\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{\dd}{\mathcal{D}} \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*}} \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.4 Mayer-Vietoris Sequence} \textbf{$\mathcal{U}$-small homology and sd-operator}\\ 1. Define a chain map $sd : S_p(X) \rightarrow S_p(X)$ naturally : use induction\\ $p=0 \hspace{3em}\cdots \rightarrow S_0(X)\overset{\epsilon}{\rightarrow} \mathbb{Z}\\ \hspace*{8em} \downarrow sd_0 = id \parallel\\ \hspace*{7em}\rightarrow S_0(X)\overset{\epsilon}{\rightarrow} \mathbb{Z}\\ p>0 : $ 1st for $i_p \in S_p(\bigtriangleup^p ) \hspace{1em},\forall p$\\ Define $\bet_p : {\bigtriangleup '}_k (\bigtriangleup^p )\rightarrow{\bigtriangleup '}_{k+1} (\bigtriangleup^p )$ as $\bet(v_0,\cdots,v_k ) = (b,v_0,\cdots,v_k)$ where $b$ is a barycenter of $ \bigtriangleup^p$ and ${\bigtriangleup '}_k (\bigtriangleup^p )$ is a subgroup of $S_k(\bigtriangleup^p)$ generated by affine singular $k$-simplices denoted by $(v_0 , \cdots , v_k)$.\\ Then $\bd\bet + \bet\bd =1 :$\\ Given $\sigma = (x_0,\cdots,x_k)\in {\bigtriangleup '}_k(\bigtriangleup^p ),\\ \hspace{2em} \partial \sigma = \partial (\sigma_\sharp i_k ) = \sigma_\sharp (\partial i_k ) = \sigma_\sharp (\Sigma (-1)^i (e_0,\cdots, \hat{e_i},\cdots,e_k)) = \Sigma (-1)^i (x_0,\cdots,\hat{x_i},\cdots,x_k)$\\ Now apply the formula for join operator in the proof of Lemma9.(3.Homotopy invariance of homology)\\ 1 st : Define $sd $ for $i \in \bigtriangleup_p (\bigtriangleup^p)$ as follows.\\ \[ \xymatrix @=3em @*[r] { % i \ar[r]^{\bd} \ar@{.>}[d]^{sd} & \bd i \ar[d]^{sd}\\ \bet sd \bd i \ar[r]^{\bd} & sd \bd i \ar[l]^{\bet}\in \bigtriangleup_{p-1}(\bigtriangleup^p)} \]\\ $sd i := \bet sd \bd i$\\ Then $\bd sd i = \bd\bet sd \bd i = sd \bd i - \bet \bd sd \bd i = sd \bd i- \bet sd \bd\bd i = sd \bd i$,\\ i.e., $sd i $ is a chain map. \\ 2nd on $X$ : \[ \xymatrix @M=1ex @C=1ex @R=3ex @*[c] { % % 윗줄 & i_p \ar[dl]^{\sigma_{\sharp}} \ar[rr]^{\bd} \ar[dd]^<{sd} & & \bd i \ar[dl]^{\sigma_{\sharp}} \ar[dd]^{sd}\\ % 가운뎃 줄1 \sg{p} \ar[rr] \ar@{.>}[dd]^{sd} & & \sg{p-1} \ar[dd]& \\ % 가운뎃 줄 2 & \ar[dl]^{\sigma_{\sharp}} \ar[rr] & & \ar[dl]^{\sigma_{\sharp}} \\ % 아래줄 S_{p}(X) \ar[rr] & & S_{p-1}(X)& } \] 전과 같이 왼쪽 면이 commute하게 $sd_X$를 정의\\ 그러면 앞면도 commute하게 되고 따라서 chain map으로 정의된다.\\ Naturality of $sd_p$ : same as before\\ 2. $sd \simeq id $: chain homotopic.\\ \begin{proof} use AMT.\\ $sd, id :S(X)\rightarrow S(X)$ natural chain maps\\ $\Rightarrow \exists $ natural chain homopoty $D: sd \simeq id.$\end{proof}\\ \textbf{Note.} Cor. of AMT $\Rightarrow sd$ is a natural chain homopoty equivalence.\\ 3. Let $\mathcal{U} = \{A\}$ be s.t. $X = \bigcup \{\AA | A\in \mathcal{U}\}$. Then $\forall \sigma :$ singular $p$-simplex in $X, \exists m>0 $ s.t. $sd^m(\sigma) \in S_p^{\mathcal{U}} (X) $ where $S_p^{\mathcal{U}} (X) = \{\sigma \in S_p(X) | \sigma (\bigtriangleup^p ) \subset \AA$ for some $A\subset\mathcal{U}\}$\\ \begin{proof} $X=\bigcup\{\AA| A\in \mathcal{U}\}\Rightarrow \bigtriangleup^p = \bigcup \{\sigma^{-1}(\AA)| A\in \mathcal{U}\}$\\ Let $\epsilon$ be a Lebesque number for this open cover. \\Choose $m$ s.t. $(\frac{p}{p+1})^m mesh (\bigtriangleup^p ) < \epsilon$\end{proof}\\ 4. Let $\mathcal{U} = \{A\}$ be s.t. $X = \bigcup \{\AA | A\in \mathcal{U}\}.$ Then $i : S^{\mathcal{U}}(X) \hookrightarrow S(X)$ is a chain homotopy equivalence.\\ \begin{proof} Construct a chain homotopy inverse $\tau: S(X) \rightarrow S^{\mathcal{U}}(X)$ s.t. $\tau\circ i = id$ and $i\circ\tau\simeq id$:\\ Let $m(\sigma)$ be the smallest integer s.t. $sd^{m(\sigma)}({\sigma})\in S_p^{\mathcal{U}}(X)$\\ Note $m(\sigma) \geq m(\sigma^{(i)})$\\ Want $T$ s.t. $\partial T + T\partial = i\circ \tau - id.\\ \overset{2.}{\Rightarrow} \exists D$ s.t. $\partial D + D\partial = sd-id\\ \Rightarrow \partial D sd + D sd \bd = sd^2 - sd\\ \hspace*{2em}\vdots\\ \partial D sd^{k-1} + D sd^{k-1}\partial = sd^k - sd^{k-1}$\\ 모두 더하면\\ $\partial D(1+\cdots + sd^{k-1}) +D(1+\cdots +sd^{k-1})\partial = sd^k - id$\\ Define $T(\sigma) = D(1+\cdots +sd^{m(\sigma)-1})(\sigma)$\\ Then $(\partial T + T\partial )(\sigma) =\partial T(\sigma) + T\partial(\sigma)\\ = sd^{m(\sigma)}(\sigma)-\sigma-D(1+\cdots + sd^{m(\sigma)-1})\partial \sigma + \Sigma(-1)^i D(1+\cdots +sd^{m(\sigma^{(i)})})^{-1}\sigma^{(i)}\\ = sd^{m(\sigma)}(\sigma) -\sigma -\Sigma(-1)^i D(sd^{m(\sigma^{(i)})}+\cdots + sd^{m(\sigma)-1})\sigma^{(i)}$\\ $\therefore$ Define $\tau(\sigma) = sd^{m(\sigma)}(\sigma) - \Sigma (-1)^i D (sd^{m(\sigma^{(i)})} + \cdots + sd^{m(\sigma)-1})(\sigma)$\\ Then $(\partial T + T\partial ) = i \circ \tau - id$.\\ Now note that $\tau\circ i = id$ on $S^{\mathcal{U}}(X)\\ (\because$ 이미 $\mathcal{U}$-small 하니까 $m(\sigma) = 0, m(\sigma^{(i)})=0$)\end{proof} \newpage {\bf Snake lemma(Zig - Zag lemma)}\\ (1) Given a short exact sequence of chain complexes,\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & \cc^{'} \ar[r]^i & \cc \ar[r]^q & \cc^{''} \ar[r] & 0} \] \hspace*{1.0em} where $\cc = \{C_{p},\bd\}, \cdots $ etc. and $i$ and $q$ are chain maps. Then we can obtain a long exact sequence\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r]^{\bd_{*}} & H_{p}(\cc^{'}) \ar[r]^{i_{*}} & H_{p}(\cc) \ar[r]^{q_{*}} & H_{p}(\cc^{''}) \ar[r]^{\bd_{*}} & H_{p-1}(\cc{}') \ar[r] & \cdots} \] 여기서 $\bd_{*}$는 다음과 같이 정의된다. 우선 $c^{''} \in Z_{p}^{''}$에 대해서 $\exists c \in Z_{p}$ such that $q(c)=c^{''}$ 하고 $q$가 chain map 이므로 $q \circ \bd(c)= \bd \circ q(c)=\bd(c^{''})=0$이 성립해야 한다. 따라서 exactness에 의해서 $i(c^{'})= \bd(c)$를 성립하는 $c^{'} \in Z_{p-1}^{'}$가 존재해야 하고 $\bd_{*}[c^{''}] := [c^{'}]$로 정의한다.\\ Well-definedness를 check하기 위해서 다음 두가지를 살펴보자. 먼저 $c$ 대신에 다른 $c_1 \in Z_{p}$를 선택했다고 가정하면, $\exists a \in Z_{p}^{'}$ such that $i(a)=c-c_{1}$가 되어 $c_{1}$에 의해서 골라진 $c_{1}^{'} \in Z_{p-1}^{'}$에 대해서 $\bd a = c^{'} - c_{1}^{'}$가 성립한다. 즉 $\bd_{*}$의 정의는 $c$의 선택과 무관하게 잘 정의된다. 또한 $c^{''}$ 대신에 $c^{''} + \bd d^{''}$을 선택해도 잘 정의된다. 여기서 $\bd d^{''}$는 $[c^{''}] = [c^{''} + \bd d^{''}]$를 성립하는 $Z_{p}^{''}$의 다른 원소이다.\\ 이제 long exact sequence의 exactness를 check 해보자.\\ 먼저 $H_{p}(\cc)$에서 exact한지를 살펴보면, $q \circ i = 0$이므로 $q_{*} \circ i_{*} =0$이 성립하고 따라서 $im \,\, i_{*} \subset ker \,\, q_{*}$이 성립한다. 또한 $q(c)=\bd d^{''}$인 $c$에 대해서 $q(d)=d^{''}$라 두면, $q(c- \bd d)=q(c)-\bd q(d) = q(c)-\bd d^{''} = 0$ 이므로 $i(c')=c- \bd d$인 $c' \in Z_{p}^{'}$가 존재한다. 따라서 $[c]=i_{*}[c^{'}]$이 성립해서 $ker \,\, q_{*} \subset im \,\, i_{*}$ 가 되어 $H_{p}(\cc)$에서는 exact하다. 같은 방법으로 diagram chasing을 통해서 $H_{p}(\cc^{''})$와 $H_{p}(\cc^{'})$에서 exact함을 쉽게 알 수 있다.\\ (2) 위와 같이 얻어진 long exact sequence는 functorial이다. 즉, 주어진 commutative diagram에 대해서 \\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & \cc{}' \ar[r]^i \ar[d]^{\alp'} & \cc \ar[r]^q \ar[d]^{\alp} & \cc^{''} \ar[r] \ar[d]^{\alp^{''}} & 0\\ 0 \ar[r] & \dd{}' \ar[r]_j & \dd \ar[r]_p & \dd^{''} \ar[r] & 0 } \] 아래 diagram이 commute한다.\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r]^{\bd_{*}} & H_{p}(\cc{}') \ar[r] \ar[d]^{\alp_{*}^{'}} & H_{p}(\cc) \ar[r] \ar[d]^{\alp_{*}} & H_{p}(\cc^{''}) \ar[r]^{\bd_{*}} \ar[d]^{\alp_{*}^{''}} & H_{p-1}(\cc{}') \ar[r] \ar[d] & \cdots\\ \cdots \ar[r]^{\bd_{*}} & H_{p}(\dd{}') \ar[r] & H_{p}(\dd) \ar[r] & H_{p}(\dd^{''}) \ar[r]^{\bd_{*}} & H_{p-1}(\dd{}') \ar[r] & \cdots } \] \newpage {\bf Mayer-Vietoris sequence}\\ 1. Let $\mathcal{U}=\{A,B\}$ be such that $\overset{\circ}{A} \cup \overset{\circ}{B} = X$. Consider $S_{p}(A), S_{p}(B) \subset S_{p}^{\mathcal{U}}(X)$. Then we have a short exact sequence\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% && \ar[d]^{\bd} & \ar[d]^{\bd \oplus \bd} && \\ (\star) & 0 \ar[r] & S_{p}(A \cap B) \ar[r]^>>>{\phi} \ar[d]^{\bd} & S_{p}(A) \bigoplus S_{p}(B) \ar[r]^>>>{\psi} \ar[d]^{\bd \oplus \bd} & S_{p}^{\mathcal{U}}(X) \ar[r] & 0\\ &&&&&& } \] \hspace*{2.0em} where $\phi(c) = (c,-c)$ and $\psi(a,b) = a+b$.\\ (Check) $\phi$ is one-to-one.\\ \hspace*{3.2em} $\psi$ is onto.\\ \hspace*{3.2em} $\psi \circ \phi = 0$ and $ker \,\, \psi \subset im \,\, \phi$\\ {\bf Note} $\phi$ and $\psi$ are chain maps. Furthermore, $(\star)$ is functorial,\\ \hspace*{1.0em} i.e., for any $f : (X,A,B) \to (X^{'}, A^{'}, B^{'})\\ \Rightarrow$ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & S(A \cap B) \ar[r]^>>>{\phi} \ar[d]^{f_{\sharp}} & S(A) \bigoplus S(B) \ar[r]^>>>{\psi} \ar[d]^{f_{\sharp} \oplus f_{\sharp}} & S^{\mathcal{U}}(X) \ar[r] \ar[d]^{f_{\sharp}} & 0 &\\ 0 \ar[r] & S(A^{'} \cap B^{'}) \ar[r]_>>>{\phi} & S(A^{'}) \bigoplus S(B^{'}) \ar[r]_>>>{\psi} & S^{\mathcal{U}}(X^{'}) \ar[r] & 0 & \textrm{commutes}\\ } \] Hence, by snake lemma, $(\star)$ induces a functorial long exact sequence\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & H_{p}(A \cap B) \ar[r]^>>>{\phi_{*}} \ar[d]^{f_{*}} & H_{p}(A) \bigoplus H_{p}(B) \ar[r]^>>>{\psi_{*}} \ar[d]^{f_{*} \oplus f_{*}} & H_{p}^{\mathcal{U}}(X) \ar[r]^>>>{\bd_{*}} \ar[d]^{f_{*}} & H_{p-1}(A \cap B) \ar[r] \ar[d]^{f_{*}} & \cdots \\ \cdots \ar[r] & H_{p}(A^{'} \cap B^{'}) \ar[r] & H_{p}(A^{'}) \bigoplus H_{p}(B^{'}) \ar[r] & H_{p}^{\mathcal{U}}(X^{'}) \ar[r] & H_{p-1}(A^{'} \cap B^{'}) \ar[r] & \cdots \\ } \] 여기서 이 long exact sequence 를 {\bf Mayer-Vietoris sequence} 라고 부른다.\\ {\bf Note} $\phi_{*}(\gam) = (\gam, - \gam)$ and $\psi_{*}(\alp, \bet) = \alp + \bet$.\\ 이제 $\bd_{*}\alp$를 계산해보자. $\alp \in H_{p}(X) \cong H_{p}^{\mathcal{U}}(X)$는 $\alp = \{a\} = \{a_{1}+a_{2}\}$, for $a_{1} \in S_{p}(A)$ and $a_{2} \in S_{p}(B)$로 생각할 수 있고, 아래의 diagram을 보면,\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r]^{\phi} & (a_1 , a_2) \ar[r]^{\psi} \ar[d] & a_{1}+a_{2} \ar[d]^{\bd}\\ \bd a_{1} \ar[r] & (\bd a_{1}, \bd a_{2}) \ar[r] & \bd a_{1} + \bd a_{2} = 0 } \] 따라서 $\bd_{*}\alp = \bd_{*}\{a\} = \bd_{*}\{a_1 + a_2\} = \{\bd a_{1}\}$가 된다.\\ 2. Mayer-Vietoris sequence holds also for reduced homology if $A \cap B \neq \emptyset$ ; It suffices to check that $(\star)$ holds for augmented chain complex.\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & S_{0}(A \cap B) \ar[r]^>>>{\phi} \ar[d]^{\eps} & S_{0}(A) \bigoplus S_{0}(B) \ar[r]^>>>{\psi} \ar[d]^{\eps \oplus \eps} & S_{0}^{\mathcal{U}}(X) \ar[r] \ar[d]^{\eps} & 0\\ 0 \ar[r] & \zb \ar[r]^{\phi} & \zb \bigoplus \zb \ar[r]^{\psi} & \zb \ar[r] & 0 \\ } \] 위의 diagram에서 diagram commutativity 를 check 할 수 있으므로 우리는 reduced homology의 funtorial long exact sequence를 얻게 된다.\\ 3.{\bf(Remark)} Mayer-vietoris sequence holds in general for an excisive couple $A, B \subset X (A \cup B = X)$.\\ \hspace*{1.0em} $\{A,B\}$ is an $excisive \,\, couple$ if $S(A) + S(B) \hookrightarrow S(X)$ induces an isomorphism in homology.\\ 4. {\bf Mayer-Vierotis sequence for simplicial homology}\\ $\bigtriangleup(K_{1}) + \bigtriangleup(K_{2}) = \bigtriangleup(K_{1} \cup K_{2})$ and $C(K_{1}) + C(K_{2}) = C(K_{1} \cup K_{2})$가 성립하므로 simplicial homology에 대한 Mayer-Vietoris sequence가 성립한다.\\ 5.{\bf Example}\\ \begin{floatingfigure}[l]{3cm} \begin{pspicture}(-0.5,0)(2,1.5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] % S^2% \rput(1,1){\psellipse(0,0)(.7,.2) \pspolygon[fillstyle=solid,fillcolor=white,linestyle=none](-1,0)(1,0)(1,0.5)(-1,0.5)(-1,0) \psellipse[linestyle=dashed](0,0)(.7,0.2) \pscircle(0,0){.7}% \psellipse[linecolor=blue](0,-0.3)(.6,.12) \psellipse[linecolor=blue](0,0.3)(.6,.12) \pscurve[linecolor=blue](0.69,0.3)(0.72,-0.2)(0.69,-0.7)\rput(1,-0.2){$B$} \pscurve[linecolor=blue](-0.69,-0.3)(-0.72,0.2)(-0.69,0.7)\rput(-1.2,0.2){$A$} }% \end{pspicture} \end{floatingfigure} (1) {\bf $H_{*}(S^{n})$}\\ 그림과 같이 $A$를 북반구와 남반구의 약간을 포함한 것으로 하고 $B$를 남반구와 북반구의 약간을 포함한 것으로 하면, $A$와 $B$는 $\overset{\circ}{D^{n}}$ 즉, point와 homotopy type이 같고, $A \cap B$는 $S^{n-1}$와 homotopy type이 같다. 따라서 Mayer-vietoris sequence를 적용해보면, \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & \widetilde{H_{p}}(A \cap B) \ar[r] & \widetilde{H_{p}}(A) \bigoplus \widetilde{H_{p}}(B) (=0) \ar[r] & \widetilde{H_{p}}(S^{n}) \ar[r]^>>>{\bd} & \widetilde{H_{p-1}}(A \cap B) \ar[r] & \cdots } \] $\therefore \widetilde{H_{p}}(S^{n}) \cong \widetilde{H_{p-1}}(S^{n-1}) \cong \cdots \cong $가 성립해서 정리하면 다음과 같다.\\ $n>0 \hspace{1.0em} p=n>0 \Rightarrow \widetilde{H_{n}}(S^{n}) \cong \cdots \cong \widetilde{H_{0}}(S^{0}) \cong \zb \\ \hspace*{2.5em} p>n \Rightarrow \widetilde{H_{p}}(S^{n}) \cong \cdots \cong \widetilde{H_{p-n}}(S^{0}) \cong 0\\ \hspace*{2.5em} p}(0,0){0.8}{0}{95} \psarc{*->}(0,0){0.8}{180}{275} \rput(0,1.1){$a_{1}$} \rput(0,-1.1){$a_{2}$} } \end{pspicture} \end{floatingfigure} 우선 $ \cdots \overset{\bd}{\rightarrow} S_{0}(S^{0}) (=\zb \bigoplus \zb) \overset{\eps}{\rightarrow} \zb \rightarrow 0$을 살펴보면 $ker \eps = (a,-a)$이고 따라서 $S_{0}(S^{0})$의 cycle은 $\langle (1,-1)\rangle$ 이다. 결국 $\widetilde{H_{0}}(S^{0})$의 generator는 $x-y$가 된다.\\ 이제 옆의 그림과 같이 1-chain $a_{1}, a_{2}$를 정의하자. 그러면 앞의 논의에 의해서 $\bd_{*}(a_{1}+a_{2}) = \bd a_{1} = x-y$가 성립한다. 따라서 $H_{1}(S^{1})$의 generator는 $ \{a_{1}+a_{2}\}$가 된다.\\ {\bf $H_{2}(S^{2}) = \zb$} generator는 무엇인가?(check)\\ (2)Figure eight(=$X$) 에 대한 homology를 계산해보자. \begin{center} \begin{pspicture}(-0.5,-0.5)(2.5,2.5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] % S^2% \rput(1,1){\pscurve(0,0)(-0.7,0.5)(-1.0,0)(-0.7,-0.5)(0,0) \pscurve(0,0)(0.7,0.5)(1.0,0)(0.7,-0.5)(0,0) \pscircle[linestyle=dashed](-0.6,0){0.75} \pscircle[linestyle=dashed](0.6,0){0.75} \rput(-0.6,0.65){$A$} \rput(0.6,-0.65){$B$} } \end{pspicture} \end{center} 그림과 같이 $A$와 $B$를 선택하면, $A$와 $B$는 $S^{1}$과 $A \cap B$는 point와 homotopy type이 같다. 이제 Mayer-Vietoris sequence에 적용시켜보면,\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & \widetilde{H_{p}}(A \cap B) \ar[r] & \widetilde{H_{p}}(A) \bigoplus \widetilde{H_{p}}(B) \ar[r] & \widetilde{H_{p}}(X) \ar[r] & \widetilde{H_{p-1}}(A \cap B) \ar[r] & \cdots} \] $\widetilde{H}(pt) = 0 $이므로 $\widetilde{H_{p}}(A) \bigoplus \widetilde{H_{p}}(B) \cong \widetilde{H_{p}}(X)$가 되어 \begin{displaymath} \widetilde{H_{p}}(X)=\{\begin{array}{ll} \zb \bigoplus \zb & , p=1\\ 0 & , \textrm{otherwise}\end{array}. \end{displaymath} 가 성립한다. 이 방법을 이용하면, 일반적으로 \\ $X=\underset{r}{S^{1} \vee \cdots \vee S^{1}}$ 일 경우 \begin{displaymath} \widetilde{H_{p}}(X)=\{\begin{array}{ll} \zb^{r} & , p=1\\ 0 & , \textrm{otherwise}\end{array}. \end{displaymath} 가 성립한다.\\ (3) {\bf Closed surface} \begin{center} \begin{pspicture}(-1.0,0.3)(4,1.7)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \rput(-1.0,1){$X$} \rput(-0.5,1){=}% % Torus% \rput(1,1){\psellipse(0,0)(1,.7) } % genus \rput(1,1){\psarc(0,0.9){1}{240}{300}\psarc(0,-0.95){1}{70}{110} } \rput(2.4,1){=}% \rput(3,0.5){\pspolygon(0,0)(1,0)(1,1)(0,1) \psline{->}(0,0)(0,0.55) \psline{->}(0,0)(.55,0) \psline{->}(0,1)(.55,1) \psline{->}(1,0)(1,0.55) \rput(1.1,.5){$b$} \rput(.5,-.1){$a$} \rput(-.1,.5){$b$} \rput(.5,1.1){$a$} % 구멍 \rput(.8,.8){$B$} \rput(1.2,.1){$A$} \psline{->}(1.1,.1)(.6,.4) \rput(.5,.5){\pscircle[fillstyle=solid,fillcolor=lightgray](0,0){0.15}\pscircle(0,0){0.03}} } \end{pspicture} \end{center} 오른쪽 그림에서와 같이 Torus에서 한점을 빼준 open set을 $B$, 그 점을 포함하는 open neighborhood를 $A$라고 하면 $A$는 point와 homotopy type이 같고, $B$는 figure eight 그리고 $A \cap B$는 $S^{1}$와 homotopy type이 같다. 따라서 Mayer-Vietoris sequence에 적용하면, \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & \widetilde{H_{p}}(A \cap B) \ar[r] & \widetilde{H_{p}}(A) \bigoplus \widetilde{H_{p}}(B) \ar[r] & \widetilde{H_{p}}(X) \ar[r] & \widetilde{H_{p-1}}(A \cap B) \ar[r] & \cdots } \] 가 되고(여기서 $\widetilde{H_{p}}(A) = 0$이다.), 특별히 \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & \widetilde{H_{2}}(X) \ar[r] & \widetilde{H_{1}}(S^{1}) \ar[r] & \widetilde{H_{1}}(B) \ar[r] & \widetilde{H_{1}}(X) \ar[r] & \widetilde{H_{0}}(S^{1}) \ar[r] & \cdots } \] 를 살펴보면 $(\widetilde{H_{2}}(B) =0) , \widetilde{H_{1}}(S^{1}) \to \widetilde{H_{1}}(B)$가 zero map 이 되므로 $\widetilde{H_{2}}(X) \cong \widetilde{H_{1}}(S^{1}) = \zb, \widetilde{H_{1}}(X) \cong \widetilde{H_{1}}(B) = \zb \bigoplus \zb$가 된다. 또한, $p \geq 3$이면, $\widetilde{H_{p}}(B) = 0, \widetilde{H_{p-1}}(S^{1})=0$이므로 \[ H_{p}(X)= \begin{cases} \zb & , p=0\\ \zb \bigoplus \zb & ,p=1 \\ \zb & ,p=2 \\ 0 & ,\textrm{otherwise} \end{cases} \] 가 된다.\\ \begin{center} \begin{pspicture}(-2.5,-0.5)(10.5,3.5)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \rput(-1.0,1.5){$X$} \rput(-0.5,1.5){=} \rput(0,1.5){$P^{2}$} \rput(0.5,1.5){=} %그림하나 \rput(1.0,0){ \pscircle[fillstyle=solid,fillcolor=lightgray](1.5,1.5){1.2}% \psarc{*-*}(1.5,1.5){1.2}{0}{180} \psarc{*->}(1.5,1.5){1.2}{0}{95} \psarc{*->}(1.5,1.5){1.2}{180}{275} \rput(1.5,3.1){$a$} \rput(1.5,-0.1){$a$}} %그림둘 \rput(4.0,0){ \pscircle(1.5,1.5){1.2} \psarc{*-*}(1.5,1.5){1.2}{0}{180} \psarc{*->}(1.5,1.5){1.2}{0}{95} \psarc{*->}(1.5,1.5){1.2}{180}{275} \pscircle(1.5,1.5){0.15} \pscircle[fillstyle=solid,fillcolor=lightgray](1.5,1.5){0.4} \rput(2.6,1.5){$B$} \rput(1.3,1.3){$A$} } %그림셋 \rput(7.0,0){ \pscircle(1.5,1.5){0.8} \psarc{->}(1.5,1.5){0.8}{0}{135} \rput(1.2,2.3){$a$} \rput(2.5,1.5){$\cong$} \rput(3.0,1.5){$B$} } \end{pspicture} \end{center} $P^{2}$에서 두번째 그림과 같이 한점을 빼준 open set을 $B$, 그점을 포함하는 open neighborhood는 $A$라고 하면 $A$는 point, $B$와 $A \cap B$는 $S^{1}$과 homotopy type이 같다. Mayer-Vietoris sequence에 적용하면, \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & \widetilde{H_{p+1}}(X) \ar[r] & \widetilde{H_{p}}(S^{1}) \ar[r] & \widetilde{H_{p}}(B) \ar[r] & \widetilde{H_{p}}(X) \ar[r] & \widetilde{H_{p-1}}(S^{1}) \ar[r] & \cdots } \] 특히, \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & \widetilde{H_{2}}(X) \ar[r]^f & \widetilde{H_{1}}(S^{1}) \ar[r]^g & \widetilde{H_{1}}(B) \ar[r]^h & \widetilde{H_{1}}(X) \ar[r] & \widetilde{H_{0}}(S^{1}) \ar[r] & \cdots } \] 를 보면, $g : \widetilde{H_{1}}(S^{1}) (=\zb) \to \widetilde{H_{1}}(B) (=\zb)$는 1을 $2a$로 보내는 map이므로 (세번째 그림 참조) injective이다. 따라서 $f$가 zero map 이 되어, $\widetilde{H_{2}}(X) = 0$이 된다. 또한 $\widetilde{H_{0}}(S^{1}) = 0$이므로 $h$가 surjective 이고, $\widetilde{H_{1}}(X) \cong \widetilde{H_{1}}(B) / ker \,\, h = \widetilde{H_{1}}(B) / im \,\, g \cong \zb / 2\zb$가 된다. 즉, \[ H_{p}(X)=\begin{cases} \zb & , p=0\\ \zb / 2\zb & , p=1 \\ 0 & , \textrm{otherwise} \end{cases} \] 이다.\\ {\bf Note} \begin{floatingfigure}[l]{4cm} \begin{pspicture}(-1.0,0)(3.0,3.0)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] % S^2% \rput(1.5,1.5) { \psellipse[fillstyle=solid,fillcolor=lightgray](0,0)(1.5,1.2) \psarc{->}(0,0){1.5}{0}{10} \rput(1.4,0){$c$} \pscircle[fillstyle=solid,fillcolor=white](-0.5,0){0.3} \psarc{->}(-0.5,0){0.3}{0}{40} \rput(-0.2,0.3){$a$} \pscircle[fillstyle=solid,fillcolor=white](0.5,0){0.3} \psarc{->}(0.5,0){0.3}{0}{50} \rput(0.7,0.3){$b$} \rput(0,0.8){$D$} \rput(0,-1.0){$\circlearrowleft$} } \rput(-1.0,1.5){$X$} \rput(-0.5,1.5){=} \end{pspicture} \end{floatingfigure} 그림과 같이 $a,b,c$ 그리고 $D$를 정의하면, $0 = \{\bd D\} = \{c\} - \{a\} - \{b\}$가 성립하므로, $H_{1}(X)$안에서 $\{c\} = \{a\} +\{b\}$가 된다.\\ 마지막으로 $P^{2}$에서 coefficient를 $\zb$대신 $\rb$ 또는 $\zb / 2$를 생각해보자. 우선 $\rb$의 경우 $im \,\, g =\rb$이므로 $H_{1}(P^{2},\rb) = 0$이 된다. 마찬가지로 $\zb /2$의 경우에는 $g$가 zero map 이 되므로 $H_{2}(P^{2},\zb /2) \cong H_{1}(S^{1},\zb /2) = \zb /2$, 그리고 $H_{1}(X,\zb/2) \cong H_{1}(B,\zb/2) = \zb / 2$가 된다.\\ {\bf 숙제23} Compute $H_{*}$(closed surface).\\ \end{document}