So I've been reading about derived categories recently (mostly via Hartshorne's *Residues and Duality* and some online notes), and while talking with some other people, I've realized that I'm finding it difficult to describe what a "triangle" is (as well as some other confusions, to be described below).


Let $\mathcal{A}$ be an abelian category, and let $K(\mathcal{A})$ and $D(\mathcal{A})$ be its homotopy category and derived categories respectively.


By definition, in either $K(\mathcal{A})$ or $D(\mathcal{A})$,


(1) A triangle is a diagram $X\rightarrow Y\rightarrow Z\rightarrow X[1]$ which is isomorphic to a diagram of the form
$$X\stackrel{f}{\rightarrow}Y\rightarrow Cone(f)\rightarrow X[1]$$


This is the definition, though I don't really understand the motivation.


Somewhat more helpful for me, is the definition:


(2) A cohomological functor from a triangulated category $\mathcal{C}$ to an abelian category $\mathcal{A}$ is an additive functor which takes triangles to long exact sequences.


Since taking cohomology of a complex (in either $K(\mathcal{A})$ or $D(\mathcal{A})$) is a cohomological functor, this definition tells me that I should think of a triangle as being like "a short exact sequence" (in the sense that classically, taking cohomology of a short exact sequence results in a long exact sequence). This idea is also supported by the fact:


(3) If $0\rightarrow X^\bullet\rightarrow Y^\bullet\rightarrow Z^\bullet\rightarrow 0$ is an exact sequence of chain complexes, then there is a natural map $Z^\bullet\rightarrow X[1]^\bullet$ in the derived category $D(\mathcal{A})$ making $X^\bullet\rightarrow Y^\bullet\rightarrow Z^\bullet\rightarrow X[1]^\bullet$ into a triangle (in $D(\mathcal{A})$).


This leads to my first precise **question**: Can there exist triangles in $D(\mathcal{A})$ which don't come from exact sequences? If so, is there a characterization of them? Is (3) false in the homotopy category $K(\mathcal{A})$? (certainly the same proof doesn't work).


I sort of expect that the answers to the first and last questions above to both be "Yes", which makes the comparison between triangles and "exact sequences" a bit weird. Of course, $K(\mathcal{A})$ and $D(\mathcal{A})$ are almost never abelian categories, and so it's "weird" to talk about exact sequences there.


I suppose at a fundamental level, I find the "homotopy category" somewhat mysterious. I don't completely understand it's role in the construction of the derived category, since after all homotopy equivalences *are* quasi-isomorphisms. I also find it difficult to internalize this notion of "homotopic morphisms of chain complexes". To me, it's just a "technical trick" which allows one to do all this magic with mapping cones which allows $K(\mathcal{A})$ to be a triangulated category, whereas the normal abelian category of chain complexes is not. I can prove things with it, but whenever I do, I sort of feel unsettled - as if I'm playing with something 'magical' that could, at any moment, turn on me unexpectedly.


In addition to my specific questions above, I suppose I was hoping that someone would be able to articulate in a nice way how one should think of triangles, why this notion of a triangulated category is so successful, and hopefully alleviate some of my unsettlement regarding homotopy.



What is a triangle?
To answer your first precise questions:


* Yes, every distinguished triangle in $D(A)$ comes from a short exact sequence. For every distinguished triangle $X \to Y \to \mathrm{Cone}(f) \stackrel{+1}\to $ there is a short exact sequence 
$$ 0 \to Y \to \mathrm{Cone}(f) \to X[-1] \to 0$$
of complexes in $A$, and our distinguished triangle arises from this one by rotation.
* By the same argument, every distinguished triangle in $K(A)$ comes from a short exact sequence (at least up to a rotation). However, *not* every short exact sequence gives rise to a distinguished triangle in $K(A)$. If 
$$ 0 \to X \stackrel f \to Y \to Z \to 0$$ is a short exact sequence of complexes in $A$, then there is a natural map $\mathrm{Cone}(f) \to Z$ which in general is only a quasi-isomorphism, not a homotopy equivalence. If for example the short exact sequence splits degree-wise, then it is always a homotopy equivalence, and we get a triangle in $K(A)$.


Alright. About your question "Why $K(A)$?". You are right that homotopy equivalences are quasi-isomorphisms. So in principle one could construct $D(A)$ in "one step" by taking the category of complexes and inverting quasi-isomorphisms. But there are some technical reasons for preferring the construction via the category $K(A)$. First off, the category $K(A)$ is already triangulated, and this is easy to prove. This means that to construct $D(A)$ we are in the situation of *Verdier localization*: we have a triangulated category, and we localize it at the class of morphisms whose cone is in a specific thick triangulated subcategory. In particular $D(A)$ becomes triangulated. In general, localizations of categories can be complicated, and in the "one-step" construction it is not even obvious that $D(A)$ is a category, i.e. that the morphisms from one object to another form a set. 


To motivate why we should care about triangles at all, note that it doesn't make sense to talk about kernels or images of morphisms in $D(A)$, so that we can't talk about exactness. Given that we want to be able to do homological algebra we need some sort of substitute for this. Triangles, encoding short exact sequences, turn out to be enough to develop much of the theory, and a posteriori one could want to axiomatize the theory only in terms of triangles. One reason we could expect the notion of a "distinguished triangle" to really be intrinsic to $D(A)$ (even though the class of distinguished triangles need to be specified in the axioms for a triangulated category) is that any triangle 
$$ X \to Y \to Z \stackrel{+1}\to$$
gives rise to a long sequence of abelian groups after applying $[W,-]$ for any object $W$; this long sequence will be a long exact sequence when the triangle is distinguished, for any $W$, and this is really a very special property! A remark is that nowadays people will tell you that "stable $\infty$-categories" are for all purposes better than triangulated categories, and in a stable $\infty$-category, the class of distinguished triangles does *not* need to be specified in advance, so to speak: equivalent stable $\infty$-categories will have the same distinguished triangles.


