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\begin{document}
 \parindent=0cm


\section*{II.4 Existence of covering spaces}
{\bf 1. Existence of a covering space assuming the existence of a
universal covering space.}

\vspace*{1.0em}
\framebox{\hspace*{1em}\parbox[b]{14cm}{$\hspace*{1em} \tx$ \\
$\hspace*{1.5em} \downarrow p$ : a universal covering with a deck
transformation group $G \overset{\theta (\cong)}
 \leftarrow \pi_1 \x$
\\ $\hspace*{1em} \x $}}

\begin{thm}
If $H < G$, then \hspace{1.0em} $\tx$ \\ \hspace*{14.5em}
$\searrow
q$\\ \hspace*{12.0em} $p \downarrow \hspace{3.0em} \hx$\\
\hspace*{14.5em} $\swarrow r$ \\ \hspace*{11.0em} $\x$\\

\hspace*{1.0em} where $\widetilde{X}_H = H \setminus \Xt$ with
$r_{\sharp}\pi_{1}\hx =
\theta^{-1}(H)$ and the quotient map $q$ and the induced map $r$ become covering maps.\\
\end{thm}

\begin{proof}

\begin{figure}[htb]
\centerline{\includegraphics*[scale=0.5,clip=true]{no1.eps}}
\end{figure}

왼쪽 그림에서와 같이 $p$에 의해 evenly cover되는 $U$를 잡으면
$H$는 $p^{-1}(U)$에 permutation으로 작용하므로 $H$ action에 대한
quotient map $q$가 covering map이 되는 것은 분명하고 $r$을 $q$에
의해 $p$로 부터 induced되는 $\bar{p}$로 정의하면 $r$은 당연히
onto이고, $U$가 역시 $r$에 의해서 evenly covered 되므로 $r$은
covering 이 된다. 그리고 \\

$\alp \in \theta^{-1}(H) \subset \pi_1\x \hspace{1.0em}
\Leftrightarrow \hspace{1.0em} \theta(\alp) \in H \hspace{1.0em}
\Leftrightarrow \hspace{1.0em} \tilde{x} \cdot \alp \in H \cdot
\tilde{x}$ \\ $\hspace{1.0em} \Leftrightarrow \hspace{1.0em}
\bar{x}:=f(\tilde{x}) = q(\tilde{x} \cdot \alp) = \bar{x} \cdot
\alp \hspace{1.0em} \Leftrightarrow \hspace{1.0em} \alp \in
\Pi_{\bar{x}} =
r_{\sharp}\pi_1\hx$ 가 성립하므로, \\
$r_{\sharp}\pi_1 \hx = \theta^{-1}(H)$ 이다.

\end{proof}

{\bf 2. Existence of a universal covering space.}\\

\framebox{\hspace*{1em}\parbox[b]{14cm}{Idea : the set of homotopy
classes of paths from $x$ to $y$ \\ $\underset{1-1}
\leftrightarrow p^{-1}(y) \subset \widetilde{X}$ , universal
covering , via $[\alp] \leftrightarrow \tilde{\alp}(1)$}}

\begin{defn}
A space $X$ is said to be {\bf semilocally simply connected} if
for each $x \in X$, there is a neighborhood $U$ of $x$ such that
the homomorphism\\
\hspace*{8.0em} $i_{\sharp}: \pi_1(U,x) \to \pi_1(X,x)$\\
induced by inclusion is trivial.\\
\end{defn}

Assume that $X$ is path-connected, locally path-connected and semilocally simply connected.\\
Let $(P,x)=\{ \alp : I \to X \ | \alp(0)=x\}$\\ Define
$\widetilde{X} = (P,x)/\sim \hspace{1.0em} ($ recall $\alp \sim
\bet \Leftrightarrow \alp \simeq \bet $ rel $\partial )$ and
$p:\widetilde{X} \to X$ by $p([\alp]) = \alp(1)$\\

{\bf Topology of $\widetilde{X}$}\\
For $\alp \in (P,x)$ and open set $U$ with $\alp(1) \in U$, let
$(\alp,U) := \{\alp*\alp' | \alp' : I \to U\ $ with $\alp'(1)=\alp(0)\}/\sim  \hspace{1.0em} \subset \widetilde{X}$.\\
이때 주어진 $y=\alp(1) \in X$에 대해, $U$를 path-connected ,
semilocally simply connected neighborhood of $y(=\alp(1))$ 라고
하면 $p|:(\alp,U) \to
U$는 onto 이고 one-to-one 이다.\\

이제 $p$가 evenly covered 임을 보이자. 즉, $p^{-1}(U)=\underset{\alp(1)=y}\coprod (\alp,U)$:\\
$(\alp,U)\cap(\bet,U) \neq \varnothing$ 라고 하자.\\ $\gam \in
(\alp,U)\cap(\bet,U)$에 대하여 $\alp'$와 $\bet'$가 존재해서 \\
\hspace*{1.0em}$\alp*\alp' \sim \gam \sim \bet*\bet'$\\
$\Rightarrow \hspace{1.0em} \alp*\alp'*\bar{\alp'} \sim
\bet*\alp'*\bar{\alp'} (\because U$가 semilocally simply connected이므로 $\bet*\bet' \sim \bet*\alp')$\\
$\Rightarrow \hspace{1.0em} \alp \sim \bet$\\

$\therefore (\alp,U)\cap(\bet,U) \neq \varnothing \hspace{2.0em}
\Rightarrow \hspace{2.0em} (\alp,U) = (\bet,U)$\\

Take $\{(\alp,U) | U$ : open neighborhood of $\alp(1)$, $\alp \in
(P,x)\}$ as a base for a topology of $\widetilde{X}$.\\
{\bf Check}\\
1. $\bigcup (\alp,U) = \widetilde{X}$(obvious) \\
2. $\gam \in (\alp,U)\cap(\bet,U)$\\
\hspace*{1.0em}$\Rightarrow \exists W \subset U \cap V $ such that
$(\gam,W) \subset (\alp,U)\cap(\bet,U)$ 여기서 $W$는 path-connected neighborhood of $\gam(1)$이고 $U$가 semilocally simply connected 이므로 subset $W$도 마찬가지(obvious)\\

$\widetilde{X}$가 path-connected 임을 보이자.\\
임의의 $[\alp] \in \widetilde{X}$에 대해서 $\alp_{s}(t) :=
\alp(st)$라고 하면 $\alp_s$는 $\alp$와 $x$(constant path)를 잇는
path이다. 여기서 $\tilde{\alp}(s) := [\alp_s]$라고 정의하면
$\tilde{\alp}(0)=[x], \tilde{\alp}(1)=[\alp]$가 되어 $[\alp]$와
$[x]$는 path 로 연결된다. ({\bf exercise} : $\tilde{\alp}$ is
continuous)\\

마지막으로 $\widetilde{X}$가 simply connected임을 보이자.\\
Let $\tau$ be a loop in $\tx$, where $\tilde{x}=[x]$\\
\hspace*{1.0em} $\Rightarrow \alp := p \circ \tau$ is a loop in
$X$ and $\tilde{\alp} = \tau \\
\hspace*{1.0em} \Rightarrow [\alp] = \tilde{\alp}(1) = \tau(1) =
[x] \hspace{1.0em} \Rightarrow \hspace{1.0em} \alp \sim x
\hspace{1.0em} \Rightarrow \hspace{1.0em} \tau \sim \tilde{x} \hspace{1.5em} \blacksquare $\\

\begin{cor}
$\forall H < \pi_1 \x, \exists$ a covering space $\tx$
corresponding to $H$, i.e. $p_{\sharp}\pi_{1}\tx = H$
\end{cor}

{\bf Remark}\\
1. Universal covering is "universal", i.e. it covers every other
covering by lifting theorem and universal covering is clearly unique up to isomorphism.\\

2. $X$ has a universal covering.\\
\hspace*{1.0em} $\Rightarrow X$ is semilocally simply
connected. \\
$\hspace*{2.0em} \pi_1(\tilde{U},\tilde{x}) \hspace{1.0em}
\overset{i_{\sharp}}\longrightarrow \hspace{1.0em} \pi_1\tx$ =0\\
\hspace*{2.0em} $\cong \downarrow p_{\sharp} \hspace{1.5em}
\circlearrowright \hspace{2.5em} \downarrow p_{\sharp}$\\
\hspace*{2.0em} $\pi_1(U,x) \hspace{1.0em}
\overset{i_{\sharp}}\longrightarrow \hspace{1.0em} \pi_1\x$\\
로 부터 당연하다.\\

\newpage

{\bf 숙제 8}\\
Let $(G,e)$ be a topological group and $p:(\tilde{G},\tilde{e})
\to (G,e)$ be a covering. Then we can lift the group structure of
$G$ to $\tilde{G}$ so that $p$ becomes a homomorphism unique up to
the choice of identity $\tilde{e} \in p^{-1}(e)$.\\


\end{document}
