Definition 4.1.
The group \(G\) acts evenly on \(Y\) if every point \(y \in Y\) has a neighbourhood \(V\) such that \(g \cdot V \cap h \cdot V = \emptyset\) whenever \(g \neq h\text{.}\)
Section 4.1 G-Coverings
Let \(G\) be a group acting on a space \(Y\text{.}\) When is the quotient map \(p \colon Y \to Y/G =: X\) a covering?
Remark 4.2.
Note that such an action is always free, i.e., \(g \cdot x = x \implies g=e\text{.}\) Indeed, suppose otherwise, then taking a neighborhood \(U\) of \(x\text{,}\) we see that \(x \in g \cdot U \cap U\text{,}\) a contradiction.
Example 4.3.
Let \(\mathbb{Z}\) act on \(\mathbb{R}\) via translation: \(n \cdot r = n+r\text{.}\) For \(r \in \mathbb{R}\text{,}\) let \(V_r = (r-1/2,r+1/2)\text{,}\) so that
\begin{equation*}
m \cdot V_r = (r+m-1/2,r+m+1/2), \quad n \cdot V_r = (r+n-1/2,r+n+1/2).
\end{equation*}
These intersect if and only if \(r+n-1/2<r+m+1/2\) and \(r+m-1/2<r+n+1/2\text{,}\) i.e., \(n-m < 1\) and \(m-n<1\text{,}\) i.e., \(| m-n | <1\text{,}\) i.e., \(m = n\text{.}\) Thus, the action is even.
Exercise 4.4.
Show that a finite group acting freely on a Hausdorff space acts evenly.
Proposition 4.5.
If \(G\) acts evenly on \(Y\text{,}\) then \(p \colon Y \to Y/G\) is a covering map.
Proof.
The map \(p\) is continuous by definition of the quotient topology, and it is open: for any open \(V \subseteq Y\text{,}\)
\begin{equation*}
p^{-1}(p(V)) = \bigcup_{g \in G} g \cdot V
\end{equation*}
is open in \(Y\text{,}\) so \(p(V)\) is open in \(Y/G\text{.}\)
Let \(y \in Y\) and choose an open neighbourhood \(V \ni y\) as in the definition of an even action, so that \(g\cdot V \cap h\cdot V = \emptyset\) for all \(g \neq h\) in \(G\text{.}\) Set \(U := p(V)\text{,}\) an open neighbourhood of \(p(y)\text{.}\) Then
\begin{equation*}
p^{-1}(U) = \bigcup_{g \in G} g\cdot V = \coprod_{g \in G} g\cdot V,
\end{equation*}
the union being disjoint by evenness. It remains to show that \(p|_{g\cdot V} \colon g\cdot V \to U\) is a homeomorphism for each \(g \in G\text{.}\)
\(p|_V \colon V \to U\) is a homeomorphism. Surjectivity is immediate from \(U = p(V)\text{.}\) For injectivity, suppose \(y_1, y_2 \in V\) satisfy \(p(y_1) = p(y_2)\text{;}\) then \(y_2 = h\cdot y_1\) for some \(h \in G\text{,}\) so \(y_2 \in V \cap h\cdot V\text{.}\) Evenness forces \(h = e\text{,}\) hence \(y_1 = y_2\text{.}\) Thus \(p|_V\) is a continuous bijection. It is also open: any \(W \subseteq V\) open in \(V\) is open in \(Y\) (as \(V\) is open in \(Y\)), so \(p(W)\) is open in \(Y/G\) by openness of \(p\text{,}\) and \(p(W) \subseteq
U\text{.}\) Hence \(p|_V\) is a homeomorphism onto \(U\text{.}\)
The remaining sheets. Fix \(g \in G\) and let \(g \colon V \to g\cdot V\) denote the (homeomorphism given by the) action of \(g\text{.}\) Since \(p(g\cdot y') = p(y')\) for all \(y' \in
V\text{,}\) we have
\begin{equation*}
p|_{g\cdot V} \circ g = p|_V,
\qquad\text{so}\qquad
p|_{g\cdot V} = p|_V \circ g^{-1}.
\end{equation*}
As a composite of homeomorphisms, \(p|_{g\cdot V} \colon g\cdot V \to U\) is a homeomorphism.
Thus \(U\) is evenly covered by \(p\text{.}\) Since \(y \in Y\) was arbitrary, \(p\) is a covering map.
Example 4.6.
Returning to the previous example, we have \(\mathbb{R}/\mathbb{Z} \cong S^1\text{,}\) and the map \(p \colon \mathbb{R} \to S^1\) is the usual covering map of \(S^1\) by \(\mathbb{R}\text{.}\)
Definition 4.7.
A covering arising as in Proposition 4.5 is called a \(G\)-covering. An isomorphism of \(G\)-coverings \(Y_1\) and \(Y_2\) is a covering isomorphism \(\varphi \colon Y_1 \to Y_2\) that is also \(G\)-equivariant: \(\varphi(g \cdot y) = g \cdot \varphi(y)\text{.}\)
Exercise 4.8.
Let \(p \colon Y \to X\) be a \(G\)-covering. If \(X\) is connected and two isomorphisms of \(G\)-coverings \(\varphi_1,\varphi_2 \colon Y \to Y\) over \(X\) agree at one point of \(Y\text{,}\) show that \(\varphi_1=\varphi_2\text{.}\)
Proposition 4.9.
For any \(G\)-covering \(p \colon Y \to X\text{,}\) the map
\begin{equation*}
G \to \operatorname{Deck}(p), \quad g \mapsto (y \mapsto g \cdot y)
\end{equation*}
is a homomorphism. If \(Y\) is connected, it is an isomorphism.
Proof.
Injectivity. If \(g \cdot y = y\) for all \(y\text{,}\) the even-action condition forces \(g = e\text{.}\)
Surjectivity. Let \(\varphi\) be an automorphism and fix \(y_0 \in Y\text{.}\) Since \(p(y_0) = p(\varphi(y_0))\text{,}\) both points lie in the same \(G\)-orbit, so there exists \(g \in G\) with \(g \cdot y_0 = \varphi(y_0)\text{.}\) The automorphism \(y \mapsto g \cdot y\) and \(\varphi\) agree at \(y_0\text{,}\) so they agree everywhere by the uniqueness of lifts (Lemma 2.8), since \(Y\) is connected.
