\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} \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{\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{\bd}{\partial} \newcommand{\cb}{\mathbb{C}} \newcommand{\cc}{\mathcal{C}} \newcommand{\sbb}{\mathbb{S}} \newcommand{\rb}{\mathbb{R}} \newcommand{\zb}{\mathbb{Z}} \newcommand{\key}[1]{\emph{\textbf{#1}}} \newcommand{\Hom}{\textrm{Hom}} \newcommand{\hqc}{H^q_c} \newcommand{\hcq}{H^q_c} \newcommand{\chq}{\check{H}^q} \newcommand{\hcqq}{H^{q+1}_c} \newcommand{\hq}{H^q} \newcommand{\hnq}{H_{n-q}} \newcommand{\hqq}{H^{q+1}} \newcommand{\ad}{\ar@{.*{\downarrow}}[d]} \newcommand{\dlim}[1][K]{\underset{#1}{\underrightarrow{\lim}}} \newcommand{\ndlim}{\underrightarrow{\lim}} \newcommand{\ch}{\check{H}} \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]{}} \newenvironment{pf1}{{\bf 정리의 증명}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \psset{unit=2cm} \section*{VI. Other duality} \section*{VI.1 Alexander duality} 1. \begin{thm} Let $(X,A)$ be a compact pair. Then there exists a long exact sequence \[ \xymatrix @=2em { % \cdots \ar[r] & \hcq(X-A)\ar[r]^-i &\hq(X)\ar[r]^-j&\chq(X)\ar[r]^-{\del} &\hcqq(X-A)\ar[r]&\cdots. }\] \end{thm} 먼저 map들이 어떻게 정의되었는지 살펴보자.\\ (1) $i$ : $U^\textrm{open}\subset X \Rightarrow \exists$a canonical homomorphism $i: \hqc(U)\to\hq(X)$: \\ 임의의 compact set $K\subset U$에 대하여 아래 diagram이 commute하므로 $i$가 induce된다. \[ \xymatrix @=3em @M+1ex @R=2em { % \hq(U,U-K)\ar[dr]^{\hspace{3em}\displaystyle{\circlearrowleft}} \ar@{.*{\downarrow}}[d] & \hq(X,X-K)\ar[l]^\cong_{\textrm{excision}} \ar[d]^{i^*} \\ \hqc(U) \ar@{.>}[r]_-{\displaystyle{\exists ! "i"}}& \hq(X) }\] (2) $j$: \[ \xymatrix @=2em @M+1ex @*[c] { % \hq(X)\ar[r]^{j^*, j=\textrm{inclusion}} \ar[dr]^{\hspace{3em}\displaystyle{\circlearrowleft}}_-{\displaystyle{\exists ! "j"}}& \hq(V)\ar@{.*{\downarrow}}[d]\\ & \dlim[V^\textrm{open}\supset A]\hq(V):=\chq(A) }\] (3) $\del$: \\ $K=X-V$, $U=X-A$라 두면 다음 diagram을 얻는다. \[ \xymatrix @=2.5em @M+1ex @R=1.5em @*[c] { % \hq(V)\ar[r]^-{\del} \ar@{.*{\downarrow}}[d]& \hqq(X,V) \ar[r]^-{\textrm{excision}}_-\cong& \hqq(U,U-K)\ar[d]\\ \chq(A) \ar@{.>}[rr]_-{\displaystyle{\exists ! "\del"}}&&\hcqq(U) }\] 이 때, $\del$가 natural하므로 $"\del"$가 induce된다.\\ \begin{pf} \[ \xymatrix @=1.2em @M+.7ex @*[c] { % &\hq(X,V)\ar[d]^\cong&&&\hqq(X,V)\ar[d]^\cong\\ \cdots\ar[r] & \hq(U,U-K)\ar[r] \ad & \hq(X)\ar[r] \ad & \hq(V)\ar[r] \ad & \hqq(U,U-K)\ar[r] \ad & \cdots\\ \cdots \ar[r] & \hcq(X-A)\ar[r]^-i &\hq(X)\ar[r]^-j&\chq(X)\ar[r]^-{\del} &\hcqq(X-A)\ar[r]&\cdots }\] $\ndlim$은 exact functor이므로 위 diagram과 같이 $(X,V)$의 cohomology long exact sequence로부터 원하는 exact sequence를 얻는다. \end{pf}\\ 2. (Alexander duality) $X=M^n$: $R$-orientable compact manifold, $A^\textrm{cpt}\subset M$.\\ Consider \xymatrix @=2em @C=2.5em @M+.7ex {\hq(V)\ar[r]^-{\zeta_A\cap} \ad \ar[dr] & \hnq(V,V-A)\ar[d]^{\textrm{excision}}_\cong \\ \chq(A) \ar[r]_-{{\exists ! D_A}} & \hnq(M,M-A)} \\ where $\zeta_A\in H_n(V,V-A)$, restriction of orientation class $\zeta_A\in H_n(M,M-A)$.\\ Then \[ \xymatrix @=2.5em { D_A: \chq(A) \ar[r]^-\cong & \hnq(M,M-A) \hspace{2em} \forall q} \] \begin{pf} 다음 diagram에 5-lemma를 적용하면 증명이 끝난다. \[ \xymatrix @=1.2em @R=2.5em @*[c] { % \cdots\ar[r] & \hcq(U)\ar[r] \ar[d]^\cong_{D_U}\ar@{}[dr]|{\displaystyle{(1)}} & \hq(M)\ar[r] \ar[d]^\cong_{D_M}\ar@{}[dr]|{\displaystyle{(2)}} & \chq(A)\ar[r]^\del \ar[d]^{??}_{D_A} \ar@{}[dr]|{\displaystyle{(3)}}& \hcqq(U)\ar[r] \ar[d]^\cong_{D_U} & \cdots\\ \cdots \ar[r] & \hnq(U)\ar[r] &\hnq(M)\ar[r]&\hnq(M,U)\ar[r]^\bd &H_{n-q-1}(U)\ar[r]&\cdots }\] $U$는 앞서와 마찬가지로 $M-A$이고 $D_U$와 $D_M$은 Poincar\'{e} duality map들이다. \\ 따라서 $(1),(2),(3)$이 commute한다는 것만 보이면 된다. (1)은 cap product의 naturality에 의하여 commute한다. \\ (2)는 다음 diagram에서 \[ \xymatrix @=2em @M+.7ex @C=3em @*[c] { % \hq(M)\ar[r] \ar[d]^\cong_{D_M} \ar@{}[dr]|{\displaystyle{\circlearrowleft}}& \hq(V) \ar[d]^{\zeta_A\cap} \ar@{.*{\rightarrow}}[r] & \chq(A) \ar[d]_{D_A} \\ \hnq(M)\ar[r]&\hnq(V,V-A)\ar[r]^\cong_{\textrm{excision}}&\hnq(M,U) }\] cap product의 naturality에 의하여 왼쪽 사각형이 commute하고 이것의 direct limit이 (2)이므로 commute한다. \\ (3)은 up to sign으로 commute하는데 다음 diagram에서 확인할 수 있다. \[ \xymatrix @=1.2em @M+.7ex @C=-1em @*[c] { % \hspace*{10em}&&&\hqq(M,V) \ar[dr]^-{\textrm{excision}}_-\cong& \\ &&\hq(V) \ar[ur]^\del \ar[ddll]_{\zeta_A\cap} \ad & & \hqq(U,U-K) \ar@/^3em/[dd]^{\zeta_K\cap} \ad \\ &&\chq(A) \ar[d]_{D_A}\ar[rr]^\del & &\hcqq(U) \ar[d]_{D_U}& \\ \hnq(V,V-A)\ar[rr]_-{\cong} &&\hnq(M,U) \ar[rr]^\bd && H_{n-q-1}(U) } \] 여기서 $V=M-K$이다. chain level에서 $\bd(\zeta\cap a)=(-1)^q(\bd\zeta\cap a+\zeta\cap \del a)=(-1)^q(\zeta\cap \del a)$이므로 up to sign으로 commute함을 확인할 수 있다. \end{pf}\\ 3.(Taut embedding) When is $\chq(A)=\hq(A)$?\\ e.g. non-taut embedding $$ A=\{(x,\sin\frac{1}{x}\}\cup \{(1,y)~|~ |y|\leq 1 \}$$ \begin{center} \begin{pspicture}(2,1.5) \psset{xunit=5cm,yunit=1.2cm} \psline(0,0)(0,2) \rput(0,1){ \psline (0.065,0.31774272123529) (0.07,0.988987116719693) (0.08,-0.0663218973512007) (0.09,-0.993333042454911) (0.1,-0.54402111088937) (0.11,0.327700708813498) (0.12,0.887294108094694) (0.13,0.986959210970907) (0.14,0.75762841539272) (0.15,0.37415123057122) (0.16,-0.0331792165475568) (0.17,-0.390184867177364) (0.18,-0.665101514978822) (0.19,-0.852122368368379) (0.2,-0.958924274663138) (0.21,-0.998774344144139) (0.22,-0.98609877449093) (0.23,-0.934279696804283) (0.24,-0.85475260723884) (0.25,-0.756802495307928) (0.26,-0.647699566344023) (0.27,-0.532973610967149) (0.28,-0.416721651753499) (0.29,-0.301898272335703) (0.3,-0.190567962875485) (0.31,-0.0841142931130684) (0.32,0.0165918922293479) (0.33,0.11106003812413) (0.34,0.199077200627792) (0.35,0.280629399514357) 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Then there exists $U^\textrm{open}$ such that \xymatrix{A \ar[r]_i & U \ar@/_/[l]_r} with $r i=id$.\\ $\Longrightarrow$ \xymatrix @M+.7ex {\hq(A) \ar[r]_{r^*} & \hq(U) \ar@/_/[l]_{i^*}\ad \\& \chq(A)\ar@{.>}[ul]^{\displaystyle{\exists !\kappa}} }\hspace{3em} and $\kappa$ is onto. \\ Furthermore, let $X$ be a binormal ANR. (i.e. $X\times I $ is also normal.)\\ $A\subset U^\textrm{open}\subset X^\textrm{normal} \Rightarrow \exists U'$ open such that $A\subset U'\subset \overline{U'}\subset U $.\\ Consider \[ \xymatrix @=3em @M+.7ex @C=6em @*[c]{ \mbox{ \begin{pspicture}(2,1.3) %\psgrid % \rput(0,0.3){\psline(0,0)(2,0) \pspolygon[fillstyle=solid,fillcolor=lightgray] (.7,0)(1.3,0)(1.3,1)(.7,1) \psline(0,1)(2,1) \pscurve[linestyle=dashed](0,0)(1,-.25)(2,0)\rput(0,0.5){$I$} %\pscurve[linestyle=dashed](0.7,0)(1,-.1)(1.3,0) } \rput(1,0.2){$A$} \rput(1.7,0.05){$\overline{U'}$} \end{pspicture}} \ar@(u,u)@<2ex>[r]^-r \ar[d]^{\textrm{inclusion}} \ar@<1.5ex>[r]^{F} \ar@(dr,d)[r]_-{\textrm{inclusion}}="a" & \mbox{ \begin{pspicture}(2,0) %\psgrid % \rput(0,0.1){\rput(0,0){\psline(0,0)(2,0)\psline(.3,.1)(.3,-.1)\psline(1.8,.1)(1.8,-.1) %\pscurve[linestyle=dashed](0.3,0)(1,-.25)(1.8,0) } \rput(1.2,-.2){$\overline{U'}$} \psarc(.2,0){.2}{150}{210}\rput(2.2,0){$U$} \psarc(1.8,0){.2}{-30}{30}} \end{pspicture}}\\ \mbox{ \begin{pspicture}(2,1.5) %\psgrid % \rput(0,0.3){\psline(0,0)(2,0) %\pspolygon[fillstyle=crosshatch,fillcolor=lightgray,linestyle=none](0,0)(2,0)(2,1)(0,1) \pspolygon[fillstyle=solid,fillcolor=lightgray] (.7,0)(1.3,0)(1.3,1)(.7,1) \pscurve[fillstyle=solid,fillcolor=white,linestyle=dashed](0,0)(.4,.5)(0,1) \pscurve[fillstyle=solid,fillcolor=white,linestyle=dashed](2,0)(1.6,.5)(2,1) \psline[linecolor=white](0,0)(0,1) \psline(0,1)(2,1) \psline[linestyle=dashed](.55,0)(.55,1) \psline[linestyle=dashed](1.45,0)(1.45,1) %\pscurve[linestyle=dashed](0,0)(1,-.25)(2,0) %\pscurve[linestyle=dashed](0.7,0)(1,-.1)(1.3,0) } \rput(1,0.1){$V$}\rput(1.7,0.1){$N$} \psline{->}(1.7,.15)(1.65,.5)%\psline{->}(.6,.15)(.65,.5) \pscurve[linestyle=dashed](.55,.3)(1,.2)(1.45,.3) \end{pspicture}}\ar@{.>}[];"a"_-{\displaystyle{\exists \overline{F}}\ \textrm{ on }\displaystyle{N} } & }\] ,where $N$ is an open neighborhood of \mbox{\psset{unit=.5cm} \begin{pspicture}(2,1.3) %\psgrid % \rput(0,-.3){\psline(0,0)(2,0) \pspolygon[fillstyle=solid,fillcolor=lightgray] (.7,0)(1.3,0)(1.3,1)(.7,1) \psline(0,1)(2,1) } \end{pspicture}}$=:C$\\ $F$를 위의 그림과 같이 $C$에서 $U$로 가는 map으로서 0-level에서는 inclusion으로 1-level에서는 $r$로, $A\times I$ 에서는 $A$로의 projection으로 정의한다. $\overline{U'}\times I$는 normal space $X\times I$의 closed subset이므로 normal이고 $C$는 $\overline{U'}\times I$의 closed subset이다. $X$가 ANR이므로 $U$도 ANR이다. 따라서 $N$에서 정의된 $F$의 extension $\overline{F}$가 존재한다.\\ %Choose $V^\textrm{open}\supset A$ such that $V\times I \subset N$. %Then $$ \overline{F}|_{V\times I} : \{j:V\hookrightarrow %U\}\simeq\{ ir' : V\overset{r'}{\to} A\overset{i}{\hookrightarrow} %U, r'=r|_V\}$$ $V^\textrm{open}\supset A$를 $V\times I \subset N$이 되도록 잡으면 $ \overline{F}|_{V\times I}$는 $j:V\hookrightarrow U$와 $ir' : V\overset{r'}{\to} A\overset{i}{\hookrightarrow} U, r'=r|_V $ 사이에 homotopy를 준다. 따라서 $r'^*i^*=j^*$이다.\\ 이제 $\kappa$가 1-1임을 보이자. $\kappa(x)=0$이라 하면, 어떤 $U$에 대하여 $x=\{x_u\},x_u\in \hq(U)$이고, $i^*(x_u)=0$이다. 따라서 $j^*(x_u)=r'^*i^*(x_u)=0$이고 결국 $x=\{j^*(x_u)\}=0$임을 알 수 있다.\\ Note. The product of Paracompact space and compact Hausdorff is paracompact.\\ (Munkres, p.259)\\ In particular $\kappa$ becomes an isomorphism for $A^\textrm{closed, ANR}$ in paracompact manifold $X$. (eg. $A$ is a closed submanifold of $X$.)\\ {\bf Remark.} The proof of 3 shows in particular that if $\exists V\supset A$ such that $A$ is a deformation retract of $V$ then $A$ is taut.\\ 4. If $X$ is compact and $\ch^q(A)\cong H^q(A)$, then we have \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% &&H^q(U,U-K)\ar[r]^{\cong}\ar[d]&H^q(X,X-K)\ar[ddl]^<<<{i^*}&\save[]+<1cm,0cm>*\txt<8pc>{where $X-K=V$ and $ X-A = U$} \restore \\ \cdots \ar[r]&\ch^{q-1}(A)\ar[r]^{\delta}\ar[d]^{\kappa:\cong}& H^q_c(X-A)\ar[r]\ar@{.>}[d]^{\exists !}&H^q(X)\ar[r]\ar[d]^{id:\cong}&\ch^q(A)\ar[d]^{\kappa:\cong} \ar[r]&\cdots\\ \cdots\ar[r] &H^{q-1}(A)\ar[r]^{\delta}&H^q(X,A)\ar[r]&H^q(X)\ar[r]&H^q(A)\ar[r]&\cdots } \] Diagram commutes since all the maps involved are induced by inclusion and $\delta$ is natural. Then by 5 lemma, $H_c^q(X-A)\cong H^q(X,A).$\\ \textbf{숙제 25.} Show that $H_c^q(\mathbb{R}^n)\cong H^q(S^n,\infty)=\widetilde{H^q}(S^n)$.\\ 5. Let $M$ be a compact $n$-manifold, $A^{closed\hspace{0.2em} ANR}\subset M$. Then \[\xymatrix@M=1ex @C=1em @R=1em @*[l]{% H^q_c(M-A) \ar[rr]^{\cong}\ar[d]^{P.D :\cong} && H^q(M,A)\ar@{.>}[lld]^{\textrm{"Lefschetz duality"}:\cong}\\ H_{n-q}(M-A) &}\] e.g. $H^q(M,\bd M)\cong H_{n-q}(M-\bd M)\cong H_{n-q}(M)\hspace{0.5em}(\bd M$ has a collar)\\ \textbf{Remark.} $(K,L)$: compact pair in $M$, a compact manifold.\\ relative A.D. : $\ch^q(K,L)\overset{\cong}{\rightarrow}H_{n-q}(M-L,M-K)$\\ In particular, if $M=K$, then $\ch^q(M,L)\overset{\cong}{\rightarrow} H_{n-q}(M-L)$.\textbf{(숙제 26.)}\\ \underline{\textbf{Application of Alexander Duality}}\\ \textbf{6.} $A^{compact}\subset \mathbb{R}^n\subset S^n\\ \Rightarrow (1) \ch^q(A)\cong \widetilde{H}_{n-q-1}(\mathbb{R}^n-A)\\ \hspace*{1.3em}(2)\check{\widetilde{H}^q}(A)\cong \widetilde{H}_{n-q-1}(S^n-A)$, where $\check{\widetilde{H}^q}(A) = \underset{\underset{V}{\rightarrow}}{lim}\tilde{H}^q(V)$\\ \begin{pf}(1) and (2) : \[\xymatrix@M=1ex @C=2em @*[c]{% \ch^q(A)\ar[r]^-{A.D :\cong}& H_{n-q}(S^n, S^n-A)\ar[r]^{excision : \cong }\ar[d]^{(*):\cong}& H_{n-q}(\mathbb{R}^n, \mathbb{R}^n-A)\ar[r]^{\cong}& \widetilde{H}_{n-q-1}(\mathbb{R}^n-A)\\ &\widetilde{H}_{n-q-1}(S^n-A)&& }\] (*) : reduced long exact sequence of pair where $q \neq 0$\\ $q=0$ case $ : \\ \rightarrow \widetilde{H}_n(S^n-A)\rightarrow \widetilde{H}_n(S^n)\rightarrow H_n(S^n, S^n-A)\overset{\bd}{\rightarrow}\widetilde{H}_{n-1}(S^n-A)\rightarrow \widetilde{H}_{n-1}(S^n)\\ \hspace*{2em}(\overset{open}{=}0)\hspace{3em}(=R)\hspace{20em}(=0)\\ \Rightarrow \widetilde{H}_{n-1}(S^n-A)\cong H_n(S^n, S^n-A)/R$.\\ Now note that $\ch^0(A)/R\cong\check{\widetilde{H}^0}(A)$.\\ Then by A.D.,$H_n(S^n, S^n-A)\cong \ch^0(A)$ and $\widetilde{H}_{n-1}(S^n-A)\cong \breve{\widetilde{H}^0}(A)$.\end{pf}\\ \textbf{7.} $A\subset\mathbb{R}^n\subset{S^n},\hspace{0.5em}A:$ compact $(n-1)$-manifold.\\ $\Rightarrow$ the number of components of $\mathbb{R}^n-A$= (the number of components of $A$)+1 \begin{pf} $\widetilde{H}_0(\mathbb{R}^n-A;\mathbb{Z}/2)\cong H^{n-1}(A;\mathbb{Z}/2)\overset{P.D.}{=}H_0(A,\mathbb{Z}/2) = (\mathbb{Z}/2)^k$ where $k$= the number of components of $A$ \end{pf}\\ \textbf{8.} A non-orientable closed $M^n$ can not be embedded in $\mathbb{R}^{n+1}.$\\ \begin{pf} May assume $M$ is connected.\\ $M$ is non-orientable.\\ $\Rightarrow H_n(M) =0 \Rightarrow rk(H^n(M;\mathbb{Z}))=0$ and hence $ H^n(M;\mathbb{Z})\cong\widetilde{H}_0(\mathbb{R}^{n+1}-M;\mathbb{Z})=0\\ \therefore \mathbb{R}^{n+1}-M$ has 1 component. This is a contradiction to 7.\end{pf}\\ \textbf{9.} $A$ = a link with $k$ components in $\mathbb{R}^3\\ \Rightarrow H_*(\mathbb{R}^3-A)=H_*(\mathbb{R}^3$- trivial link with $k$ component)\\ \begin{pf} $\widetilde{H}_{n-q-1}(\mathbb{R}^3-A)=H^q(A) = H^q($trivial link)=$\widetilde{H}_{n-q-1}(\mathbb{R}^3-$trivial link with $k$ component). \end{pf}\\ \end{document} %\ar@{.*{\rightarrow}}[r] & \chq(A) %\ar@{}[dr]|{\displaystyle{\circlearrowleft}}