I already know the 'Homotopy lifting property' in Algebraic topology which is, "Given covering map $p:\tilde{X}\to X$, a homotopy $f_t: Y\to X$ and a lift $\tilde{f_0}:Y\to \tilde{X}$ of $f_0$, there is a unique lift $\tilde{f_t}:Y\to \tilde{X}$ of $f_t$ extending $\tilde{f_0}$. It seems that this can be restated as If $f_0=p\circ\tilde{f_0}$ with $f_0=p\circ\tilde{f_0}\simeq f_1$ then there is a unique $\tilde{f_1}$ such that $\tilde{f_0}\simeq\tilde{f_1}$". Is this fine? Well, in Theorem 54.3 in Munkres' textbook, it talks somewhat similar statement as above: Let $p:(\tilde{X},\tilde{x_0})\to (X,x_0)$ be a covering map. "Let $f\simeq_{p} g$. Then, if we take a lift $\tilde{f},\tilde{g}$ of $f,g$ such that $\tilde{f}(0)=\tilde{g}(0)=\tilde{x_0}$, then $\tilde{f}\simeq_p\tilde{g}$". Ok, I wonder why we need the assumption $\textit{$\tilde{g}(0)=\tilde{x_0}$}$. Of course that if we state in terms of paths, we need to initiate the starting point of the paths to lift, so we need that assumption. But, they are homotopic. In Homotopy lifting property, we do not assume anything about $\tilde{g}(0)$. So I think we don't need that assumption and simply state as: "Let $p:(\tilde{X},\tilde{x_0})\to (X,x_0)$ be a covering map. Let $f\simeq_{p} g$. Then, if we take a unique lift $\tilde{f}$ of $f$ so that $f=p\circ\tilde{f}$, then there is a unique lift $\tilde{g}$ of $g$ such that $\tilde{f}\simeq_p\tilde{g}$". So what I'm thinking is that Theorem 54.3 is just a restatement of Homotopy lifting property. What am I misunderstanding?


1 Answer 1


The homotopy lifting property says that if $f_0$ has a lift $\tilde f_0$, then the homotopy $f_t$ has a lift $\tilde f_t$ starting with the given $\tilde f_0$.

In 54.4 Munkres considers two paths $f,g : I \to X$ starting at $x_0$ and ending at $x_1$ which are path homotopic. We know that they have unique lifts $\tilde f, \tilde g : I \to \tilde X$ such that $\tilde f (0) = \tilde g (0) = \tilde x_0$. The homotopy lifting property shows that any path homotopy $h_t : f \simeq g$ lifts to a homotopy $\tilde h_t$ starting with $\tilde f$. But it is not a priori clear that that $\tilde h_t$ is a path homotopy $\tilde f \simeq \tilde g$. That is what Munkres proves.

  • $\begingroup$ Thanks for the comment. But I'm still in wonder. Yes, the homotopy lifting property shows the lift of homotopy $h_t$ to $\tilde{h_t}$ such that $\tilde{h_0}=\tilde{f}$. But what about $\tilde{h_1}$? Since they are path homotopies, $\tilde{h_1}(0)=\tilde{h_0}(0)$ and $\tilde{h_1}(1)=\tilde{h_0}(1)$. Doesn't it implie that (path) homotpy lifting property ensures that $\tilde{h_1}(0)=\tilde{x_0}$ which is actually $\tilde{g}(0)$? Since clearly, $\tilde{h_1}$ is a lift of $g$ so Ok to assume $\tilde{g}=\tilde{h_1}$. $\endgroup$ Commented Sep 15, 2020 at 7:11
  • $\begingroup$ Of course the proof is not difficult. But you need the homotopy lifting property and an additional ingredient: Each path in $X$ starting at $x_0$ has a unique lift starting at a given $\tilde x_0 \in p^{-1}(x_0)$. This shows that the constant paths $h_0(s), h_1(s)$ lift to constant paths in $\tilde X$. And this implies $\tilde f(0) = \tilde h_1(0), \tilde f(1) = \tilde h_1(1)$. But only the requirement $\tilde g (0) = \tilde x_0$ implies what you want to show. If you drop that, then $\tilde g$ may start at another point of $\tilde X$ and cannot be path homotopic to $\tilde f$. $\endgroup$
    – Paul Frost
    Commented Sep 15, 2020 at 8:33

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