I am reading 1.3 Covering space of Hatcher's 'Algebraic Topology' and I cannot find out what I am missing.

In the book, he says

"The advantage of this is that, by the homotopy lifting property, homotopy classes of paths in $\widetilde{X}$ starting at $\widetilde{x_0}$ are the same at homotopy classes of paths in $X$ starting at $x_0$."

where $\widetilde{X}$ is a universal covering of $X$ ($p:\widetilde{X}\rightarrow X$).

I don't really get it. For example, I think, we can build two different lifts, $f_1$ and $f_2$ of a path $f$ in $X$ such that $f_1\not\simeq f_2$.

Thanks in advance!

  • 2
    $\begingroup$ Well you think so, but that doesn't mean it's true. It actually isn't, that's the point; Hatcher must prove it at some point $\endgroup$ – Maxime Ramzi Feb 12 '19 at 23:12

Let $p : \tilde{X} \to X$ be any covering (it may be the universal covering, but we do not require it). Hatcher proves that for any homotopy $h : Y \times I \to X$ and any lift $\tilde{h}_0 : Y \to \tilde{X}$ of $h_0$ (where $h_t : Y \to X, h_t(y) = h(y,t)$) there exists a unique lift $H : Y \times I\to \tilde{X}$ of $h$ such that $H_0 = \tilde{h}_0$.

This implies the unique path lifting property: Given a path $u : I \to X$ starting at $x_0 \in X$, then for any $\tilde{x}_0 \in p^{-1}(x_0)$ there exists a unique lift $U : I \to \tilde{X}$ such that $U(0) = \tilde{x}_0$.

You are right, a path $u$ in $X$ can have lifts which are not homotopic as paths ( a homotopy of paths is one keeping the end-points fixed). However, if both lifts start at the same point, then they are identical.

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  • $\begingroup$ Thank you! I got what I was missing! $\endgroup$ – Lev Bahn Feb 12 '19 at 23:36

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