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This is a question from "A First Course in Algebraic Topology" by C. Kosniowski

Let $p_1$: $X_1 \to X$, $p_2$: $X_2 \to X$ be covering maps with $X$ connected and locally path connected. (i) Prove that if there is a continuous surjection $f$: $X_1 \to X_2$, then $f$: $X_1 \to X_2$ is a covering map. (ii) Prove that if $X_2$ is path connected and if there is a continuous map $f$: $X_1 \to X_2$, then $f$: $X_1 \to X_2$ is a covering map.

Not much clue on how to prove (i). One difficulty is to show $f$ is a homeomorphism "locally". For (ii), even if assuming (i), I can't show $f$ must be a surjection.

Can anyone give me some help or clues?

In case this exercise is wrong, counterexamples are also welcome.

Note

It is NOT assumed that $f\circ p_2 = p_1$. So it is different from:

[1] Exercise 1.3.16 in Hatcher

[2] math.stackexchange.com/q/109695

Yan Zhu
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  • You are missing some assumptions about your maps. – Moishe Kohan Dec 08 '22 at 04:03
  • @MoisheKohan If you can give counterexamples, it is also good for me. – Yan Zhu Dec 08 '22 at 04:19
  • If $X, Y$ are connected compact manifolds of positive dimension, then there is always a continuous surjection $X\to Y$ that factors though a map from $X$ to $[0,1]$. – Moishe Kohan Dec 08 '22 at 04:31
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    Your question has been asked quite often in this forum. The first occurrence seems to be this: https://math.stackexchange.com/q/109695 – Paul Frost Dec 08 '22 at 10:19
  • Also observe that Kosniowski's definition of a covering is a bit sloppy: "The continuous map $p: \tilde X \to X$ is said to be a covering map if each point $x \in X$ has an open neighbourhood evenly covered by $p$." But the preceding definition of $U \subset X$ being evenly covered does not exclude $p^{-1}(U) = \emptyset$ so that $p$ is not necessarily onto. What he writes after "In other words ..." is the adequate definition. – Paul Frost Dec 08 '22 at 10:27
  • @PaulFrost Thanks for bringing these answered questions. But I believe the assumption here is different. It doesn't say $f \circ p_2 = p_1$. It does make me feel the exercise may be wrong. – Yan Zhu Dec 08 '22 at 18:01
  • @YanZhu You are right. I reopened the question. – Paul Frost Dec 08 '22 at 18:06

1 Answers1

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If we require $f \circ p_2 = p_1$, then it is true. See the linked questions.

Without this assumption it is not true. Here is a counterexample.

Take $X_1 = X_2 = X = [0,1]$ and $p_1 = p_2 = id$. The map $f : [0,1] \to [0,1], f(x) = 2x$ for $x \le 1/2$, $f(x) = 1$ for $x \ge 1/2$, is surjective, but no covering map.

More generally, if the claim in the exercise were true, then each continuous surjection $f : X \to X$, where $X$ is an arbitrary connected and locally path connected space, would be a covering map. This is of course nonsense.

Paul Frost
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