Lifts of relative homotopic paths to the same base point $\Rightarrow$ The lifts homotopic and agree on the end point.

algebraic-topology

Let $p:X\rightarrow Y$ a covering map, $a,b : I\rightarrow Y$ are relatively homotopic to each other: $a \simeq_{\partial I} b $ (denoting $F:I\times[0,1] \rightarrow Y$ the homotopy between $a$ and $b$). The lifts $\tilde{a} , \tilde{b}$ satisfies $\tilde{a}(0)=\tilde{b}(0)$. I wish to prove that $\tilde{a}(1) = \tilde{b}(1)$ and $\tilde{a} \simeq \tilde{b}$.

For $\tilde{a}(1) = \tilde{b}(1)$ I tried to show by using the homotopy lifting property of $F$, and use the homotopy $G:I\times [0,1] \rightarrow X$ satisfies: $G_0 = \tilde{a}$ and $p\circ G_t =F_t$. Then I took two approaches:

  1. Show the existence of a lift $\gamma$ of $b$ such that $\gamma(0) = \tilde{a}(0)$ and $\gamma(1) = \tilde{a}(1)$ then from path lifting uniqueness $\gamma$ must be equal to $\tilde{b}$.
  2. Show that $G_1$ and $\tilde{b}$ agree in one point and then by theorem they agree on all points. Yet this approach (which I didn't succeeded to prove), just satisfies the fact $\tilde{a} \simeq \tilde{b}$ but as fat as I understand doesn't promise $\tilde{a}(1) = \tilde{b}(1)$.

This post is very similar, relating to loops which are lifted to the same base point:
Are the lifts of homotopic curves necessarily homotopic?

Best Answer

Let $\tilde a,\tilde b:I\rightarrow X$ be lifts of $a,b:I\rightarrow Y$ satisfying $p\circ \tilde a=a$, $p\circ \tilde b=b$, and let $F_t:a\simeq_{\partial I}b:I\rightarrow Y$ be a relative homotopy satisfying $F(s,0)=a(s)$, $F(s,1)=b(s)$, $F(0,t)=a(0)=b(0)$ and $F(1,t)=a(1)=b(1)$. Now apply the HELP lemma in the diagram

$\require{AMScd}$ \begin{CD} 0\times I\cup I\times \partial I@>G'>> X\\ @Vi V V @VVpV\\ I\times I @>F>> Y \end{CD}$\require{AMScd}$

where $G'(s,0)=\tilde a(s)$, $G'(s,1)=\tilde b(s)$, $G'(0,t)=\tilde a(0)=\tilde b(0)$. Note that $p\circ G'=F\circ i$. From the data the HELP lemma gives a map $G:I\times I\rightarrow X$ satisfying

$$G(s,0)=\tilde a (s),\quad G(s,1)=\tilde b(s),\quad G(0,t)=\tilde a(0)=\tilde b(0),\quad p\circ G(s,t)=F(s,t).$$

In particular $p(G(s,1))=p(\tilde a(1))=p(\tilde b(1))=F(s,1)=a(0)=b(0)$, so the assignment $s\mapsto G(s,1)$ is a path in $X$ contained in the fibre over $a(0)=b(0)$, going from $\tilde a(0)$ and ending at $\tilde b(0)$. Since the covering projection has discrete fibres such a path must be constant.

Hence $\tilde a(1)=\tilde b(1)$, and $G:\tilde a\simeq_{\partial I}\tilde b$ is a relative homotopy.

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