I am currently learning about covering spaces and the Homotopy Lifting Property for a covering space. As of now, I'm having some trouble giving the proof for that property over arbitrary topological spaces.

More precisely, the Theorem I want to prove is this:

Let $\pi\colon E\to X$ be any covering map, and $Y$ a topological space (with no further assumptions). Given a homotopy $H\colon Y\times [0,1] \to X$, suppose that the map $f$ defined by $f(y)=H(y,0)$ admits a lift $\tilde{f}\colon Y\to E$. In that case, there exists a lift $\tilde{H}\colon Y\times[0,1]\to E$ such that $\pi \circ \tilde{H}=H$ and $\tilde{H}(y,0)=\tilde{f}(y)$ for all $y\in Y$.

So far, I've procceded in this way:

Fix any $y\in Y$. The homotopy $H$ defines a path $H^{y}(t)=H(y,t)$ on $X$. Because of this, using the Path Lifting Property for covering spaces, there is a unique lift $\tilde{H}^{y}\colon [0,1]\to E$ s.t. $\pi(\tilde{H}^{y}(t))=H^{y}(t)=H(y,t)$ for every $t\in [0,1]$, and $\tilde{H}^{y}(0)=\tilde{f}(y)$.

Define $\tilde{H}\colon Y\times [0,1]\to E$ by $\tilde{H}(y,t)=\tilde{H}^{y}(t)$ for every $(y,t)\in Y\times [0,1]$. By construction, it's immediate that $\tilde{H}(y,0)=\tilde{f}(y)$ and $\pi \circ \tilde{H}=H$. It remains to check that $\tilde{H}$ is continuous.

At this point, I believe I managed to prove continuity when $Y$ is locally connected. For every $y\in Y$, by local connectedness of $Y$ and compactness of $[0,1]$, it's possible to find an open connected neighbourhood of $y$, $N_{y}$, and a natural number $N$ such that $H(N_{y}\times[\frac{k-1}{N},\frac{k}{N}])$ lies in an evenly covered subset of $X$ for every $k=1,...,N$. Using the same argument that is used in proving the Path Lifting Property (where I needed to use the connectedness of $N_{y}\times\{{\frac{k}{N}}\}$), we can define a continuous lift $L\colon N_{y}\times [0,1]\to E$ of $H$ such that $L(\cdot,0)=\tilde{f}$ in $N_{y}$. Finally, for every $z\in N_{y}$, $L(z,\cdot)$ and $\tilde{H}^{z}$ are (continuous) lifts of $H^{z}$ for which $L(z,0)=\tilde{H}^{z}(0)=\tilde{f}(z)$. Therefore, $L(z,t)=\tilde{H}^{z}(t)=\tilde{H}(z,t)$ for all $(z,t)\in N_{y}\times [0,1]$, so $L=\tilde{H}$ in their common domain, which implies that $\tilde{H}$ is continuous in $N_{y}\times [0,1]$. Since $y$ was arbitrary, we conclude that $\tilde{H}$ is continuous.

From here, I have two questions:

$(1)$ Is this proof that $\tilde{H}$ is continuous correct, when $Y$ is a locally connected space?

$(2)$ When $Y$ is an arbitrary topological space (not neccesarily locally connected), is the statement still true? How can one prove it without the local connectedness assumption?

Edit: I've seen some proofs in the case that $Y=[0,1]$ (i.e. the Path Homotopy Lifting Property), and it seems that I can define $\tilde{H}$ locally and then extend the local pieces via the Pasting Lemma, skipping the first part of my proof. Nevertheless, for me it's a little clearer having $\tilde{H}$ globally defined from the start and then checking continuity, even if it's not really neccesary.

Thank you in advance!


I thought for a while now and can not detect any flaws in the following, so here we go:

Pick $y \in Y$. We may, as you said, pick some open neighbourhood of $y$ say $N_y$ and a natural number $n$ s.t. $H(N_y \times [ \frac{k-1}{n},\frac{k}{n}])$ lies inside an evenly covered neighbourhood $U_k$. Say $(V_{k,i})_{i\in I}$ are the disjoint open sets that map homeomorphically to $U_k$ via $\pi$.

Now here comes my reasoning why I think we can skip the connectedness of $N_y \times \{ \frac{k}{n} \}$: $\tilde{f}(y)$ lies in one of the $V_{1,i}$, and after replacing $N_y$ by $\tilde{f}^{-1}(N_y)$ we may assume that so does all of $\tilde{f}(N_y)$. However in this case one can define a continous lift $H'_{1}$ of $H|_{N_y \times [0, \frac{1}{n}]}$ simply by composing with $\pi^{-1}$. Since by construction the whole image of this lift lies in one of the $V_{1,i}$ we can repeat this process (with $\tilde{f}$ replaced by $H'_{1}(-,\frac{1}{n})$) to construct a continous lift $H'$ of $H|_{N_y \times [0, 1]}$. Moreover as you wrote this continous lift has to coincide with $\tilde{H}|_{N_y \times [0, 1]}$. After all for every $z \in N_y$, $H'(z, -)$ and $\tilde{H}(z,-)$ both provide continous lifts of $H(z, -)$ with starting point $\tilde{f}(z)$ and path-lifting is always unique.

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  • $\begingroup$ Seems perfect to me. So the idea is basically to shrink the neighbourhood $N_{y}$ finitely many times for the "gluing" to work, I understand? Thank you! $\endgroup$ – Darth Lubinus May 1 at 12:32
  • $\begingroup$ Yes, that's what I meant :) $\endgroup$ – Photographer May 2 at 16:20

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