\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*}} \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.6 Relative Homology and Excision} \begin{defn} $A \underset{i}{\hookrightarrow}X$, a subspace of a topological space $X \Rightarrow S_p(A)\underset{i_{\sharp}}{\hookrightarrow} S_p(X) $\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \Rightarrow & 0 \ar[r] & S_p(A) \ar[r]^i \ar[d]^{\bd} & S_p(X) \ar[r]^q \ar[d]^{\bd} & S_p(X) / S_p(A) \ar[r] \ar@{.>}[d]^{\bar{\bd}} & 0\\ & 0 \ar[r] & S_{p-1}(A) \ar[r] & S_{p-1}(X) \ar[r] & S_{p-1}(X)/ S_{p-1}(A)\ar[r] & 0 } \] $\hspace{3em}$,where $\bar{\bd}$ is an induced boundary homomorphism s.t. ${\bar{\bd}} ^2 = \bar{{\bd}^2} = 0$.\\ $\Rightarrow \{ S_p (X,A) , "\bd" = \bar{\bd}\} = S(X,A) :$ relative chain complex or singular chain complex for a pair $(X,A).\\ \Rightarrow H_p (S(X,A)) =: H_p (X,A) $: singular homology for a pair $(X,A)$(or relative homology X mod A). \end{defn} Remark 1. This is a general construction for $\forall$ chain complexes $\mathcal{C}' \subset \mathcal{C}$, and hence can talk about $H_p (K,A)$ for a simplicial pair $A}[d]^{\bar{f_{\sharp}}="f_{\sharp}"} & 0& \textrm{and $"f_{\sharp}"$ is a chain map.}\\ &0\ar[r] & S_p(B)\ar[r] & S_p(Y)\ar[r] & S_p(Y)/S_p(B)\ar[r] & 0&} \] $\Rightarrow f_* : H_p(X,A) \rightarrow H_p(Y,B).\\ \hspace*{5.5em} \{c\}\mapsto \{f\circ c\}$\\ (i) $id : (X,A)\circlearrowleft \Rightarrow id_* = id.$\\ (ii) $(X,A) \overset{f}{\rightarrow} (Y,B) \overset{g}{\rightarrow} (Z,C) \Rightarrow (g\circ f)_* = g_* \circ f_*$\\ Example.\\ (i) $\{X_{\alp}\} :$ path-components of $X$ and $A_{\alp} = A \cap X_{\alp} \Rightarrow H_p(X,A) = \underset{\alp}{\oplus}H_p(X_{\alp},A_{\alp})$\\ (ii) $X$ : path-connected and $A \neq \phi \Rightarrow H_0(X,A) = 0$.\\ \textbf{2. Homotopy Invariance}\\ $f \underset{F}{\simeq} g : (X,A)\rightarrow (Y,B) \hspace{1em}(F(A\times I) \subseteq B) \Rightarrow f_* = g_* .$\\ \begin{proof} Let $i_0 , i_1 :(X,A)\rightarrow(X\times I, A\times I)$ be level maps as before.\\ Need a chain homotopy $D_{X,A} : S_p(X,A)\rightarrow S_{p+1}(X\times I, A\times I)$\\ between $i_{0\sharp}$ and $i_{1\sharp}$: \[ \xymatrix @=2em @*[c]{% 0 \ar[r] & S_p(A)\ar[r]\ar[d]^{D_A = D_X |} & S_p(X)\ar[r]\ar[d]^{D_X} & S_p(X,A)\ar[r]\ar@{.>}[d]^{\bar{D}_X = D_{X,A}} & 0\\ 0 \ar[r] & S_{p+1}(A\times I)\ar[r] & S_{p+1}(X\times I)\ar[r] & S_{p+1}(X\times I, A\times I)\ar[r] & 0} \] Check $\bar{\bd}\bar{D}\bar{\sigma} + \bar{D}\bar{\bd}\bar{\sigma} = \bar{i_{0\sharp}}\bar{\sigma} - \bar{i_{1\sharp}}\bar{\sigma}$: trivial.\\ $(X,A)\overset{i_0}{\underset{i_1}{\rightrightarrows}}(X\times I, A\times I)\overset{F}{\rightarrow} (Y,B)$ s.t. $f= F\circ i_0$ and $g = F\circ i_1$.\\ $\bar{D} : i_{0\sharp} \cong i_{1\sharp} \Rightarrow i_{0*} = i_{1*} : H(X,A)\rightarrow H(X\times I, A\times I)\\ \therefore f_* = (F\circ i_0 )_* = F_* \circ i_{0*} = F_* \circ i_{1*} = (F\circ i_1 )_* = g_*$\end{proof}\\ \textbf{3.} $0\rightarrow S(A)\rightarrow S(X)\rightarrow S(X,A)\rightarrow 0\\ \overset{\textrm{snake}}{\Rightarrow} \exists$ a functorial long exact sequence \[ \xymatrix @=2em @*[c]{% \cdots \ar[r]& H_p(A) \ar[r]^{i_*} & H_p(X) \ar[r]^{q_*} & H_p(X,A) \ar[r]^{\bd _*} & H_{p-1}(A)\ar[r] & \cdots} \] $\hspace*{20em}\{c\}\mapsto \{\bd c\}$\\ Remark. If $A \neq \phi $, define $\widetilde{H}(X,A) = H(X,A).$ Then 3 holds for reduced case also. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% & \ar[d]& \ar[d] & \ar[d] & \\ 0 \ar[r]& S_0(A)\ar[r]\ar[d]^{\epsilon} & S_0(X)\ar[r]\ar[d]^{\epsilon} & S_0(X,A)\ar[r]\ar[d] & 0\\ 0 \ar[r]& \mathbb{Z}\ar[r]^{id}\ar[d] & \mathbb{Z} \ar[d]\ar[r]& 0\ar[r]\ar[d] & 0\\ & 0 & 0 & 0 &} \] Example. $H_p(B^n, \bd B^n )$ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & \widetilde{H_p} (\bd B^n) \ar[r] & \widetilde{H_p}(B^n) \ar[r] & \widetilde{H_p} (B^n , \bd B^n) \ar[r]^{\bd} & \widetilde{H}_{p-1}(\bd B^{n-1}) \ar[r] & \cdots } \] $\widetilde{H_p}(B^n) =0 \hspace{0.5em},\forall p\hspace{0.3em}\Rightarrow\hspace{0.3em} \bd : \cong \hspace{0.3em}\Rightarrow \\H_p (B^n , \bd B^n)=\widetilde{ H}_p (B^n , \bd B^n) \cong \widetilde{H}_{p-1}(S^{n-1}) = \mathbb{Z}\hspace{0.8em} p=n\\ \hspace*{19em} = 0 \hspace{0.8em} p\neq n$\\ \textbf{4. Long Exact Sequence for Triple}\\ Let $A\subset B\subset X$. Then we have a functorial long exact sequence \\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & H_p (B,A)\ar[r] & H_p(X,A) \ar[r] & H_p (X,B) \ar[r]^{\bd} & H_{p-1}(B,A) \ar[r] & \cdots} \] \begin{proof} \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & S(B)/S(A) \ar[r] & S(X)/S(A) \ar[r] & S(X) /S(B) \ar[r] & 0 & \textrm{: s.e.s.}} \] and apply snake lemma.\end{proof} \textbf{5. Five Lemma} \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% A_1 \ar[r] \ar[d]^{f_1} & A_2 \ar[r] \ar[d]^{f_2} & A_3 \ar[r] \ar[d]^{f_3} & A_4 \ar[r] \ar[d]^{f_4} & A_5 \ar[d]^{f_5 \hspace{2em} \textrm{all} \curvearrowright} & \textrm{:exact}\\ % & & & & & \textrm{all} \curvearrowright \\ B_1 \ar[r] & B_2\ar[r] & B_3\ar[r] & B_4\ar[r] & B_5 & \textrm{:exact}} \] $f_1 $: onto , $f_5 : 1-1$ and $f_2 , f_4 :\cong \hspace{0.5em}\Rightarrow \hspace{0.5em}f_3 :\cong$.\\ (In particular, $f_i : \cong \hspace{0.5em} i = 1,2,4,5. \hspace{0.5em} \Rightarrow \hspace{0.5em} f_3 : \cong$.)\\ \begin{proof} by diagram chasing. \end{proof} Example. Show $ j : (B^n , S^{n-1}) \rightarrow (B^n, B^n \setminus 0)$ induces an isomorphism\\ $j_* : H(B^n , S^{n-1}) \rightarrow H(B^n, B^n\setminus 0) $. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & H_p (S^{n-1}) \ar[r]\ar[d]^{j_* : \cong} & H_p(B^n)\ar[r]\ar[d]^{j_* : = } & H_p(B^n , S^{n-1})\ar[r]\ar[d]^{\textrm{five lem.에 의해} j_* :\cong} & H_{p-1}(S^{n-1}) \ar[r] \ar[d]^{j_* : =}& H_{p-1}(B^n )\ar[r]\ar[d]^{j_* : =}& \cdots\\ \cdots \ar[r] & H_p (B^n\setminus 0) \ar[r] & H_p(B^n)\ar[r] & H_p(B^n , B^n\setminus 0) \ar[r] & H_{p-1}(B^n\setminus 0)\ar[r] & H_{p-1}(B^n )\ar[r]& \cdots} \] \textbf{6. Split Short Exact Sequence}\\ Given short exact sequence $0\rightarrow A \overset{i}{\rightarrow} B \overset{p}{\rightarrow} C \rightarrow 0, $ TFAE :\\ (i) $\exists A \overset{j}{\leftarrow} B $ s.t. $j\circ i = id$.\\ (ii) $\exists B\overset{s}{\leftarrow} C$ s.t. $p\circ s = id$.\\ In this case, $B = A\oplus C$ and the short exact sequence is said to be split.\\ \begin{proof} (i)$\Rightarrow$(ii):\\ $0\rightarrow A \overset{j}{\underset{i}{\leftrightarrows}} B \underset{p}{\rightarrow} C \rightarrow 0 \Rightarrow B = i(A) \oplus ker(j) : b = ij(b) +(b-ij(b))\\ (\because j(b) -jij(b) = 0 \Rightarrow b-ij(b) \in ker(j)$ and $ji(a) = 0 \Rightarrow a=0\Rightarrow i(a) =0$.)\\ $p|_{ker(j)}: ker(j) \rightarrow C : \cong \hspace{0.3em}$ since $ker (p) = i(A) \Rightarrow \exists s = (p|)^{-1} $ s.t. $p\circ s = id.$\\ (ii)$\Rightarrow$(i) : exactly same.\end{proof} Example. $A \underset{\underset{r}{\leftarrow}}{\overset{i}{\hookrightarrow}}X , $ a retract $\Rightarrow H_p (X) = H_p(A) \oplus H_p(X,A)$. \newpage \textbf{Excision} \begin{thm} $(X,A) : $ a pair of spaces and let $U\subset X$ be s.t. $\bar{U} \subset \AA.$\\ Then the inclusion $i :(X-U, A-U) \hookrightarrow(X,A)$ induces an isomorphism $i_* : H_* (X-U, A-U) \overset{\cong}{\rightarrow} H_*(X,A)$. (i.e., $U$ may be excised without altering relative homology.)\end{thm} \begin{proof} Let $\mathcal{U} = \{X-U , A\}$ so that $\{(X-U)^{\circ}$ ,$ \AA\}$ covers X. (See the footnote.\footnote{$(X-U)^{\circ}= (U^C)^{\circ} = \bar{U}^C$})\\ Then $i : S^{\mathcal{U}}(X)\hookrightarrow S(X)$ induces an isomorphism on homology. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r]& S(A)\ar[r]\ar[d]^{i=i|} & S^{\mathcal{U}}(X)\ar[r]\ar[d]^i & S^{\mathcal{U}}(X)/S(A)\ar[r]\ar@{.>}[d]^j & 0 \\ 0 \ar[r]& S(A)\ar[r] & S(X)\ar[r] & S(X)/S(A)\ar[r] & 0} \]where $j$ is induced as quotient map and becomes a chain map as before. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \Rightarrow & \cdots \ar[r]& H_p(A)\ar[r]\ar[d]^{i_* =id} & H_p(S^{\mathcal{U}}(X))\ar[r]\ar[d]^{i_* \cong} & H_p (S^{\mathcal{U}}(X)/S(A))\ar[r]\ar[d]^{j_* : \cong \textrm{by five lem.}} & H_{p-1}(A)\ar[r]^{\bd_*}\ar[d]^= & \cdots\\ & \cdots\ar[r] & H_p(A)\ar[r] & H_p(X)\ar[r] & H_p(X,A)\ar[r] & H_{p-1}(A)\ar[r]^{\bd_*} & \cdots} \] Now consider $S^{\mathcal{U}}(X)/S(A) : S_p ^{\mathcal{U}} = S_p(X-U) + S_p(A)$ \\ $ \Rightarrow S_p ^{\mathcal{U}}/S_p(A) = \frac{S_p(X-U)+S_p(A)}{S_p(A)} \underset{\underset{i}{\leftarrow}}{\cong} S_p(X-U)/ (S_p(X-U)\cap S_p(A)) \\ \hspace*{5em}= S_p(X-U)/S_p(A-U) = S_p(X-U, A-U)$\\ $\Rightarrow H_p(X-U, A-U ) \underset{\underset{i_*}{\rightarrow}}{\cong} H_p(S^{\mathcal{U}}(X)/S(A))\underset{\underset{j_*}{\rightarrow}}{\cong} H_p(X,A)$\end{proof}\\ Remark. Recall $X_1 , X_2 \subset X , X = X_1 \cup X_2.\\ \{X_1 , X_2\}$ : an excisive couple if $S(X_1) +S(X_2) \hookrightarrow S(X_1 \cup X_2 = X)$ induces an isomorphism on homology.\\ \newpage \textbf{숙제 24.} $\{X_1, X_2\}$ is an excisive couple. $\Leftrightarrow (X_1 , X_1\cap X_2 )\hookrightarrow (X_1 \cup X_2 , X_2)$ is an excision map, i.e., induces an isomorphism on homology.\\ \begin{proof} Hint: Consider the following commutative diagram. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% S(X_1) /S(X_1\cap X_2) \ar[rr]^j \ar[dr]^i &&S(X_1 \cup X_2 )/S(X_2)\\ &S(X_1)+ S(X_2) / S(X_2)\ar[ur]^k &} \]\end{proof} Example.\\ \begin{floatingfigure}[l]{3cm} \begin{pspicture}(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) \pscurve[linecolor=blue](0.69,-0.3)(0.72,-0.5)(0.69,-0.7)\rput(1,-0.5){$V$} }% \end{pspicture} \end{floatingfigure} $X = S^n, E^n_+$: 북반구, $E^n_- = A $: 남반구, $U = \AA$라 두면 $U$는 excision theorem을 이용하여 $X$에서 바로 excise할 수 없다. ($\because \bar{U}\nsubseteq \AA)$ 그렇지만 $V$를 그림과 같이 잡아서 $V$를 excision theorem 에 의해 excise한 다음 deformation retract를 이용하여 $U$를 "excise"할 수 있다.\\ $H_p(S^n, E^n_-)\underset{\underset{i_*}{\leftarrow}}{\cong} H_p(S^n -V, E^n- V)$\\ Note $j : (E_+^n , S^{n-1})\underset{\underset{\exists r}{\leftarrow}}{\hookrightarrow}(S^n -V, E_-^n -V) $ where $r $ is a deformation retraction\\(i.e., $rj = id, jr \simeq id) \Rightarrow j_* : \cong \Rightarrow H_p(S^n, E^n_-) \underset{\underset{i_*}{\leftarrow}}{\cong} H_p(E_n^+ , S^{n-1})$\\ $\therefore \{E_+^n , E_-^n\}: $ excisive couple and hence can apply Mayer-Vietori sequence directly.\\ In general, $V\subset U \subset A$ and suppose that $V$ can be excised. If $(X-U, A-U)$ is a deformation retract of $(X-V,A-V)$, then $U$ can be excised by the exactly same argument as above. Hence in this case, $\{X-U, A\}$ is an excisive couple. \newpage {\bf Equivalence of simplicial and singular homology}\\ Let $K$ be a simplicial complex. Recall that we have 3 homology theories.\\ (1) $C(K) \rightsquigarrow H(K)$ ; simplicial homology\\ (2) $\bigtriangleup(K) \rightsquigarrow H^{\triangle}(K)$ ; ordered simplicial homology\\ (3) $S(|K|) \rightsquigarrow H(|K|)$ ; singular homology\\ And we showed $\mu : \bigtriangleup(K) \to C(K)$ is a natural chain equivalence,\\ $\hspace*{7.0em} (v_{0}, \cdots v_{p}) \mapsto [v_{0}, \cdots, v_{p}]$\\ so that $\mu_{*} : H^{\triangle}(K) \to H(K)$ is a natural isomorphism.\\ 이제 (2)와 (3)을 비교해보자.\\ Let $\theta : \bigtriangleup(K) \to S(|K|) \\ \hspace*{1.0em} (v_{0}, \cdots v_{p}) \mapsto l$\\ \hspace*{1.0em} where $l : \triangle^{p} \to |K|$ is an affine map determined by $l(e_{i}) = v_{i} , i=0,1, \cdots , p$.\\ $\Rightarrow \theta$ is a natural(augmentation-preserving) chain map so that $\theta_{*} : H^{\triangle}(K) \to H(|K|)$ is a natural transformation.\\ If $K_{0} < K$, a subcomplex, then \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% 0 \ar[r] & \bigtriangleup(K_{0}) \ar[r]^{i} \ar[d]^{\theta |} & \bigtriangleup(K) \ar[r] \ar[d]^{\theta} & \bigtriangleup(K)/\bigtriangleup(K_{0}) \ar[r] \ar@{.>}[d]^{\bar{\theta}="\theta"} & 0\\ 0 \ar[r] & S(|K_{0}|) \ar[r] & S(|K|) \ar[r] & S(|K|)/S(|K_{0}|) \ar[r] & 0 } \] $\Rightarrow \theta_{*} : H^{\triangle}(K,K_{0}) \to H(|K|,|K_{0}|)$ is naturally well-defined.\\ \framebox{\hspace*{1em}\parbox[b]{14cm}{$\theta_{*} : \widetilde{H^{\triangle}}(K) \to \widetilde{H}(|K|)$ is a natural isomorphism. (Hence so is $\theta_{*} : H^{\triangle}(K) \to H(|K|)$ and $H_{*} : H^{\triangle}(K,K_{0}) \to H(|K|,|K_{0}|)$ if $K_{0} < K$ by 5-lemma.)}}\\ {\bf step 1} Assume that $K$ is finite.\\ Prove by induction on $n$(the number of simplices in $K$).\\ $n$=1 \hspace{2.0em} $K=\{v\} \Rightarrow \widetilde{H^{\triangle}}(K) = \widetilde{H}(|K|) = 0$.\\ $n > 1$\\ Let $\sigma$ be a simplex of $K$ of maximal dimension so that $K_{1} := K - \{\sigma\}$ is a subcomplex of $K$.\\ $\underline{\sigma} := \sigma$ as a subcomplex of $K$\\ $bd \underline{\sigma}$ := subcomplex of $\underline{\sigma}$ consisting of proper faces of $\sigma$\\ \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \widetilde{H_{p}^{\triangle}}(K) \ar[r]^{\theta_{*}} \ar[d]^{i_{*}} & \widetilde{H_{p}}(|K|) \ar[d]^{i_{*}}\\ H_{p}^{\triangle}(K,\underline{\sigma}) \ar[r]^{\theta_{*}} & H_{p}(|K|,\sigma)\\ H_{p}^{\triangle}(K_{1}, bd \underline{\sigma}) \ar[r]^{\theta_{*}} \ar[u]^{j_{*}} & H_{p}(|K_{1}|,bd \sigma) \ar[u]^{j_{*}} } \] 위의 diagram에서 $j_{*}$는 excision theorem에 의해서 isomorphism이고, 가장 아래의 $\theta_{*}$는 induction hypothesis에 의해 isomorphism이다. 마지막으로 $i_{*}$는 각각 $(|K|,\sigma), (K, \underline{\sigma})$ pair에 대한 long exact sequence에 의해서 isomorphism이다.($ \because \sigma$에 대한 homology는 항상 0 이므로) 따라서, \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \widetilde{H_{p}^{\triangle}}(K) \cong \widetilde{H_{p}}(|K|)} \] {\bf step2} Assume that $K$ is infinite.\\ (1) $\theta_{*}$ is onto.\\ Given $\{z\} \in \widetilde{H_{p}}(|K|), |z|:=$ support of $z$. Then $|z|$ is compact, and hence is contained in $|L|$, a finite subcomplex of $K$. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \widetilde{H_{p}^{\triangle}}(L) \ar[r]^{\theta_{*}}_{\cong} \ar[d]^{i_{*}} & \widetilde{H_{p}}(|L|) \ar[d]^{i_{*}} \ni & \{z\} \ar[d] \\ \widetilde{H_{p}^{\triangle}}(K) \ar[r]^{\theta_{*}} & \widetilde{H_{p}}(|K|) \ni & \{z\}} \] step 1 으로 부터 $L$에 대한 $\theta_{*}$가 $\cong$이고 diagram이 commute하므로 $K$에 대한 $\theta_{*}$는 onto이다.\\ (2) $\theta_{*}$ is 1-1.\\ Suppose $ \{z\} \in \widetilde{H_{p}^{\triangle}}(K)$ and $\theta_{*}\{z\} = \{\theta(z)\} = 0.$ Then $\exists c \in \widetilde{S_{p+1}}(|K|)$ such that $\bd c = \theta(z)$. $|c|$ is compact, hence contained in $L$, a finite subcomplex of $K$. By above diagram, $\theta_{*}$ is 1-1.\\ {\bf Note} Given a continuous map $f : |K| \to |L|$, let $g : K' \to L$ be a simplicial approximation of $f$. 그러면 \[ \xymatrix @M=1ex @C=1ex @R=3ex @*[c] { % % 윗줄 & \ar[dl]_{\mu_{*}(\cong)} H^{\triangle}(K') \ar[rr]^{g_{*}} \ar[dd]^<<<<{\theta_{*}(\cong)} & & \ar[dl]^{\mu_{*}(\cong)} H^{\triangle}(L) \ar[dd]^{\theta_{*}(\cong)}\\ % 가운뎃 줄 H(K') \ar[rr]^{g_{*}} \ar[dr]_{\eta_{*}=\theta_{*} \circ \mu_{*}^{-1}} & & H(L) \ar[dr]_{\eta_{*}}& \\ % 아래줄 & H(|K'|) \ar[rr]_{g_{*}=f_{*}} && H(|L|) } \] 위의 그림에서 모든 diagram이 commute하고, $\mu_{*} , \theta_{*}$가 natural isomorphism이므로 $\eta_{*}$도 natural isomorphism이 된다.\\ \newpage {\bf Relative Mayer-Vietoris sequence} \psset{unit=2cm} \begin{center} \begin{pspicture}(4,2)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \rput(2,1){ \psellipse(0,0)(1.5,0.8) \psellipse(-0.6,0)(0.8,0.5) \psellipse(0.6,0)(0.8,0.5) \psellipse(-0.3,0)(0.5,0.2) \psellipse(0.3,0)(0.5,0.2) \rput(0,0.9){$X$} \rput(-0.6,0.6){$X_{1}$} \rput(0.6,0.6){$X_{2}$} \rput(-0.3,0.3){$A_{1}$} \rput(0.3,0.3){$A_{2}$} } \end{pspicture} \end{center} 위와 같은 경우 아래와 같은 sequence들을 생각할 수 있다. \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% & 0 \ar[d] & 0 \ar[d] & 0 \ar[d] && \\ 0 \ar[r] & S(A_{1} \cap A_{2}) \ar[r]^{\phi|} \ar[d]^{i} & S(A_{1}) \bigoplus S(A_{2}) \ar[r]^{\psi|} \ar[d]^{i_{1}\oplus i_{2}} & S(A_{1}) + S(A_{2}) \ar[r] \ar[d]^{j} & 0 & \textrm{s.e.s}\\ 0 \ar[r] & S(X_{1} \cap X_{2}) \ar[r]^{\phi} \ar[d] & S(X_{1}) \bigoplus S(X_{2}) \ar[r]^{\psi} \ar[d] & S(X_{1}) + S(X_{2}) \ar[r] \ar[d] & 0 & \textrm{s.e.s}\\ 0 \ar[r] & \frac{S(X_{1} \cap X_{2})}{S(A_{1} \cap A_{2})} \ar@{.>}[r]^{\bar{\phi}} \ar[d] & \frac{S_(X_{1})}{S(A_{1})} \bigoplus \frac{S(X_{2})}{S(A_{2})} \ar@{.>}[r]^{\bar{\psi}} \ar[d] & \frac{S(X_{1}) + S(X_{2})}{S(A_{1}) + S(A_{2})} \ar[r] \ar[d] & 0 & \textrm{$(\star)$}\\ & 0 & 0 & 0 & & } \] (1) $\bar{\phi}$ and $\bar{\psi}$ are chain maps. ($\because$ chain map의 quotient는 chain map)\\ (2) $(\star)$ is exact by 9-lemma.\\ Suppose $\{X_{1},X_{2}\}$ is an excisive couple of $X_{1} \cup X_{2}$ and $\{A_{1},A_{2}\}$ is an excisive couple of $A_{1} \cup A_{2}$. Then $(\star)$ induces an long exact sequence \[ \xymatrix @=1em @*[c]{% (\ast) \cdots \ar[r] & H_{p}(X_{1} \cap X_{2}, A_{1}\cap A_{2}) \ar[r] & H_{p}(X_{1},A_{1}) \bigoplus H_{p}(X_{2},A_{2}) \ar[r] & H_{p}(X_{1} \cup X_{2}, A_{1}\cup A_{2}) \ar[r] & \cdots } \] 여기서 $H_{p}(X_{1} \cup X_{2}, A_{1} \cup A_{2})$이 되는 이유는 \\ $S(X_{1}) + S(X_{2}) \overset{i}{\hookrightarrow} S(X_{1} \cup X_{2})$ 와 $S(A_{1}) + S(A_{2}) \overset{i}{\hookrightarrow} S(A_{1} \cup A_{2})$이 homology에서 isomorphism을 induce하고, 5-lemma 에 의해서 $\frac{S(X_{1})+S(X_{2})}{S(A_{1})+S(A_{2})} \hookrightarrow \frac{S(X_{1} \cup X_{2})}{S(A_{1} \cup A_{2})}$ 도 homology에서 isomorphism을 induce하기 때문이다. $(\ast)$를 {\bf Relative Mayer-Vietoris sequence} 라고 부른다.\\ {\bf Special cases}\\ (1) $X=X_{1} \cup X_{2} \Rightarrow H_{p}(X_{1} \cup X_{2} , A_{1} \cup A_{2}) = H_{p}(X, A_{1} \cup A_{2})$ \\ (2) Furthermore, if $X_{1} = X_{2} = X$, then \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & H_{p}(X, A_{1} \cap A_{2}) \ar[r] & H_{p}(X, A_{1}) \bigoplus H_{p}(X,A_{2}) \ar[r] & H_{p}(X, A_{1} \cup A_{2}) \ar[r] & \cdots } \] (3) If $X = X_{1} \cup X_{2}$ and $A_{1} = A_{2} = A$, then \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% \cdots \ar[r] & H_{p}(X_{1} \cap X_{2}, A) \ar[r] & H_{p}(X_{1}, A) \bigoplus H_{p}(X_{2},A) \ar[r] & H_{p}(X,A) \ar[r] & \cdots} \] \end{document}