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I know that, if $X$ is a locally arcwise connected and locally simply connected topological space, then the restrictions of any locally constant sheaf $\mathcal{F}$ on $X$ corresponding to inclusions $U\subseteq V$ of open subsets which are homotopy equivalences are isomorphisms (this is clear because a locally constant sheaf on $X$ corresponds to a functor $F: \mathbf{ \Pi }_1(X)\rightarrow\mathbf{Set}$).

I was wondering: is it possible to prove the other implication, i.e. whenever $\mathcal{F}$ is a sheaf on $X$ such that $\mathcal{F}(V)\rightarrow\mathcal{F}(U)$ is an isomorphism each time $U\subseteq V$ is an homotopy equivalence, then $\mathcal{F}$ must be locally constant?

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  • $\begingroup$ Yes, you're right! Thank you, I edited my question. $\endgroup$ – mfox Sep 28 '18 at 22:48
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This is certainly not true in this generality. For instance, let $X=\mathbb{R}^3\setminus\mathbb{Q}^3$. Then $X$ is locally path connected and locally simply connected, but I'm pretty sure no nontrivial inclusion of open subsets of $X$ is a homotopy equivalence. So, any sheaf at all on $X$ would satisfy your condition.

If you assume $X$ is locally contractible, then it is true. Indeed, the restriction of $\mathcal{F}$ to any contractible open set $U$ is then constant, since we can identify $\mathcal{F}(V)$ with $\mathcal{F}(U)$ for any contractible open $V\subseteq U$ via the restriction map and such $V$ form a basis for the topology.

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  • $\begingroup$ Thank you for your answer. I don't have a clue on how to prove that no nontrivial inclusion of open subsets of $\mathbb{R}^3\setminus\mathbb{Q}^3$ is a homotopy equivalence, could you give me at least some hint? $\endgroup$ – mfox Sep 28 '18 at 22:57
  • $\begingroup$ I haven't worked out a full proof myself. The idea though is that there is nontrivial homotopy no matter how close you zoom in on any point, and so no point can be removed without losing some homotopy class of maps into your space. $\endgroup$ – Eric Wofsey Sep 28 '18 at 23:01
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    $\begingroup$ In particular, let $Y$ be a 2-dimensional version of the "Hawaiian earring" (so, a union of infinitely many $S^2$ spheres which converge on a common intersection point). For any open $U\subseteq \mathbb{R}^3\setminus\mathbb{Q}^3$ and any $x\in U$, I think you can construct a map $f:Y\to U$ which maps the common point of the spheres to $x$, such that this $f$ is not homotopic to any map $Y\to U\setminus\{x\}$. This would show that for any $V\subseteq U\setminus\{x\}$, the inclusion $V\to U$ is not a homotopy equivalence. $\endgroup$ – Eric Wofsey Sep 28 '18 at 23:04
  • $\begingroup$ (The idea is that $f$ is a sequence of spheres in $U$ closing in on $x$. Any map homotopic to $f$ must have its spheres enclose the same sets of rational points as the spheres of $f$ do, and so must also map the common point to $x$.) $\endgroup$ – Eric Wofsey Sep 28 '18 at 23:05

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