Covering map with a right inverse is a homeomorphism

Question

I am stuck on the following problem. I believe I am very close.

Let $p:E\to B$ be a covering map. Suppose that there exists a continuous $f: B\to E$ so that $p\circ f = id_B$ (that is, $p$ has a right inverse). Prove that if $E$ is path connected, then $p$ is a homeomorphism.

Since $p$ is covering map, I know that it suffices to show $p$ is injective. I am guessing I am to show that for any $b\in B$, $|p^{-1}(\{ b\} )|=1$.

Let $b\in B$. Let $e\in p^{-1}(\{ b\} ) $. I know that the monodromy action is transitive since $E$ is path connected. This, and the fact that $p$ is surjective, gives $1 \leq |p^{-1}(b)| \leq | \pi _1 (B,b)| $. (This comes from the correspondence $\pi _1(B,b) \to p^{-1}(b) $ via $[\gamma ] \mapsto e \cdot [\gamma ] $ being surjective.)

From this, I predict that I will need to show that $\pi _1(B,b)$ is trivial (i.e. $B$ is simply connected). However I have no idea how to show this.

How do we use the right inverse? The most I can get out of it is that $f$ is a lift of the identity function on $B$. This gives the induced homomorphism $p_*: \pi _1 (E,e) \to \pi _1 (B,b)$ is an isomorphism. Does this help me?

Any suggestion would be great.

Answer 1

You are correct that it suffices to show that $p$ is injective. This is because $p \circ f \circ p = p \circ id_E$, so if $p$ is injective then $f \circ p = id_E$ since injective maps can be cancelled from the left.

To show that it is injective, consider $a, b \in E$ such that $p(a) = p(b)$. Since $E$ is path-connected, let $\gamma : [0, 1] \to E$ be a path such that $\gamma(0) = a$ and $\gamma(1) = b$. Now consider the map $p \circ \gamma : [0, 1] \to B$. Note that $p \circ (f \circ p \circ \gamma) = (p \circ f) \circ (p \circ \gamma) = id_B \circ (p \circ \gamma) = p \circ \gamma$.

By the path lifting property of covering spaces, this implies that $(f \circ p \circ \gamma) = \gamma$. In that case, $a = \gamma(0) = f(p(\gamma(0))) = f(p(a)) = f(p(b)) = f(p(\gamma(1))) = \gamma(1) = b$.

Therefore, $a = b$.