# Locally constant sheaves and homotopy equivalences

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?

• Yes, you're right! Thank you, I edited my question. – mfox Sep 28 '18 at 22:48

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.
• 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? – mfox Sep 28 '18 at 22:57
• 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. – Eric Wofsey Sep 28 '18 at 23:04
• (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 $x$.) – Eric Wofsey Sep 28 '18 at 23:05