MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.)

share|cite|improve this question
up vote 3 down vote accepted

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, 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.

share|cite|improve this answer
Yes this is a very nice paper! For the tensor case one knows at least for morphisms between geometric stacks that every isomorphism $f^* \cong g^*$ of tensor functors comes from a (unique) isomorphism $f \cong g$ (i.e. $f=g$ in the scheme case), see Lurie's Tannaka duality for geometric stacks (in my thesis there will be a simpler proof, though). – Martin Brandenburg Sep 20 '13 at 23:56
That result of Lurie's is an interesting counterpoint to mine. Clearly it is false that there is only one automorphism of functors of $f^*$, unless (as in my paper) you assume the automorphism is of a very restricted class. This alternative restriction (of being a tensor functor) seems like a different direction than mine because it explicitly refers to quasicoherent sheaves; i.e. it is rather more geometric. That's usually a good thing, of course! But I was hoping to get a tensor-related theorem that applies to, say, perverse sheaves. – Ryan Reich Sep 24 '13 at 3:15
Also, I want to thank you for reading and apparently enjoying my paper. – Ryan Reich Sep 24 '13 at 4:43

The isomorphism $f^* q^* \cong p^*$ induced by $q_* f_* \cong p_*$ is defined to be the composition

$f^* q^* \to f^* q^* p_* p^* \cong f^* q^* q_* f_* p^* \to f^* f_* p^* \to p^*$.

Now it is easily checked that the two definitions of the morphism agree. In fact, the first one is more natural, whereas the second one is somehow artificial since it coindices with the first one plus an expansion of the isomorphism $f^* q^* \cong p^*$, which is then removed again.

More generally, if $p: X \to S,~ p' : X' \to S,~ q' : X' \to X, ~q : S' \to S$ are morphisms (in your case $p=id$) such that the obvious diagram commutes, then there is a natural transformation $p'^* q_* \to q_* p'^*$. It is studied, for example, in EGA I, 9.3. Remark that we only use here that we have a weakly commutative diagramm of categories and functors such that the vertical functors $p_*,p'_*$ have left adjoints. See this paper by Calmes, Hornbostel (Lemma 1.2.6) for a generalization to categories.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.