\documentclass[12pt ]{article} \setlength{\textwidth}{14 true cm} \setlength{\textheight}{20 true cm} \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} \usepackage{floatflt} \usepackage[arrow,matrix]{xy} \usepackage{graphicx} \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{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^m}} \newcommand{\rb}{\mathbb{R}} \newcommand{\zb}{\mathbb{Z}} \newcommand{\bd}{\partial} \newcommand{\key}[1]{\emph{\textbf{#1}}} \newcommand{\rh}[1]{\widetilde{H}(#1)} \newcommand{\hh}{\widetilde{H}} \newcommand{\hhp}{\widetilde{H}_p} \newcommand{\ssx}[1][p]{\triangle^{#1}} \newcommand{\iosh}{i_{0\sharp}} \newcommand{\ish}{i_{1\sharp}} \newcommand{\ios}{i_{0*}} \newcommand{\is}{i_{1*}} \newcommand{\sm}{S^n-s^r} \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{pf}{{\bf 증명}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \section*{I. Adjunction Space (Attaching space)} \textbf{I.1 Construction} \begin{defn} $X, Y$: disjoint topological spaces , $A \overset{closed}{\subset} X$ and $ f: A\rightarrow Y$, a map.\\ Define an equivalence relation $\sim$ on $X\coprod Y$ generated by $a\sim f(a),\hspace{1em} \forall a\in A$.\\ The quotient space $X\underset{f}{\cup} Y = X\coprod Y/\sim$ is called the adjunction space determined by $f$ and $f$ is called an attaching map.\end{defn} \begin{thm}(Extension principle)\\ Let $g: X\rightarrow Z$ and $h: Y\rightarrow Z$ s.t. $g(a)=hf(a),\hspace{1em} \forall a\in A\Rightarrow$ \[ \xymatrix @M=1ex @C=3em @R=1em @*[c]{% X\coprod Y \ar[r]^{g\coprod h} \ar[dd]^{p}&Z&\\ & & \textrm{ $\exists ! k$ s.t. the diagram commute}\\ X\underset{f}{\cup} Y \ar@{.>}[uur] & &} \] \end{thm} \begin{thm} Let $X\coprod Y \overset{p}{\rightarrow} X\underset{f}{\cup} Y$ be the quotient map.\\ (1) $Y$ is embedded as a closed subset of $X\underset{f}{\cup}Y:\\ \hspace*{1.5em}p|_Y : Y\rightarrow p(Y)$ is a homeomorphism.\\ (2) $X-A$ is embedded as an open subset of $X\underset{f}{\cup}Y:\\ \hspace*{1.5em}p|_{X-A} : X-A \rightarrow p(X\setminus A) $ is a homeomorphism.\end{thm} \begin{pf}(1) $p|_Y$ is continuous and 1-1. \\ Show $p|_Y$ is a closed map:\\ $C\subset Y$ a closed subset and show $p(C)$ is closed in $X\underset{f}{\cup}Y$,\\ i.e., $p^{-1}p(C)= f^{-1}(C)\coprod C$ is closed. And the assertion clearly holds.\\ (2) $p|_{X-A}$ is continuous and 1-1. Show it is an open map:\\ $U\subset X-A$ open $\Rightarrow U$ open in $X \Rightarrow p(U)$ is open since $p^{-1}p(U)=U$ is open is $X\coprod Y$.\end{pf} \newpage \begin{thm} (Separation Axiom)\\ $X ,Y : T_1 \Rightarrow X\underset{f}{\cup}Y : T_1\\ X, Y :$ normal $\Rightarrow X\underset{f}{\cup} Y$: normal\\ Ref. See Munkres p.210\end{thm} \begin{defn}(Collared pair) $(X,A)$ is called a collared pair if\\ (1)$ A\subset X$ is closed.\\ (2)$ X$ is Hausdorff.\\ (3) Points in $X-A$ can be separated from $A: \forall x\in X-A, \hspace{0.5em} \exists U, V:$ disjoint open sets s.t. $x\in U $ and $ A\subset V.$\\ (4) $A$ has a collaring $B$ in $X$ : $\exists$ open $B\supset A$ s.t. $A$ is a strong deformation retract of $B$.\end{defn} \begin{prop} $(X, A)$: a collard pair, $Y$: Hausdorff $\Rightarrow (X\underset{f}{\cup} Y , Y)$ : a collard pair.\\ In fact, $B$: a collaring of $A \Rightarrow Y\cup p(B)$: a collaring of $Y$.\end{prop} \begin{pf} (1) : clear from 정리 2(1)\\ (2) $X\underset{f}{\cup} Y$ is Hausdorff: \\ Case 1. $z_1 , z_2 \in X\underset{f}{\cup}Y -Y \cong X-A \Rightarrow $clear.\\ Case 2. $z_1 \in Y , z_2\notin Y \overset{\textrm{정의 2(3)}}{\Rightarrow} \exists U \ni z_2 , V\supset A\\ \Rightarrow p(U):$ open neighborhood of $z_2$ and $p(V)\cup Y$: open neighborhood of $z_1$ gives a separation. (Note $p^{-1}(p(V)\cup Y) = V\coprod Y$: open in $X\coprod Y.$)\\ Case 3. $z_1 , z_2 \in Y :$ Let $z_1 \in V_1 , z_2 \in V_2 $ be a separation and \\ $r :B\rightarrow A$ a strong deformation retract. Let $U_i = r^{-1}f^{-1}(V_i )$: open in $X.\\ \Rightarrow p(U_1 )\cup p(V_1 )$ and $p(U_2) \cup p(V_2)$ give a separation for $z_1$ and $z_2$\\ (Note $p^{-1}(p(U)\cup p(V)) = U\coprod V$.)\\ (3) $ z\notin Y$. Then use 정의 2(3) to get disjoint open sets $U\ni z$ and $ V\supset A \Rightarrow p(U)$ and $p(V)\cup Y$ give a separation for $z$ and $Y$.(cf. Case2.)\\ (4) Let $D: id \simeq i\cdot r$(rel $A$) be a strong deformation retract :\\ $D:B\times I \rightarrow B$ s.t. $\left\{\begin {array}{cc} D(a,t)=a ,& \forall a\in A\hspace{1em} t\in I\\ D(b,0) = b , & \forall b\in B\\ D(b,1) = r(b) \in A ,& \forall b\in B \end{array}\right\}$\\ Define $\bar{D} : p(B)\cup Y \times I \rightarrow p(B)\cup Y$ by $\bar{D}(z,t) = \left\{\begin {array}{ccc} z ,& z\in Y&\\ p(D(b,t)) ,& z = p(b) ,& b\in B-A \end{array}\right\} $\\ \[ \xymatrix @M=1ex @C=3.5em @R=1em @*[c]{% (B\coprod Y)\times I\ar[rr]^{D\coprod p_1 }\ar[dd]^{p\times id} && B\coprod Y\ar[dd]^{p}\\ & \curvearrowright & \\ (p(B)\cup Y)\times I\ar[rr]^{\bar{D}} & & p(B)\cup Y} \] $\Rightarrow \bar{D}$ is continuous by the following fact.\\ Fact. $p:X\rightarrow Y$ quotient , $C:$ locally compact Hausdorff.\\ $\Rightarrow p\times id :X\times C \rightarrow Y\times C$ is a quotient map.\\ \begin{pf} Ref. Munkres p.113\end{pf} \end{pf} \end{document}