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?

  • $\begingroup$ Covering theory won't necessarily apply unless you add more hypotheses: I think that a hypothesis of local path connectedness on all spaces should suffice. $\endgroup$
    – Lee Mosher
    Apr 17, 2018 at 23:47

2 Answers 2


$p^{\ast}$ may not be surjective without extra assumptions. For example, you can check out Zeeman's example in Spanier's textbook of non-equivalent coverings corresponding to the same subgroup of $\pi_1$.

However, you can get a little ways by only making assumptions about $Y$. For instance, $p^{\ast}$ is certainly a homeomorphism if $\tilde{X}$ is path-connected and $Y=[0,1]^n$. For $n=1$ see Proposition 3.7 of this paper and $n\geq 2$ follows from inductively playing around with the exponential law. At the moment, it is unclear to me how far the same argument will get you for more general $Y$.


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 $\alpha$.


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