# Canonicity of the push-pull map for sheaves of sets

This question is Vakil 2.7.F (in the December 2022 version of the notes).

Suppose we have the following commutative diagram of topological spaces:

$$\require{AMScd} \begin{CD} W @>{\beta'}>> X \\ @V{\alpha'}VV @VV{\alpha}V \\ Y @>>{\beta}> Z \end{CD}$$

Suppose also we have a sheaf $$\mathscr{F}$$ on $$X$$. Then, we define two maps from $$\operatorname{hom}(\mathscr{F}, \mathscr{F})$$ to $$\operatorname{hom}(\beta^{-1}\alpha_*\mathscr{F}, \alpha'_*\beta'^{-1}\mathscr{F})$$. Given $$f: \mathscr{F} \to \mathscr{F}$$, we get $$\beta'^{-1}f: \beta'^{-1}\mathscr{F} \to \beta'^{-1}\mathscr{F}$$. We may then take the transpose of this map, apply $$\alpha_*$$, apply commutativity, and finally take the transpose again to get the desired map (see Vakil for more details). Dually, you may also apply $$\alpha_*$$ first to get a second map. Call these maps $$h_1, h_2$$.

Vakil 2.7.F asks us to show that $$h_1(\operatorname{id})=h_2(\operatorname{id})$$. You could obviously do this by passing to components and keeping track of all the functor applications and transposes, but this is super messy. Is there a nice, abstract nonsense proof of this fact?

Here is an abstract nonsense proof, based on the observation that we are constructing a natural transformation of functors from $$\beta^{-1}\alpha_*$$ to $$\alpha’_*\beta’^{-1}$$, usually called the base change homomorphism. The component of this natural transformation for $$\mathscr{F}$$ is exactly the image of the identity under either of the $$h_i$$.

Build the natural transformation as the composition of: $$\beta^{-1}\alpha_* \rightarrow \beta^{-1}\alpha_*\beta’_*\beta’^{-1}\rightarrow \beta^{-1}\beta_*\alpha’_*\beta’^{-1}\rightarrow \alpha’_*\beta’^{-1}.$$

Here the first map is the unit, middle is the commutativity isomorphism, and last is counit map. Then to see that the component on $$\mathscr{F}$$ is either of the maps you described applied to $$\text{Id}$$, one needs to sit and stare at this diagram for a while.

I’ll do one half of this, viewing things as natural transformations of functors, imagine we are building this transformation up via pieces. Start with the identity natural transformation of the identity, apply $$\beta’^{-1}$$ then take mates. This gives the unit for $$\beta’$$, then apply $$\alpha_*$$, and apply the commutativity iso. Then apply $$\beta^{-1}$$, and take mates again to compose with the final counit. This shows your first map applied to the identity is the component of this natural transformation. For the other one, interpret the other map, and get to the same composition, showing that these are equal.

• I have tried working through this, and I am afraid I don't quite understand. I see why the component of the natural transform you give is $h_1(\operatorname{id})$, but I don't see why it should be $h_2(\operatorname{id})$. The construction of $h_2$ goes through a completely different commutativity isomorphism, namely $\beta'^{-1}\alpha^{-1} \to \alpha'^{-1}\beta^{-1}$ and I don't see any way to connect this to your construction. Commented Jan 4, 2023 at 6:57
• Sorry you are absolutely right, this lets us see that our first map is the component of this “base change morphism”, but the second one is the component of a (a priori) different base change morphism, which starts with the adjoint natural transformation of the other one. Then, you need to show that these two base change maps are actually the same, which is straightforward (but long) by writing out the definition of the adjoint natural transformation and expanding. This is a large diagram but only uses the definitions and naturality, and is categorical, but I think onriv’s approach is better. Commented Jan 8, 2023 at 9:54
• If you’re still looking to prove that these two base changes agree and are struggling, I can try help further. This proof is slick but not enlightening in my view though. Commented Jan 8, 2023 at 9:55

This is equivalent to showing that the functoral morphism $$\beta^{-1}\alpha_*\to\alpha_*'(\alpha')^{-1}\beta^{-1}\alpha_*=\alpha_*'(\beta')^{-1}\alpha^{-1}\alpha_*\to\alpha_*'(\beta')^{-1}$$ is equal to the functoral morphism $$\beta^{-1}\alpha_*\to\beta^{-1}\alpha_*\beta_*'(\beta')^{-1}=\beta^{-1}\beta_*\alpha_*'(\beta')^{-1}\to\alpha_*'(\beta')^{-1}$$. This follows from the fact that the diagram of sheaf functors above commutes, where each square/parallelogram is the naturality of a unit or counit, and the triangles are the definition of unit-counit conjunction.

Here is an argument from the cases of sets to presheaves to sheaves. It should not be counted as category-theoretic proof. But I believe it catches some essential part of a category-theoretic proof. A generalized statement in category theory may contain some more construction like locale or other structures. (An reference may be Borceus's Handbook of Categorical Algebra Vol 3, Categories on Sheaves Section 2.12. It's named Change of Base, as ChirsH hinted.)

First observe that one has a similar setting: given a map $$f: A \to B$$ of sets. The map $$f$$ induces direct image and inverse image functors $$f_∗: PA → PB$$ and $$f^{−1}: PB → PA$$. Here $$PA$$ and $$PB$$ are the power sets of $$A$$ and $$B$$ respectively, treated as posets. $$f_*$$ takes any subset $$S$$ of $$A$$ to $$f(S)$$ of $$B$$ and $$f^{-1}$$ takes any subset $$T$$ of $$B$$ of $$f^{-1}(T)$$ of $$A$$. Then $$(f_*, f^{-1})$$ forms an adjoint pair: $$f(S) \subset T$$ if and only if $$S \subset f^{-1}(T)$$. (cf. Riehl’s Category in Context Example 4.18).

With this similar setting, one can perform Vakil’s definition of the push-pull map similarly: Given a commutative diagram in the category sets as follow and $$F$$ is a subset of $$X$$.

$$\begin{CD} W @>{\beta'}>> X \\ @V{\alpha'}VV @VV{\alpha}V \\ Y @>>{\beta}> Z \end{CD}$$

one can define the push-pull map $$\alpha^\prime_*(\beta^\prime)^{-1} F \to \beta^{-1}\alpha_*F$$ in two ways(here the map is in fact just the inclusion of two subsets of $$Y$$)

1. as the definition in 2.7.4 of Vakil’s FOAG, starting with the identity $$(\beta^\prime)^{-1}F \to (\beta^\prime)^{-1}F$$ on $$W$$, which is just the identity map on the inverse image $$(\beta^\prime)^{-1}F$$. By adjointness of $$(\beta^\prime_*, (\beta^\prime)^{-1}))$$, this is the map as the map $$\beta^\prime_*(\beta^\prime)^{-1} F \to F$$, which in fact says that $$\beta^\prime_*(\beta^\prime)^{-1} F \subset F$$. Apply $$\alpha_*$$ to get a map $$\alpha_*\beta^\prime_*(\beta^\prime)^{-1} F \to \alpha_*F$$. By the commutative diagram, this is $$\beta_*\alpha^\prime_*(\beta^\prime)^{-1} F \to \alpha_*F$$. By adjointness of $$(\beta_*, (\beta)^{-1}))$$, this yields a map $$\alpha^\prime_*(\beta^\prime)^{-1} F \to \beta^{-1}\alpha_*F$$. In fact, it’s just that $$\alpha^\prime_*(\beta^\prime)^{-1} F \subset \beta^{-1}\alpha_*F$$.

2. as in exercise 2.7.F, the dual way starts with $$\alpha_*F \to \alpha_*F$$, and gets $$F \to \alpha^{-1}\alpha_{*}F$$ using the adjointness of $$(\alpha_*, \alpha^{-1})$$, and gets $$(\beta^\prime)^{-1}F \to (\beta^\prime)^{-1}\alpha^{-1}\alpha^{*}F$$. By the commutative diagram, it’s $$(\beta^\prime)^{-1}F \to (\alpha^\prime)^{-1}\beta^{-1}\alpha_{*}F$$. Using the adjointness of $$(\alpha^\prime_*, (\alpha^\prime)^{-1})$$, it’s $$\alpha^\prime_*(\beta^\prime)^{-1}F \to \beta^{-1}\alpha_{*}F$$.

For posets, there is at most one morphism between two objects. Hence for this similar setting, the two construction is the same.

Now move to the case of the category of presheaves. For a continuous map $$\pi: X\to Y$$, one still has that $$(\pi^{-1}_{pre}, \pi_*)$$ forms an adjoint pair. Here for a presheaf $$\mathscr{G}$$ of $$Y$$, the presheaf $$\pi^{-1}_{pre}\mathscr{G}$$ is defined by $$\pi_{\text {pre }}^{-1} \mathscr{G}(\mathrm{U})=\operatorname{colim}_{V \supset \pi(\mathrm{U})} \mathscr{G}(\mathrm{V})$$ for any open set $$U$$ of $$X$$. (cf 2.7.2 in Vakil’s FOAG).

Perform two definitions of the push-pull map in the case of the category of presheaves. There are two definitions for the map $$\beta^{-1} \alpha_{*} \mathscr{F} \longrightarrow \alpha_{*}^{\prime}\left(\beta^{\prime}\right)^{-1} \mathscr{F}$$, i.e., two definitions for the map $$(\beta^{-1} \alpha_{*} \mathscr{F})(U) \to(\alpha_{*}^{\prime}\left(\beta^{\prime}\right)^{-1} \mathscr{F})(U)$$ for any open set $$U$$ on $$Y$$. By definition, $$(\beta^{-1} \alpha_{*} \mathscr{F})(U) = \operatorname{colim}_{V \supset \beta(U)} (\alpha_{*}\mathscr{F})(U) = \operatorname{colim}_{V \supset \beta(U)} \mathscr{F}(\alpha^{-1}(V))$$ and $$(\alpha_{*}^{\prime}\left(\beta^{\prime}\right)^{-1} \mathscr{F})(U) = (\left(\beta^{\prime}\right)^{-1} \mathscr{F})((\alpha^\prime)^{-1}(U)) = \operatorname{colim}_{V \supset \beta^{\prime}((\alpha^\prime)^{-1}(U))} \mathscr{F}(V)$$. Hence there are two definitions for the map $$\operatorname{colim}_{V \supset \beta(U)} \mathscr{F}(\alpha^{-1}(V)) \to \operatorname{colim}_{V \supset \beta^{\prime}((\alpha^\prime)^{-1}(U))} \mathscr{F}(V)$$. Since $$V \supset \beta(U)$$ if and only if $$\alpha^{-1}(V) \supset \alpha^{-1}(\beta(U))$$, $$\operatorname{colim}_{V \supset \beta(U)} \mathscr{F}(\alpha^{-1}(V))$$ is the same as $$\operatorname{colim}_{V \supset \alpha^{-1}(\beta(U))} \mathscr{F}(V)$$ . By the previous argument on the setting of sets one knows that $$\beta^{\prime}((\alpha^\prime)^{-1}(U)) \subset \alpha^{-1}(\beta(U))$$. Hence $$\{V \supset \alpha^{-1}(\beta(U))\}$$ is a sub-filtered set of $$\{V \supset \beta^{\prime}((\alpha^\prime)^{-1}(U))\}$$ . Hence there is a unique map $$\operatorname{colim}_{V \supset \alpha^{-1}(\beta(U))} \mathscr{F}(V) \to \operatorname{colim}_{V \supset \beta^{\prime}((\alpha^\prime)^{-1}(U))} \mathscr{F}(V)$$ making the two colimit cones commute. Since both two definitions make it commute, they are the same. The details are:

Back to how $$(\pi^{-1}_{pre}, \pi)$$ being an adjoint pair, i.e., given $$\pi: X \to Y$$, the isomorphism between $$PSh_X(\pi_{pre}^{-1}\mathscr{G}, \mathscr{F})$$ and $$PSh_Y(\mathscr{G}, \pi_* \mathscr{F})$$:

Given morphism of presheaves $$\pi_{pre}^{-1}\mathscr{G} \to\mathscr{F}$$, one has $$\operatorname{colim}_{V \supset \pi(U)} \mathscr{G}(V) \to \mathscr{F}(U)$$ for every open subset $$U \subset X$$. Replacing it with $$\pi^{-1}(U)$$ for every open subset $$U \subset Y$$, one has $$\operatorname{colim}_{V \supset \pi(\pi^{-1}(U))} \mathscr{G}(V) \to \mathscr{F}(\pi^{-1}(U))$$. Since $$U \supset \pi(\pi^{-1}(U))$$, one has the colimit component $$\mathscr{G}(U) \to \operatorname{colim}_{V \supset \pi(\pi^{-1}(U))} \mathscr{G}(V)$$. This two morphisms compose to $$\mathscr{G}(U) \to \operatorname{colim}_{V \supset \pi(\pi^{-1}(U))} \mathscr{G}(V)\to \mathscr{F}(\pi^{-1}(U))$$ for every open subset $$U \subset Y$$. Hence a morphism of presheaves $$\pi_{pre}^{-1}\mathscr{G} \to\mathscr{F}$$ corresponds to a morphism of presheaves $$\mathscr{G} \to \pi_*\mathscr{F}$$.

Conversely, given a morphism of presheaves $$\mathscr{G} \to \pi_*\mathscr{F}$$, one has $$\mathscr{G}(U) \to \mathscr{F}(\pi^{-1}(U))$$ for every open subset $$U \subset Y$$. Hence, for the filtered set $$\{V \supset \pi(U)\}$$, here $$U$$ switches to the notation for an open subset $$U \subset X$$, one has $$\{\mathscr{G}(V) \to \mathscr{F}(\pi^{-1}(V))\}_{V \supset \pi(U)}$$. Since $$\pi^{-1}(V) \supset \pi(U)$$ for every $$V \supset \pi(U)$$, the restriction of sections $$\operatorname{res}: \mathscr{F}(\pi^{-1}(V)) \to \mathscr{F}(U)$$ gives a cone: $$\{\mathscr{G}(V) \to \mathscr{F}(\pi^{-1}(V))\stackrel{\operatorname{res}}{\rightarrow}\mathscr{F}(U)\}_{V \supset \pi(U)}$$. It corresponds to a morphism of $$\operatorname{colim}_{V \supset \pi(U)} \mathscr{G}(V) \to \mathscr{F}(U)$$. Hence a morphism of presheaves $$\mathscr{G} \to \pi_*\mathscr{F}$$ corresponds to a morphism of presheaves $$\pi_{pre}^{-1}\mathscr{G} \to\mathscr{F}$$.

Then for the case in the definition of the push-push map, the identity $$\beta^{\prime,-1} \mathscr{F} \to\beta^{\prime,-1} \mathscr{F}$$ corresponds to the map $$\mathscr{F} \to \beta^{\prime}_*\beta^{\prime, -1}\mathscr{F}$$ defined by $$\mathscr{F}(U) \to \operatorname{colim}_{V \supset(\beta^{\prime}(\beta^{\prime,-1}(U)))}\mathscr{F}(V) \to (\beta^{\prime, -1}\mathscr{F})(\beta^{\prime,-1}(U))$$. Here the second map is just identity, hence it is $$\mathscr{F}(U) \to \operatorname{colim}_{V \supset(\beta^{\prime}(\beta^{\prime,-1}(U)))}\mathscr{F}(V)$$, i.e., the colimit component map. Then $$\alpha_*\mathscr{F} \to \alpha_*\beta^{\prime}_*\beta^{\prime, -1}\mathscr{F}$$ is defined by $$\mathscr{F}(\alpha^{-1}(U)) \to \operatorname{colim}_{V \supset(\beta^{\prime}(\beta^{\prime,-1}(\alpha^{-1}(U))))}\mathscr{F}(V)$$ for every open subset $$U \subset Z$$. The identity $$\alpha_*\beta^{\prime}_* =\beta_*\alpha^{\prime}_*$$ corresponds to the fact that $$\beta^{\prime,-1}(\alpha^{-1}(U))) = \alpha^{\prime}(\beta^{-1}(U))$$. And $$\alpha_*\mathscr{F} \to \beta_*\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F}$$ is defined by $$\mathscr{F}(\alpha^{-1}(U)) \to \operatorname{colim}_{V \supset(\beta^{\prime}(\alpha^{\prime}(\beta^{-1}(U))))}\mathscr{F}(V)$$. It corresponds to $$\beta^{-1}\alpha_*\mathscr{F} \to\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F}$$ defined by the cone $$\{(\alpha_*\mathscr{F})(V) \to (\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F})(\beta^{-1}(V)) \stackrel{\operatorname{res}}{\to} (\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F})(U)\}_{V \supset \beta(U)}$$. Here $$(\alpha_*\mathscr{F})(V) \to (\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F})(\beta^{-1}(V))$$ is $$\mathscr{F}(\alpha^{-1}(V)) \to \operatorname{colim}_{V^\prime \supset(\beta^{\prime}(\beta^{\prime,-1}(\alpha^{-1}(V))))}\mathscr{F}(V^\prime)$$ and $$(\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F})(U)$$ is $$\operatorname{colim}_{V\supset \beta^{\prime}(\alpha^{\prime, -1}(U))} \mathscr{F}(V)$$ as in the above discussion. And $$\beta^{-1}\alpha_*\mathscr{F} \to\alpha^{\prime}_*\beta^{\prime, -1}\mathscr{F}$$ is defined by the cone $$\{ \mathscr{F}(\alpha^{-1}(V)) \to \operatorname{colim}_{V^\prime \supset(\beta^{\prime}(\beta^{\prime,-1}(\alpha^{-1}(V))))}\mathscr{F}(V^\prime) \to \operatorname{colim}_{V^\prime\supset \beta^{\prime}(\alpha^{\prime, -1}(U))} \mathscr{F}(V^\prime)\}_{V \supset \beta(U)}$$. And as in the above discussion, $$V \supset \beta(U)$$ if and only if $$\alpha^{-1}(V) \supset \alpha^{-1}(\beta(U))$$, it is same with the cone $$\{ \mathscr{F}(V) \to \operatorname{colim}_{V^\prime \supset(\beta^{\prime}(\beta^{\prime,-1}(V)))}\mathscr{F}(V^\prime) \to \operatorname{colim}_{V^\prime\supset \beta^{\prime}(\alpha^{\prime, -1}(U))} \mathscr{F}(V^\prime)\}_{V \supset \alpha^{-1}(\beta(U))}$$, i.e., the same with $$\operatorname{colim}_{V \supset \alpha^{-1}(\beta(U))} \mathscr{F}(V) \to \operatorname{colim}_{V^\prime\supset \beta^{\prime}(\alpha^{\prime, -1}(U))} \mathscr{F}(V^\prime)$$.

The situation the dual argument is similar: the identity of $$\alpha_*\mathscr{F} \to \alpha_*\mathscr{F}$$ corresponds to the map $$\alpha^{-1}\alpha_*\mathscr{F} \to \mathscr{F}$$ defined by the cone $$\{\mathscr{F}(\alpha^{-1}(V)) \to \mathscr{F}(\alpha^{-1}(V))\stackrel{\operatorname{res}}{\rightarrow}\mathscr{F}(U)\}_{V \supset \alpha(U)}$$. The first map is just identity, hence it’s just the cone $$\{ \mathscr{F}(\alpha^{-1}(V))\stackrel{\operatorname{res}}{\rightarrow}\mathscr{F}(U)\}_{V \supset \alpha(U)}$$. Then $$\beta^{\prime, -1}\alpha^{-1}\alpha_*\mathscr{F} \to \beta^{\prime,-1}\mathscr{F}$$ is defined by $$\operatorname{colim}_{V \supset \beta^{\prime}(U)}(\alpha^{-1}\alpha_*\mathscr{F})(V) \to \operatorname{colim}_{V \supset \beta^{\prime}(U)}\mathscr{F}(V)$$. It corresponds to map of cones $$\{(\alpha^{-1}\alpha_*\mathscr{F})(V) \to \mathscr{F}(V)\}_{V \supset \beta^{\prime}(U)}$$. And unraveling the definition for $$\alpha^{-1}\alpha_*\mathscr{F} \to \mathscr{F}$$, one has the diagram

$$\begin{CD} \mathscr{F}(\alpha^{-1}(V^\prime)) @>{\operatorname{res}}>> \mathscr{F}(V) \\ @V{}VV @|{}V \\ (\alpha^{-1}\alpha_*\mathscr{F})(V) = \operatorname{colim}_{V' \supset \alpha(V)} \mathscr{F}(\alpha^{-1}(V^\prime)) @>>{}> \mathscr{F}(V) \\ @V{}VV @VV{}V \\ \operatorname{colim}_{V \supset \beta^{\prime}(U)}(\alpha^{-1}\alpha_*\mathscr{F})(V) @>>{}> \operatorname{colim}_{V \supset \beta^{\prime}(U)}\mathscr{F}(V) \end{CD}$$

. It’s $$\operatorname{colim}_{V \supset \beta^{\prime}(U)}(\alpha^{-1}\alpha_*\mathscr{F})(V) = \operatorname{colim}_{V \supset \beta^{\prime}(U)}\operatorname{colim}_{V' \supset \alpha(V)} \mathscr{F}(\alpha^{-1}(V^\prime)) = \operatorname{colim}_{V' \supset \beta(\alpha^{\prime}(U))}\mathscr{F}(\alpha^{-1}(V^\prime))$$. The map $$\beta^{\prime, -1}\alpha^{-1}\alpha_*\mathscr{F} \to \beta^{\prime,-1}\mathscr{F}$$ is defined by the map of cones $$\{\mathscr{F}(\alpha^{-1}(V^\prime)) \to \operatorname{colim}_{V \supset \beta^{\prime}(U)}\mathscr{F}(V) \}_{V' \supset \alpha(\beta^{\prime}(U))}$$. It switches to $$\alpha^{\prime,-1}\beta^{-1}\alpha_*\mathscr{F} \to \beta^{\prime,-1}\mathscr{F}$$ just via the relation $$\alpha(\beta^{\prime}(U)) = \beta(\alpha^{\prime}(U))$$ and defined by $$\{\mathscr{F}(\alpha^{-1}(V^\prime)) \to \operatorname{colim}_{V \supset \beta^{\prime}(U)}\mathscr{F}(V) \}_{V' \supset \beta(\alpha^{\prime}(U))}$$, i.e., the map $$\operatorname{colim}_{V' \supset \beta(\alpha^{\prime}(U))}\mathscr{F}(\alpha^{-1}(V^\prime)) \to \operatorname{colim}_{V \supset \beta^{\prime}(U)}\mathscr{F}(V)$$ .Then it moves to $$\beta^{-1}\alpha_*\mathscr{F} \to \alpha_{*}^\prime\beta^{\prime,-1}\mathscr{F}$$ by replacing $$U$$ to $$\alpha^{\prime,-1}(U)$$ in the last colimit maps: $$\operatorname{colim}_{V' \supset \beta(\alpha^{\prime}(\alpha^{\prime, -1}(U)))}\mathscr{F}(\alpha^{-1}(V^\prime)) \to \operatorname{colim}_{V \supset \beta^{\prime}(\alpha^{\prime,-1}(U))}\mathscr{F}(V)$$ , and then compose $$\operatorname{colim}_{V \supset \alpha^{-1}(\beta(U))} \mathscr{F}(V) \to \operatorname{colim}_{V \supset \alpha^{\prime}(\alpha^{\prime,-1}(U))} \operatorname{colim}_{V' \supset \alpha^{-1}(\beta(V))}\mathscr{F}(V') = \operatorname{colim}_{V' \supset \alpha^{-1}(\beta(\alpha^{\prime}(\alpha^{\prime,-1}(U)))} \mathscr{F}(V') = \operatorname{colim}_{V' \supset \beta(\alpha^{\prime}(\alpha^{\prime, -1}(U)))}\mathscr{F}(\alpha^{-1}(V^\prime)).$$ Hence it’s still the map $$\operatorname{colim}_{V \supset \alpha^{-1}(\beta(U))} \mathscr{F}(V) \to \operatorname{colim}_{V^\prime\supset \beta^{\prime}(\alpha^{\prime, -1}(U))} \mathscr{F}(V^\prime)$$.

Back to the case of category of sheaves, first one can prove that the following diagram commute:

$$\begin{CD} PSh_W(\beta^{\prime, -1}_{pre}\mathscr{F},\beta^{\prime, -1}_{pre}\mathscr{F}) @>{(\beta^{\prime,-1}_{pre},\beta^{\prime}_*)}>{\sim}> PSh_X(\mathscr{F},\beta^\prime_*\beta^{\prime, -1}_{pre}\mathscr{F}) \\ @V{(-)^{sh}}VV @VV{\beta^{\prime}_*sh_*}V \\ Sh_W(\beta^{\prime, -1}\mathscr{F},\beta^{\prime, -1}\mathscr{F}) @>{(\beta^{\prime,-1},\beta^{\prime}_*)}>{\sim}> Sh_X(\mathscr{F},\beta^\prime_*\beta^{\prime, -1}\mathscr{F}) \end{CD}$$

Since adjoint pair is natural in both two variables, given the map $$sh:\beta^{\prime, -1}_{pre}\mathscr{F} \to \beta^{\prime, -1}\mathscr{F}$$(cf. exercise 2.4.I), treating it as a morphism in the category of presheaves, one have

$$\begin{CD} PSh_W(\beta^{\prime, -1}_{pre}\mathscr{F},\beta^{\prime, -1}_{pre}\mathscr{F}) @>{(\beta^{\prime,-1}_{pre},\beta^{\prime}_*)}>{\sim}> PSh_X(\mathscr{F},\beta^\prime_*\beta^{\prime, -1}_{pre}\mathscr{F}) \\ @V{sh_*}VV @VV{\beta^{\prime}_*sh_*}V \\ PSh_W(\beta^{\prime, -1}_{pre}\mathscr{F},\beta^{\prime, -1}\mathscr{F}) @>{(\beta^{\prime,-1}_{pre},\beta^{\prime}_*)}>{\sim}> Sh_X(\mathscr{F},\beta^\prime_*\beta^{\prime, -1}\mathscr{F}) \end{CD}$$

Then $$PSh_W(\beta^{\prime, -1}_{pre}\mathscr{F},\beta^{\prime, -1}\mathscr{F}) \cong Sh_W(\beta^{\prime, -1}\mathscr{F},\beta^{\prime, -1}\mathscr{F})$$ by the sheafification and forgetful adjoint pair and $$Sh_W(\beta^{\prime, -1}\mathscr{F},\beta^{\prime, -1}\mathscr{F}) \cong Sh_X(\mathscr{F},\beta^\prime_*\beta^{\prime, -1}\mathscr{F})$$ by the $$(\beta^{\prime,-1},\beta^{\prime}_*)$$ adjoint pair. I believe that they composing to the isomorphism $$PSh_W(\beta^{\prime, -1}_{pre}\mathscr{F},\beta^{\prime, -1}\mathscr{F}) \cong Sh_X(\mathscr{F},\beta^\prime_*\beta^{\prime, -1}\mathscr{F})$$ has to be used directly(hence there should be some condition on this about the three adjoint pairs in the general setting here). Using this composition, one get the commutative diagram in the begining of the paragraph.

Then the commutative diagram induced by applying $$\alpha_*$$ is easier to see. A third commutative diagram using the adjointness of $$(\beta_*, \beta^{-1})$$ is similar to the argument of the above diagram.

The dual definition gives similar results. Composing all these diagrams transforms the result to the category of sheaves.

• I believe your last statement is not strictly correct. You need to apply sheafification, then the post-compose with the universal map $\operatorname{sh}\alpha'_* \beta'^{-1}_{\text{pre}} \to \alpha'_*\beta'^{-1}$ . From there we then need to show that both definitions of the sheaf push-pull factors as this composition. The details of showing this seem to be fairly grisly to me. Commented Jan 6, 2023 at 6:53
• Additionally, you appear to be missing a step showing that both definitions make the relevant colimit cones commute. Commented Jan 6, 2023 at 7:01
• @Josiah yeah, I was very vague in the two steps you pointed out. I am trying to add more details. Commented Jan 6, 2023 at 8:28
• @Josiah btw, why you said the dual definition is a natural transformation $\beta^{\prime -1}\alpha \to \alpha^{\prime -1} \beta$ in the comment under ChrisH’s answer? It seems the dual definition still gives a natural transformation with the same domain and codomain functors. Commented Jan 6, 2023 at 8:34
• @Josiah Hi, I changed the argument for moving the result from presheaves to sheaves. I believe that two points are essential: 1. the naturality of adjointness. 2. The three adjoint pairs, here the presheaf inverse image and pushforward adjoint pair, the sheaf inverse image and pushforward adjoint pair, the sheafification and forgetful adjoint pair, has to satisfy some condition (some kind of commutivity) in general categorical settings. Commented Jan 6, 2023 at 14:04

I've come up with a fully categorical approach. Since it includes a LOT of large diagrams, I've typeset the answer in LaTeX and hosted the pdf on my blog.

To simplify the notation a bit I use $$\alpha$$ for $$\alpha_*$$. The sheaf $$\mathcal{F}$$ is not written. Here below, we get the commutative diagram that gives a simple categorical proof in the presheaf category, that extend trivially to the sheaf category.

$$\require{AMScd} \begin{CD} \beta^{-1}\alpha @>\eta^{\alpha'}_{\beta^{-1}\alpha}>>=\cases{\alpha'\alpha'^{-1}\beta^{-1}\alpha\\ \alpha'\beta'^{-1}\alpha^{-1}\alpha} @>\alpha'\beta'^{-1}\epsilon^\alpha>>\alpha'\beta'^{-1} \\ @V\beta^{-1}\alpha\eta^{\beta'}VV @V\alpha'\alpha'^{-1}\beta^{-1}\alpha\eta^{\beta'}VV @V\alpha'\beta'^{-1}\eta^{\beta'}VV\\ =\cases{\beta^{-1}\alpha\beta'\beta'^{-1}\\ \beta^{-1}\beta\alpha'\beta'^{-1}}@>\eta^{\alpha'}_{\beta^{-1}\alpha\beta'\beta'^{-1}}>> =\cases{\alpha'\alpha'^{-1}\beta^{-1}\alpha\beta'\beta'^{-1}\\\alpha'\beta'^{-1}\alpha^{-1}\alpha\beta'\beta'^{-1}\\ \alpha'\alpha'^{-1}\beta^{-1}\beta\alpha'\beta'^{-1}}@>\alpha'\beta'^{-1}\epsilon^\alpha_{\beta'\beta'^{-1}}>> \alpha'\beta'^{-1}\beta'\beta'^{-1}\\ @V\epsilon^\beta_{\alpha'\beta'^{-1}} VV @V\alpha'\alpha'^{-1}\epsilon^\beta_{\alpha'\beta'^{-1}} VV @V\alpha'\epsilon^{\beta'}_{\beta'^{-1}} VV\\ \alpha'\beta'^{-1}@>\eta^{\alpha'}_{\alpha'\beta'^{-1}}>> \alpha'\alpha'^{-1}\alpha'\beta'^{-1}@>\alpha'\epsilon^{\alpha'}_{\beta'^{-1}}>>\alpha'\beta'^{-1} \end{CD}$$

0/ The first column and the first row express the two composed morphisms we are asked to show they are equal

1/ The last column and the last row actually express two composed morphisms that are equal to the identity by the counit-unit identities

2/ The top two squares and the first bottom one are commutative because of the naturality of the unit and counit natural transformations

3/ The last bottom square is commutative because the counit of the composition of two adjoint functors is the composition of the counits, and we use the commutation relation $$\alpha\beta'=\beta\alpha'$$