\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{graphicx} \usepackage{picins} %\usepackage{pstricks} %\usepackage{pst-node} %\usepackage{pst-coil} %\usepackage{pstcol} %\usepackage{pst-eps} %\newpsobject{showgrid}{psgrid}{subgriddiv=1,griddots=10,gridlabels=6pt} \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{\bM}{\textbf{M}} \newcommand{\bbU} {\mathcal{U}} \newtheorem{thm}{정리} \newtheorem{cor}[thm]{따름정리} \newtheorem{lem}[thm]{보조정리} \newtheorem{prop}[thm]{명제} \newtheorem{cl}{Claim} {\theorembodyfont{\rm} \newtheorem{ex}{Example} \newtheorem{que}{질문} \newtheorem{notation}{Notation}[section] \newtheorem{defn}{정의} \newtheorem{rem}{remark} \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*{Basic topological property} \textbf{1}. $M$ : n-manifold. \\ $\Rightarrow$ M \phantom{ }is Hausdorff, locally compact, locally connected, 1st countable.\\ \\ \textbf{2}. $M$ is connected $\Leftrightarrow$ M\,\, is path-connected \\ \begin{proof} At first, path-connected space is connected clearly. Conversely, since $M$ is locally homeomorphic to Euclidean space, every point in $M$ has a path-connected neighborhood. Recall that $X$ is path-connected if and only if $X$ is connected and every point in $X$ has a path-connected neighborhood. (Proof: Each path-component of $X$ is open because each point in path-component has a path-connected neighborhood inside its component by maximality. A path-component is also closed since the compliment is open. By connectedness, open and closed set is only $X$ itself.) \end{proof} \begin{thm} Let $M$ be a compact n-manifold. \\ 1) Let $\bbU$ be an open cover of $M$. Then there exists a \textbf{partition of unity} subordinate to $\bbU$ .\\ 2) There exists $\phi : M \hookrightarrow \mathbb{R}^N$, an \textbf{embedding} into a Euclidean space. \end{thm} \begin{proof} 1) We can choose a finite coordinate refinement of $\bbU$ because $M$ is compact. In fact, we can choose $\{V_1 , V_2 , \ldots , V_k \}$, an open covering of $M$ such that $V_j \subset \overline{V_j} \subset U _j$ for a coordinate chart $\{ U _j , \phi _j \} $, where $\{ U_1, \ldots, U_k \} $ is an open refinement of $\bbU$. $M$ is a normal space since it is compact Hausdorff. \\ Using Urysohn lemma, we can construct $g_j : M \rightarrow [0,1]$ by \\ $$g_j (x) = \left\{ \begin{array}{c} 1 , \,\, x \in \overline{ V_j} \\ 0 , \,\, x \in U_j^c \\ \end{array} \right. \,.$$ Define \begin{displaymath} f_j (x) := \frac{g_j(x)}{\sum_{j=1}^k g_j(x)} \,\,. \end{displaymath} Then we can easily check $\{f_j\}$ is a partition of unity.\\ 2) For each $i=1,\ldots,k,$ define $\psi_i : M \rightarrow \mathbb{R}^{n+1}$ by $\psi_i(x) = (g_i(x)\phi_i(x),g_i(x))$. then $\psi_i$ is well-defined on $M$. Let $\psi : M \rightarrow \mathbb{R}^{n+1} \times \cdots \times \mathbb{R}^{n+1} = \mathbb{R}^{k(n+1)} = \mathbb{R}^N$ be given by $\psi = ( \psi_1,\ldots,\psi_k)$. We claim $\psi$ is one-to-one. Suppose that $\psi(x)= \psi(y)$, then $\psi_i(x)=\psi_i(y)$ and $g_i(x)=g_i(y)$ for every $i=1,\ldots,k$. And if $x \in V_j$ then $g_j(y)=g_j(x)=1$ and hence $y \in U_j$. Now $\phi_j(x)=\phi_j(y)$ implies $x=y$ as $\phi_j$ is a homeomorphism. Hence $\psi$ is an embedding since $M$ is compact. \end{proof} \begin{thm} If M is $2^{nd}$ countable, then M is paracompact.\\ In fact, each open cover has a countable locally finite refinement consisting of open sets with compact closures. \\ \end{thm} \begin{proof} At first, we show that there exists a countable basis $\mathcal{A}$ consisting of relatively compact open sets. Since $M$ is $2^{nd}$ countable, there exists a countable basis $\mathcal{B}$. Let $\mathcal{A} = \{B \in \mathcal{B} : \overline{B}$ is compact$\}$. %\parpic[l]{ % \includegraphics[width=0.4\textwidth]{figure2.eps} % } %\begin{center} %\begin{pspicture}(0,0)(4,4) %\showgrid% %\pscircle(2,2){2}% %pscircle[linestyle=dotted](2,2){1.5}% %\pscircle[linestyle=dashed](2,2){1}% %\pscircle(2,2){0.5}% %\rput(2,2){$x$}% %\rput[lb](-0.4,0){$\mathbf{B} \in \mathcal{B}$} %\rput*(2,4){$\mathbf{M}$} %\rput*(3,3){$U$}% %\rput*(3,2){$V$} %\psline(-0.2,0.3)(1.8,1.8) %\end{pspicture} %\end{center} $M$ 은 locally compact 이기때문에 $x$를 포함하는 임의의 neighborhood $U$에 대하여 relatively compact open set $V$를 선택할 수 있고, $V$또한 $x$를 포함하는 open set이므로 $x \in B \subset V$ 인 basis element $B$를 찾을수 있다. $\overline{B} \in \overline{V}$ 즉 compact set 안의 닫힌 집합이므로 $B$는 relatively compact 이다. 따라서 $\mathcal{A}\,$ is also a countable basis and denote $\mathcal{A} = \{ A_1,A_2,\ldots \}$.\\\\ Secondly, there exists a compact exhaustion, i.e., $\{G_i : i=1,2,\ldots \}$ such that $$G_1 \subset \overline{G_1} \subset G_2 \subset \overline{G_2} \subset G_3 \subset \ldots$$ and $M = \bigcup_{i=1}^\infty G_i$, $G_i$ is open and $\overline{G_i}$ is compact. Indeed define $G_k$ inductively with $G_1 = A_1$ and $G_k=A_1 \bigcup \cdots \bigcup A_{j_k}$. Then $\overline{G_k} \subset A_1 \bigcup \cdots \bigcup A_{j_k} \bigcup A_{j_k+1} \bigcup \cdots \bigcup A_{j_{k+1}}$ for some $j_{k+1} > j_k$ since $\overline{G_k}$ is compact. Now let $G_{k+1} = A_1 \bigcup \cdots \bigcup A_{j_{k+1}}$.\\ Let $\bbU = \{U_\alpha : \alpha \in \mathcal{I} \}$ be a given cover of $M$. For each fixed index $i$, let $V_\alpha = U_\alpha \bigcap ( G_{i+2}- \overline{G_{i-1}})$. Then $\{V_\alpha : \alpha \in \mathcal{I} \}$ is an open cover of a compact set $\overline{G_{i+1}} - G_i $ and hence there exists a finite subcover $\mathcal{V}_i$. Now $\mathcal{V} = \bigcup_{i=1}^\infty \mathcal{V}_i$ is the desired refinement. \end{proof} \begin{cor} If $M$ is $2^{nd}$ countable, then there exists a partition of unity subordinate to an arbitrarily given open cover. \end{cor} \textbf{*} 일반적으로 manifold $M$ 은 $2^{nd}$ countable 을 가정한다.\\\ HW2. (1) 1학기때 배운 필요한 topology를 review할것.(제출할 필요없음.)\\ (2) smooth bump function을 construct할 수 있나?\\ \end{document}