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Let $p:E\to B, e_0\to b_0$ be a covering map. Let $f,g$ be two paths from $b_0$ to $b_1$ and let $\tilde f,\tilde g$ be their liftings to $B$ starting at $e_0$. Suppose $f,g$ are path homotopic. I'm trying to understand why the liftings must be homotopic.

First, let $F(s,t)$ be a homotopy between $f$ and $g$: $$F(s,0)=f(s)\\F(s,1)=g(s)\\F(0,t)=b_0,\ F(1,t)=b_1$$

Note that $F(0,0)=b_0$. By Lemma 54.2 in Munkres:

Let $p:E\to B$ be a covering map; let $p(e_0)=b_0.$ Let the map $F:I\times I\to B$ be continuous, with $F(0,0)=b_0$. There is a unique lifting of $F$ to a continuous map $$\tilde F:I\times I\to E$$ such that $\tilde F(0,0)=e_0$. If $F$ is a path homotopy, then $\tilde F$ is a path homotopy.

there exists a unique lifting $\tilde F:I^2\to E$ such that $\tilde F(0,0)=e_0$. Moreover, the lemma guarantees that $\tilde F$ is a path homotopy. So we know that $$\tilde F(0,t)=e_1,\ \tilde F(1,t)=e_2$$ for all $t$ and for some $e_1,e_2\in E$.

Munkres says that $\tilde F(0,t)=e_0$. Why is that so? By commutativity of the diagram we only know $p(\tilde F(0,t))=F(0,t)=b_0$, so $\tilde F(0,t)\in p^{-1}(b_0)$.

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    $\begingroup$ You choose that first, then turn the crank on the proof. There's not a unique lift and that lift happens to satisfy $\tilde{F}(0,0)=e_0$. There is a unique lift that satisfies the constraint $\tilde{F}(0,0)=e_0$. In other words, a lift exists and it is unique once you specify $\tilde{F}(0,0)=e_0$. $\endgroup$ – Randall Oct 26 '18 at 2:33
  • $\begingroup$ @Randall I still don't get it. I understand that we consider the specific lift $\tilde F$ such that $\tilde F(0,0)=e_0$. But the claim is that for all $t\in I$, $\tilde F(0,t)=e_0$. The lemma I mentioned does not say that we can choose $\tilde F$ with that property, we can only choose what $\tilde F(0,0)$ is. The values $\tilde F(0,t)$ for $t\in(0,1]$ are uniquely determined, but I don't understand why all of them equal $e_0$. $\endgroup$ – user531587 Oct 26 '18 at 2:43
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    $\begingroup$ Ah, the leg $\{0\} \times [0,1]$ is connected and its image lies in the fiber. $\endgroup$ – Randall Oct 26 '18 at 2:45
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Munkres spell out a reason in his proof of the Lemma. The inverse image of $b_0$ has the discrete topology, therefore any subset of $p^{-1}(b_0)$ with cardinality greater than $1$ is the union of two disjoint open sets, i.e., separated. The map $\tilde{F}$ is continuous, hence since $0\times I$ is connected $\tilde{F}(0\times I)$ is connected, therefore since non empty it must have cardinality $1$. The same argument applies to $1\times I$.

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    $\begingroup$ I guess for completeness it should be noted that $\tilde F(\{0\}\times I)$ is $e_0$ (and not some other point in $p^{-1}(b_0)$) because the image of $(0,0)\in \{0\}\times I$ lies in $\{e_0\}$, so the image of the whole image of $\{0\}\times I$ under $\tilde F$ should lie in $\{e_0\}$. $\endgroup$ – user531587 Oct 26 '18 at 3:07

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