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Let $\mathsf{Top} \downarrow X$ denote the slice category of topological spaces over some base space $X$. Then the functor $\Gamma: \mathsf{Top} \downarrow X \to \mathsf{Sh}(X)$ which assigns to any map of spaces $p: E \to Y$ the sheaf of sections has a left adjoint, which is the etale space (i.e. geometric realization) construction $\mathrm{Et}: \mathsf{Sh}(X) \to \mathsf{Top} \downarrow X$. The latter functor is fully faithful, and hence the unit $\eta: 1 \Rightarrow \Gamma \mathrm{Et}$ is a natural isomorphism. The counit $\varepsilon: \mathrm{Et} \Gamma \Rightarrow 1$ is not a natural isomorphism. However, by the triangle identities the whiskered transformations $\Gamma \varepsilon : \Gamma \mathrm{Et} \Gamma \Rightarrow \Gamma$ and $\varepsilon \mathrm{Et}: \mathrm{Et} \Gamma \mathrm{Et} \Rightarrow \mathrm{Et}$ are both natural isomorphisms.

For a continuous map $f: Y \to X$, often the inverse image of sheaves $f^{-1}: \mathsf{Sh}(Y) \to \mathsf{Sh}(X)$ is defined to be the composite $$f^{-1} := \Gamma f^* \mathrm{Et}$$ where $f^*: \mathsf{Top} \downarrow Y \to \mathsf{Top} \downarrow X$ is the pullback or base change.

The wikipedia page states that "the pullback of bundles then corresponds to the inverse image of sheaves." I would like to show more generally that the pullback between slice categories over topological spaces corresponds to the inverse image of sheaves. In other words, I would like to prove the following:

Proposition: $f^{-1} \Gamma \cong \Gamma f^*$.

Attempted proof: By definition, $f^{-1}\Gamma = \Gamma f^* \mathrm{Et} \Gamma .$ The (whiskered) counit gives us a natural transformation $$\Gamma f^* \varepsilon: f^{-1}\Gamma = \Gamma f^* \mathrm{Et} \Gamma \Rightarrow \Gamma f^* $$ How can I verify that this is a natural isomorphism? Or am I going about this the wrong way? I'd prefer an "abstract" proof, in this sort of spirit, if possible.

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Nice question, but I think this is false even in the case of vector bundles.

Take for example $f:\{0\}\rightarrow\mathbb{R}$ the inclusion and $p:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ the first projection so we can see $\mathbb{R}\times\mathbb{R}$ as an object in $\mathrm{Top}/\mathbb{R}$.

We have $f^*p=\mathbb{R}$ seen as a toplogical space over the point, and $\Gamma f^*p=\mathbb{R}$ seen as a sheaf on the point.

On the other hand $\Gamma p$ is the sheaf of continuous function on $\mathbb{R}$, and $i^{-1}\Gamma p$ is the set of germs of continuous function at $\{0\}\subset\mathbb{R}$ (seen as a sheaf over the point).

Note that the natural transformation you defined is the evaluation map. But the latter is much larger than the former.

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  • $\begingroup$ Hmm...that's pretty unfortunate. I felt so sure it would be true. I guess it should hold, at the very least for pulling back along the inclusion of an open subset $i: U \hookrightarrow X$? $\endgroup$ – ಠ_ಠ Mar 2 '17 at 9:34
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    $\begingroup$ Yes for open subset it works. In fact, you can prove that your natural transformation is an isomorphism in this case using stalks. Maybe if you like formal argument, you can use the fact that $i ^{-1}$ as an exact left adjoint if $i$ is the inclusion of an open subset. You can also use ideas from these nice notes maths.ed.ac.uk/~tl/sheaves.pdf. $\endgroup$ – Roland Mar 2 '17 at 9:47

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