# Converse of the path lifting lemma

We know that covering spaces have the path lifting property i.e. if $p:E \rightarrow B$ is a covering map and $u:I \rightarrow B$ is a path wish intial point $a$, then for each $w \in p^{-1}(a)$, there is a path $p_w:I \rightarrow E$ which is a lifting of $p$. Clearly the converse is not true : $p$ has path lifting property does not imply that $p$ is a covering map. E.g. $p:\mathbb{R}_{indiscrete}\rightarrow \mathbb{R}$ which takes $x \mapsto x$ has the path lifting property but is not a covering map. My question is

Suppose $p:E \to B$ is a surjective map and both E,B are path connected. Then is it true that $p$ has the path lifting property ?

I am unable to find a proof or a counter example. Also,

Are there any weaker assumptions (than assuming that $p$ is a covering projection) that ensure the path lifting property of $p$ ?

Thanks !

• There is a generalization of a covering map that you may find interesting. A map that has the homotopy lifting property with respect to any space is called a fibration. – André 3000 Jan 31 '15 at 0:43

A counterexample follows: let $$\begin{matrix} p: &[0,2\pi] &\longrightarrow &S^1 \\ & x & \mapsto & e^{ix} \end{matrix}$$
which is a surjective map, and $u: I \longrightarrow S^1$ a double loop $$u(t) = e^{4\pi it}$$
This path $u$ can not be lifted to a continuous path in $[0,2\pi]$. In fact, $p$ can be extended to $\mathbb{R}$: you get a covering map $\mathbb{R} \longrightarrow S^1$ and the unique lift of $u$ inside $\mathbb{R}$ is not inside the interval $[0,2\pi]$.