\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{\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{\dis}{\displaystyle} \newcommand{\disi}{\displaystyle{\sum_{i=1}^n}} \newcommand{\disj}{\displaystyle{\sum_{j=1}^m}} \newcommand{\bd}{\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{\key}[1]{\emph{\textbf{#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{pf}{{\bf 증명}}{\hfill\framebox[2mm]{}} \newenvironment{pf1}{{\bf 정리의 증명}}{\hfill\framebox[2mm]{}} \begin{document} \parindent=0cm \psset{unit=2cm} \section*{III.2 $H_q (M^n , M-A ) ,\hspace{0.5em} q\geqslant n$} idea : Compare $H_n(M, M-A)$ with $\Gamma A.$\\ 6. Suppose $U^{open} \subset M. $ Then $\bet_U \in H_n(M, M-U)$ can be viewed as a section as before and we have a homomorphism $j_U : H_n(M, M-U) \rightarrow \Gamma U.\\ \hspace*{23em} \bet_U\mapsto j_U(\bet_U ) : x\mapsto \bet_U |_x$ \\ In general, $\forall A\subset M, $ does $j_A : H_n(M, M-A)\rightarrow \Gamma A$(defined by $"\bet" = j_A(\bet) : x\mapsto \bet |_x )$ define a homomorphism, i.e., is $j_A(\bet)$ continuous section on $A$?\\ \begin{pf} Want $"\bet "$ is locally constant, i.e., $\forall a\in A , \hspace{0.5em} \exists V$ and $\bet_V$ s.t. $\bet|_x = \bet _V|_x\hspace{0.5em} ,\forall x\in A\cap V.$\\ Recall : Can represent $\bet = \{b\}$ with $\bd b \subset M-A .\\ |\bd b| :$ compact $\Rightarrow U = M- |\bd b|$ is open and choose $V$, a coordinate ball neighborhood of $a$ with $V\subset U.$ \[ \xymatrix @C=1em @*[c]{% (M, M-U)\ar[r]\ar[d] & (M, M-V)\ar[d]^{\cong on H_n} &\Rightarrow& \bet '= \{b\} \ar@{.>}[r]\ar[d] & \exists \bet_V \ar[d]^{\cong} \\ (M, M-A)\ar[r] & (M, M-a) & & \bet ' |_A = \bet \ar[r]& \bet |_a \ar[u] } \] And $\forall x\in A\cap V, \hspace{0.5em} \bet_V|_x = \bet'|_V|_x = \bet'|_x = \bet'|_A|_x = \bet|_x.$\end{pf}\\ 7. When is $j_A : H_n (M, M-A) \rightarrow \Gamma A$ an isomorphism?\\ Know : true if $A = U$, a coordinate ball.\\ \[ \xymatrix @C=3em @*[c]{% H_n (V, V-U )\ar[r]^{excision :\cong} \ar[d]^{\psi_* : \cong} & H_n(M, M-U)\ar[r]^{j_U} &\Gamma U \ar[d]^{\psi_* :\cong}\\ \mathbb{Z} = H_n (\mathbb{R}^n, \mathbb{R}^n\backslash D )\ar[rr]^{j_D : \cong} &&\Gamma D = \mathbb{Z} } \] Also note $H_q(M, M-U ) \cong H_q(\mathbb{R}^n , \mathbb{R}^n\backslash D ) \cong \widetilde{H_{q-1}}(S^{n-1}) = 0 $ if $ q>n.$\\ Let $M = \mathbb{R}^n.$\\ If $A$ is a "nice " compact set, then $j_A$ is $\cong$.\\ e.g. $ A =D , [0,1]\times [0,1] , [0,1], $ point, $\cdots$ etc.\\ But note that if $A = M =\mathbb{R}^1, \hspace{0.5em} H_1(\mathbb{R}^1, \mathbb{R}^1\backslash \mathbb{R}^1 ) = H_1 (\mathbb{R}^1 ) = 0$ but $\Gamma \mathbb{R}^1 = \mathbb{Z}$.\\ \textbf{Note.} $A$: closed. Then\[ \xymatrix @C=1em @*[c]{% H_n(M, M-A)\ar[rr]^{j_A} \ar[dr]^{j_A}& & \Gamma A \\ & \Gamma_c A\ar[ur]^{\subset} & } \]i.e., $j_A(\bet) \in \Gamma_c A \hspace{0.5em} ,\forall \bet$ where $\Gamma_c A$ consists of sections with compact support.\\ \begin{pf}Let $\bet = \{b\}$: relative cycle $\Rightarrow |b| :$ compact. \\Then $\forall x\in (M-|b|)\cap A, \bet|_x = 0 $ since $|b| \subset M-x.$ Think of this in chain level. \[ \xymatrix @C=1em @*[c]{% 0\ar[r] & S_n(M-A)\ar[d]\ar[r] & S_n(M) \ar[r]\ar[d]^{=}& S_n(M)/S_n(M-A)\ar[r]\ar[d] & 0\\ 0 \ar[r]& S_n(M-x)\ar[r] & S_n(M)\ar[r] & S_n(M)/S_n(M-x)\ar[r] & 0 } \] $\therefore \bet$ has a support $\subset |b|\cap A:$ compact\end{pf}\\ So the right statement is $j_A : H_n(M, M-A) \rightarrow\Gamma_c A .$ \\Furthermore, $j_A$ is natural:\\ \[ \xymatrix @C=2em @*[c]{% B\subset A\subset M \Rightarrow & H_n(M, M-A)\ar[r]^<<<{j_A} \ar[d]^{\rho_B^A} &\Gamma A\ar[d]^{restriction}&\textrm{commute.}\\ &H_n(M, M-B) \ar[r]^<<<{j_B}&\Gamma B } \] Exercise. \textbf{숙제 11.} \[ \xymatrix @C=2em @*[c]{% f : (M, A)\ar[r]^{\cong} & (N, B)\Rightarrow &H_n(M, M-A)\ar[r]^<<<{j_A} \ar[d]^{f_* : \cong} & \Gamma A\ar[d]^{"f_*"}&\textrm{commute.}\\ & & H_n(N, N-B)\ar[r]^<<<{j_B} & \Gamma B& } \]\\ {\bf 8. (Theorem)} Let $M$ be an $n$-dimensional manifold and $A^{\textrm{closed}}\subset M$. Then\\ (1) $H_q(M,M-A)=0$ for $q>n$\\ (2) $H_n(M,M-A)\cong \Gam_cA$\\ \begin{pf} \begin{lem} (MV) $A, B$ closed $\subset M.$ If the theorem is true for $A, B$ and $A\cap B,$ then so is for $A\cup B$.\end{lem} \begin{pf} $A, B$ : closed $\Rightarrow M-A, M-B$ open with $(M-A)\cap(M-B) = M-A\cup B\\ \hspace*{20em}(M-A)\cup(M-B)=M-A\cap B$\\ relative MV: \[\scriptsize{ \xymatrix @C=1em @*[c]{% \textrm{hypothesis}\ar[r]&0\ar[r]\ar[d]^{\cong}& H_n(M, M-A\cup B)\ar[r]\ar[d]^{j}&H_n(M, M-A)\oplus H_n(M, M-B)\ar[d]^{\cong}\ar[r] & H_n(M, M-A\cap B)\ar[d]^{\cong} \ar[r] &0\\ \textrm{exact :}&0\ar[r]& \Gamma(A\cup B)\ar[r] & \Gamma(A)\oplus\Gamma(B)\ar[r] &\Gamma(A\cap B)\ar[r]&0\\ && s\ar@{|->}[r]&(s|_A, s|_B ) &&\\ &&&(a,b)\ar@{|->}[r]& a|_{A\cap B}-b|_{A\cap B}& } }\] (i) follows from relative MV-sequence.\\ (ii) follows from 5-lemma. \end{pf} \begin{lem} If $M=\rb^n$ and $A$ is a compact subset of $\rb^n$, then the theorem is true. \end{lem} \begin{pf} Know : The theorem is true for a "nice" compact set $A\subset \rb^n$, for instance $A=$rectangle.\\ By lemma 1, the theorem is true if $A$ is a finite union of rectangles by induction on number of rectangles.\end{pf} \\ 이제 compact set $A$에 대하여 정리가 성립함을 보이자. \\ 먼저 $j_A: H_n(M,M-A)\to \Gam A =\Gam_c A$가 onto임을 보인다. \\ (i) For $s\in \Gam A $, there exists an open set $U$ containing $A$ such that $s$ can be extended to $\bar{s}$ on $U$.\\ pf) $s(A)$ is compact $\Rightarrow$ $s(A)$ lies in finitely many sheets of $\rb^n_\mathcal{O} \cong \rb^n\times \zb$.\\ Let $A_i$=$s^{-1}(\rb^n\times i)$ for $i\in \zb$. $A_i$ is compact.\\ So there exists an open set $U_i$ containing $A_i$ such that $U_i$'s are pairwise disjoint and $s$ can be extended to $\bar{s}$ over $U=\bigcup U_i$.\\ (ii) Cover $A$ by finitely many rectangles in $U$ and let $A'$ be the union of rectangles.\\ Then the following diagram commutes. \[ \xymatrix @=2em @C=4em @*[c] { % H_n(M,M-A') \ar[r]^<<<<<<<{j_{A'}}_<<<<<<<{\cong} \ar[d] & \Gam A'\ni\bar{s}|_{A'} \ar[d]^{\textrm{restriction}}\\ H_n(M,M-A) \ar[r]^<<<<<<<{j_{A}} & \Gam A \ni s}\] 임의의 $s\in \Gam A$에 대하여 (i)에 의하여 $\bar{s}$가 존재하므로 $\bar{s}|_{A'}$은 $s$에 대응되는 $\Gam A'$의 원소이다. 또한 $j_{A'}$은 isomorphism이므로 $j_A$는 onto이다. \\ 이제 $j_A$이 1-1이고 $H_q(M,M-A)=0$ for $q>n$임을 보이자. \\ Let $\alp\in H_q(M,M-A)$ and assume $j_A(\alp)=0$ if $q=n$.\\ Suppose $\alp=\{a\}$ with $\bd a\subset M-A$. Since $|\bd a|$ is compact, $V=M-|\bd a| $ is open. Let $\alp'=\{a\}\in H_q(M,M-V)$. \\ ($q=n$) : Let $U\subset V $ be an open set containing $A$ each of whose components intersects $A$. Let $A'$ be a finite union of rectangles which covers $A$ and is contained in $U$.\\ Then $j_U(\alp'|_U)$ is a section on $U$ which has a zero on each of its component.\\($\because$ Each component intersects $A$ and $j_A(\alp)=0$.)\\ By uniqueness of sections on connected sets, $j_U(\alp'|_U)=0$. \[ \xymatrix @=1.5em @C=4em @*[c] { % H_n(M,M-U) \ar[r]^<<<<<<<{j_{U}} \ar[d] & \Gam U \ar[d]&\alp'|_{U} \ar@{|->}[r] \ar@{|->}[d] &0\ar@{|->}[d]\\ H_n(M,M-A') \ar[r]^<<<<<<<{j_{A'}}_<<<<<<<{\cong} \ar[d] & \Gam A' \ar[d]&\alp'|_{A'} \ar@{|->}[r] \ar@{|->}[d]&0\ar@{|->}[d]\\ H_n(M,M-A) \ar[r]^<<<<<<<{j_{A}} & \Gam A &\alp \ar@{|->}[r] &0}\] Therefore by the above diagram, $\alp'|_{A'}=0$ and hence $\alp=0$. ($q>n$) : The theorem is true for $A'$. So $\alp'|_{A'}=0$ and hence $\alp=0$. \end{pf}\\ \begin{pf1}\\ {\bf Step 1.} $A^{\textrm{compact}}\subset M$:\\ $A$ is a finite union of compact sets each of which is contained in a coordinate ball neighborhood($\approx \rb^n$) and apply lemma 1 and lemma 2.\\ {\bf Step 2.} $A\subset U^{\textrm{open}}\subset \overline{U}^{\textrm{compact}}$ $\Rightarrow$ The theorem is true for $A\subset U(=M)$\footnote{manifold의 open subset은 manifold이므로 $U=M$으로 하여도 무방하다.}\\ \hspace*{4em}where $A$ is a closed subset of $U$:\\ $\bd U =\overline{U}-U$ is compact and $U-\overline{A}=U-A$. \\ Consider $(M,M-\bd U, M-(\bd U\cup \overline{A} ))$.\\ ($q>n$): $$ H_{q+1}(M,M-\bd U) \to H_q(M-\bd U,M-(\bd U \cup \overline{A}))\to H_q(M,M-(\bd U \cup \overline{A})) \to \cdots $$ By excision theorem, $H_q(M-\bd U,M-(\bd U \cup \overline{A}))=H_q(U,U-A)$. \\By step 1, $H_{q+1}(M,M-\bd U)=0$ and $H_q(M,M-(\bd U \cup \overline{A}))=0$.\\ Therefore $H_q(U,U-A)=0$.\\ ($q=n$): \[ \xymatrix @=1.5em @*[c] { % 0 \ar[r]&H_n(U,U-A) \ar[r] \ar[d]^{j_A} &H_n(M,M-(\bd U \cup \overline{A})) \ar[r]\ar[d]^{\cong(\textrm{step 1})} &H_n(M-\bd U) \ar[d]^{\cong(\textrm{step 1})}\\ 0 \ar[r]&\Gam_cA \ar[r]_<<<<<<<<<<<<{\parbox{5em}{\scriptsize extension by 0 outside support}}&\Gam(\bd U\cup \overline{A}) \ar[r]_<<<<<<<<<<{\textrm{restriction}} &\Gam(\bd U) }\] So, $j_A$ is an isomorphism by 5 lemma.\\ {\bf Step 3.}(general case)\\ Show $j_A: H_n(M,M-A)\to \Gam_cA$ is onto:\\ $\forall s\in \Gam_cA$, let supp $s=K$. \\ Then K is compact and $\exists U$ such that $K\subset U \subset \overline{U}^\textrm{compact}$ \\ Let $A'=A\cap U$. \[ \xymatrix @=1.5em @C=4em @*[c] { % H_n(U,U-A') \ar[r]^<<<<<<<<{\cong\textrm{(step 2)}} \ar[d]^{i_*} & \Gam_cA'\ni s|_{A'} \ar[d]^{\parbox{4.5em}{\scriptsize extension by 0 outside {\it K}}} \\ H_n(M,M-A) \ar[r]^{j_A} & \Gam_cA\ni s }\] So, $j_A$ is onto.\\ Show $j_A$ is 1-1 and $H_q(M,M-A)=0$ if $q>n$: \\ $\alp \in H_q(M,M-A) $ and assume $j_A(\alp)=0$ if $q=n$.\\ If $\alp=\{a\}$ with $\bd a \subset M-A$, $|a|$: compact $\Rightarrow |a| \subset U \subset \overline{U}^\textrm{compact}$.\\ Let $A'=A\cap U$. Apply the above diagram.\\ $q=n$ : Since $|a|\subset U$, $\alp'=\{a\}\in H_n(U,U-A')$. So in the above diagram, $i_*(\alp')=\alp$. Since $j_A(\alp)=0$ and $0|_{A'}=0$, $j_{A'}(\alp')=0$. So $\alp'=0$ and hence $\alp=0$.\\ $q>n$ : By step 2, $\alp'=\{a\}\in H_q(U,U-A')=0$. So $\alp=0$. \end{pf1}\\ {\bf 9. Consequences of the theorem}\\ (1) Let $A$ be connected and closed but not compact. Then $H_n(M,M-A)=0$. In particular if $M$ is connected but not compact then $H_n(M)=0$. \\ \begin{pf} If $s\in \Gam_cA$, $\nu\circ s : A\to \zb^{\geq 0}$ is continuous and $=0$ at some $a\notin $ supp $s$. So $\Gam_cA=0$ \end{pf}\\ (2) If $M$ is orientable along $A$, $A$ is compact and has $k$ components, then $H_n(M,M-A)\cong\zb^k$.\\ \begin{pf} $\Gam_cA=\Gam A \cong \zb^k$ by 5. (2). \end{pf}\\ (3) If $A\subset \rb^n$ is compact and has $k$ components, then $\widetilde{H}_{n-1}(\rb^n-A)\cong\zb^k$.\\ \begin{pf} From homology sequence of pair $(\rb^n, \rb^n-A)$, we get $$ 0\to {H}_{n}(\rb^n, \rb^n-A) \to \widetilde{H}_{n-1}(\rb^n-A)\to 0$$ So $\widetilde{H}_{n-1}(\rb^n-A)\cong {H}_{n}(\rb^n, \rb^n-A) \cong\zb^k$ by (2). \end{pf}\\ (4) If $M$ is connected and closed\footnote{A manifold $M$ is closed if it is compact without boundary.}, then $H_n(M)=\left\{ \begin{array}{cl} \zb, & \textrm{if }M \ \textrm{ is orientable} \\ 0, &\textrm{if }M \ \textrm{ is non-orientable} \end{array} \right. $\\ \begin{pf} Clear from 5. (3) and (4). \end{pf}\\ {\bf Remark.} For a PID $R$, $H_n(M)=\left\{ \begin{array}{cl} R, & \textrm{if }M \ \textrm{ is } R-\textrm{orientable} \\ 0, &\textrm{if }M \ \textrm{ is not }R-\textrm{orientable} \end{array} \right. $\\ {\bf 10. Fundamental class of $M$ and degree}\\ A choice of generating section is an orientation and the corresponding homology class $\zeta$\footnote{$[M]$으로 쓰기도 한다. } is called the \key{fundamental (orientation) class} of $M$, i.e., $\zeta|_x\in H_n(M,M-x)$ is the preferred orientation at $x$ for all $x\in M$.\\ Let $M^n, N^n$ be oriented closed connected manifolds. For $f: M\to N$, if $f_*(\zeta_M)=k\cdot \zeta_N$, $k$ is called the \key{degree} of $f$. \\ {\bf 숙제 14} $\forall y\in N$, "regular value" i.e., $\forall x\in f^{-1}(y),$ $f$ is a homeomorphism on a neighborhood $U_x$ ($f|_{U_x}:U_x\overset{\cong}{\to} V_y)$. Then $$ deg f =\sum_{x\in f^{-1}(y)}deg_x f$$ where $deg_xf$ is defined by $f|_{U*}:H_n(U,U-x)\to H_n(V,V-y)$.\\ In particular, if $p:M\to N$ is a $k$-fold covering, $deg p =k$ (with respect to the induced orientation on M).\\ {\bf 숙제 15} (22.43) (22.49) (22.50)* \end{document}