# Lifting homotopic paths

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 $$\tilde F(0,t)\in p^{-1}(b_0)$$.

• 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 $\tilde{F}(0,0)=e_0$. – Randall Oct 26 '18 at 2:33
• @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 $e_0$. – user531587 Oct 26 '18 at 2:43
• Ah, the leg $\{0\} \times [0,1]$ is connected and its image lies in the fiber. – Randall Oct 26 '18 at 2:45

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 $$1\times I$$.
• 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 $\{e_0\}$. – user531587 Oct 26 '18 at 3:07