In the following all spaces $C^0(X,Y)$ are spaces of base point preserving maps with the compact-open topology.Furthermore all spaces I consider in the following are locally pathwise connected.

Under which assumptions can we prove the following statement:

If $Y,X,\widetilde{X}$ are pointed topological spaces, $p: \widetilde{X} \rightarrow X$ is a covering, $\widetilde{X} $ is connected and and $Y$ is simply connected, then the map

$ p^* : C^0(Y, \widetilde{X}) \rightarrow C^0(Y,X) $

is a homeomorphism?

I can prove that the map is bijective by standard covering theory and that it is continuous by the definition of the compact open topology (this only uses $p$ continuous).

I am mainly interested in finding sufficient assumptions on $X, \widetilde{X},Y$ to make this statement true, in particular, if it holds for locally compact Hausdorff spaces, but more general statements seem interesting as well.

Edit: Added the assumption, that all spaces are locally path connected to ensure, that the universal lifting theorem applies.

Edit2: I think I can prove this under the following additional assumptions:

$Y$ is compact, $X$ and $ \widetilde{X}$ are metric spaces, such that:

$\varepsilon := \inf _ { \{x,y \in \widetilde{X} : x \ne y \land p(x)=p(y) \} }(d_{ \widetilde{X} }(x,y)) < \infty $

and $p$ induces the metric on $X$, meaning

$d_X(x,y)= \inf_ { \{ \tilde x , \tilde y \in \widetilde{X} : p(\tilde x) =x \land p( \tilde y ) =y \}} (d_{\widetilde{X}} ( \tilde x, \tilde y)) $

Sketch of Proof: The compact open topology is metrizable with the metric

$d_{c^0(Y,X)} ( f,g) = \sup_{y \in Y } ( d_X (f(x), g(x)))$

and the same for $\widetilde{X}$. Now for $\delta < \varepsilon /2 $, we should get:

$\forall f \in C^0(Y,\widetilde{X}): p^*( B _ \delta (f)) = B_ \delta (p \circ f)$

This implies $p^*$ open. Hence $p^*$ is a homeomorphism.

I am not 100% sure, if this proof holds and I do not know how to remove the infimum assumption at all.

If the decktransformation group acts tranisitivley and preserves the metric, the infimum condition should hold I guess.

And I am new to this forum: Should I have posted everything under Edit2 rather as an answer then an edit?