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
Best Answer
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.