6
$\begingroup$

I'm currently reading Hatcher's Algebraic topology book. In page 68 he says:

A consequence of the lifting criterion is that a simply-connected covering space of a path-connected, locally path-connected space $X$ is a covering space of every other path-connected covering space of $X$ A simply-connected covering space of $X$ is therefore called a universal cover. It is unique up to isomorphism, so one is justified in calling it the universal covering.

Let's precise that a little bit. Let $X$ be a path-connected, locally path-connected space and $x_0\in X$. Let $p_1:(\tilde{X}_1,\tilde{x}_1)\to (X,x_0)$ be a simply-connected covering space and $p_2:(\tilde{X}_2,\tilde{x}_2)\to (X,x_0)$ be any other path-connected covering space. Using the fact that $\tilde{X}_1$ is simply connected and the lifting criterion one can find a map $\tilde{p}_1:(\tilde{X}_1,\tilde{x}_1)\to(\tilde{X}_2,\tilde{x}_2)$ such that $p_2\tilde{p}_1=p_1$ and $\tilde{p}_1$ must be the desired covering map. Note that this is where one uses the hypothesis $X$ path connected and locally path-connected and $\tilde X_1$ and $\tilde X_2$ both path-connected

However I'm having some trouble proving it. Let $x\in \tilde{X}_2$ then $p_2(x)\in X$ and there is a nbd $U$ of $p_2(x)$ such that $p^{-1}_1(U)=\cup_i V_i$ where the $V_i$ are open and disjoint subsets of $\tilde{X}_1$ and $p|V_i:V_i\to U$ is an homeomorphism. The desired nbd of $x$ that makes $\tilde{p}_1$ into a covering map should be $p^{-1}_2(U)$. We have $\tilde{p}_1^{-1}(p^{-1}_2(U))=p^{-1}_1(U)$ and then one expects that $\tilde{p}_1|V_i:V_i\to p_2^{-1}(U)$ is an homeomorphism to make this work. But I can't prove that last statement, I'm not even sure it's surjective (Hatcher doesn't assume covering spaces to be surjective).

Is this the right approach? Or maybe there is a simpler way to do it.

$\endgroup$
5
$\begingroup$

Lemma: Given the commutative diagram

$$\begin{array}{ccccccccc} \widetilde{X} & \\ \downarrow{\small{p}} & {\searrow}^{q} \\ X_1 & \!\!\!\!\! \xleftarrow{p_1} & \!\!\!\! X_2\end{array}$$

where $p_1, p$ are covering maps, then so is $q$, where $X_1, X_2, \tilde{X}$ are all path-connected and locally path-connected.

Proof:

  1. $q$ is surjective: $\sigma$ be a path in $X_2$ from $x_0$ and $x$. Pushforward by $p_1$ to get a path $p_1 \circ \sigma$ in $X_1$ from $p_1(x_0) = x_0'$ and $q(x)$. Lift to $\widetilde{X}$ to get a path $\widetilde{\sigma}$ starting from some point $x_0''$ in the fiber over $x_0'$. Pushforward by $q$ to get path $q \circ \widetilde{\sigma}$ starting at $x_0$. Uniqueness of path-lifing says $q \circ \widetilde{\sigma} \simeq \sigma$, so that $q$ maps the endpoint of $\tilde{\sigma}$ to the endpoint $x$ of $\sigma$. As $X_2$ is path connected, we can apply this argument for all $x \in X_2$ to prove $q$ is surjective.

  2. $q$ is a covering map: Pick $x \in X_2$. Pushforward by $p_1$ to get $p_1(x)$ in $X_1$. There is a path-connected neighborhood $\mathscr{U}$ of $p_1(x)$ evenly covered by $p_1$ and $p$ (take neighborhoods evenly covered by $p_1$ and $p$ and take intersection). $\mathscr{V}$ be the slice in $p_1^{-1}(\mathscr{U})$ containing $x$. $\{\mathscr{U}_\alpha\}$ be the slices in $p^{-1}(\mathscr{U})$. $q$ maps each slice $\mathscr{U}_\alpha$ to distinct slices in $q^{-1}(\mathscr{U})$. $q^{-1}(\mathscr{V})$ is then union of slices in $\{\mathscr{U}_\alpha\}$ which are mapped homeomorphically onto $\mathscr{V}$. I claim all $\mathscr{U}_\alpha$ are mapped homeomorphically on $\mathscr{V}$ by $q$. This can be proved slicewise, recalling that given a commutative diagram with any two arrows as homeomorphism, so is the third. $\blacksquare$


If $\widetilde{X}$ is simply connected, $p : \widetilde{X}\to X$ the universal cover, $p_1 : X_2 \to X_1$ a covering map, then as $p_*(\pi_1(\widetilde{X}))$ fits inside ${p_1}_*(\pi_1(X_2))$, being the trivial group, we can lift $p$ to $\tilde{p} : \widetilde{X} \to X_1$. By previous discussion, $\tilde{p}$ is a covering map, since it fits inside a commutative diagram like above. Thus, $\widetilde{X}$ covers $X_2$, as desired.

$\endgroup$

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.