# Example of a map of coverings which is not a covering map, if the base space is not locally path-connected

Consider two coverings $$X \to Y$$ and $$X^\prime \to Y$$. A surjective map $$X \to X^\prime$$, which fits in a commutative triangle with the covering maps, is again a covering, given that $$Y$$ is locally path-connected. (See here.) If one drops that last assumption, does someone know any counterexamples?

Take $$Y=\mathbb{Q}$$ and $$X=X'=\mathbb{Q}\times\mathbb{N}$$, both covering $$Y$$ by the projection map. Now define $$f:X\to X'$$ as follows. Let $$(\alpha_n)$$ be a decreasing sequence of irrational numbers converging to $$0$$. Define $$f(x,2n)=(x,0)$$ if $$x<\alpha_n$$ and $$f(x,2n)=(x,n)$$ if $$x>\alpha_n$$, and define $$f(x,2n+1)=(x,n)$$. Then $$f$$ is surjective and commutes with the projection maps down to $$Y$$. However, $$f$$ is not a covering map because no neighborhood of $$(0,0)\in X'$$ is evenly covered: if $$x>0$$ then the fiber of $$f$$ over $$(x,0)$$ is finite (there are only finitely many $$n$$ such that $$x<\alpha_n$$) but the fiber of $$f$$ over $$(0,0)$$ is infinite.
Let me also remark that for this to be impossible, it suffices for $$Y$$ to be just locally connected, not necessarily locally path connected. In that case, given any $$y\in Y$$, we can find a connected neighborhood $$U$$ of $$y$$ that is evenly covered by both the covering maps $$X\to Y$$ and $$X'\to Y$$. It then follows that each of the copies of $$U$$ in $$X'$$ must be evenly covered by the map $$X\to X'$$, since by connectedness that map can only map entire copies of $$U$$ to other entire copies of $$U$$ (it cannot split them up and map part to one copy of $$U$$ and part to another).