Given a topological space $X$ and a continuous map $\pi\colon X \to Y$ , there is an adjunction between the categories of sheaves on $X$ and sheaves on $Y$ . The adjunction is somewhat mysterious to me as I begin to write this, so the hope is that by writing about it, I’ll begin to understand.

The functors in question are the pushforward and inverse image functors, which arise naturally in algebraic geometry (so I am led to believe). What makes the adjunction difficult to understand is that the definitions of the functors in question don’t seem to play nicely with each other.

The proof (with a little setting the stage) appears below. The method of proof, like most category-theoretic proofs, is a story of careful bookkeeping with functors and natural transformations, along with simple calculations to check. The idea essentially seems to be that the unit $\eta_\mathscr{F} \colon \mathscr{F} \to \pi_\ast\pi^{-1}\mathscr{F}$ “shreds a section $f$ into pieces,” while the counit $\epsilon_{\mathscr{G}} \colon \pi^{-1}\pi_\ast\mathscr{G} \to \mathscr{G}$ recognizes that these “shredded up” sections assemble into sections of $\mathscr{G}$ .

Adjunctions

Remember, given categories $C$ and $D$ , a pair of functors $F \colon C \to D$ and $G \colon D \to C$ form an adjoint pair when there exists an isomorphism

D(Fc,d) \cong C(c,Gd)

which is natural in $c$ and $d$ . We say $F$ is left adjoint to $G$ , or equivalently that $G$ is right adjoint to $F$ . In particular, setting $d = Fc$ , we have

D(Fc,Fc) \cong C(c,GFc).

which gives us a natural transformation $\eta \colon 1_C \Longrightarrow GF$ . Similarly, we have $\epsilon\colon FG \Longrightarrow 1_D$ . The natural transformation $\eta$ is called the unit of the adjunction, and $\epsilon$ is the counit. In fact, the existence of these natural transformations could be taken to be the definition of an adjunction, provided the natural transformations satisfy the identities

\epsilon F \circ F\eta = 1_F \quad\text{and}\quad G\epsilon \circ \eta G = 1_G,

which says that for all $c \in C$ and $d \in D$ , the compositions

\begin{CD} Fc @>{F\eta_c}>> FGFc @>{\epsilon_{Fc}}>> Fc \end{CD}

and

\begin{CD} Gd @>{\eta_{Gd}}>> GFGd @>{G\epsilon_d}>> Gd \end{CD}

are the identity.

My goal in this blog post is to show that the functors I am interested in satisfy this latter definition of an adjunction.

Sheaves

Given a pair of spaces $X$ and $Y$ , and a continuous map $\pi \colon X \to Y$ , we have functors $\pi_\ast \colon \operatorname{Sh}_X \to \operatorname{Sh}_Y$ and $\pi^{-1}\colon \operatorname {Sh}_Y \to \operatorname{Sh}_X$ from the category of sheaves on $X$ to the category of sheaves on $Y$ and vice versa. The former is simple to describe: if $\mathscr{G}$ is a sheaf on $X$ , the pushforward sheaf $\pi_\ast\mathscr{G}$ on $Y$ is the assignment

U \mapsto \mathscr{F}(\pi^{-1}(U))

for all open subsets $U$ of $Y$ . If $\psi \colon \mathscr{G} \to \mathscr{G}'$ is a morphism of sheaves on $X$ , the morphism $\pi_\ast\psi$ is defined by the rule

(\pi_\ast\psi)(U) = \psi(\pi^{-1}(U)).

The latter is slightly more complicated. Recall that if $\mathscr{F}$ is a sheaf on $Y$ , there is a space $F$ equipped with a local homeomorphism $\rho \colon F \to Y$ . The sheaf $\mathscr{F}$ is the sheaf of local sections of $\rho$ ; $F$ is called the espace étalé for the sheaf $\mathscr{F}$ . The set of points of $F$ above $y \in Y$ is the stalk $\mathscr{F}_y$ .

As usual, a section $s\colon U \to F$ of the projection $\rho \colon F \to X$ is an element $(s_y)_{y \in U}$ of $\prod_{y \in U} \mathscr{F}_y$ . In our case, continuity of $s$ is the following condition

$$\begin{equation}\label{star}\tag{$\ast$} \text{for all } y \in U, \text{ there exists an open neighborhood } U_y \text{ of } y\\ \text{ and } f\in\mathscr{F}(U_y) \text{ such that } f_x = s_x \text{ for all } x \in U_y. \end{equation}$$

Given that $F$ and $X$ admit maps to $Y$ , we can form their pullback,

\begin{CD} \pi^{-1}F @>>> F \\ @VVV @VV{\rho}V \\ X @>{\pi}>> Y \end{CD}

which as a set is $\{ (x,f_y) \in X \times F : \pi(x) = y\}$ . The topology on $\pi^{-1}F$ is induced from the product topology on $X \times F$ . The inverse image sheaf $\pi^{-1}\mathscr{F}$ is the sheaf of sections of $\pi^{-1}F \to X$ . The universal property of the pullback ensures that this really defines a functor from the category of sheaves on $Y$ to the category of sheaves on $X$ .

Condition $\eqref{star}$ for continuity of a section $s\colon V \to \pi^{-1}F$ —that is, a tuple $(s_{\pi(x)})_{x \in V}$ in $\prod_{x \in V}\mathscr{F}_{\pi(x)}$ —says that for all $x \in V$ there exists an open neighborhood $V_x$ of $x$ and an open neighborhood $U_x$ in $Y$ such that $\pi(V_x) \subset U_x$ . Furthermore there exists a section $f \in \mathscr{F}(U_x)$ such that $f_{\pi(v)} = s_{\pi(v)}$ for all $v \in V_x$ .

If $\phi\colon \mathscr{F} \to \mathscr{F}'$ is a morphism of sheaves on $Y$ , the resulting morphism $\pi^{-1}\phi \colon \pi^{-1}\mathscr{F} \to \pi^{-1}\mathscr{F}'$ is defined by

(\pi^{-1}\phi)(V)\left[(s_{\pi(x)})_{x \in V}\right] = \left(\phi_{\pi(x)}(s_{\pi(x)})\right)_{x \in V}.

One needs to demonstrate that the right-hand side satisfies condition $\eqref{star}$ ; the check is simple, so I’ll leave it to the reader.

The claim is that $\pi^{-1}$ is left adjoint to $\pi_\ast$ . Therefore we should expect natural transformations $\eta \colon 1_{\operatorname{Sh}_Y} \Longrightarrow \pi_\ast\pi^{-1}$ and $\epsilon \colon \pi^{-1}\pi_\ast \Longrightarrow 1_{\operatorname{Sh}_X}$ .

The Unit of the Adjunction

So let $\mathscr{F}$ be a sheaf on $Y$ , and let $U$ be an open set in $Y$ . Since $\pi_\ast\pi^{-1}\mathscr{F}(U) = \pi^{-1}\mathscr{F}(\pi^{-1}(U))$ , we can think of $s \in \pi_\ast\pi^{-1}\mathscr{F}(U)$ as a section $s \colon \pi^{-1}(U) \to \pi^{-1}F$ , i.e. a tuple $(s_{\pi(x)})_{x \in \pi^{-1}(U)}$ satisfying the compatibility condition above.

Note that in particular if $U \subset Y$ is open and $f \in \mathscr{F}(U)$ , the continuity condition $\eqref{star}$ for a section $\pi^{-1}(U)\to \pi^{-1}F$ is satisfied by the section $(f_{\pi(x)})_{x \in \pi^{-1}(U)}$ , and thus defines an element of $\pi_\ast\pi^{-1}\mathscr{F}(U)$ .

Let us thus define a map $\eta_{\mathscr{F}}(U)\colon \mathscr{F}(U) \to \pi_\ast\pi^{-1}\mathscr{F}(U)$ as

\eta_{\mathscr{F}}(U)(f) = (f_{\pi(x)})_{x\in\pi^{-1}(U)}.

(Let’s parse the notation: a natural transformation $\eta$ yields a map of sheaves $\eta_{\mathscr{F}}$ for each sheaf $\mathscr{F}$ . This is the definition of that map on the open subset $U \subset Y$ .)

To prove the claim that $\eta$ is a natural transformation, we need to show that given a map $\phi \colon \mathscr{F} \to \mathscr{F}'$ of sheaves on $Y$ , the following diagram commutes

\begin{CD} \mathscr{F} @>{\eta_{\mathscr{F}}}>> \pi_\ast\pi^{-1}\mathscr{F} \\ @V{\phi}VV @VV{\pi_\ast\pi^{-1}\phi}V \\ \mathscr{F}' @>{\eta_{\mathscr{F}'}}>> \pi_\ast\pi^{-1}\mathscr{F}'. \end{CD}

Given $f \in \mathscr{F}(U)$ , commutativity of the square is the claim that the following equality holds

\left((\phi(U)(f))_{\pi(x)}\right)_{x \in \pi^{-1}(U)} = \left(\phi_{\pi(x)}(f_{\pi(x)})\right)_{x \in \pi^{-1}(U)}.

This is in fact true, since taking the stalk at $x$ is functorial.

The Counit of the Adjunction

Now suppose that $\mathscr{G}$ is a sheaf on $X$ . Let $\pi_\ast G$ denote the espace étalé over $Y$ for the sheaf $\pi_\ast\mathscr{G}$ .

A section $s \colon V \to \pi^{-1}\pi_\ast G$ is a tuple $(s_{\pi(x)})_{x \in V}$ taking values in $\pi_\ast\mathscr{G}_{\pi(x)}$ satisfying the continuity condition $\eqref{star}$ , which in our particular case asserts the existence, for each $x \in V$ , of an open neighborhood $V_x$ of $x$ and an open neighborhood $U_x$ in $Y$ satisfying $\pi(V_x) \subset U_x$ . Furthermore there exists $g \in \pi_\ast\mathscr{G}(U_x) = \mathscr{G}(\pi^{-1}(U_x))$ such that the image of $s_{\pi(x)}$ in $\mathscr{G}_x$ under the natural map $\pi_\ast\mathscr{G}_{\pi(x)} \to \mathscr{G}_x$ is equal to $g_x$ .

Since the sets $\pi^{-1}(U_x)$ cover $V$ by assumption, sheafiness of $\mathscr{G}$ implies the existence of a unique $g \in \mathscr{G}(V)$ such that $g_x$ is equal to the image of $s_{\pi(x)}$ under the natural map $\pi_\ast\mathscr{G}_{\pi(x)} \to \mathscr{G}_x$ for all $x \in V$ . Let us thus define a map $\epsilon_{\mathscr{G}}(V) \colon \pi^{-1}\pi_\ast\mathscr{G}(V) \to \mathscr{G}(V)$ sending $(s_{\pi(x)})_{x\in V}$ to $g$ .

We claim that $\epsilon$ defines a natural transformation from $\pi^{-1}\pi_\ast\mathscr{G}$ to $\mathscr{G}$ . To prove the claim, we need to show that given a map $\psi\colon \mathscr{G} \to \mathscr{G}'$ of sheaves on $X$ , the following diagram commutes

\begin{CD} \pi^{-1}\pi_\ast\mathscr{G} @>{\epsilon_{\mathscr{G}}}>> \mathscr{G} \\ @V{\pi^{-1}\pi_\ast\psi}VV @VV{\psi}V \\ \pi^{-1}\pi_\ast \mathscr{G}' @>{\epsilon_{\mathscr{G}'}}>> \mathscr{G}'. \end{CD}

Given $(s_{\pi(x)})_{x \in V} \in \pi^{-1}\pi_\ast\mathscr{G}(V)$ , the upper right path of the square sends us first to some $g \in \mathscr{G}(V)$ such that $g_x$ is the image of $s_{\pi(x)}$ in $\mathscr{G}_x$ for all $x \in V$ , and then sends $g$ to $\psi(V)(g) \in \mathscr{G}'(V)$ . On the other hand, the lower left path of the square sends $(s_{\pi(x)})_{x \in V}$ first to $\left((\pi_\ast\psi)_{\pi(x)}(s_{\pi(x)})\right)_{x \in V}$ in $\pi^{-1}\pi_\ast\mathscr{G}'(V)$ , and then to some $g' \in \mathscr{G}'(V)$ such that $g'_x$ is the image of $(\pi_\ast\psi)_{\pi(x)}(s_{\pi(x)})$ in $\mathscr{G}'_x$ for all $x \in V$ .

If $U \subset Y$ is an open set containing $\pi(x)$ such that $s \in \pi_\ast\mathscr{G}(U) = \mathscr{G}(\pi^{-1}(U))$ represents $s_{\pi(x)}$ for $\pi(x) \in U$ , we have $(\pi_\ast\psi)_{\pi(x)}(s_{\pi(x)}) = \psi(\pi^{-1}(U))(s)_{\pi(x)}$ . Furthermore we know that $g \in \mathscr{G}(V)$ is such that the image of $s_{\pi(x)}$ in $\mathscr{G}_x$ is $g_x$ for all $x \in V$ . Therefore we have that the image of $\psi(\pi^{-1}(U))(s)_{\pi(x)}$ in $\mathscr{G}'_x$ is equal to $\psi(V)(g)_x$ , as desired.

The inverse image–pushforward adjunction

To conclude, we just need to show that two double compositions of natural transformations are the identity. The first, translated into our particular case, says that if $\mathscr{F}$ is a sheaf on $Y$ , we have that

\begin{CD} \pi^{-1}\mathscr{F} @>{\pi^{-1}\eta_{\mathscr{F}}}>> \pi^{-1}\pi_\ast\pi^{-1}\mathscr{F} @>{\epsilon_{\pi^{-1}\mathscr{F}}}>> \pi^{-1}\mathscr{F} \end{CD}

is equal to the identity. The difficulty here seems to be largely notational. Let $V \subset X$ be an open set. Suppose $(s_{\pi(x)})_{x \in V}$ is a section in $\pi^{-1}\mathscr{F}(V)$ . We have

(\pi^{-1}\eta_{\mathscr{F}})(V)\left[(s_{\pi(x)})_{x\in V}\right] = \left((\eta_{\mathscr{F}})_{\pi(x)}(s_{\pi(x)})\right)_{x \in V} \in \pi^{-1}\pi_\ast\pi^{-1}\mathscr{F}(V).

First observe that the continuity condition $\eqref{star}$ tells us that there is, for each $x \in V$ , an open neighborhood $V_x$ of $x$ and an open set $U_x \subset Y$ containing $\pi(V_x)$ . On this open set we have $s \in \mathscr{F}(U_x)$ such that $\left((\eta_{\mathscr{F}})(U_x)(s)\right)_{\pi(x)} = (\eta_{\mathscr{F}})_{\pi(x)}(s_{\pi(x)})$ .

We saw above that $(\eta_{\mathscr{F}})(U_x)(s)$ is the section $(s_{\pi(x)})_{x \in \pi^{-1}(U_x)} \in \pi_\ast\pi^{-1}\mathscr{F}(U_x) = \pi^{-1}\mathscr{F}(\pi^{-1}(U))$ . The map $\epsilon_{\pi^{-1}\mathscr F}$ recognizes that the sections $(s_{\pi(x)})_{x \in \pi^{-1}(U_x)}$ satisfy the compatibility conditions to glue up to form a section $(s_{\pi(x)})_{x \in V}$ in $\pi^{-1}\mathscr{F}(V)$ , demonstrating that the double composition is indeed the identity. So far so good.

The second double composition we need to show is equal to the identity is defined for a sheaf $\mathscr{G}$ on $X$ as follows

\begin{CD} \pi_\ast\mathscr{G} @>{\eta_{\pi_\ast\mathscr{G}}}>> \pi_\ast\pi^{-1}\pi_\ast\mathscr{G} @>{\pi_\ast\epsilon_{\mathscr{G}}}>> \pi_\ast\mathscr{G}. \end{CD}

If $U \subset Y$ is an open set and $g \in \pi_\ast\mathscr{G}(U)$ is a section, the first map sends $g$ to the section $(g_{\pi(x)})_{x \in \pi^{-1}(U)} \in \pi_\ast\pi^{-1}\pi_\ast\mathscr{G}(U)$ .

The second map recognizes that the above section and $g \in \mathscr{G}(\pi^{-1}(U)) = \pi_\ast\mathscr{G}$ agree on each stalk, and so sends our section back to $g$ .

This completes the proof that inverse image and pushforward form an adjoint pair of functors.