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 $p$.)
[Math] Lifts across covering maps
at.algebraic-topologygn.general-topology
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My answer to this question gives an example of a locally path connected (but non-semilocally simply connected) space $HA\subset\mathbb{R}^3$ called the Harmonic archipelago: draw the Hawaiian earring on a disk and between each hoop push the surface straight up to make a hill of height 1.
The fundamental group is uncountable but the only covering of $HA$ is itself so it is the unique object in its category of coverings. This weird covering behavior happens because you can deform any loop as close as you want to the basepoint (over finitely many hills) but cannot actually contract it all the way to the basepoint because the homotopy would have to take the loop over infinitely many of hills (contradicting compactness).
Topologies on the fundamental group detect this behavior. For instance the quotient topology from the loop space (the quasitopological fundamental group, $\pi_{1}^{qtop}(X)$) satisfies: If $p:Y\to X$ is a covering map, then $p_{\ast}:\pi_{1}^{qtop}(X)\to \pi_{1}^{qtop}(Y)$ is an open embedding of quasitopological groups. So if $\pi_{1}^{qtop}(X)$ is an indiscrete group there are no proper open subgroups so the only covering of $X$ is itself.
This works for all spaces including non-locally path connected ones. Really, I'd say it is easier to produce non-locally path connected examples. Here is a compact space $X$ with $\pi_{1}^{qtop}(X)\cong \mathbb{Z}_n$ indiscrete.
Let $T\subset \mathbb{R}^2$ be the closed topologists sine curve, $a\in T$ be the endpoint in the open path component and $b$ be a point in the other path component - say the origin. Now let $$Y=\frac{T\times S^1}{T\times \{1\}\cup \{a\}\times S^1}$$
In other words, if $a$ is the basepoint of $T$, $Y$ is the reduced suspension $\Sigma T$.
You have $\pi_1(Y)\cong \mathbb{Z}$ where the loop $L:S^1\to Y$, $t\mapsto (b,t)$ represents a generator, but every neighborhood of $L$ contains a trivial loop so the quotient topology on $\pi_1(Y)$ is the indiscrete topology. If you attach a 2-cell to $Y$ using $L^n$ as an attaching map, you get a space $X$ where $\pi_{1}^{qtop}(X)\cong \mathbb{Z}_n$ is indiscrete. Both $Y$ and $X$ spaces have no nontrivial coverings.
Constructing locally path connected examples with countable indiscrete fundamental groups is much harder. In fact, I think it is an open question whether or not there exists a Peano continuum with finite, indiscrete fundamental group. For such spaces it is more efficient to talking about coverings via the "shape topology" and Spanier groups (the last section of this paper shows the shape topology consists precisely of the data of the category of coverings). Applied to your question: locally path connected $X$ has a categorical universal covering iff there is a minimal open subgroup in $\pi_1(X)$ with the shape topology. If the minimal open subgroup is the trivial subgroup you get back a classical universal covering and a discrete group.
In general there is an obstruction living in $H^3(X,\pi_2Y)$. Choose a CW structure on $X$ and $Y$ with only one 0-cell. Then you can use $\varphi$ to define a map at the level of 1-skeleta (just by sending every 1-cell $e$ to a cellular representative of $\varphi([e])$). Since $\varphi$ is a map of fundamental groups it respects homotopies between paths so you can extend it to the 2-skeleton (the border every 2-cell has trivial class in $\pi_1X$ so it gets sent to a loop whose class in $\pi_1Y$ is itself trivial). So you have a continous map $f:X^2\to Y$ realizing $\varphi$ as a map of fundamental groups. To extend it you need that the cohomology class of the map sending every 3-cell $e$ of $X$ to $[f(\partial e)]\in\pi_2Y$ is 0 (the addition of a boundary correspond to a modification of $f$ at the 2-skeleton that doesn't change its behaviour on $\pi_1X$). If this obstruction is 0 analogously you can find an obstruction living in $H^4(X,\pi_3Y)$ and so on and so forth. If all groups $H^{n+1}(X,\pi_nY)$ are 0 you can realize your map. A special case is when the universal cover of $Y$ is contractible (i.e. $\pi_iY=0$ for all $i>1$), for example for any hyperbolic manifold.
Best Answer
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.