Lifts across covering maps

Question

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 $p$.)

Answer 1

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.

Answer 2

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.

Answer 3

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.