Lifting homotopic paths

Question

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

Answer 1

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 $1\times I$.