Assume $C$ is a $1$-category and is $M : C \to Cat$ a $1$-functor, such that for every morphism $f : x \to y$, the functor $M(f) : M(x) \to M(y)$, which is denoted by $f\_\\*$, has a left-adjoint functor, which is denoted by $f^\* : M(y) \to M(x)$. Now I expect that $x \to M(x), f \mapsto f^\*$ defines a contravariant pseudo-functor $M \to Cat$. For example, it should be possible to conclude by this method that $X \to \text{Mod}\_X, f \mapsto f^\*$ is a pseudo-functor on the category of ringed spaces.


Now here is what I have tried: Using the units and counits of the adjunction, we get morphisms


$\epsilon\_x : (id\_x)^\* = (id\_x)^\* (id\_x)\_\* \to id\_{M(x)}$,


$\alpha\_{f,g} : f^\* g^\* \to f^\* g^\* (gf)\_\* (gf)^\* = f^\* g^\* g\_\* f\_\* (gf)^\* \to f^\* f\_\* (gf)^\* \to (gf)^\*$,


and these turn out to be isomorphisms. I have already done some calculations resp. diagram chases (which you probably don't want to see), but the necessary compatibility conditions ([Vistoli's Descent notes](http://homepage.sns.it/vistoli/descent.pdf), Def. 3.10) don't seem to be true automatically. Or am I missing something?


If not: Where can I find a nice exposition when these adjoint functors $f^\*$ may be chosen in such a way that we get a pseudo-functor? Or at least, why does everything works out fine in the example with modules, without tedious calculations?



Pointwise left-adjoint yields a pseudo-functor
Here is a way of proving the following (condition b) of Vistoli in your question). I'm assuming condition a) to be strict (the $id$ is the $id$) in here for the moment.


Let us state condition b)



> 
> Consider a pseudo-functor $(-)\_\*$ with structural morphisms in the direction $(fg)\_\* \to f\_\* g\_\*$, >that is associative, meaning that
> 
> 
> $$\begin{matrix} (fgh)\_\* & \to & f\_\* (gh)\_\* \\ 
> > \downarrow & & \downarrow \\ 
> > (fg)\_\* h\_\* & \to & f\_\* g\_\* h\_\* \end{matrix}$$
> 
> 
> commutes. Then if $f\_\*$ has a left adjoint $f^\*$ for any $f$, there is a unique way of defining a >pseudo-functor structure $(-)^\*$ such that $(f)^\*=f^\*$ and such that the two following diagrams >commute:
> 
> 
> $$\begin{matrix}
> >Id & \to & (gf)\_\*(gf)^\* \\
> >\downarrow & & \downarrow \\
> >g\_\* f\_\* f^\* g^\* & \to & g\_\*f\_\* (gf)^\*
> >\end{matrix}
> >$$ and 
> 
> 
> $$\begin{matrix}
> >f^\*g^\*(gf)\_\* & \to & f^\*g^\* g\_\* f\_\* \\
> >\downarrow & & \downarrow \\
> >(gf)^\* (gf)\_\* & \to & Id
> >\end{matrix}
> >$$
> 
> 
> Furthermore, if $(-)\_\*$ is associative, meaning the following diagram is commutative:
> 
> 
> $$\begin{matrix}
> >(fgh)\_\* & \to & (fg)\_\*h\_\* \\
> >\downarrow & & \downarrow \\
> >f\_\*(gh)\_\* & \to & f\_\*g\_\*h\_\*
> >\end{matrix}
> >$$
> 
> 
> Then $(-)^\*$ will be associative too, meaning the following diagram is commutative:
> 
> 
> $$\begin{matrix}
> >h^\*g^\*f^\* & \to & (gh)^\*f^\* \\
> >\downarrow & & \downarrow \\
> >h^\*(fg)^\* & \to & (fgh)^\*
> >\end{matrix}
> >$$
> 
> 
> 


Ok, so now, let's prove this statement.


Now here is a way of looking at this type of diagrams obtained from adjunctions.


From now on, some diagonal arrows will appear in the diagrams. They are the natural transformations you are asking about such as $(fg)\_\* \to f\_\*g\_\*$ (or 2-maps, if you want), whereas the other arrows (vertical and horizontal) in the diagrams will simply be usual functors (or 1-maps), such as $f\_\*$.


First of all, you have to convince yourself that when you put together a bunch of diagrams to make a bigger one, the order in which you apply the 2-maps (when you have a choice) is irrelevant, because they act on 1-maps that are in different parts of the formulas giving objects. For example, in the following diagram,
$$\begin{matrix}
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet
\end{matrix}
$$
you have to apply first the lower left 2-map, but then you can do either the lower right or the middle left, and so on. It does not matter, in the end, you get the same 2-map from the lower left part of the perimeter to the upper right one. 


All this had nothing to do with adjunctions, so far. Now let's start using them.


You have to convince yourself of the following lemma (see <http://arxiv.org/abs/0806.0569>, lemma 1.2.6).


Given a diagram
$$\begin{matrix}
a & \to & a' \\
\downarrow & \nearrow & \downarrow \\
b & \to & b'
\end{matrix}
$$


such that the vertical maps have left adjoints. Then, there is a diagram


$$\begin{matrix}
a & \to & a' \\
\uparrow & \nwarrow & \uparrow \\
b & \to & b'
\end{matrix}
$$


such that two diagrams commute (see loc. cit.). The point is the existence of the 2-map given by the diagonal arrow, and it is defined exactly as you do it in your question, in the particular example you are considering.


**Whenever this lemma will be applied in what follows, the maps with adjoints will be vertical. The horizontal maps will never be changed.**


Now convince yourself of the following fact: if I put two such diagrams beside each other


$$\begin{matrix}
\bullet & \to & \bullet & \to & \bullet \\
\downarrow & \nearrow & \downarrow & \nearrow & \downarrow \\
\bullet & \to & \bullet & \to & \bullet
\end{matrix}
$$


you can compose the 2-maps. Now assuming all vertical maps have left adjoints, you can apply the lemma to both squares and then compose the resulting 2-maps. You can also apply the lemma to the rectangle (forgetting the middle vertical arrow) with the composition as diagonal 2-map. The point is that you get the same result.


You can prove a similar lemma, but with two squares stacked one above each other instead of besides each other.


Finally, here is the relevant point. It helps you getting some commutative diagrams involving left adjoints from commutative diagrams involving right adjoints. I'm not going to draw this here, because it is too painful, but look at Lemma 1.2.7 p. 8 of [loc. cit.](http://arxiv.org/pdf/0806.0569v1). It is stated in the opposite way of what you want: it starts with left adjoints, and says something about right ones. But you can reverse the statement (so start with the second cube and get the first one). The proof of this lemma is left as an exercise in the reference, but the point is that you have no choice as to how to prove it, because it is very general. Write both compositions, and try to complete with smaller commutative diagrams, you'll have no choice as to which diagram to add at each step.


Now, at last, let's apply all this to your own problem. Here is a file with a description of the cube you have to use.


<http://www.math.uni-bielefeld.de/~bcalmes/fichiers/diagram.pdf>


The 2-maps appear as double arrows. I've called $\xi$ and $\zeta$ the pseudo-functor morphisms you are interested in. All unlabeled single arrows are identities. Unlabeled double arrows are either identity or some obvious adjunction morphism (unit or counit).


You have to check that the 2-maps you are getting by adjunction are what you want, but it is completely straightforward.


Now, for condition a) of Vistoli on identities, you can use the same kind of tricks.


