Continuous maps between covering spaces are surjective

This is Exercise 10.16 from Rotman's An Introduction to Algebraic Topology:

Let $(\widetilde{Y}, q)$ and $(\tilde{X}, p)$ be covering spaces of $X$. If there exists a continuous $h: \widetilde{Y} \rightarrow \tilde{X}$ with $ph = q$, then $h$ is a surjection. (Hint: Use unique path lifting.)

Here is what I have so far:

Given $\tilde{x} \in \tilde{X}$, let $c_{\tilde{x}}$ be the constant map at $\tilde{x}$. Then $pc_{\tilde{x}}$ is the constant path at $x \in X$ and so lifts to the constant path $c_{\tilde{y}} \subseteq \widetilde{Y}$ for some $\tilde{y} \in \widetilde{Y}$. Further, $c_{x} = qc_{\tilde{y}} = phc_{\tilde{y}}$, so $hc_{\tilde{y}}$ lies in the fiber over $x$.

However, I'm not really sure how to control the starting point of $hc_{\tilde{y}}$, so I'm not sure how to incorporate unique path lifting.

• You must have some connectivity assumption. – Amitai Yuval May 21 '17 at 19:37
• @AmitaiYuval Rotman assumes covering spaces to be path connected. – Jacob Bond May 21 '17 at 19:41

Let $y\in \tilde Y$ and $x\in \hat X$. There exists a path $c$ such that $c(0)=h(y)$ and $c(1)=x$. The unique lifting property implies that there exists a unique path $d_t:[0,1]\rightarrow \tilde Y$ such that $d(0)=y$ and $q(d)=p(c)$. The path $h(d)$ verifies $h(d)(0)=c(0)$ and $p(h(d))=p(c)$ The unique path lifting property implies that $h(d)=c$, and henceforth that $h(d(1))=x$.
There is at least one point in the image of $h$, call it $x$. You want to prove that any other point is also in the image. So let $x'$ be another point in $\tilde{X}$. Take a path connecting $x$ to $x'$ (we assume path connectivity). Push this path forward to the base $X$, then pull it back to $\tilde{Y}$. The path you get in $\tilde{Y}$ is pushed forward by $h$ to the path you took in $\tilde{X}$ to begin with. In particular, the endpoint $x'$ is in the image.