Suppose $X$ is a graph (a $1$-dimensional CW complex), and $X \to Y$ is a covering map. Then is $Y$ a graph?
I think the answer is yes, but I haven't been able to think of a proof.
Edit: After posting my question, I came up with an idea that works when $X \to Y$ is a normal cover and $X$ has finitely many vertices. In this case, we can view $Y$ as the quotient of $X$ by the deck group $G$ associated to this cover. Since $X$ has finitely many vertices, we can refine its graph structure so that the orbit of every vertex consists only of vertices. Note that this also means that the orbit of any point that is not a vertex consists only of points that are not vertices. In particular, if $e$ is an edge and $g \in G$ is a deck transformation, then the image $g(e)$ consists of exactly two vertices, so it is also an edge! Quotienting by the deck group therefore involves identifying edges with one another, the result of which is a graph.
Note that this argument does not work (at least in its current form) when $X$ is an infinite graph. For instance, we could take $X = \mathbb{R}$ with a vertex at $n + 1 / (|n| + 1)$ for each integer $n$. Then there is no way to "refine" the graph structure of $X$ so that the orbit of each vertex under the translation action by $\mathbb{Z}$ is a vertex.