An exhaustive continuous map is a covering map.

Question

$p_1:\tilde X_1 \rightarrow X \, ; \, p_2:\tilde X_2 \rightarrow X$ two coverings maps, where $X$ connected and locally path-connected, and suppose that $f:\tilde X_1 \rightarrow \tilde X_2$ is an exhaustive continuous map, then $f$ is a covering map.

I have been thinking but I don´t see nothing until now. Thanks.

Answer 1

You need to have a commutative diagram, i.e, $p_2\circ f = p_1$. Then, $f$ is just a covering homomorphism which is a covering map by Lee, Proposition 11.36, b).