Is sheaf direct image exact for covering maps?

Question

Let $\pi\colon X\to Y$ be a covering map. Is $\pi_*$ an exact functor of abelian sheaves? It's straightforward to show that this is true if $\pi$ is a finite cover, so I'm going to assume that $\pi$ is infinite.

Answer 1

I hope I didn't made a mistake, here is a try for a counter-example : let $X = D \times \Bbb N$ where $D$ is the unit disk. Write $D_n := D \times \{n\}$. $Y = D$ and $\pi$ is the projection.

Define a sheaf $F_n$ on each $D_n$ by $F_n(U) = \Bbb Z$ if $U \subset D(0,1/n)$ and $0$ else. Let $F = \prod_{n \geq 0} F_n$. Now, let $G = \prod_{n \geq 0} G_n$ where $G_n$ is the skyscraper sheaf at $0 \in D_n$ with stalk $\Bbb Z$. Clearly, we have a natural morphism $f : F \to G$ and such morphism is surjective.

On the other hand, if we look at the direct image we should have a surjection $(\pi_*F)_0 \to (\pi_*G)_0$ if our functor was exact. Let $s_0 \in (\pi_*G)_0$, if there is another germ $t_0 \in (\pi_*F)_0$ with $f_0(t_0) = s_0$ then this means by definition that we should have a small open $U$ with $t \in (\pi_*F)(U)$, $s \in (\pi_*G(U))$ and $f(t) = s$, and the stalk at $0$ of $t$ (resp. $s$) is $t_0$ (resp $s_0$.) Such open $U$ contains some ball of radius $\delta > 1/N > 0$ around zero. If we look at $0 \in D_N$, we have $F_N(\pi^{-1}(U)) = \{0\}$ by definition, but $G_N(\pi^{-1}(U)) = \Bbb Z$, in particular we don't have a surjection $(\pi_*F)_0 \to (\pi_*G)_0$. So this functor is not exact.