Discontinuous lifts of continuous maps While recapitulating the theory of covering maps, I thought that it would a most inconvenient thing if the lift of a continuous map defined by continuation along paths to a covering space should be discontinuous under certain assumptions.
Unfortunately, I failed to prove the continuity of those lifts of a continuous function in general, and that local path-connectedness of the domain of the function is a requirement for the classical theorem about the existence of a continuous lift strongly suggests that unfortunately, there might be a continuous function with such a discontinuous lift.
Is this true? Is there such an example? Are there conditions on the covering (or the covered) space that preclude such calamities?
Thank you very much in advance for any help.
 A: The standard lifting theorem says that the following:
Let $p : (X,x_0)  \to (Y,y_0)$ be  a covering map and $Z$ be a path connected locally path connected space. Then $f : (Z,z_0) \to (Y,y_0)$ has a (continuous!) lift $\tilde f : (Z,z_0) \to (X,x_0)$ if and only if $f_*(\pi_1(Z,z_0)) \subset p_*(\pi_1(X,x_0))$.
The condition on fundamental groups is clearly necessary for the existence of a lift (even without any assumptions on $Z$). For the converse let us consider a path connected $Z$. One proceeds as follows:
For each $z \in Z$ choose a path $u : I \to Z$ such that $u(0) = z_0$ and $u(1)= z$. The path $u' = f \circ u : I \to Y$ (which satsfies $u'(0) = y_0$) has a unique lift $\tilde u' : I \to X$ such that $\tilde u'(0) = x_0$. Define
$$\tilde f(z) = \tilde u'(1) .$$
It is easy to see that the condition on fundamental groups assures that $\tilde f(z)$ does not depend on the choice of $u$. Thus we obtain a well-defined function $\tilde f : (Z,z_0) \to (X,x_0)$ such that $p \circ \tilde f = f$.
This works for any path connected $Z$. Let us emphasize that if $f$ has a continuous lift $g = (Z,z_0) \to (X,x_0)$, then it must have necessarily have the form $g = \tilde f$ as above. In fact, given a path $u$ from $z_0$ to $z$, the path $g \circ u$ is a lift of the path $u' = f \circ u$ with $(g \circ u)(0) = g(z_0) = x_0$ and by unique path lifting we conlude $g  \circ u = \tilde u'$ and hence $g(z) = (g \circ  u)(1) =  \tilde u'(1) =  \tilde f(z)$.
As you know, the continuity of  $\tilde f$ can easily be proved for locally path connected $Z$. But even if $Z$ is not locally path connected, some maps can have continous lifts (for example, all maps with image in an evenly covered set).
Here is an example showing that if $Z$ is not local path connected, then we can get a discontinous $\tilde f$.
Let $p : (\mathbb R,0) \to (S^1,1), p(t) = e^{2\pi i t}$. This is a covering map with extremely nice base and total space. Let $Z$ be the Warsaw circle (see e.g. here and here). This is a simply connected space. Let $L = \{ (0,y):-1\le y \le 1\} \subset Z$ and $q : Z \to Z' = Z/L$ be the quotient map. There exists an obvious homeomorphism $h : S^1 \to Z'$ such $h(-1) = [L]$. Combining it with a rotation on $S^1$, we get a homeomorphism $H : Z' \to S^1$ such that $H([L]) = 1$. Now consider $f = H \circ q : Z \to  S^1$ and take $z_0 = (0,0)$. The condition on fundamental groups is satisfied because $Z$ is simply connected, thus we get $\tilde f$ as above. But this map is discontinuous: We have $\tilde f(z_0) = 0$ and $\tilde f(\zeta_n) \to 1$, where the sequence $(\zeta_n)$ has the form $\zeta_n = (t_n,0) \in Z$ with $t_n \to 0$. Look at the graphical representation of $Z$ to get the intuition; there are no "short paths" from $z_0$ to the $\zeta_n$ although they are arbitrarily close to $z_0$.
