Help in my understanding of Lemma $54.2$ from Munkres Topology concerning path homotopy lifting.

Question

I am looking for a step by step explanation of what is going on in this proof. I am lost from the second line of the proof onward! I just can't wrap my head around what is going on in the proof technically.

Answer 1

Since $p : E \to B$ is a covering map, there exists an open cover $V_\alpha $ for $B$ such that, over a given $V_\alpha$, $E$ looks like a disjoint union of open sets, each homeomorphic to $V_\alpha$ via the projection $p$.

You are given a map $F : I \times I \to B$ and you want to lift this to a map $\tilde F : I \times I \to E$. The strategy is to break up the domain $I \times I$ into lots of little squares, small enough that $F$ maps each little square inside one of the $V_\alpha$'s.

Why is this useful? Because once you have mapped your little square into a $V_\alpha$, there is a very natural way to lift your little square up to $E$. Remember that $p^{-1}(V_\alpha)$ looks like a disjoint union of open sets homeomorphic to $V_\alpha$. Let's write $$p^{-1}(V_\alpha) = \tilde V_{\alpha, 1} \cup \dots \cup \tilde V_{\alpha, n}$$ You can lift your little square from $V_\alpha$ to any one of the $\tilde V_{\alpha, i}$'s by taking its preimage under the homeomorphism $p : \tilde V_{\alpha, i} \to V_{\alpha}$.

The idea is to do this lift one square at a time, working from left to right then top to bottom in $I \times I$.

But there is a problem: Once you have mapped a little square to a $V_\alpha$, how do you chose which of the $\tilde V_{\alpha, i}$'s to lift it to?

The answer is to lift your little square to the $\tilde V_{\alpha, i}$ such that the image of your little square joins continuously with the images of the previous little squares that you have already put in place. It's like solving jigsaw puzzle piece by piece.

I hope this gives you enough intuition to tackle to proof.