11
$\begingroup$

Let $X$ be a stratified topological space (in my case $X$ is a compact space presented as a finite union of locally closed topological manifolds of finite dimension (strata) such that the closure of any stratum is a union of some other strata). Let $\mathcal{F}^\bullet\in D^b(Sh_X)$ is an object of the bounded derived category of sheaves of vector spaces on $X$ whose cohomology sheaves are constructible along the strata of the given stratification.

Let $Y$ be a 'nice' topological space. Let $f\colon Y \times [0,1]\to X$ be a continuous map such that for any $y\in Y$ the image $f(\{y\}\times [0,1])$ is contained in some stratum.

Question. Is it true that $f^*(\mathcal{F}^\bullet)|_{Y\times\{0\}}$ and $f^*(\mathcal{F}^\bullet)|_{Y\times\{1\}}$ are isomorphic in $D^b(Sh_Y)$?

Remark. If $\mathcal{F}^\bullet$ is a local system, then the answer is positive regardless of any stratification.

A precise statement, ideally a reference, would be helpful (if it exists).

$\endgroup$
4
$\begingroup$

Here are two comments:

1) I suspect the answer is "yes," so long as your homotopy has the property that your pullback is locally constant (and hence, by triviality of the interval, constant) along the homotopy parameter, which I believe your condition implies. You might also want to add the hypothesis that the homotopy is stratified appropriately so that you can compare these sheaves against a common refinement.

2) Although this doesn't directly answer your question, I think your type of question is probably easier to understand with the equivalence between constructible sheaves and functors from the exit path category in mind.

The exit path category has points for objects and homotopy classes of exits paths for morphisms. An exit path is a continuous map from the interval with the property that the dimension of the containing stratum is non-decreasing. Homotopies are through exit paths.

Here is a short treatment of this equivalence done in the language of constructible cosheaves, although everything dualizes to the classical statement:

http://arxiv.org/abs/1603.01587

However, we abstract away the question whether every homotopy can be stratified and broken up into elementary homotopies.

We also give a new definition of "constructible," which I believe is better for many reasons.

As a caveat, I've heard it said that the derived category of constructible sheaves (take the category of constructible sheaves, then derive it) is not the same as the category of constructible derived sheaves (derived complexes of sheaves with constructible cohomology sheaves), but a theorem of Beilinson says they are the same for a triangulation.

$\endgroup$
1
$\begingroup$

The above question has positive answer. It immediately follows from the following slightly more general statement.

Proposition. Let $X$ be a compact topological space. Let $\mathcal{F}^\bullet\in D^b(Sh_{X\times [0,1]})$ has the property that for any point $x\in X$ the restriction $\mathcal{F}^\bullet\Big|_{\{x\}\times[0,1]}$ has constant cohomology sheaves (of finite rank). Then the adjunction morphism $$p^*p_*\mathcal{F}^\bullet\to \mathcal{F}^\bullet$$ is an isomorphism, where $p\colon X\times [0,1]\to X$ is the obvious projection.

Proof. For any $x\in X$ we denote by $I_x\colon \{x\}\times [0,1]\to X\times [0,1]$ and $i_x\colon \{x\}\to X$ the obvious imbeddings. Let $p_x\colon \{x\}\times[0,1]\to \{x\}$ be the obvious projection. It suffices to show that the restriction of the adjunction morphism $$I_x^*(p^*p_*\mathcal{F}^\bullet)\to I_x^*\mathcal{F}^\bullet$$ is an isomorphism. We have \begin{eqnarray*} I_x^*(p^*p_*\mathcal{F}^\bullet)=p_x^*i_x^*(p_*\mathcal{F}^\bullet)=p^*_x((p_x)_*I_x^*\mathcal{F}^\bullet), \end{eqnarray*} where the last equality is by the proper base change theorem.Thus we may replace $\mathcal{F}^\bullet$ with $I_x^*\mathcal{F}^\bullet$ and $X$ by a single point and prove the proposition in this situation. But if $X$ is a point the result is (almost) obvious. QED.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.