\documentclass[12pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} %\usepackage{graphicx} \usepackage{hangul} \usepackage{amscd,amsmath} \usepackage{amsfonts} \usepackage{amssymb,theorem} \usepackage{longtable} %\usepackage{floatflt} %\usepackage{texdraw, epic, pstcol ,pstricks ,pst-3d, pst-poly,pst-grad,pst-node, pst-text} \newcommand{\Aff}{\mbox{\it Aff}} \newcommand{\aff}{\mbox{\it aff}} \newcommand{\alp}{\alpha} \newcommand{\bet}{\beta} \newcommand{\del}{\delta} \newcommand{\gam}{\gamma} \newcommand{\vep}{\varepsilon} \newcommand{\eps}{\epsilon} \newcommand{\lam}{\lambda} \newcommand{\kap}{\kappa} \newcommand{\sig}{\sigma} \newcommand{\ome}{\omega} \newcommand{\Gam}{\Gamma} \newcommand{\Ome}{\Omega} \newcommand{\Sig}{\Sigma} \newcommand{\Del}{\Delta} \newcommand{\Lam}{\Lambda} \newcommand{\dis}{\displaystyle} \newcommand{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^m}} \newcommand{\pa}{\partial} \newcommand{\cb}{\mathbb{C}} \newcommand{\cc}{\mathcal{C}} \newcommand{\ccn}{\mathcal{C}^{\infty}} \newcommand{\sbb}{\mathbb{S}} \newcommand{\rb}{\mathbb{R}} \newcommand{\tx}{(\widetilde{X}, \widetilde{x})} \newcommand{\hx}{(\widetilde{X}_H, \bar{x})} \newcommand{\txx}{(\widetilde{X}_1, \widetilde{x}_1)} \newcommand{\txxx}{(\widetilde{X}_2, \widetilde{x}_2)} \newcommand{\x}{(X,x)} \newcommand{\xt}{\widetilde{x}} \newcommand{\yt}{\widetilde{y}} \newcommand{\Xt}{\widetilde{X}} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{예} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{주} \newtheorem{note}{Note} } \renewcommand{\thenote}{} \renewcommand{\therem}{} \newenvironment{proof}{{\bf 증명}}{\hfill\framebox[2mm]{}} \newenvironment{proof1}{{\bf 정리증명}}{\hfill\framebox[2mm]{}} \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} \\ \vspace{20.0em} 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 \\ $\alp \in \theta^{-1}(H) \subset \pi_1\x \hspace{1.0em} \Leftrightarrow \hspace{1.0em} \theta(\alp) \in H \hspace{1.0em} \Leftrightarrow \hspace{1.0em} \tilde{x} \cdot \alp \in H \cdot \tilde{x}$ \\ $\hspace{1.0em} \Leftrightarrow \hspace{1.0em} \bar{x}:=q(\tilde{x}) = q(\tilde{x} \cdot \alp) = \bar{x} \cdot \alp \hspace{1.0em} \Leftrightarrow \hspace{1.0em} \alp \in \Pi_{\bar{x}} = r_{\sharp}\pi_1\hx$ 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 $\alp'(0)=\alp(1)\}/\sim \hspace{1.0em} \subset \widetilde{X}$.\\ 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. $\gam \in (\alp,U)\cap(\bet,V)$\\ \hspace*{1.0em}$\Rightarrow \exists W \subset U \cap V $ such that $(\gam,W) \subset (\alp,U)\cap(\bet,V)$ 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}