Constructing Base Spaces

# Constructing Base Spaces

Given a topological space $Y$ we would like to construct another topological space for which $Y$ is a cover.

 Definition: Let $Y$ be a topological space and let $G \subseteq \mathrm{Hom}(Y)$. Define an equivalence relation $\sim_G$ on $Y$ for all $y_1 ,y_2 \in Y$ by saying that $y_1 \sim_G y_2$ if there exists an $f \in G$ such that $f(y_1) = y_2$. The Quotient of $Y$ by $G$ is denoted by $Y/G$ and is the quotient space $Y / \sim_G$.

Here, the notation "$\mathrm{Hom}(Y)$" is used to denote the set of homeomorphisms from $Y$ to $Y$.

If $G \subseteq \mathrm{Hom}(Y)$ and if $p : Y \to Y/G$ is the quotient map then we would like to determine under what conditions $(Y, p)$ is a covering space of $Y/G$. Fortunately there are few conditions for this to be true. We define the main one below.

 Definition: Let $Y$ be a topological space and let $G \subseteq \mathrm{Hom}(Y)$. Then $G$ is said to Act Properly Discontinuously on $Y$ if for every $y \in Y$ there exists an open neighbourhood $U$ of $y$ such that for all $f_1, f_2 \in G$ with $f_1 \neq f_2$ we have that $f_1(U) \cap f_2(U) = \emptyset$.

In other words, a subgroup of the group of all homeomorphisms from $Y$ to $Y$ is said to act properly discontinuously on $Y$ if for every point $y$ in $Y$ we can find an open neighbourhood $U$ of $y$ such that the images of $U$ under every homeomorphism in [[4 G $]] are disjoint. The following theorem tells us that if$G$acts properly discontinuously on$Y$then$(Y, p)$will be a covering space of$Y/G$.  Theorem 1: Let$Y$be a topological space and let$G \subseteq \mathrm{Hom}(Y)$act properly discontinuously on$Y$. Let$p : Y \to Y/G$be the quotient map. Then$(Y, p)$is a regular covering space of$Y/G$. Furthermore,$A(Y) = G$, that is, the group of all covering transformations of$Y$is precisely$G\$.