I wish to prove the following statement:

Let $X$ be a simply connected and locally connected space, and let $p:Y\to Z$ be a covering map. Then given $f:X\to Z$ continuous, $x_0\in X$, $y_0\in Y$ such that $p(y_0)=f(x_0)$, there is a unique continuous function (a "lift") $\tilde f:X\to Y$ such that $p\circ\tilde f=f$ and $\tilde f(x_0)=y_0$.

This question has gone through many iterations here on MSE. So far, I have shown that a lift is unique if it exists, and I have proven the theorem in the case $X=[0,1]$ and $X=[0,1]^2$. This general version was also attempted but specialized to an incorrect lemma.

Here is what I have so far: Define $\tilde f$ such that $\tilde f(x)=\tilde \phi(1)$ for any path $\phi:[0,1]\to X$ from $x_0$ to $x$ (where the lifting sets $\tilde\psi(0)=y_0$). To show that $\tilde f$ is well-defined, suppose that $\phi$ and $\psi$ are paths from $x_0$ to $x$. Then they are path-homotopic by simple connectedness, and the homotopy $h$ lifts to $\tilde h$; since $\tilde h(0,x)$ is a lift of $\phi$ and $\tilde h(1,x)$ is a lift of $\psi$, uniqueness of lifts implies $\tilde\phi(x)=\tilde h(0,x)$ and $\tilde\psi(x)=\tilde h(1,x)$, so $\tilde\phi$ and $\tilde\psi$ are path-homotopic and in particular have the same endpoint $\tilde\phi(1)=\tilde\psi(1)$.

It is easy to check that $p\circ\tilde f=f$ and $\tilde f(x_0)=y_0$, and so the main remaining task is the proof that $\tilde f$ is continuous. "Following my nose" leads to the following proof skeleton: Take a point $x\in X$ and $\tilde f(x)\in U\subseteq Y$ open; we wish to show that there is a neighborhood $x\in V\subseteq X$ with $\tilde f[V]\subseteq U$. By the covering map property and since $f(x)=p(\tilde f(x))\in p[U]$ is open ($p$ is an open map), there is a neighborhood $f(x)\in A\subseteq p[U]$ that is evenly covered, so suppose that $W$ is the slice of $Y$ that contains $\tilde f(x)$, such that $p\restriction W$ is a homeomorphism onto $p[W]=A$. We also know that $X$ is locally connected, so there is an $x\in M\subseteq f^{-1}[A]$ that is open and connected. My guess is that $M$ is the desired open set, but I don't see how to make the connection.

There is a puffy cloud at this point, but let's suppose that we could find a $V\ni x$ (probably $M$ or a subset) such that $\tilde f[V]\subseteq W$. Then since $V$ is a neighborhood of $x$ it is sufficient to show that $\tilde f\restriction V:V\to W$ is continuous in the subspace topology, which is to say $(f\restriction V)^{-1}[W']$ is open in $V$ for every open $W'\subseteq W$:

\begin{align} (\tilde f\restriction V)^{-1}[W']&=(\tilde f\restriction V)^{-1}[(p\restriction W)^{-1}[(p\restriction W)[W']]]\\ &=((p\restriction W)\circ(\tilde f\restriction V))^{-1}[(p\restriction W)[W']]\\ &=(p\circ(\tilde f\restriction V))^{-1}[(p\restriction W)[W']]\\ &=((p\circ \tilde f)\restriction V)^{-1}[(p\restriction W)[W']]\\ &=(f\restriction V)^{-1}[(p\restriction W)[W']]. \end{align}


You need to assume $X$ is locally path-connected, not just locally connected (otherwise the result is not true; the "long circle" that gave a counterexample to your previous question can also give a counterexample to this one). Your idea to take $V=M$ is a good one. If $M$ is path-connected, fix a path $\phi$ from $x_0$ to $x$, and then for every other point $x'\in M$, consider a path from $x_0$ to $x'$ that is obtained by concatenating $\phi$ with a path from $x$ to $x'$ that is entirely in $M$. Can you use this to show that $\tilde{f}[M]\subseteq W$?

  • 1
    $\begingroup$ For future reference, the "long circle" counterexample sets $X$ to be the one point compactification of the long ray, $Y=\Bbb R$ and $Z=\Bbb R/\Bbb Z$, $p$ the quotient map, and $f(x,0)=x,f(x,\alpha)=0$ for $0<\alpha<\omega_1$, and $f(\infty)=0$. Then the only possibility for $\tilde f$ with $\tilde f(\infty)=1$ is the definition in the OP, but this sets $\tilde f(x,\alpha)=0$ so it is not continuous at $\infty$. $\endgroup$ – Mario Carneiro Jul 8 '15 at 5:24
  • $\begingroup$ @MarioCarneiro: Yep, exactly. $\endgroup$ – Eric Wofsey Jul 8 '15 at 5:27
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    $\begingroup$ To complete the proof, if $\psi$ is the path from $x$ to $x'$ and $\phi\ast\psi$ is the path join, since $\tilde\phi\ast\tilde\psi$ (where $\tilde\psi(0)=\tilde\phi(1)$) is a lift of $\phi\ast\psi$, $\tilde f(x')=\tilde\phi\ast\tilde\psi(1)$. Thus $\tilde\psi$ is a continuous function from $\tilde f(x)$ to $\tilde f(x')$ which lies in $p^{-1}[A]$, and $W$ is clopen in $p^{-1}[A]$ by the covering map property, so $\tilde\psi:[0,1]\to W$ and $\tilde f(x')\in W$. $\endgroup$ – Mario Carneiro Jul 8 '15 at 5:53

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