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?



covering theory with compact open topology
I now have a proof under very different assumptions:


Let Y have the following property:


$ \forall K \subset Y $ compact: $\exists \alpha : [0,1] \rightarrow Y : \mathop {Im}( \alpha ) \supset K$


Define $ \Omega(K,U) = \{ f \in C^0(Y, X ) : f(K) \subset U \}$


Then we can do the following. Let $ (U\_i) \_{i \in I}$ be a path connected base of the topology of $X$, over which $p$ is trivial. Hence we can lift it to a base $(U\_{ij})\_{i \in I, j \in F} $ of the topology of $ \widetilde{X}$. Assume furthermore, that for each $i, i', j$, there is at most one $j'$, such that $U\_{ij} \cap U\_{i'j'}$ is non empty. What we now need to show, is, that for every $K\_1 ... K\_n $ compact, $ U\_{i\_1j\_1}, ...U\_{i\_nj\_n}$, the image under $p^\*$ of


$ \Omega:= \{ f \in C^0(Y, \widetilde X ) : f(K\_l) \subset U\_{i\_lj\_l}, l=1...n \}$


is open. Now we define $K = \bigcup \limits \_{l=1}^n K\_l$ and $\alpha$ as in the property $Y$ has. Now we define


$A = \{\gamma \in C^0([0,1],X): \forall l=1,..n : \tilde \gamma ( \alpha^{-1} (K\_l)) \subset U\_{i\_lj\_l} \}$


where $ \tilde \gamma$ denotes the unique lift (as $\gamma (0)=x\_0$). By the Lesbegue's number lemma, we get:


$\exists n\_\gamma , (i^\gamma\_l,j^\gamma\_l)\_{l=1...n\gamma}: \tilde \gamma ([\frac{l-1}{n\_\gamma} , \frac l{n\_\gamma} ]) \subset U\_{i\_l^\gamma j\_l^\gamma} $ .


Finally, we conclude


$p^\*( \Omega)=\bigcap \limits \_{l=1} ^n \Omega(K\_l, U\_{i\_l}) \cap (\bigcup \limits \_{ \gamma \in A } \bigcap \limits \_{l=1} ^{n\_\gamma} \Omega (\alpha([\frac{l-1}n , \frac ln ]), U\_{i\_l^\gamma}) $


Which proves the claim.


PS: The condition about the topology base on $ \widetilde X$ can be reformulated to the existence of a path connected topology base on $X$, such that every loop not in the image of $p\_\*$ (the induced map on $\pi \_1$) does not lie in any union of two open base sets.


The condition on Y is fullfilled for any CW complex Y, as every compact subset is contained in a finite union of cells. Now using space filling curves and the path connectedness of Y, we can find such an $\alpha$.


