1
$\begingroup$

I'm studying covering maps and homotopy lifting and I would like to clarify a few things which my lecture notes doesn't seem to make clear.

A lemma in my lecture notes says:

Let $p: \tilde Y \to Y$ be a covering map, and let $f:X \to Y$ be a continuous map, with $X$ connected. Suppose that $\tilde f_1 : X \to \tilde Y$ and $\tilde f_2 : X \to \tilde Y$ are lifts of $f$. If $\tilde f_1(X) = \tilde f_2(x)$ for some point $x \in X$, then $\tilde f_1$ and $\tilde f_2$ agree everywhere.

But it true given a covering map $p: \tilde Y \to Y$, and given a continuous map $f:X \to Y$, you can always find a lift of $f$ to $\tilde Y$?

I know this is true for a path in $Y$ but does this generalize?

Thanks!

$\endgroup$
  • 2
    $\begingroup$ No, this is the homotopy lifting criterion. You need $f_*(\pi_1X)\subseteq p_*(\pi_1\tilde{Y})$. $\endgroup$ – Chris Gerig Mar 22 '13 at 20:20
  • 1
    $\begingroup$ For example, if $X=Y=S^1$ and $\bar Y=\mathbb R$ with the normal map $p:\mathbb R\to S^1$, then $\text{id}:S^1\to S^1$ can't lift. $\endgroup$ – Thomas Andrews Mar 22 '13 at 20:27
  • 1
    $\begingroup$ Yes, and the lifting criterion requires $X$ to be locally and globally path-connected. $\endgroup$ – Stefan Hamcke Mar 22 '13 at 20:27
  • 1
    $\begingroup$ In general, if $\text{id}_Y$ lifts, then $\bar Y\cong Y\times D$ for some discrete space $D$. $\endgroup$ – Thomas Andrews Mar 22 '13 at 20:30

Your Answer

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

Browse other questions tagged or ask your own question.