# 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.
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