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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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1 Answer 1

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
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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
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@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

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