# Homotopy lifting property for covering spaces

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.

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.
• 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! – Darth Lubinus May 1 at 12:32