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).



Homotopy property of constructible sheaves on stratified spaces
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.


