*I asked this question on [math.SE](https://math.stackexchange.com/q/2789364/), 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?**



Path-lifting property: function space interpretation
Yes, $p$ is continuous. In proposition 3.7 of [this paper](https://projecteuclid.org/download/pdf_1/euclid.hha/1355321064) it is shown that the lifting map $Map((I,0),(B,b))\to Map((I,0),(E,e))$, $f\mapsto \tilde{f}$ is continuous (and therefore a homeomorphism) when you fix $p(e)=b$. Essentially the same proof will tell you that your map $p$ is continuous, you just don't fix the initial point. Recall the compact open topology is generated by subbasic sets $\langle K,U\rangle=\{f|f(K)\subseteq U\}$ where $K$ is compact and $U$ is open. The key in the proof is to use basic open neighborhoods of the form $$\bigcap\_{j=1}^{n}\Big\langle \left[\frac{j-1}{n},\frac{j}{n}\right],U\_j\Big\rangle\cap \bigcap\_{j=1}^{n-1}\Big\langle \left\{\frac{j}{n}\right\},U\_j\cap U\_{j+1}\Big\rangle$$ where $n\geq 3$ and $U\_j$ evenly covers $\pi(U\_j)$. 


By using the exponential law, your generalization also has a "yes" answer when $Y=[0,1]^n$ but I think it is not so clear if it holds more generally, for instance, if $Y$ is not a compact metric space.


