Let $p : E \to X $ be a covering map, with $X$ path-connected and locally path-connected, and $E$ path-connected. (The local path-connectedness of $E$ follows.) Also, let $f$ be a homeomorphism of $X$. I want to determine whether there is a "homeomorphism" $g$ of $E$ such that $p \circ g = f \circ p$, that is, $g$ lifts $f$.
In my intuition, I think this is false (I first tried to prove the statement assuming true, but I got stuck.) However, I am having a hard time finding a counterexample. How do I have to proceed?
Best Answer
Well $f\circ p$ factors through $g$ if and only if the image of $\pi_1(E)$ by the induced morphism is included in that of $\pi_1(E)$.
So you are right that it doesn't always happen. It does happen if $E$ is a universal covering (because $\pi_1(E) =0$), so in our example we have to chose a nonuniversal covering.
Here's one example : let $\mathbb R\times S^1 \to \mathbb T = S^1\times S^1$ be the standard covering and $f:\mathbb{T\to T}$ be the homeomorphism that switches coordinates : $f(z_1,z_2) = (z_2,z_1)$.
Then if it lifted, we would get a map $g:\mathbb R\times S^1\to \mathbb R\times S^1$ that satisfied $p\circ g = f\circ p$. Now look at $h(z) := pr_{\mathbb R}\circ g (0,z)$ : it's a map $S^1\to \mathbb R$ and it satisfies $\exp \circ h (z) = z$, where $\exp : \mathbb R \to S^1$ is the standard covering map.
This is of course absurd.