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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?

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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).

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