# Functoriality in $f$ of the inverse image $f^{-1}\mathscr{F}$ presheaf

Let $$f:X\to Y$$ and $$g:Y\to Z$$ be continuous maps between topological spaces and $$\mathscr{H}$$ be a presheaf on $$Z$$. For now, lets define the presheaf inverse image $$g^{-1}\mathscr{H}$$ as $$\Gamma(V,g^{-1}\mathscr{H}):=\operatorname*{colim}_{g(V)\subset W}\mathscr{H}(W).$$

I want to prove $$f^{-1}\circ g^{-1}=(g\circ f)^{-1}$$ as functors. In other words, I want to show that $$f^{-1}(g^{-1}\mathscr{H})=(g\circ f)^{-1}\mathscr{H}$$ and that, if $$\varphi:\mathscr{H}_1\to\mathscr{H}_2$$ is a morphism of presheaves, $$f^{-1}(g^{-1}\varphi)=(g\circ f)^{-1}\varphi$$.

As for the first statement, it boils down to proving that $$\operatorname*{colim}_{g(f(U))\subset W}\mathscr{H}(W)=\operatorname*{colim}_{f(U)\subset V}\operatorname*{colim}_{g(V)\subset W}\mathscr{H}(W).$$ I think this follows from the fact that an open set $$W\subset Z$$ contains $$g(f(U))$$ if and only if it contains a subset of the form $$g(V)$$, where $$V\subset Y$$ is an open set containing $$f(U)$$. But it is not clear to me.

As for the second statement, I have no idea about how to prove it.

• There is actually a very simple proof using Yoneda lemma and the fact that $(g\circ f)_*=g_*\circ f_*$. Are you interested or do you prefer to stick with the colimit manipulations ? – Roland Oct 22 at 21:45
• @Roland: do you mean using adjunction? I thought it worked for sheaves only... – Mindlack Oct 22 at 21:55
• @Mindlack Yes using adjunction, and it also work for presheaves since $f^{-1}$ for presheaves (that is, without sheafification) is the left adjoint of $f_*$ seen as a functor from presheaves to presheaves. The same proof carries over. – Roland Oct 22 at 22:00
• @Roland, I was primarily interested in colimit manipulations but I am surely curious about your proof! Please write about it! – Gabriel Oct 22 at 22:13

## 3 Answers

As you correctly pointed out, the first part is indeed a purely topological problem – given $$W \supset (g \circ f)(U)$$, $$V=g^{-1}(W)$$ should work. So this means that your colimits are essentially over the same set of indices so they’re canonically equal.

For the second part, you should just note that if $$\phi: \mathcal{F} \rightarrow \mathcal{G}$$, then $$(g^{-1}\phi)(V) : (g^{-1}\mathcal{F})(V) \rightarrow (g^{-1}\mathcal{G})(V)$$ is the colimit of the $$\phi(W): \mathcal{F}(W) \rightarrow \mathcal{G}(W)$$ over the $$W \supset g(V)$$. Then it really is (admittedly painful) abstract nonsense.

This is another proof of the isomorphism $$(g\circ f)^{-1}=f^{-1}\circ g^{-1}$$, hence not really answer of the original question, but this was requested in the comments.

You have an adjunction $$\operatorname{Hom}_{PSh(X)}(f^{-1}\mathcal{F},\mathcal{G})=\operatorname{Hom}_{PSh(Y)}(\mathcal{F},f_*\mathcal{G})$$ and an isomorphism of functor (in fact a true equality here) $$(g\circ f)_*=g_*\circ f_*$$. Using this, we have : \begin{align*} \operatorname{Hom}_{PSh(X)}((g\circ f)^{-1}\mathcal{F},\mathcal{G})&=\operatorname{Hom}_{PSh(Z)}(\mathcal{F},(g\circ f)_*\mathcal{G})\\ &=\operatorname{Hom}_{PSh(Z)}(\mathcal{F},g_*f_*\mathcal{G})\\ &=\operatorname{Hom}_{PSh(Y)}(g^{-1}\mathcal{F},g_*f_*\mathcal{G})\\ &=\operatorname{Hom}_{PSh(X)}(f^{-1}g^{-1}\mathcal{F},\mathcal{G})\\ \end{align*} This is natural in $$\mathcal{F}$$ and $$\mathcal{G}$$, hence by Yoneda, we have an isomorphism $$(g\circ f)^{-1}\mathcal{F}\simeq f^{-1}g^{-1}\mathcal{F}$$. Since this is natural in $$\mathcal{F}$$, we have an isomorphism of functors $$(g\circ f)^{-1}\simeq f^{-1}g^{-1}$$.

• Is there a proof for this adjunction that uses less abstract nonsense than the proof of the claim? I remember that the proof of the adjunction is rather technical in the case of sheaves – but perhaps I haven’t found the right proof, or it’s better for presheaves? – Mindlack Oct 23 at 9:07
• @Mindlack a first observation is that, since the sheafification functor is a left-adjoint of the inclusion functor, the statement about sheaves follows from the one about presheaves. Now, there are natural maps between the two sides of the adjunction. I think perhaps it is boring to show that they are inverses of eachother? I'll try to do it in detail and then I'll get back to you! – Gabriel Oct 23 at 9:19
• Perhaps using the presheaf case is indeed simpler, but I really remember a lot of abstract nonsense (more, and of a harder sort than the question you were asking). Perhaps I was doing it wrong, though. – Mindlack Oct 23 at 9:26
• @Mindlack Well, I prefer this proof for two reasons : the adjunction property also has colimits manipulation, but a single colimit instead of two, so it is easier in my opinion. But also, and mostly, because the adjunction property is really an important property (you can derived the adjunction for sheaves, but this also has very important implications in the case of presheaves), so it has to be done anyway, the composition quickly follows, as shown. – Roland Oct 23 at 9:50
• @Mindlack I wrote an answer with my proof of the adjunction below. – Gabriel Oct 23 at 13:34

This is not really an answer to my question, which has already great answers but my attempt to proving the adjunction in the case of presheaves, as discussed in the comments.