g-squature. So, the answer is when you use that. So, I'll say that. And then... This is a map in here, and you have a training model M, a g-module M, and a list of maps. So, here it can apply space change factor, I think it's the right... And... Okay, because when I say we think together, this one is the fiber of this one, and we can also do that. We can also... When we're in B, here, this one is in B, whereas identity, and here we have the communication. This is a location of the g-squature on B, and we can do that on B, G, 1, B. And so... So, I think these are two sequences, actually are the same when my autism is flat Y. Well, you have to prove that one lose may have maps like that. And it's enough to prove that one lose is an equivalent. And this one is, I'm flat Y is that in the situation that I'm at, that HB is flat. That's when I say that overall M, a g-module, is equal to the normal Y. Another big rule of today's change is this... Today's change is a normal theory. Okay? So, to prove this lemma, I mean, one side is easy, is you take M, which one is it? If it's flat, then it satisfies this condition, because you take... No, it's not very right. If you add that, and you want to show that this basic formula is as five, okay, I have five, I write a long exact sequence of homology of this fiber, and because I have that, I can prove that every time, there's another multiple case. I might write it. So, say I have M1, M0, M2, and M... and this is a fiber, okay? So, that's a Gaussian diagram. I have a long exact sequence of my mat, and the one... and the one... and then I'm going to... plus one, I guess. Okay, so... you write it in case, you write it in case, you take it into base change, and you write the same thing with the fiber of the base change, and you see that the homology groups, because of this relation here, you put the two long exact sequence, one side to the other, and then you have isomorphisms, which implies that actually isomorphisms are the fibers, when you have change base and change base, and then you take the fiber, and I'll... sort of write it. So, there's this side of the lemma, this side of the lemma is the monoclonal, try to prove. So, using that, it's really easy to know this one is an equivalent, okay? And so, this one is due. So, what's important in that? Yeah, so, this means that flat morphism preserves a T-structure, okay? And in fact, it's an equivalent, okay? So morphism is flat, and you have to preserve the T-structure, right? Preserving the T-structure, I mean, exactly that. Okay, so zoos are exactly the same. So, those two sequences are equivalent. Okay. Okay, so, the flat, yeah. Right back down. That means it preserves the structure. Okay, I'm going to... There's one. There's a flat wall to the relative structure, and I can take, now, x be the right stack, and x is the alternate. So, why do we need x to be on? I need x to have a flat address, okay? So I can use... Because the quasi-quadric, the category of quasi-quadric machine on x is just gluing of a AG module for A, cloud, or X. And then, when I have a relation to A to B, I can do it, so... Can I ask you a question? There's proof about that preserving the structure. It seems you only needed some proof that it preserved push-outs, is that right? No, it always preserved push-outs. Right. So, no. But how do you... But isn't that enough to deduce that to not? I need to prove it. So, I think one of the things, the equivalence, and I don't know if my model is not flat, then there should not be... There may not be equivalences. There's no reason for them to be. You're saying that the co-molg would be the slider there by not being... Yeah....or the former there by not being. Yeah, that's what I'm saying. But there's all lemmings, or is it... You're saying you can't do them without a frat ass? Well, there needs to be a few co-molg groups of this thing. And to do so, I need to slam up. And to prove them, since then, I'd actually exactly say that the co-molg group of this one is a co-molg group of this one. Well, a co-molg group of lower than you. And what's important is also that, intentionally by being, you're still got something which is in degree, not in degree. Okay. And this one will make you prove that you're a learned. We did. I don't know. Okay. So, yes, we have a flat ass. And so, well, I'm just saying that we can glue T-structure. I'll just do the structure test. One thing about it, why is it not of a T-structure over X? So, I have... I have three four Xs on number three. And why is that? This is just the limit over the A to X flat. And morphology and screen therapy, flat T, are a more pre-hand. So, I'm going to call it... What? Call it? Call it. No, yes. I mean, X is a little bit... So, this one is going to be a bit funny to you. Yeah. So, 10-4 is the same thing. And so, the art on the T-structure... So, it's a section of this one and all of this one is zero. And so, the tilt of this. And it's actually... so that's quasi-coherent sheets over X. But it's just a definition of A. But this is actually the same thing as a category of quasi-coherent sheets of truncation of X. So, if you take the thought of this T-structure, you forget the geratin structure. So, you're having an enhancement of a stack. And if you take the geratin structure, you don't change the part of your geratin category, but you change the geratin category. Okay, so... I'm going to take the 20 minutes to try and explain the Bayesian formula. So, to try it on Bayesian formula, I think you can play through... and see. So... Um... So, Bayesian formula. Well, if you have a diagram like that, so we take X, Y, Z, and... So, the way that you see that is the hardest, is that you can actually take the limited preserved intersection to categories and how I should figure this out. I don't understand what you said. So, if I were to figure out what the intersection between these two T-structure are, then I would go to column X. Because I take a limit, I'd be just trying to take intersection... Yeah, yeah, yeah. Because the limit preserved on the intersection, that's how I know the two categories. Yeah, that's good. That's what makes you happy. And so, if I take something like that, so, Cartesian diagram in derived stats, we don't make these structure are, but we have, so to say, these, these, these, you. We have two ways of getting from the derived category of the algebraic category of X. So, these two ways are basically... Is this the book? I'll just have some of those cool back-up elements over here. Yeah, yeah, cool. So, what do you do with Z and......the last one. So, one way is to first......move forward and go back, basically. This one and the second one is......to show you through that one. And that's... And you have......for the information that you can use to learn that, which is just given by the adjunction. The adjunction can be......backed as this. So, what... The question is, the basic formula is, when is this transformation in coordinates? Okay? So, we're going to......to prove, to catch a clue. So, let's call the transformation alpha. And so, F, M, S, Y, Z......and......and......and the complex. Rides...... etmeüne......men. Alumnus and men. Okay, so is it the same statement as the one you derived from? Okay. You can make a very good base change. There are all skips and one with a very good base change. Oh yeah, yeah, here we don't need, that's the important part, we don't need actually flat, which we need in non-derived case. Okay, we don't need actually flat and why that? Because our category is stable, so the base change always presents the products we need. Okay, so I mean are you going to define the stable case and the term in that case? You use the same statement. Well, the difference is that the base change always preserves cofibers, okay, so they're joined. So because it's stable it also preserves spiders. That's all. I mean I think maybe the more important point is that the fiber product is derived. Yeah, yeah, yeah. So I mean if F is flat then we could just naively compute the fiber product and that would be the correct thing. If F is not flat then is that different from the classical fiber product? That's what's making this perfect. So if it would be real different, this is not what we're trying to do in this situation. So you guys have any point to say about this? Okay, so you can't say that you have to use a product that you can't use. Yeah, I'm going to give that to you. Okay. So we'll try to prove that when X and Y are fine. And then we'll see how, we'll see if not creative. But we'll test how to, to, to, to generalize from X and Y are fine too. And then you, doesn't come back crazy crazy crazy. You're right. So first X and Y are fine. You're right. So this one is F. F is FK. So because Z is fairly compact, okay, this means that I have to write down this diagram again. So Z is, uh, as the compact and well separated. So in fact, in fact, and well separated. So in the right category of the right stack, the problem is the final problem is the fine scheme. The right scheme. And what should I do with that? Well, so we have to prove that. Um, we cannot go to Z. Um, since that, this is true, okay. So that's alpha and equivalence. It's first stable by an angle in it, and then it's the same as the fine. So first case for one one, let's say, one whole point is Z. So Z is a spec C. Okay, so C B is just a spec of tensor to drive tensor five. And the right is pure. Usually fine. Okay, so Y is, um, first Y is just a, um, a spec of a particular module M, uh, C D G module. Then this thing is just a definition and, um, I don't know if you like that. So the problem with Y is that the problem with Y is that it's not a good thing. So, um, what should I say? I don't know. And then the other thing is, yeah, the other thing is this one. So it's a H A of D, which is A of D. So this is the definition of the homology. What we need to, this is just a forgetful function from D G module to D G module. So this is just M C, there's an A D G module. Yeah, yeah, yeah, yeah. Okay. Um, so, uh, yeah, this thing is the same thing. It's a homology. Well, let's say it's just, I first restrict you to the same, okay, so I can take M and I transfer it with C and then I forget that it is a B tensor C module but I just say it to the B module. Okay, so, I, yeah, it's also a B, that's why that's the problem. Yeah, yeah, yeah, that's the problem. So, I'm going to take the same thing, okay, and then I'm going to transfer it to the C module. No, no, please leave it over. B is C, okay. What is the number? What is the C? What is the C? You can't say the C. The C is over C. C is over C. What's that? Mine is M, that's M, and those two are equivalent. Okay? Yeah, okay. So, it's time to pull it back in. So, now I've proved that this kind of isomorphism is stable when I'm moving a finite number of Zs, okay? And the important thing is that the only thing you can do is understand things. This is just......the number of Zs. And so, the thing is, you can understand what this map is. Find here, and then pull this diagram. And so, this......what is that? Well, if I just E, in the diagram, the integral of the graph, the graph can go quite pretty cheap of Z. Then I have projections, only need to find, for this finite number of finds, which Z is the probability. So, I have projections in some... let's say Z-alpha, even more, but Z is the probability of......the spec of Z-alpha. I'm going to start to find it. Okay, so I have these projections, and then I can go and forget about Z-alpha in knowledge. I can go to the knowledge picture, so I get into this thing, and so for any alphas, works well in harmony. And this one, this one are compatible, and so in here, I have a diagram indexed by this same alphas, okay? So, I take the limits of this diagram, and that's my P start, and I take the limit of P. So, I use one alphas, but let's say Z is quite alphas, and I take the limit of P. So, now, by predicting this, and I want to use some data, this one will just be the limit of everything, because, in fact, again, the right-hand side product deserves the finite limits. So... Sorry, Professor, so the alphas are the morphine from... Yeah, yeah, so this is from the spec of... Sorry. Yeah. My alphas are morphine from spec of the alphas, please. Um... Yeah. When I take it, so my finite limit is here, and so the finite limit will just go out of the tensor, drag the subproduct, so it will be here, I mean, okay? So, just will still be an increment if Z is such a finite form. Okay, you just write and write that the tensor product, the right-hand product, can use the finite limit, and you get much more. Okay? I think that the whole step is to say that the full value of Z is the limit of C. Yeah, yeah. That's why it makes it more... Yeah, yeah. It's a category of zero-ass tags to the top of the... the top of us, so... um, the problem is how you do those all, so if Z is the finite form of C alpha, then ZB is the finite form of C alpha's tensor, drag tensor, we go right. Okay. In the... in the spite of the... in spite of the... Yeah, yeah, yeah. Yeah. Um, yeah, so I'm going to write that a unit of Z alpha and this one is set for the coordinate of the aspect of C alpha and so it's anything. Okay. So, you're doing that, doing that, and you're doing this ability to get that when the... you can't lose quality. Now, the question I wanted to... try and look at that is how to... to get the basis to do finite. I mean, here, you said, okay, when you had a diagram like that, and... and you also said, okay, when X, Y, and Z are the finite, this works. Now, when X and Y are the finite, and Z are the finite, then this works. So, check basis formula first. Now, seeing that X could be a finite is... it's easy because... you write that, I mean, we go into the right category of X, so here we can just test the equivalence on every... on every aspect A or X. Okay? So, seeing that X can be a finite, you ought to see that Y can be a finite. So, how... this is the part that we need... we need to see that... so, first, we write every of non-quadrant equations about every sheet of the scene, which I did also find that... I'm just naming them... write a p star, just to remember, this is not possible to compute. And... so, we said earlier that this one does not compute. Actually, in our case, it does when we do an R-close compact, it was separated, then this can be used, and that's the reason why... what we have to do, actually, is like testing whether or not, if I take E, P, U, I go there, I do that, and I have to test whether or not it's quite coherent or not. But this can be done in... on any... a finite or a Y, and then, in something like that, that future of state, you think, and then, L, R, it's there somewhere. Yeah, so if I want to... this is Y. Okay. So this is one I know it can use. So to test at this one, can use... I just have to tell you that this one can use for any state A over Y. Okay. So that's how I get to Y as a fine. Okay, and then, X as a fine, and Z as a fine. We go back and get... and I think I'm... okay. Here. APPLAUSE Questions or comments? Can we observe where the state is last step? It is... So the base change formula always works in the non-cosqlian formula, just by definition. If you write down what's the first formula of non-cosqlian formula, it doesn't say base change just by definition. So if you know that the first formula preserves cosqlian G, you actually prove the base change formula. And now what does it mean for the first formula to preserve cosqlian G? A cosqlian G is just something that flips on a fine and it has to be stable by base change. So it's actually step 1.2. It tells you that the first formula will preserve cosqlian G. Okay. I have a question. So as you said, there's a space for the polar bear from the quasi-perherent modules and some stack x into just usual OX modules. So now I said I have an OX module. How do I actually check the real amount of this quasi-perherent? Well, that's kind of what you just said. You could try to... I mean, you have the challenge like that. Okay. And you should be like... On the fine side of the whole section. Okay. So... What's really... What you need to do is just... Okay, if I take my sheet here, E, and I want to know if it's quite coherent, I take every... I can do it. We don't need Z. We can restrict to Z at 5, okay? Because it's quite coherent. If I'm only here, there's a push forward pullback. It's fine. Oh, really? Yeah. It's quite coherent. Yeah. So let's say this is Z. E is quasi-perherent. Even on a......radiified map, it's quite coherent. Ah, so that's very nice. Yeah, I should say that. That's the definition. By definition, OZ module is a sheet of module on the big side of the fine scheme to resume. It is defined on the fine scheme. Checking this quasi-perherent is just saying that the value over B is just the value over A and Z. Vibranatum. So it's just a definition. And so this gives a much more complete definition than what we said. Yeah, I mean, what's a G? For any of the points, spec A to Z, I have the day module, A of N. Any time I have a map from spec B to spec A, I have a map from E of A and Z of B to E of B. That's it. And, you know, combatable with this composition is Vism. And the sheet is quasi-perherent, if it's Vism, I'm sorry, what's the next? Okay, nice. Great. So I have a question. What if we, if we start using in DG categories, I mean, we can't believe that this conceptual saving that we can save all DG categories or a common stable alternative work category, and I'm wrong, but is it just a historical reason? I mean, the leaves, the leaves are DG categories for this? Why do you use GK as a word? Because you spoke about GK as a word. Yeah, I didn't do this. That's a historical reason, I mean. One and a half years. It's just a model. Now, the thing is, it's special. I mean, you shouldn't care if you use what was wrong, there's no reason except that, you know, it takes half a case to define what's a DG category, and it takes hundreds of pages to define what's a linear stable DG category, so that's the only reason. There's no reason to decide that. But this conceptual saving and characteristic heroes have proven that there's a head-eye on it. Well, it seems that the alternative is not used in the paper, but it's true already. The other doesn't know why characters exist in this paper. We were discussing with Marco before, but he doesn't know why. I guess, Christine, there's a need for a DG. If he doesn't see it. If he asks, I'll tell you because he doesn't know why. I don't know why.
thing as a category of quasi-coherent sheets of Z......and......and......and the complex. actually prove the base change formula. And now