Let $X,Y,Z$ be connected topological spaces, $f\colon X\to Y$ be a continuous map and $p\colon Z\to Y$ be a covering map. The problem is the existence of a continuous lift of $f$ across $p$. A standard result involving fundamental groups and induced homomorphisms requires that $X$ be path-connected and locally path-connected. Sufficient conditions however exist also in case of not necessarily locally path-connected spaces $X$. Say, if $X$ is contractible then the lift does exist. Could you please recommend a work containing sufficient conditions for more general spaces $X$ than the locally path-connected ones? (Possibly with further restrictions on $f$ or $p$.)

  • $\begingroup$ Do you have any specific example in mind? What kind of spaces are you thinking of? $\endgroup$ May 11, 2014 at 18:11

3 Answers 3


Suppose you have basepoints $x_0\in X$, $z_0\in Z$ and $p(z_0)=f(x_0)$. The lift $\tilde{f}:X\to Z$ such that $p\circ \tilde{f}=f$ exists and is continuous if and only if

1) $f_{\ast}(\pi_1(X,x_0))\subseteq p_{\ast}(\pi_1(Z,z_0))$ (this is equivalent to $\tilde{f}$ being a well-defined function).

2) For every evenly covered neighborhood $U\subset Y$, there is an open neighborhood $V\subset X$ such that if $\alpha,\beta:([0,1],0)\to (X,x_0)$ are paths with $\alpha(1),\beta(1)\in V$, then the lifts of $f\alpha$ and $f\beta$ starting at $z_0$ end in the same slice of $U$ in $Z$. (this is equivalent to the continuity of $\tilde{f}$)

For arbitrary spaces, this is about as good as it gets. Without more conditions on $X$ to control how paths vary with respect to their endpoints, there is no way to get around having to deal with how the given cover lifts paths. There are some conditions that do provide insight for some non-locally path connected spaces.

Here is a sufficient condition which generalizes local path-connectivity:

Suppose $(PX)_{x_0}$ is the space of paths in $X$ starting at $x_0$ with the compact-open topology and $ev:(PX)_{x_0}\to X$, $ev(\alpha)=\alpha(1)$ is endpoint-evaluation.

Theorem: If $f_{\ast}(\pi_1(X,x_0))\subseteq p_{\ast}(\pi_1(Z,z_0))$ and $ev:(PX)_{x_0}\to X$ is a quotient map, then $\tilde{f}$ exists and is continuous.

For a proof, see Lemma 2.5 and Corollary 2.6 of

J. Brazas, Semicoverings: a generalization of covering space theory, Homology Homotopy Appl. 14 (2012) 33-63.

The proof doesn't require local triviality. To see an example of this generalization in action, consider something like the suspension of a non-discrete, zero-dimensional space (like the Cantor set). Such a space is not locally path connected, but the evaluation map is quotient so lifts are guaranteed to be continuous. The endpoint-evaluation map is not continuous for Zeeman's example that ACL mentions showing that there is not going to be a nice characterization for all spaces.

  • $\begingroup$ This is basically what I was looking for, thank you. One more question. Are there any sufficient conditions for the map $ev$ to be quotient? $\endgroup$ May 13, 2014 at 12:09
  • $\begingroup$ I used a sufficient condition in that paper which is a little easier to check (I have called it wep-connectedness in a few papers for a space having enough "well ended paths"). A path $\alpha:[0,1]\to X$ is well-targeted if for every neighborhood $U$ of $\alpha$ (in the compact-open topology of $(PZ)_{z_0}$), there is an open neighborhood $V$ of $\alpha(1)$ such that for every $v\in V$, there is a path $\beta\in U$ with $\beta(1)=v$. Now $Z$ is wep-connected iff for any point $z\in Z$ there is some well-targeted path $\alpha$ from $z_0$ to $z$. $\endgroup$ May 13, 2014 at 12:25
  • $\begingroup$ It turns out that this property of $Z$ does not depend on the choice of basepoint $z_0$ and it's a straightforward exercise to show that it implies $ev$ is quotient. $\endgroup$ May 13, 2014 at 12:26
  • $\begingroup$ @JeremyBrazas Can you edit the above? \tilde{f}: X to Z,\ and f_*(X, x_0) \subseteq p_*(Z, z_0) I think are what you meant. $\endgroup$ Aug 13, 2016 at 4:18
  • $\begingroup$ A necessary and sufficient condition for $ev$ being quotient is given in Lemma 6.2 of tac.mta.ca/tac/volumes/30/35/30-35.pdf $\endgroup$ Oct 27, 2017 at 3:23

If every cover of $X$ is trivial, then the lift does exist.

The example of Zeeman presented in pp. 258-259 of Hilton-Wylie's Homology Theory: An introduction to algebraic topology makes me doubt that a more general (usable) sufficient condition can exist.

  • $\begingroup$ I think Zeeman's example rather suggests there is no reasonable condition that works for all spaces. The one I give in my answer is not just more general but is actually quite useful. It was critical in the proof of a general Nielsen-Schreier theorem for topological groups (open subgroups of free topological groups are free topological), which had been a long-standing open problem in the theory of topological groups. $\endgroup$ May 12, 2014 at 0:51

I feel it would be difficult to deal with non locally path connected spaces and covering spaces, though I may be wrong. However the existence of a covering map $f: X \to Y$ implies certain local conditions on $Y$; these are usually stated in the case $f$ is a universal cover, but the more general case is in Chapter 10 of Topology and Groupoids, Section 5, as that $Y$ is

"semilocally $\chi_f$-connected",

which means that each point $y \in Y$ has a neighbourhood $U$ such that for all $x \in X$ with $f(x)=y$ the image of $\pi_1(U,y)$ in $\pi_1(Y,y)$ is contained in the image under $f$ of $\pi_1(X,x)$. So to get a lift of $p: Z \to Y$ to all possible base points you seem to need a similar condition on $Z$ and $p$.

The proofs in above the book go by modelling covering maps of spaces by covering morphisms of groupoids. This modelling seems to me particularly convenient, as against the usual modelling in terms of actions of groups, when considering lifting of maps.


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