If $(\tilde Y, q)$ and $(\tilde X, p)$ are covering spaces of $X$ with $h : \tilde Y \rightarrow \tilde X$ and $ph=q$, then $h$ is onto.

If $(\tilde Y, q)$ and $(\tilde X, p)$ are both covering spaces of $X$ with $h : \tilde Y \rightarrow \tilde X$ and $ph=q$, then $h$ is onto.

The hint given says to use unique path lifting to solve this but I can't see any way how.

By "unique path lifting", I'm assuming that means using the unique paths $\tilde f : (I, 0) \rightarrow (\tilde X, \tilde x_0)$ and $\tilde g : (I, 0) \rightarrow (\tilde Y, \tilde y_0)$ such that $p\tilde f = f$ and $q \tilde g = g$ where $p : \tilde X \rightarrow X$ and $q : \tilde Y \rightarrow X$, but I can't figure out how to proceed.

Anyone have any ideas?

• Whoops: $p: \tilde X \rightarrow X, q: \tilde Y \rightarrow X.$ I've added their definitions in the descriptions. – Oliver G Sep 8 '17 at 14:00
• How much connectivity can I assume? – Randall Sep 8 '17 at 14:22
• @Randall I'm not sure what you mean by that, from Rotman's Algebraic topology $\tilde X, \tilde Y$ and $X$ have to be path connected in order for $q,p$ to be covering projections. – Oliver G Sep 8 '17 at 15:25
• Oh, well that is negotiable for some authors. I'll assume path connectivity, then. – Randall Sep 8 '17 at 15:29

1 Answer

We need to assume $\tilde X$ path connected, otherwise it may not work.

First, choose a base point in eacH of the spaces, $x_0 \in X$, $\tilde y_0 \in \tilde Y$, $\tilde x_0 \in \tilde X$, that map accordingly (...). Now let $\tilde x \in \tilde X$ arbitrary. Take a path $\tilde \gamma$ from $\tilde x_0$ to $\tilde x$ in $\tilde X$. The path $\gamma \colon = p \circ \tilde \gamma$ in $X$ can be lifted in $\tilde Y$ starting from $\tilde y_0$. Let's call this path $\eta$. Then $h \circ \eta$ is a path in $\tilde X$ starting at $\tilde x_0$ and is a lift for $\gamma$. By the uniqueness of lifts we have $h \circ \eta = \tilde \gamma$. In particular, $h(\eta(1)) = \tilde x$.