Let $p:X \to S$ and $q:Y\to S$ be two objects in the category of ringed spaces over the ringed space
$S$, and let $f:X \to Y$ be a morphism over $S$.


Given a sheaf $\mathcal{F}$ of $\mathcal{O}\_Y$-modules, there are at least
two different ways to produce a morphism
$$
q\_\*\mathcal{F} \to p\_\*f^\*\mathcal{F}
$$
of sheaves of $\mathcal{O}\_S$-modules by just using canonical operations. We could for instance
apply the adjunction unit for $(f^\*, f\_\*)$ to $\mathcal{F}$, push the map forward to $S$
and then use the natural isomorphism $q\_\*f\_\* \simeq p\_\*$. That is
$$
\mathcal{F} \to f\_\*f^\*\mathcal{F}
$$
$$
q\_\*\mathcal{F} \to q\_\*f\_\*f^\*\mathcal{F}
$$
$$
q\_\*\mathcal{F} \to p\_\*f^\*\mathcal{F}.
$$
On the other hand, we could start by applying the adjunction co-unit for $(q^\*, q\_\*)$,
pull it back to $X$, use the natural isomorpism $f^\*q^\* \simeq p^\*$ and finally use the
adjuncton $(p^\*, p\_\*)$ to produce a morphism of $\mathcal{O}\_S$-modules. That is
$$
q^\*q\_\*\mathcal{F} \to \mathcal{F}
$$
$$
f^\*q^\*q\_\*\mathcal{F} \to f^\*\mathcal{F}
$$
$$
p^\*q\_\*\mathcal{F} \to f^\*\mathcal{F}
$$
$$
q\_\*\mathcal{F} \to p\_\*f^\*\mathcal{F}.
$$


In this case we get the same map (something which at least I find tedious to check;
but I might be thinking of it in the wrong way).


My question is:


Is there a general coherence result which frees us from checking such equalities case by case,
just as in the case of for instance symmetric monoidal categories. I'm thinking of something like:
Start with a commutative diagram of ringed spaces and a map of sheaves of modules over one of
the spaces. Is any map produced from this map, by just applying push-forwards and pull-backs and using
adjunctions, uniquely determined by the strings of symbols in the domain and co-domain of the new map?


(One could probably state a similar question where we throw the tensor product into the mix of canonical operations.)



Coherence for pull-backs and push-forwards
I would like to revive this old question by answering it with a pointer to my recent preprint [Obvious natural morphisms of sheaves are unique](http://arxiv.org/abs/1307.4678), which addresses precisely this problem. In particular, I give an example in the first section that discuses the kind of natural morphism you ask about. I don't yet know how to add tensor products to the mix in general, though there are examples there as well that do concern it when it enters the problem in a non-invasive kind of way.


The summary answer, by the way is: yes, such morphisms are always unique, as long as the start and end functors are both of the form $f\_\*$ or $\mathrm{id}$ (both $f^\*$ or $\mathrm{id}$ also works), or can be reduced to such by means of adjunctions and isomorphisms like $(fg)\_\* \cong f\_\* g\_\*$. If they necessarily have more then you need some conditions ensuring that certain base-change maps are isomorphisms or else there are counterexamples.


I should say that despite the intricacy of the morphisms I consider in that paper, it seems hard in practice to find a map constructed from more than a handful of units and counits and isomorphisms such as $(fg)\_\* \cong f\_\* g\_\*$, so although it is a labor-saving device, it seems so far not to save *unbounded* labor. I would like to be proven wrong on that assertion, however; I know that when you add the $!$ functors the labor really ramps up, but alas, I haven't been able to add those to the mix either.


