\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[all]{xy} \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{\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{\bd}{\partial} \newcommand{\key}[1]{\emph{\textbf{#1}}} \newcommand{\td}[1]{\widetilde{#1}} \newcommand{\ssx}[1][p]{\triangle^{#1}} \newcommand{\sg}[2][X]{S_{#2}({#1})} \newcommand{\sgi}[2][X]{S_{#2}({#1}\times I)} \newcommand{\sgp}[1][X]{S_{p}({#1})} \newcommand{\sgip}[1][X]{S_{p}({#1}\times I)} \newcommand{\iosh}{i_{0\sharp}} \newcommand{\ish}{i_{1\sharp}} \newcommand{\ios}{i_{0*}} \newcommand{\is}{i_{1*}} \newcommand{\bigt}{\bigtriangleup} \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*{VII.2 Definition of singular homology} Let $e_{0}=(0,0,\cdots,0)$ in $\rb^{\infty}$ ( or $\rb^{N})$.\\ \hspace*{1.5em} $e_{1}=(1,0, \cdots,0)$\\ \hspace*{1.5em} $e_{2}=(0,1, \cdots,0)$\\ \hspace*{4.5em} \vdots\\ \hspace*{1.5em} etc.\\ $\bigt^{p}$ = the simplex spanned by $\{e_{0},e_{1}, \cdots , e_{p}\}$ = the standard $p$-simplex.\\ The $i$-th face map $f_{p}^{i} : \bigt^{p-1} \rightarrow \bigt^{p}$ is an affine map given by\\ \begin{displaymath} f_{p}^{i}(e_{j}) = \{ \begin{array}{ll} e_{j} & j < i \\ e_{j+1} & j \geq i \end{array}.\\ \end{displaymath} i.e. "$f_{p}^{i}(\bigt^{p-1}) = f_{p}^{i}(e_{0}, \cdots, e_{p-1}) = (e_{0}, \cdots, \widehat{e_{i}}, \cdots, e_{p})$" $< \bigt^{p}$.\\ {\bf 예}\\ \psset{unit=1.5cm} \begin{pspicture}(-1.0,-1.0)(4.0,2.0) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psline{|-|}(0,0.5)(1.0,0.5)\psline{->}(0,0.5)(0.5,0.5) \rput(0,0){$e_{0}$}\rput(1.0,0){$e_{1}$} \pspolygon(2.0,0)(3.0,0)(2.0,1.0)(2.0,0) \psline{->}(2.0,0)(2.5,0) \psline{->}(2,0)(2,0.5) \psline{->}(3,0)(2.5,0.5) \rput(1.8,-0.5){$e_{0}$} \rput(3.2,-0.5){$e_{1}$} \rput(1.8,1.2){$e_{2}$} \pscurve{->}(0.5,0.5)(1.5,1.0)(2,0.5) \pscurve{->}(0.6,0.5)(1.5,0)(2.5,0) \pscurve{->}(0.4,0.5)(2.0,-0.5)(3.2,0)(2.5,0.5) \rput(1.5,0.8){$f^{1}$} \rput(1.5,0.2){$f^{2}$} \rput(3.5,0){$f^{0}$} \end{pspicture} \begin{center} $f^{0}(e_{0},e_{1}) = (e_{1}, e_{2})$ \\ $f^{1}(e_{0},e_{1}) = (e_{0}, e_{2})$ \\ $f^{2}(e_{0},e_{1}) = (e_{0}, e_{1})$ \end{center} Let $X$ be a topological space.\\ A singular $p$-simplex in $X$ is a map $\sigma : \bigt^{p} \rightarrow X$.\\ $i$-th face $\sigma^{(i)}$ of $\sigma$ is a map : $\bigt^{p-1} \overset{f^{i}}{\rightarrow} \bigt^{p} \overset{\sigma}{\rightarrow} X$,i.e., $\sigma^{(i)} = \sigma \circ f_{p}^{i}$\\ $S_{p}(X)$를 singular $p$-simplices in $X$에 의해서 generated되는 free abelian group이라고 하자.(more generally, the free $R$-module generated by singular $p$-simplices in $X$) 또는, the group(module) of singular $p$-chains이라고도 한다.\\ 그러면, $c \in S_{p}(X)$는 finite sum, $\sum_{i=1}^{k} n_{i}\sigma_{i}$로 유일하게 표현된다.\\ 이제 boundary operator를 정의하자.\\ Define a homomorphism $\partial : S_{p}(X) \rightarrow S_{p-1}(X)$ by \begin{displaymath} \partial\sigma = \sum^p_{i=0} (-1)^{i}\sigma^{(i)} = \sum^p_{i=0} (-1)^{i} \sigma \circ f_{p}^{i} \end{displaymath} 또는 informally, \begin{displaymath} \partial\sigma(e_{0}, \cdots, e_{p-1}) = \sum^{p}_{i=0} (-1)^{i} \sigma(e_{0}, \cdots, \widehat{e_{i}}, \cdots, e_{p}) \end{displaymath} {\bf Note}\\ 1.$\partial^{2}$=0 : same as before.\\ 2. {\bf 숙제 21(check)}\hspace{0.5em} In general, $\sigma(e_{0}, \cdots, e_{i}, e_{i+1}, \cdots, e_{p}) \sim -\sigma(e_{0}, \cdots, e_{i+1}, e_{i}, \cdots, e_{p})$\\ $p$가 1인 경우에 대해서 살펴보자. \begin{center} \psset{unit=1.5cm} \begin{pspicture}(-0.5,-0.5)(3.5,1.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psline{|-|}(0,0)(1.0,0) \rput(0,-0.2){$e_{0}$} \rput(1.0,-0.2){$e_{1}$} \rput(1.5,0.4){$\rightarrow$} \pscurve(2.0,0)(2.25,0.4)(2.5,0.5)(2.75,0.6)(3.0,1.0) \pscurve{->}(2.0,0)(2.25,0.4)(2.5,0.5)(2.75,0.6) \pscurve{->}(3.0,1.0)(2.75,0.6)(2.5,0.5)(2.25,0.4) \rput (2.25,0.6){$\bar{\tau}$} \rput(2.75,0.4){$\tau$} \end{pspicture} \end{center} 위와 같이 주어진 $\tau$와 $\bar{\tau}$에 대해서 $\bar{\tau} + \tau \sim 0$임을 보이자. 다시 말해 어떤 $c \in S_{2}(X)$에 대해서 $\bar{\tau} + \tau = \partial c$임을 보이면 된다. 아래의 그림과 같이 $\sigma, \xi$를 정의하면, \begin{center} \psset{unit=1.5cm} \begin{pspicture}(-0.5,-0.5)(4.0,3.5) %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \psline{|-|}(0,0)(1.0,0) \psline{->}(0,0)(0.5,0) \psline{->}(1.0,0)(0.5,0) \rput(0,-0.2){$e_{0}$} \rput(1.0,-0.2){$e_{1}$} \rput(0.5,1.0){$\downarrow$} \rput(0.3,1.0){$\psi$} \pspolygon[fillstyle=solid, fillcolor=lightgray](0,2.0)(1.0,2.0)(0.5,2.732)(0,2.0) \psline{->}(0,2.0)(0.5,2.0) \psline{->}(1.0,2.0)(0.75,2.366) \rput(-0.2,2.0){0} \rput(1.2,1.8){1} \rput(0.5,2.8){2} \pscurve{->}(1.4,2.25)(1.6,2.4)(1.8,2.25) \rput(1.6,2.6){$\sigma$} \pscurve{*->}(2.0,2.0)(2.3,2.7)(2.7,2.5) \pscurve{*->}(3.5,3.0)(3.2,2.3)(2.7,2.5) \rput(2.0,1.7){$p$} \rput(3.5,3.2){$q$} \pspolygon[fillstyle=solid,fillcolor=lightgray](2.5,0)(3.5,0)(3.0,0.732)(2.5,0) \pscurve{->}(1.2,0.3)(1.6,1.0)(1.8,1.7) \rput(1.5,1.4){$\tau$} \pscurve{->}(3.0,0.366)(2.5,1.4)(2.0,1.9) \rput(3.5,1.0){$\xi(const.map)$} \end{pspicture} \end{center} 여기서 $\xi : \bigt^{2} \rightarrow X$는 $\xi(a)=p , \forall a \in \bigt^{2}$인 constant map이다. 또한, $\psi$는 2-simplex를 1-simplex로 그림과 같이 collapsing 시키는 map이고 $\sigma$를 $\psi \circ \tau$로 정의한다. 그러면, \\ $\partial\sigma = \sigma^{(0)} - \sigma^{(1)} + \sigma^{(2)} = \sigma(\widehat{0},1,2) - \sigma(0,\widehat{1},2) + \sigma(0,1,\widehat{2}) = \bar{\tau} - p +\tau$이고 \\ $\partial\xi = p-p+p = p$이므로,\\ $\bar{\tau} + \tau = \partial\sigma + p = \partial\sigma + \partial\xi = \partial(\sigma + \xi)$가 되어서 $\tau \sim \bar{\tau}$임을 알 수 있다.\\ 이제 $X$의 singular chain complex를 정의하자.\\ $S(X) = \{S_{p}(X),\partial\}$을 singular chain complex of $X$라고 하고, \\ \begin{displaymath} \cdots \rightarrow S_{p+1}(X) \overset{\partial}{\rightarrow} S_{p}(X) \overset{\partial}{\rightarrow} S_{p-1}(X) \rightarrow \cdots\\ \end{displaymath} $p$-th singular homology group of $X$, $H_{p}(X)$를 \begin{center} $H_{p}(X) := Z_{p}(X) / B_{p}(X)$ \\ \end{center} where $Z_{p}(X) = ker \partial_{p}$ (whose element is called a $ cycle$) and $B_{p}(X) = im \partial_{p+1}$ (whose element is called a $boundary$).\\ 라고 정의한다. 이때 $R$-module의 경우 ($H_{p}(X : R)$)와 구별하기 위해서 $H_{p}(X : \zb)$라고 쓰기도 한다.\\ 1. {\bf Functorial property for $H_{p}$}\\ Let $f : X \rightarrow Y$ be a map. Then $f$ induces a chain map.\\ Define $f_{\sharp} : S(X) \rightarrow S(Y)$ by $\sigma \mapsto f \circ \sigma$.Then\\ $\ulcorner f_{\sharp}(\partial \sigma) = f_{\sharp}(\sum (-1)^{i}\sigma^{(i)}) = \sum (-1)^{i}f_{\sharp}\sigma^{(i)} = \sum (-1)^{i}f \circ \sigma^{(i)} = \sum (-1)^{i}f\circ(\sigma\circ f^{i}) = \sum (-1)^{i}(f \circ \sigma) \circ f^{i} = \partial(f \circ \sigma) = \partial(f_{\sharp}\sigma)\\ \therefore \hspace{0.2em}f_{\sharp}\partial = \partial f_{\sharp} \lrcorner$\\ Hence $f_{\sharp}$ induces a homomorphism $f_{*}(=H_{p}(f)) : H_{p}(X) \rightarrow H_{p}(Y)$. (1) $id : X \rightarrow X \Rightarrow id_{*} = id$ \\ (2) $X \overset{f}{\rightarrow} Y \overset{g}{\rightarrow} Z \Rightarrow (g \circ f)_{\sharp} = g_{\sharp} \circ f_{\sharp} \Rightarrow (g \circ f)_{*} = g_{*} \circ f_{*}$\\ $\therefore \hspace*{0.2em} H_{p}$ is a covariant functor from the category of topological spaces to the category of abelian groups ($R$-modules).\\ 2. \begin{displaymath}{\bf \textrm{$H_p(point)$}} = \{ \begin{array}{ll} 0 & p \neq 0 \\ \zb & p = 0 \end{array} \end{displaymath} $X$가 point이므로 $S_{p}(X) = $이다. 여기서 $c_{p} : \bigt^{p} \rightarrow X$인 constant map이다. 즉\\ \begin{center} $\cdots \overset{\partial}{\longrightarrow} S_3 \overset{\partial}{\longrightarrow} S_2 \overset{\partial}{\longrightarrow} S_1 \overset{\partial}{\longrightarrow} S_0 \overset{\partial}{\longrightarrow} 0$\\ \end{center} \hspace*{12.0em} $\Arrowvert \hspace{2.5em} \Arrowvert \hspace{2.7em} \Arrowvert$ \\ \hspace*{10.0em} $ \hspace{0.5em} \hspace{0.5em} $ \\ 이다. 그런데 여기서 $\partial c_{1} = c_0 - c_0$ 이므로 $S_1 \overset{\partial=0}{\rightarrow} S_0$가 되어 $H_{0}=\zb$이다. 또한 $\partial c_{2} = c_1 - c_1 + c_1 = c_1$이므로 $S_2 \overset{\partial(\cong)}{\rightarrow} S_1$가 되어 $H_{1}=0$이다. 이와 같은 방법으로 $p>2$에 대해서 $H_{p}=0$임을 알 수 있다.\\ 3. $\{X_{\alp}\}$ : path-components of $X$\\ $\Rightarrow S_{p}(X) = \underset{\alp}{\bigoplus} S_{p}(X_{\alp})$ since $\bigt^{p}$ is connected and $\forall c \in S_{p}(X)$ can be written uniquely as $\sum c_{\alp}$. Moreover, $\partial : S_{p}(X_{\alp}) \rightarrow S_{p-1}(X_{\alp})$.\\ $\Rightarrow Z_{p}(X)= \underset{\alp}{\bigoplus}Z_{p}(X_{\alp})$ and $B_{p}(X)= \underset{\alp}{\bigoplus}B_{p}(X_{\alp})$\\ $\Rightarrow H_{p}(X)= \underset{\alp}{\bigoplus}H_{p}(X_{\alp})$\\ 4. Let $X$ be path-connected. Then $H_{0}(X) = \zb (or R)$.\\ \begin{proof} {\bf (Idea)}\\ $x_{0} \in X$를 고정시키자. 그러면 $x_{0}$와 임의의 $x \in X$ 사이에는 이 둘을 잇는 path가 존재하고, 1-simplex $\rho : \bigt^{1} \rightarrow X$의 image를 이 path라고 하면 $\partial \rho = x - x_{0}$ 가 되고 따라서 $x \sim x_{0}$이 성립한다.\\ 자세한 증명을 살펴보자.\\ 임의의 $c \in S_{0}(X)$는 $\underset{finite}{\sum} n_{i} x_{i}$로 표시할 수 있다. 여기서 $x_{i}$는 $e_{0}$를 $x_{i} \in X$로 보내주는 0-simplex이다.\\ 그러면 우선 $c=\partial c_{1} \Leftrightarrow \sum n_{i}=0$임을 보이자.\\ $\because (\Rightarrow) c_{1}=\sum k_{i} \sigma_{i} \Rightarrow c=\partial c_{1} = \sum k_{i}(\sigma_{i}(1)-\sigma_{i}(0)) \Rightarrow \sum n_{i} = \sum (k_{i}-k_{i})=0$\\ \hspace*{4.0em}여기서 $\sigma_{i}$는 1-simplex이다.\\ \hspace*{0.8em} $(\Leftarrow) c=\sum n_{i} x_{i} - (\sum n_{i})x_{0} = \sum n_{i}(x_{i} - x_{0}) = \sum n_{i} \partial \rho_{i}$\\ \hspace*{4.0em}여기서 $X$가 path-connected라는 조건이 쓰였다.\\ 이제 $S_{0}(X) \overset{\eps}{\rightarrow} \zb$ (or $R$) $\sum n_{i} x_{i} \mapsto \sum n_{i}$를 생각해보자. 그러면 $\eps$는 onto이고, $\eps \circ \partial_{1}= 0$이다. 또한 앞에서 보였던 것을 이용하면, 만약 $X$가 path-connected라면 $ker \eps = im \partial_{1}$이 되어 $S_{1} \overset{\partial_{1}}{\rightarrow} S_{0} \overset{\eps}{\rightarrow} \zb \rightarrow 0$은 $S_{0}$에서 exact이다. 따라서 $H_{0}(X)=S_{0}/ker \eps = \zb$가 된다. \end{proof} \begin{defn} Given a chain complex $\{C,\partial\}$, an epimorphism $C_{0} \rightarrow \zb \rightarrow 0$ with $\eps \circ \partial_{1} = 0$ is called an {\bf augmentation}. And the homology of $\cdots \rightarrow S_{p}(X) \rightarrow \cdots \overset{\partial}{\rightarrow} S_{0}(X) \overset{\eps}{\rightarrow} \zb \rightarrow 0$ is called a {\bf reduced homology of $X$} and denoted by $\widetilde{H_{p}}(X)$. \end{defn} {\bf Note}\\ 1. $\widetilde{H_{p}}(X) = H_{p}(X)$ if $p \geq 1$\\ 2. $H_{0}(X) \cong \widetilde{H_{0}}(X) \bigoplus \zb$, since $S_{0}(X) \cong ker \eps \bigoplus \zb$ and $im \partial_{1} \subset ker \eps$.\\ 3. $\widetilde{H_{p}}(point)= 0 \hspace{1.0em} \forall p$. \begin{defn} A chain complex $\{C_{p},\partial\}$ is called {\bf acyclic} if $H_{p}(C)=0 \hspace{1.0em} \forall p$.\\ An augmented chain complex $\{C_{p},\partial, \eps\}$ is called {\bf acyclic} if $\widetilde{H_{p}}(C)=0 \hspace{1.0em} \forall p$. i.e.,\\ \begin{displaymath} \cdots \rightarrow C_{p+1} \overset{\partial}{\rightarrow} C_{p} \overset{\partial}{\rightarrow} C_{p-1} \rightarrow \cdots \end{displaymath} is acyclic if and only if it is exact.\\ \end{defn} {\bf Example} $\{S_{p}(pt.), \partial, \eps\}$ is acyclic. \end{document}