Let $p:E\to B$ be a covering, let $b\in B$, and let $e,e'\in p^{-1}(b).$
Then a path $f:I\to B$ with $f(0)=b$ lifts uniquely to a path $g:I\to E$ such that $g(0)=e$ and $f\circ g=f.$

The proof:

subdivide $I$ into subintervals each of which maps to a fundamental neighborhood under $f$, and lift $f$ to $g$ inductively by use of the prescribed homeomorphism property of fundamental neighborhoods.

Hello,
Is it possible that the covering map $p:\tilde X \to X$ may not carry over the entire homotopy into $\tilde X$?

Where $\alpha$ is null-homotopic on $X$, and $\tilde\alpha$ lifts to $\alpha$ through covering map $p$.

sorry, I got confused by chatGPT. With this.
Given a covering map $p:\tilde X \to X$ and a loop $\alpha:[0,1]\to X$ based at $x_0$. Suppose $\tilde \alpha$ is a lift of $\alpha$ with $\tilde \alpha(0)=\tilde x_0$.

I'm trying to show that if $\alpha$ is null-homotopic in $X$, then either $\tilde \alpha$ is null-homotopic(which makes sense through $p$ sending open nbhds to open nbhds homeomorphically) in $\tilde X$ or it is a loop that does not bound in $\tilde X$

The question is: Prove that if $\alpha$ is null-homotopic in $X$, then either $\tilde \alpha$ is null-homotopic in $\tilde X$ or it is a loop that does not bound in $\tilde X$.


I don't know anything about second part of this Q.