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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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The path-lifting property for a continuous map $f:E\to B$ says that the canonical set-function $E^I\to X$ is surjective, i.e every path lifting problem has a solution (a lift in $E$). By the axiom of choice this surjectivity is equivalent to the existence of a section (not just any set-function in the opposite direction). Since such a section is furnished by the axiom of choice, nothing can be said about its continuity.

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