Let $f: X \rightarrow Y$ be a continuous map. Then a standard exercise is to show that the functors $f_*: \text{Sh}(X) \rightarrow \text{Sh}(Y)$ and $f^{-1}: \text{Sh}(Y) \rightarrow \text{Sh}(X)$ are adjoint functors, i.e. there is a natural isomorphism $$\text{Hom}_X(f^{-1}\mathcal{G},\mathcal{F}) \xrightarrow{\cong} \text{Hom}_Y(\mathcal{G},f_*\mathcal{F}).$$ To do this, one first should show the existence of natural maps $f^{-1}f_*\mathcal{F} \rightarrow \mathcal{F}$, and $\mathcal{G}\rightarrow f_*f^{-1}\mathcal{G}$. Then there is an obvious map $\Phi: \text{Hom}_X(f^{-1}\mathcal{G},\mathcal{F}) \rightarrow \text{Hom}_Y(\mathcal{G},f_*\mathcal{F})$, by applying $f_*$ to both $f^{-1}\mathcal{G}$ and $\mathcal{F}$, and then composing with $\mathcal{G}\rightarrow f_*f^{-1}\mathcal{G}$. In the other direction one gets a map $\Psi$ in a similar way. Obviously, both are natural.

Now I was able to prove that they are inverse to each other, by checking with the definitions of $f^{-1}f_*\mathcal{F} \rightarrow \mathcal{F}$ and $G \rightarrow f_*f^{-1}\mathcal{G}$, but this is a lengthy and confusing thing to do, dealing with all the colimits and sheafifications involved.

Is there an easier way to do this, using some kind of categorical argument involving the naturality, or is what I did really the way to do it?

Here's what seems to be an idea for an abstract nonsense proof.

Denote by $$f^*G (U) = \varinjlim_{V\subset Y, f(U)\subset V} G(V)$$ so that $f^{-1}G$ is $(f^*G)^a$, where $(-)^a$ denotes the sheafification.

Then $$\hom (f^{-1}G,F) = \hom((f^*G)^a,F) \cong \hom(f^*G, F) \cong \int_{U\in O(X)^{op}}\hom(f^*G(U),F(U)) = \int_{U\in O(X)^{op}} \hom(\displaystyle\varinjlim_{V\subset Y, f(U)\subset V} G(V),F(U))=\int_{U\in O(X)^{op}} \displaystyle\varprojlim_{V\subset Y, f(U)\subset V}\hom( G(V),F(U)) $$

Now let $C$ denote the category of elements of $\hom : O(X)\times O(X)^{op}\to \mathbf{Set}$, it is well known that ends on $O(X)^{op}$ in $\mathbf{Set}$ correspond to limits of functors $C\to \mathbf{Set}$ that factor through the projection $C\to O(X)\times O(X)^{op}$.

So $$\hom (f^{-1}G,F) = \lim_{(U,U',f)\in C}\displaystyle\varprojlim_{V\subset Y, f(U)\subset V}\hom( G(V),F(U')) $$.

However, $O(X)^{op}$ is a preorder category, so $C$ is equivalent to the category of pairs $(U,U')$ of opens of $X$ such that $U'\subset U$ and a there is a (unique ) morphism $(U,U') \to (V,V')$ if and only if $U\subset V, V'\subset U'$, so $V'\subset U' \subset U\subset V$. We'll still use $C$ to denote this category.

It is also well-known that limits commute (sort of Fubini's theorem) so here $$\hom (f^{-1}G,F) = \lim_{(U,U',V)\in D } \hom( G(V),F(U'))$$ where $D$ is the category of triplets $(U,U',V)\in O(X)\times O(X)\times O(Y)$ such that $U'\subset U\subset f^{-1}(V)$, and we have a (unique) morphism $(U,U',V)\to (W,W', V')$ if and only if $W'\subset U' \subset U \subset W $ and $V'\subset V$.

Thus $$\hom (f^{-1}G,F) = \lim_{(U,V)\in E}\lim_{U'\subset U, U'\in O(X)^{op}} \hom( G(V),F(U'))$$, again by Fubini's theorem for limits for the obvious $E$. But now the category of opens of $X$ included in $U$ has an initial object (when seen as a subcategory of $O(X)^{op}$), which is $U$. Thus the limit is easily computed and we get $$\hom (f^{-1}G,F) = \lim_{(U,V)\in E}\hom(G(V), F(U))$$.

We're almost there. Now $$\hom(G,f_*F) = \int_{V\in O(Y)^{op}}\hom(G(V), F(f^{-1}(V))) = \lim_{(V,V')\in Z} \hom(G(V), F(f^{-1}(V')))$$ where $Z$ is the category of pairs $(V,V')$ of opens of $Y$ which plays the same role as $C$ for $X$.

Now $V'\subset V \implies f^{-1}(V') \subset f^{-1}(V)$ which means we have a functor $R:Z\to E$ which sends $(V,V') \mapsto (f^{-1}(V'), V)$. We're now left with comparing $\lim_Z \hom\circ (G, F)\circ R$ and $\lim_E \hom\circ (G,F)$. Now te functor $R$ is be nice enough that these are the same (e.g. has coinitial image in this preorder category).

Now these are the same and all the isomorphisms are clearly natural in $G,F$, so that $\hom(f^{-1}G,F)\cong \hom(G,f_*F)$ and so $f^{-1}\dashv f_*$

  • Looks like I don't know enough category theory to understand this :D Do you have any reference where I can read up, e.g. what $\int$ means in this context? – red_trumpet Dec 7 at 18:25
  • $\int$ means (depending on where the index is) end or coend; they're a special kind of (co)limit. I've learned them while browsing through the nLab and random discussions on the internet so I don't really have any references – Max Dec 7 at 20:34

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

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