A locally constant sheaf on a locally connected space is a covering space; Proof?

As part of my hobby i'm learning about sheaves from Mac Lane and Moerdijk. I have a problem with Ch 2 Q 5, to the extent that i don't believe the claim to be proven is actually true, currently. Here is the question repeated:

A sheaf on a locally connected space $X$ is locally constant if each point $x \in X$ has a basis of open neighborhoods $\mathcal{N}_x$ such that whenever $U,V \in \mathcal{N}_x$ with $U \subset V$, the restriction $\mathcal{F}V \rightarrow \mathcal{F}U$ is a bijection. Prove that $\mathcal{F}$ is locally constant iff the associated etale space over $X$ is a covering.

I don't have a problem (i think) showing the "covering" $\rightarrow$ "locally constant" direction of the implication. My problem is in the reverse direction. In fact, in the process of trying to solve this problem i think i have come up with a very simple counterexample.

Here's my attempt at a counterexample:

Let $X = \{a,b,c\}$ be finite space with open subsets $\{\{\},\{a\},\{a,b\},\{a,b,c\}\}$. I claim this is hyper-connected and locally connected.

Let $\mathcal{F}$ be the sheaf with $\mathcal{F}(\{a,b,c\}) = \{q\}$, $\mathcal{F}(\{a,b\}) = \{r,s\}$, and $\mathcal{F}(\{a\}) = \{t\}$, where $r = q|_{\{a,b\}}$, and $t = q|_{\{a\}} = r|_{\{a\}} = s|_{\{a\}}$. I claim this defines a locally constant sheaf. Indeed we can have $\mathcal{N}_{a} = \{\{a\}\}$, $\mathcal{N}_{b} = \{\{a,b\}\}$, and $\mathcal{N}_c = \{\{a,b,c\}\}$.

Now we can calculate the corresponding etale space $p : E \rightarrow X$ as follows: $E = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$, and has these four distinct elements since $\dot{q}(a) = \dot{r}(a) = \dot{s}(a) = \dot{t}(a)$, and $\dot{q}(b) = \dot{r}(b)$. The action of $p$ follows by construction.

However, $p$ does not appear to me to be a covering space. In particular, the only neighborhoods of $b \in X$ are $\{a,b\}$and $\{a,b,c\}$ and we have $p^{-1}(\{a,b\}) = \{\dot{q}(a),\dot{q}(b),\dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b\}$ because $2 \nmid 3$. Similarly, $p^{-1}(\{a,b,c\}) = \{\dot{q}(a), \dot{q}(b), \dot{q}(c), \dot{s}(b)\}$ and this cannot be partitioned into homeomorphic images of $\{a,b,c\}$ since $3 \nmid 4$.

So i think i have a locally constant sheaf on a locally connected space whose corresponding etale space is not a covering. What have i done wrong? I have seen the claim mentioned elsewhere so i'm convinced i'm wrong somewhere. Perhaps someone knows where a correct proof is published otherwise?

Solution 1:

I am posting this just to confirm that the definition of locally constant as given in the OP (and I guess in the text that the OP is citing) is incorrect.

A sheaf is called locally constant if each point has a n.h. on which it becomes consant. This is a stronger condition than that in the OP, and it is true that a sheaf is locally constant iff the associated etale space is a covering space (perhaps under mild conditions such as local connectedness --- I haven' thought it through for a while).

The condition in the OP is much weaker: it just says that the transition maps in the definition of the stalk of $\mathcal F$ at each point $x$ are eventually constant, and many kinds of sheaves will satisfy this condition, e.g. skyscrapes, sheaves such as $j_!\mathbb Z$ (say for $j$ being the inclusion of an open set in a variety), and more generally (I think) any constructible sheaf on a variety over $\mathbb C$.

Solution 2:

I realize it's been a while since the original question was posted, but I've run into the same issue with the same question. I don't believe it's true as written, either. Here is my counterexample: let $X$ be a metric space and let $A$ be a set with more than one element. Then given $x \in X$, $\mbox{Sky}_x(A)$ is locally constant under the definition given in the problem. If $y \neq x$, then take as the neighborhood basis at $y$ the balls $$B\bigg(y;\frac{d(x,y)}{2^n}\bigg).$$ The restriction morphisms between these neighborhoods are obviously bijections, since the sets are all singletons. At $x$, we may take the neighborhood basis to be all open sets, and the restriction morphisms are again bijections, since they are all identity on $A$.

However, the bundle of sections for this sheaf is not a covering space. The fiber at $x$ consists of $|A|$ points, while the fiber at all other points consist of only one point. A covering space has fibers whose size are locally constant, so the associated bundle is not a covering map.