2
$\begingroup$

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!

$\endgroup$
1
+50
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.