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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?

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Yes, $p$ is continuous. In proposition 3.7 of this paper 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.

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  • $\begingroup$ Why do we need $n\ge3$? $\endgroup$ – Steve D May 31 '18 at 17:57
  • $\begingroup$ @SteveD Well, $n= 2$ could be included for this particular proof but this is not really an important feature. You could always take finer partitions and consider only sets of the form given where $n$ is greater than some fixed number. Allowing $n=1$ would just require an unnecessary extra case to write. $\endgroup$ – Jeremy Brazas May 31 '18 at 18:38
  • $\begingroup$ Right, I think I followed the argument, but then seeing this requirement made me second-guess it. Sure, we can always refine, but my question was more "why do we have to?" $\endgroup$ – Steve D May 31 '18 at 18:52
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    $\begingroup$ You don't have to exclude $n=1$...I guess it's a preference. But the argument absolutely requires the form above for $n\geq 2$. Why would you want to consider $n=1$ as something special when you don't have to? My instinct to start with $n\geq 3$ is related to covers of simplicial complexes...that is all. $\endgroup$ – Jeremy Brazas May 31 '18 at 23:23
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    $\begingroup$ Thank you for your answer! I think I fully understand how this works now. Do you want to post a similar answer on math.SE? I could also post my understanding of your answer there. Again, thanks! $\endgroup$ – Hempelicious Jun 7 '18 at 2:34

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