# An exhaustive continuous map is a covering map.

$p_1:\tilde X_1 \rightarrow X \, ; \, p_2:\tilde X_2 \rightarrow X$ 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.

• What does "exhaustive" mean? Does it mean "surjective"? – Zev Chonoles May 26 '15 at 3:02
• Yes, "exhaustive" mean that $f$ is surjective. – sti9111 May 26 '15 at 3:31

## 1 Answer

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).

• HI, can you be more especific with the reference, please?, because I don´t know that book by lee. – sti9111 May 26 '15 at 3:43
• Introduction to Topological Manifolds – Adam Baranowski May 26 '15 at 6:16
• Even with commutativity, isn't the statement in the OP false? Let $\tilde{X}_1=\{1,2,3\}, \tilde{X}_2=\{1,2\}, X=\{1\}$. Let $p_1$ and $p_2$ be the unique maps to $X$ which are covering maps and let $f\colon 1,2\mapsto 1, 3\mapsto 2$ which is not covering. – Dan Rust May 26 '15 at 13:31
• @DanielRust Good catch! I think that conditions on connectedness should be on $\tilde{X_i}$ instead of $X$. – user 1987 May 26 '15 at 22:06