*I asked this question on math.SE, but even with a bounty, there were no answers/comments. I hope this is not too low-level for this site.*

Suppose I have a covering map $\pi:E\rightarrow B$, and a path in $B$, which is just a map $f$ from $I=[0,1]$ to $B$; then I know I can lift this path to a map $\hat{f}:I\rightarrow E$. I was wondering about the following generalization.

If we define the subset $X\subset E\times \mathrm{Map}(I,B)$ as $$ X = \{(e,f)\mid f(0)=\pi(e)\}$$

Then path-lifting says there exists a map $p:X\rightarrow\mathrm{Map}(I,E)$. In this formulation, $p(e,f)$ is the lift $\hat{f}$ with $\hat{f}(0)=e$.

My question: **is $p$ continuous?**

I'm thinking the $\textrm{Map}$ spaces have the compact-open topology, but maybe that's not the right one for this question.

I guess this could be even further generalized: if $(Y,q)$ is a simply-connected space with basepoint, then we can still define
$$ X=\{(e,f)\mid f(q)=\pi(e)\}\subset E\times\mathrm{Map}(Y,B)$$
and we still get a map $p:X\rightarrow\mathrm{Map}(Y,E)$, and we can still ask: **is $p$ continuous?**