Definition 4.10.
A covering \(p \colon Y \to X\) with \(p(y_0) = x_0\) is called regular if \(p_*\,\pi_1(Y, y_0)\) is a normal subgroup of \(\pi_1(X, x_0)\text{.}\)
Section 4.2 Regular Coverings
Theorem 4.11.
Let \(p \colon Y \to X\) be a regular covering with \(Y\) connected, \(X\) locally path-connected, and \(p(y_0) = x_0\text{.}\) There is a canonical isomorphism
\begin{equation*}
\pi_1(X, x_0)\,/\,p_*\,\pi_1(Y, y_0) \xrightarrow{\ \cong\ } \operatorname{Deck}(p),
\end{equation*}
and \(p\) is a \(G\)-covering with \(G = \pi_1(X, x_0)/p_*\,\pi_1(Y, y_0)\text{.}\)
Proof.
We construct a left action of \(\pi_1(X, x_0)\) on \(Y\text{.}\) Given \([\sigma] \in \pi_1(X, x_0)\) and \(z \in Y\text{,}\) define \([\sigma] \cdot z\) as follows:
-
Let \(\tilde{\sigma}\) be the lift of \(\sigma\) starting at \(y_0\text{;}\) write \(y_1 = \tilde{\sigma}(1)\text{.}\)
-
Choose any path \(\gamma\) in \(Y\) from \(y_0\) to \(z\text{;}\) then \(\alpha := p \circ \gamma\) is a path in \(X\) from \(x_0\) to \(p(z)\text{.}\)
-
Let \(\tilde{\alpha}\) be the lift of \(\alpha\) starting at \(y_1\text{.}\)
Set \([\sigma] \cdot z := \tilde{\alpha}(1)\text{.}\)
The full verification that this is a well-defined continuous action and that the induced map \(\pi_1(X, x_0) \to \operatorname{Deck}(p)\) is an isomorphism with kernel \(p_*\,\pi_1(Y, y_0)\) is carried out in Exercise 4.12 below.
Exercise 4.12.
Verify the theorem in the following steps.
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Show the action is well-defined (independent of the choice of \(\gamma\)). This uses normality of \(p_*\,\pi_1(Y, y_0)\text{.}\)
-
Show it is an action: \([c_{x_0}] \cdot z = z\) and \(([\sigma]*[\tau]) \cdot z = [\sigma] \cdot ([\tau] \cdot z)\text{.}\)
-
Show \(z \mapsto [\sigma] \cdot z\) is continuous for each fixed \([\sigma]\text{,}\) using that \(X\) is locally path-connected.
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Show \(\pi_1(X,x_0) \to \operatorname{Deck}(p)\text{,}\) \([\sigma] \mapsto (y \mapsto [\sigma]\cdot y)\text{,}\) is surjective with kernel \(p_*\,\pi_1(Y, y_0)\text{.}\)
Corollary 4.13.
If \(p\colon\tilde{X} \to X\) is a universal covering, then \(\pi_1(X, x_0) \cong \operatorname{Deck}(p)\text{.}\)
Proof.
The universal cover is regular (its \(p_*\) image is trivial, which is normal). Since \(\tilde{X}\) is simply connected, \(p_*\,\pi_1(\tilde{X}) = 0\text{,}\) so the isomorphism of Theorem 4.11 gives \(\pi_1(X,x_0) \cong \operatorname{Deck}(p)\text{.}\)
Corollary 4.14.
If \(G\) acts evenly on a simply connected, locally path-connected space \(Y\text{,}\) then \(\pi_1(Y/G, x_0) \cong G\text{.}\)
Proof.
Let \(p\colon Y\to Y/G\) be the quotient map. By Proposition 4.5, \(p\) is a covering. Since \(Y\) is simply connected it is the universal cover, so \(G \cong \operatorname{Deck}(p) \cong \pi_1(Y/G, x_0)\) by Proposition 4.9, Corollary 4.13.
Exercise 4.15.
Exercise 4.16.
Compute the fundamental group of the Klein bottle.
