Are lifts continuous?

Question

Let $p:(\tilde{X}, \tilde{x}) \rightarrow (X, x)$ be a covering map, let $f:(Y, y) \rightarrow (X, x)$ be a continuous map, and let $\tilde{f}:(Y, y) \rightarrow (\tilde{X}, \tilde{x})$ be a map such that $f=p\tilde{f}$. For my question, it is important that we do not assume a priori that $\tilde{f}$ is continuous.

The answer to this question suggests that $\tilde{f}$ is automatically continuous, but I have my reservations. The setup I introduced in the first paragraph is slightly different, but couldn't we follow the exact same reasoning to argue that $\tilde{f}$ is continuous? Yet, this feels wrong.

I'll explain where my doubts stem from. It is well-known that if $X, \tilde{X},$ and $Y$ are all path-connected and locally path-connected, then $f$ admits a continuous lift if and only if $f_*(\pi_1(Y, y)) \subseteq p_*(\pi_1 (\tilde{X}, \tilde{x}))$ (see Vick's Theorem 4.9 here, or Hatcher's Proposition 1.33 here). Both proofs construct a map $\tilde{f}: (Y, y) \rightarrow (\tilde{X}, \tilde{x})$ such that $f=p\tilde{f}$. Ergo, as mentioned earlier, this should be enough to conclude that $\tilde{f}$ is continuous (see the answer to this question).

Herein lies the problem: both Vick and Hatcher demonstrate the continuity of $\tilde{f}$ in a way which seems to vitally rely on the local path-connectedness of $Y$. If the fact that $f=p\tilde{f}$ is enough to conclude that $\tilde{f}$ is continuous, then why would authors as experienced as Vick and Hatcher make their proofs unnecessarily cumbersome? I am inclined to say that there is a mistake in this answer.

Are my suspicions correct? If so, what is Jason DeVito's error?

Answer 1

Of course each $f : (Y,y) \to (X,x)$ has non-continuous lifts $\tilde f$. However, the question is more subtle. For simplicity let us assume that $Y$ is path-connected. In Continuity in the lifting lemma for a covering map a special lift is constructed by lifting the paths of the form $f_\gamma = f \circ \gamma$, where $\gamma$ is a path in $Y$ from $y$ to an arbitrary point $y' \in Y$, and defining $\tilde f(y') = $ endpoint of the lift $\tilde f_\gamma$ with $\tilde f_\gamma(0) = \tilde x$.

It has to be shown that this is well-defined (that is $\tilde f_\gamma(1) = \tilde f_{\gamma'}(1)$ for any two paths $\gamma, \gamma'$ from $y$ to $y'$). This can be done. See Hatcher.

Unforunately the result is not always continuous. See my answer to the above question.

However, it can be shown that $\tilde f$ is continuous provided $Y$ is locally path connected. No conditions on $X$ and $\tilde X$ are needed here. See my answer to Lifting criterion, where do we use "X is locally path connected" in the proof? to find an explanation where in the proof local path connectedness enters.

Why does Jason DeVito's idea to prove the continuity of the lift break down? He correctly argues that $(f|_W)^{-1}\circ g$ is a composition of continuous functions, so is continuous. But in general it is not true that $h = (f|_W)^{-1}\circ g$ near $u\in U$.

Note that $(f|_W)^{-1}\circ g$ is defined on the open neigborhood $V' = g^{-1}(V)$ of $u$ in $U$. That $h = (f|_W)^{-1}\circ g$ near $u\in U$ means that $h \mid_{V''}= (f|_W)^{-1}\circ g \mid_{V''}$ on some open neigborhood $V'' \subset V'$ of $u$. But why should we have $h(V'') \subset W$ (which would in fact imply $h = (f|_W)^{-1}\circ g$ near $u$)? Recall the construction of $h$ via lifting paths $f_\gamma$, where $\gamma$ is a path in $U$ from the basepoint $y$ to $u$. If we can find a path-connected $V''$, then $h(u') \in W$ for all $u' \in V''$ because we can take a path $\gamma_u$ from the basepoint $y$ to $u$ and prolong it in $V''$ to a path $\gamma_{u'}$ from $y$ to $u'$. But if $U$ is not locally path-connected, it will in general be impossible to find $V''$. This means that we have to take a path from $y$ to $u'$ which leaves $V'$ - and then there is no guarantee that $h(u') = \tilde f_{\gamma_{u'}} \in W$. See again my answer to Lifting criterion, where do we use "X is locally path connected" in the proof?