# When do sheaves which are constant along the fibers come from the base?

Assume we are given a map $f: X \rightarrow Y$ between topological spaces which is open, surjective and has (pathwise) connected fibers. Consider categories $\text{Sh}(X)$, $\text{Sh}(Y)$ of sheaves (with values in any given fixed sensible category) on $X$ and $Y$. The inverse image functor $f^\ast: \text{Sh}(Y) \rightarrow \text{Sh}(X)$ will then be fully faithful and hence we can consider $\text{Sh}(Y)$ up to equivalence as a full subcategory of $\text{Sh}(X)$. Obviously a necessary condition for a sheaf to lie in this subcategory is that it is constant when restricted to the fibers of $f$.

My question is: Is this condition also sufficient? Is it even sufficient for trivial fibrations $F\times Y \rightarrow Y$?

-

I found another condition: Assume that for every $x\in X$ and every open neighborhood $V$ of $x$ there exists an open neighborhood $U\subset V$ of $x$ such that $U\cap f^{-1}(y)$ is path-connected for every $y\in Y$. Then the assertion is correct. I did not look for counterexamples but I have the feeling that this extra condition is necessary.

We consider $\cal F$ as etale sheaf, that is, as topological space with a continuous map $\pi:{\cal F}\to X$ which is a local homeomorphism such that each fibre of $\pi$ carries a group structure which is continuous where defined. On the space $\cal F$ we establish an equivalence relation $\sim$ as follows: we say that $v$ and $w$ are equivalent if there exists a continuous path $\gamma$ connecting the two such that $f\circ\pi\circ\gamma$ is constant. Let $\cal G={\cal F}/\sim$ and equip this space with the quotient topology. We have a commutative diagram $${\cal F}\to\cal G$$ $$\pi\downarrow\ \ \downarrow p$$ $$X\to Y$$ where the right vertical arrow $p$ is induced. We show that $p$ is a local homeomorphism. Let $[v]\in\cal G$ and let $v\in\cal F$ be any preimage. Then there exists a neighborhood $U$ of $v$ in $\cal F$ such that $\pi(U)$ is open and $\pi$ is a homeomorphism from $U$ to $\pi(U)$. By the condition, we can also assume that $U\cap f^{-1}(y)$ is path-connected for every $y\in Y$. Then $U/\sim$ is an open neighborhood of $[v]$ in $\cal G$. Then $p(U/\sim)=f(\pi(U))$ is open in $Y$ and the same is true for any open subset of $U$, so $p$ is an open map on $U/\sim$. It also is continuous as for an open subset $W$ of the open set $p(U/\sim)$ the preimage $\pi^{-1}(f^{-1}(W))$ is open in $\cal F$ and so its image in $\cal G$ is open, which is the $p^{-1}(W)$. Finally, $p$ is injective on $U/\sim$ since if $p([v])=p([w])$ for two $u,v\in U$, then $\pi(u)$ and $\pi(v)$ lie in the same fibre of $f$. Since $\cal F$ is constant on that fibre and $U$ intersected with that fibre is path-connected, we infer $u\sim v$, so $p$ is injective, so in total, $p$ is a local homeomorphism.

-