Do covering spaces have the homotopy lifting property?

Question

I am having troubles to understand the interplay bewteen fiber bundles, fibrations and covering spaces. What I (think to) know is that:

1. A covering space is a fiber bundle
2. A fiber bundle is a fibration, i.e. it has the homotopy lifting property
3. A map $f: X \to B$ can be lifted to a covering space $p: E \to B$ if and only if $f_*(\pi_1(X)) \subseteq p_*(\pi_1(E))$

(Some additional hypotheses are needed, but I don't think they are important for the sequel)

• From 1 e 2 I would conclude that a covering space $p: E \to B$ has the homotopy lifting property. In particular we can lift every map $f: X \times I \to B$.
• From 3 this would mean that $f_*(\pi_1(X \times I)) \subseteq p_*(\pi_1(E))$.
• It is always true that $\pi_1(X \times I) \cong \pi_1(X)$
• Furthermore, if we take $E$ to be the universal cover of $B$ we will have that $\pi_1(E) \cong 0$ and so $p_*(\pi_1(E)) \cong 0$.

Finally, from the three facts stated above I would conclude that

If $p: E \to B$ is a universal covering space and $f: X \times I \to B$ is a map, then $f_*(\pi_1(X)) \cong 0$

... but this is not true!

My question is: which of the three assertions is wrong? Or is my reasoning wrong? Do covering spaces have the homotopy lifting property?

Answer 1

To apply the homotopy lifting property, you need to assume that $f: X\to B$ has a lifting. It doesn't assert that any homotopy can be lifted without assumptions. Your first bullet point is not correct.

However all 3 facts stated are correct (modulo some extra conditions on the topological spaces).

For a simple example highlighting why your argument doesn't hold, consider $f:\mathbb S^1\to\mathbb S^1$ the identity map, $\pi:\mathbb R\to \mathbb S^1$ the universal covering map. Then $f$ cannot be lifted, to $\tilde f:\mathbb S^1\to \mathbb R$, hence any homotopy between $f$ and another map cannot be lifted as well.