\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{\zb}{\mathbb{Z}} \newcommand{\tx}{(\widetilde{X}, \widetilde{x}_0)} \newcommand{\txx}{(\widetilde{X}_1, \widetilde{x}_1)} \newcommand{\txxx}{(\widetilde{X}_2, \widetilde{x}_2)} \newcommand{\x}{(X,x_0)} \newcommand{\xt}{\widetilde{x}} \newcommand{\yt}{\widetilde{y}} \newcommand{\Xt}{\widetilde{X}} \newcommand{\yx}{Y^{X}} \newcommand{\yz}{Y^{Z}} \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 \psset{unit=2cm} \section*{IV.1 Descriptions of higher homotopy groups} {\bf Notation}\\ $I^{n} = \underbrace{I \times \cdots \times I} $ = the unit $n$-cube\\ \hspace*{4.0em}$n$ \hspace{2.0em} =$\{t=(t_1, \cdots t_n) \in \rb^{n} | 0 \leq t_{i} \leq 1 , i=1,2, \cdots ,n\}$\\ $\pa I^{n} =$ boundary of $I^{n} =\{t=(t_1, \cdots t_n) \in \rb^{n} | t_{i} = 0$ or $1$ for some $i\}$\\ {\bf 1st description }\\ $F_{n}\x := \{\alp : (I^{n},\pa I^{n}) \to \x\}$\\ $\alp , \bet \in F^{n}$에 대하여, $\alp \thicksim \bet \Leftrightarrow \alp \simeq \bet$ rel $\pa I^{n}$이라고 하면,\\ \begin{center} \framebox[1.1\width]{\Large $\pi_{n}\x := F_{n}\x / \thicksim$}\\ \end{center} 이 된다. 물론 $\pi_{n}\x$ 의 원소는 $[\alp]$(eqivalence class of $\alp$)이다. \\ 먼저 $\pi_{n}\x$에 group structure 를 주자. \\ $\alp , \bet \in F_{n}\x$ 에 대해서 \\ \begin{displaymath} \alp \ast \bet(t_1, \cdots, t_n) =\left\{\begin{array}{cl} \alp(2t_{1}, t_2, \cdots ,t_{n}) & 0 \leq t_1 \leq \frac{1}{2} \\ \bet(2t_{1}-1,t_2, \cdots ,t_{n}) & \frac{1}{2} \leq t_1 \leq 1 \end{array} \right. \end{displaymath} 이라고 하고, $[\alp][\bet] := [\alp \ast \bet]$ 라고 정의하면, 이는 잘 정의된다.\\ 즉, $\alp \thicksim \alp' , \bet \thicksim \bet'$ 이면 $\alp \ast \bet \thicksim \alp' \ast \bet'$이다.(check).\\ 그리고 $\pi_1$의 경우와 똑같은 이유로 (이때 $t_2, \cdots , t_n$ 은 상수로 취급)\\ $(\alp \ast \bet ) \ast \gam \thicksim \alp \ast (\bet \ast \gam), \hspace{1.0em} \alp \ast x_0 \thicksim x_0 \ast \alp, \hspace{1.0em} \bar{\alp} \ast \alp \thicksim \alp \ast \bar{\alp}$ 가 성립하므로 $\pi_{n}\x$는 group이 된다.\\ (여기서 $\bar{\alp}(t_1, \cdots, t_{n}) := \alp(1-t_1, t_2, \cdots , t_{n})$)\\ {\bf 2nd description}\\ Observe that $\pi_{1}\x = [(S^{1},1),\x]$\\ \hspace*{10.0em} = the set of homotopy classes of maps $f:(S^{1},1) \to \x$\\ Similarly,\\ \begin{center} \framebox[1.1\width]{\Large $\pi_{n}\x = [(S^{n},e_{1}),\x]$}\\ \end{center} 이 때, $e_{1}=(1,0, \cdots ,0) \in \rb^{n+1}$이다.\\ 이 경우 $\alp , \bet :(S^{n},e_{1}) \to \x$에 대해서 $\alp \ast \bet$는 아래 그림과 같이 정의되는 $(S^{n},e_{1}) \to \x$의 map 과 대응된다.\\ \begin{figure}[htb] \centerline{\includegraphics*[scale=0.4,clip=true]{pic1.eps}} \end{figure} {\bf 3rd description}\\ Use loop space $\Ome\x$ and view $\pi_{2}\x$ as $\pi_{1}(\Ome\x ,x_{0})$.\\ 이 경우 $\pi_{1}(\Ome\x , x_{0})$를 정의하기 위해서 function space $\Ome\x$에 topology가 필요하다.\\ {\bf Topology on a function space}\\ The most commonly used and useful topology for a function space is the {\bf compact-open topology}.\\ 먼저 $\yx := \{f | f : X \to Y, $ a continuous function\}라고 하자. 그리고 주어진 $K^{compact} \subset X$ 와 $U^{open} \subset Y$에 대해서 $S(K,U) := \{f: X \to Y | f(K) \subset U\}$라고 하면 \\ $\frak{S} := \{S(K,U) | K^{compact} \subset X, U^{open} \subset Y\}$를 subbasis 로 갖는 $\yx$의 topology를 {\bf compact-open topology} 라고 한다.\\ {\bf 숙제 11}\\ 1. $Y$ is Hausdorff $\Rightarrow \yx$ is Hausdorff. \\ 2. Let $(Y,d)$ be a bounded metric space. Then Show that $id : \yx_{d} \to \yx$ is continuous.(여기서 $\yx_{d}$는 metric topology가 주어진 공간이고 $\yx$는 compact-open topology가 주어진 공간이다.)\\ 3. Let $X$ be a compact Hausdorff space and $Y$ be a bounded metric space. Then $\yx_{d} = \yx$, i.e., topology of uniform convergence is same as compact-open topology.(물론 2의 결과로 부터 topology of uniform convergence 가 compact-open topology보다 finer하다는 것을 안다.) \begin{thm} Let $\varphi : Z \times X \to Y$ and let $\hat{\varphi} : Z \to \yx$ be the induced map defined by $\hat{\varphi}(z)(x) = \varphi(z,x)$. Then\\ \hspace*{2.0em} 1) If $\varphi$ is continuous, then $\hat{\varphi}$ is continuous.\\ \hspace*{2.0em} 2) If $X$ is locally compact Hausdorff, then $\varphi$ is continuous if and only if\\ \hspace*{4.0em}$\hat{\varphi}$ is continuous. \end{thm} \begin{proof} 1) $\hat{\varphi}$가 continuous 인 것을 보이는 것은 주어진 $\hat{\varphi}_{z} \in S(K,U)$에 대해서 $\hat{\varphi}(V) \subset S(K,U)$인 $z$의 neighborhood $V$ 가 존재함을 보이는 것과 같은 것인데,\\ $\varphi$ is continuous $\Rightarrow \exists V$, a neighborhood of $z$ such that $\varphi(V \times K) \subset U$ (예전에 했던 것처럼 $K$의 compactness를 이용해서 유한개의 product neighborhood를 subcover한 후 $V$를 선택) $\Rightarrow \hat{\varphi}(V) \subset S(K,U)$ 이므로 $\hat{\varphi}$는 continuous 이다.\\ \begin{figure}[htb] \centerline{\includegraphics*[scale=0.4,clip=true]{pic2.eps}} \end{figure} 2) 주어진 $\varphi(z,x) \in U^{open} \subset Y$에 대해서 $\varphi(V_{z} \times W_{z}) \subset U$를 만족시키는 $W_{z}$와 $V_{z}$가 존재함을 보이자.\\ 먼저 $X$가 locally compact Hausdorff 이므로 $\hat{\varphi}_{z}(\bar{W_{z}}) \subset U$를 만족시키는 $x$ 의 relatively compact neighborhood $W_{z}$가 존재하는 것을 안다. 즉 $\hat{\varphi}_{z} \in S(\bar{W_{z}},U)$. 또한 $\hat{\varphi}$가 continuous이므로 $\hat{\varphi}(V_{z}) \in S(\bar{W_{z}},U)$를 만족시키는 $z$의 neighborhood $V_{z}$가 존재하는 것을 알고 따라서 $\varphi(V_{z} \times W_{z}) \subset U$이다.\\ \end{proof} \begin{cor} $X$ is locally compact Hausdorff. Then $\yx \times X \overset{ev}\longrightarrow Y$(evaluation) is continuous.\\ \end{cor} \begin{proof} $\hat{ev} : \yx \to \yx$ is an identity map and apply the above theorem. \end{proof} {\bf exercise} $X$ is a locally compact Hausdorff space. Then $Y^{X \times Z} \to (\yx)^{Z}$ is a homeomorphism.\\ 이제 다시 $\pi_{n}\x$로 돌아가자.\\ Now $\Ome \x = \x^{(I,\pa I)} \subset X^{I}.$\\ Compare $\pi_{2}\x ( = F_{2}\x / \thicksim)$ and $\pi_1(\Ome\x , x_0) ( = \Ome(\Ome\x , x_{0}) / \thicksim)$. 그러면 \\ $\alp : I_1 \times I_2 \to X \overset{1-1}\leftrightarrow \hat{\alp} :I_1 \to X^{I_{2}}$의 관계에 의해서 \\ \hspace*{10.0em} $\searrow \hspace{1.0em} \cup$\\ \hspace*{11.0em} $\Ome\x$\\ $\alp \in F_{2}\x \Leftrightarrow \hat{\alp} : (I,\pa I) \to (\Ome\x , x_{0}) \Leftrightarrow \hat{\alp} \in \Ome(\Ome\x , x_{0})$ 가 성립한다. 또한 앞의 정리에 의해서 $\hat{\alp}$는 continuous이다. 즉\\ $\wedge : F_{2}\x \overset{1-1}\leftrightarrow \Ome(\Ome\x , x_{0})$이다.\\ 또한 $\alp , \bet \in F_{2}\x$에 대해서 \\ \hspace*{1.0em}$\exists H : I^{2} \times I \to X$, homotopy between $\alp$ and $\bet$ \hspace{2.0em} ($I^{2}=I_{1} \times I_{2}$)\\ $\overset{by thm}\Longleftrightarrow \hat{H} : I_{1} \times I \to X^{I_{2}}$, homotopy between $\hat{\alp}$ and $\hat{\bet}$\\ 이 성립하므로 $\alp \thicksim \bet \Leftrightarrow \hat{\alp} \thicksim \hat{\bet}$이 된다.\\ 따라서 $\wedge : \pi_{2}\x \overset{1-1}\leftrightarrow \pi_{1}(\Ome\x , x_{0})$이다.\\ 마지막으로\\ \begin{displaymath} \hat{\alp} \ast \hat{\bet}(s)=\left\{\begin{array}{cl} \hat{\alp}(2s) = \alp(2s,-)\\ \hat{\bet}(2s-1) = \bet(2s-1,-) \end{array} \right. \end{displaymath} 이 되므로 $\hat{\alp} \ast \hat{\bet}(s) = \alp \ast \bet(s,-) = \widehat{\alp \ast \bet}(s)$가 되어 $\widehat{\alp \ast \bet} = \hat{\alp} \ast \hat{\bet}$가 성립한다.\\ \begin{center} \framebox[1.1\width]{$\wedge$ is a canonical isomorphism between $\pi_{2}\x$ and $\pi_{1}(\Ome\x,x_{0})$}\\ \end{center} \newpage In general, $\pi_{n}\x \overset{\wedge}= \pi_{n-1}(\Ome\x , x_{0})$\\ i.e. $\alp \in F_{n}\x \Leftrightarrow \alp : I^{n-1} \times I \to X$ \\ \hspace*{4.0em} $\Leftrightarrow \hat{\alp} : I^{n-1} \to X^{I} \\ \hspace*{8.0em} \searrow \hspace{1.0em} \cup\\ \hspace*{9.0em} \Ome\x$\\ 이는 $\pi_2$경우와 같은 방법으로 $\wedge$가 canonical isomorphism임을 알 수 있다.\\ 따라서 $\pi_{3}(X) = \pi_{2}(\Ome(X)) = \pi_{1}(\Ome(\Ome(X))) = \pi_{1}(\Ome^{2}X), \cdots$\\ \hspace*{3.0em} $\pi_{n}(X) = \pi_{1}(\Ome^{n-1}X)$ 이 성립한다.\\ \end{document}