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?

  • $\begingroup$ Whoops: $p: \tilde X \rightarrow X, q: \tilde Y \rightarrow X.$ I've added their definitions in the descriptions. $\endgroup$ – Oliver G Sep 8 '17 at 14:00
  • $\begingroup$ How much connectivity can I assume? $\endgroup$ – Randall Sep 8 '17 at 14:22
  • $\begingroup$ @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. $\endgroup$ – Oliver G Sep 8 '17 at 15:25
  • $\begingroup$ Oh, well that is negotiable for some authors. I'll assume path connectivity, then. $\endgroup$ – Randall Sep 8 '17 at 15:29

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, $h(\eta(1)) = \tilde x$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.