# Inverse Image as the left adjoint to pushforward

This is a repost of a question on Math stackexchange. No one is biting at it there, so I guess it is harder than I thought.

Assume $X$ and $Y$ are topological spaces, $f : X \to Y$ is a continuous map. Let ${\bf Sh}(X)$, ${\bf Sh}(Y)$ be the category of sheaves on $X$ and $Y$ respectively. Modulo existence issues we can define the inverse image functor $f^{-1} : {\bf Sh}(Y) \to {\bf Sh}(X)$ to be the left adjoint to the push forward functor $f_{*} : {\bf Sh}(X) \to {\bf Sh}(Y)$ which is easily described.

My question is this: Using this definition of the inverse image functor, how can I show (without explicitly constructing the functor) that it respects stalks? i.e is there a completely categorical reason why the left adjoint to the push forward functor respects stalks?

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Easy: the stalk at a point $x: 1 \to X$ is a functor $\text{Sh}(X) \to Set$ that may be identified with the inverse image functor

$$x^\ast: \text{Sh}(X) \to \text{Sh}(1).$$

Since we have $x^\ast \circ f^\ast \cong (f \circ x)^\ast = (f(x))^\ast$, the inverse image pulls back stalk functors to stalk functors.

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Is there any book on algebraic geometry which mentions this? I remember that I also had to prove this by myself, a couple of months after having done the "fiddly element proof". – Martin Brandenburg Apr 14 '11 at 12:31
I had assumed this was well-known, but maybe not. The direct image $x_\ast: Sh(1) \to Sh(X)$ is, pretty tautologously, the skyscraper sheaf construction, and one has the basic adjunction between taking stalks and taking skyscrapers (see e.g., Hartshorne, exercises 1.17 and 1.18, page 68, or Mac Lane & Moerdijk, lemma II.6.7, page 93). Thus the left adjoint $x^\ast$ to $x_\ast$ is canonically identified with the stalk functor. – Todd Trimble Apr 14 '11 at 12:52
I guess I'm too category-minded and insufficiently geometry-minded. I thought "inverse image along inclusion of a point" was the definition of a stalk. – Andreas Blass Apr 14 '11 at 14:44
You certainly could define it that way, but whether it's typical to define it that way... I'm not sure. Any way you define it, it's not hard to see this is an equivalent definition. :-) – Todd Trimble Apr 14 '11 at 14:48
@Martin: I can't resist to point out that Görtz' and my book (www.algebraic-geometry.de) mentions this (Section (2.8)). – Torsten Wedhorn Apr 15 '11 at 6:06

Here is a pretty messy proof that is not categorical. I recently solved this problem and decided to share my solution. I apologize if this is not what you are looking for.

Some notation:

• $X,Y$ will be topological spaces, and $f:X\to Y$ a continous map. We will denote $F$ (resp. $G$) to be a sheaf on $X$ (resp. $Y$).
• $U$ (resp. $V$) will denote an open set in $X$ (resp. $Y$).
• We will denote $\overline{f}G$ to be the presheaf on $X$ given by $\overline{f}G(U) = \lim ~ G(V)$, where $V\supseteq f(U)$.

If $P$ is a presheaf on $X$, we use the notation $(U,s)$, where $U\ni x$ and $s\in P(U)$, to denote the image of $s\in P(U)$ in the stalk $P_x$ at the point $x$. Thus, $(U,s) = (U',s')$ (where $U'\ni x$) if and only if there is an $U''\ni x$ and $s''\in P(U'')$ such that $U''\subseteq (U\cap U')$ and $s|_{U''} = s'|_{U''} = s''$. \

Let $P^+$ be the sheafification of $P$, so it comes equipped with a morphism $P\to P^+$. For $s\in P(U)$, we define $\tilde{s}$ to be the image of $s\in P(U)\mapsto \tilde{s}\in P^+(U)$. The important fact for us is that if $a\in P^+(U)$, then for any $x\in U$, there is an $U_x\ni x$, and $s\in P(U_x)$, such that $a|_{U_x} = \tilde{s}$. \

Therefore, if have a sheaf $H$ on space $X$, and two morphisms $\varphi,\varphi':P^+\to H$, such that $\varphi(U)(\tilde{s}) = \varphi'(U)(\tilde{s})$, for all $s\in P(U)$, and all $U$, then $\varphi = \varphi'$. \

Now we let $F$ be a sheaf on $X$, with $f:X\to Y$ a continuous map. For an open set $U$ in $X$, $\overline{f}f_*F(U)$ is the direct limit of $F(f^{-1}V)$, where $f^{-1}V\supseteq U$. We denote $(f^{-1}V,s)$ to be the image of $s\in F(f^{-1}V)$ in its direct limit. Hence $\widetilde{(f^{-1}V,s)}\in f^{-1}f_*F(U)$. The sheaf morphism $f^{-1}f_* F\to F$ we denote by $\alpha$. The map $\alpha(U):f^{-1}f_* F(U) \to F(U)$ satisfies the property that $\widetilde{(f^{-1}V,s)} \mapsto s|_U$, where $V$ is an open set in $Y$, and $f^{-1}V\supseteq U$, and $s\in f_*F(V) = F(f^{-1}V)$. \

For a sheaf $G$ on $Y$, we denote the sheaf morphism $G\to f_*f^{-1}G$ by $\beta$. First, we have a presheaf morphism $\beta':G\to f_*\overline{f}G$ where $\beta'(V)$ is given by $s\mapsto (V,s)$, here we think of $(V,s)$ as a representative in the direct limit $G(W)$ where $W\supseteq f(f^{-1}V)$. And we have a presheaf morphism $\mu: f_*\overline{f}G\to f_*f^{-1}G$ where, $\mu(V)(s) = \widetilde{(f^{-1}V,s)}$. Thus, $\beta(V)$ satisfies the property $s\mapsto \widetilde{(f^{-1}V,s)}$.\

If we start with a morphism $\psi:G\to f_*F$, then we obtain a morphism $f^{-1}G\to F$ by composition $\alpha\circ f^{-1}\psi$. The map $(\alpha\circ f^{-1}\psi)(U)$ satisfies the property that $\widetilde{ (f^{-1}V,s) }\mapsto \psi(V)(s)|_U$. \

If we start with a morphism $\varphi:f^{-1}G\to F$, then we obtain a morphism $G\to f_*F$ by composition $f_*\varphi\circ \beta$. The map $(f_*\varphi\circ \beta)(V)$ satisfies the property that $s\mapsto \varphi(f^{-1}V)\widetilde{(f^{-1}V,s)}$. \

Hence, given a morphism $f^{-1}G\to F$ we can obtain $G\to f_*F$, and given a morphism, $G\to f_*F$ we can obtain $f^{-1}G\to F$. We will show these two transformations are inverses of one another, completing the proof. \

Let us start with $\varphi: f^{-1}G\to F$. Construct $\psi:G\to f_*F$ and then construct its corresponding morphism $\varphi':f^{-1}G\to F$. We have to show that $\varphi = \varphi'$. Choose an open set $U$ (in $X$). We will show that $\varphi(U) = \varphi'(U)$. The map $\varphi'(U)$ sends $\widetilde{(f^{-1}V,s)} \mapsto \psi(V)(s)|_U$, where $V$ is an open set in $Y$ for which $f^{-1}V \supseteq U$ ($\iff V\supseteq f(U)$). But $\psi(V)(s) = \varphi(f^{-1}V)(\widetilde{f^{-1}V,s})$. As sheaf morphisms are consistent with restrictions and $U\subseteq f^{-1}V$ we have that, $\psi(V)(s)|_U = \varphi(U)(\widetilde{f^{-1}V,s})$. Therefore, $\varphi(U)$ and $\varphi'(U)$ both send $\widetilde{(f^{-1}V,s)}$ to the same section in $F(U)$. This is enough to complete the proof.\

Now we consider the case $\psi:G\to f_*F$. Construct $\varphi:f^{-1}G\to F$ and then construct its corresponding morphism $\psi':G\to f_*F$. We have to show that $\psi = \psi'$. Choose an open set $V$ (in $X$). We will show that $\psi(V) = \psi'(V)$. The map $\psi'(V)$ sends $s\in G(V)$ to $\varphi(f^{-1}V)\widetilde{(f^{-1}V,s)}$. But $\varphi(f^{-1}V)\widetilde{(f^{-1}V,s)} = \psi(V)(s)|_V = \psi(V)(s)$. Hence, $\psi'(V) = \psi(V)$.

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