\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{\hq}{H^q} \newcommand{\hnq}{H_{n-q}} \newcommand{\dlim}[1][K]{\underset{#1}{\underrightarrow{\lim}}} \newcommand{\htx}{\textrm{Hom}} \newcommand{\etx}{\textrm{Ext}} \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*{V.2 Poincar\'{e} Duality} Let $M$ be a $n$-dimensional manifold (without boundary), $R$-orientable. ($R$ : PID, if $M$ is non-orientable over $\zb$, take $R=\zb/2$.)\\ If $M$ is compact, we will show the following isomorphism, \[ \xymatrix @=3em @*[c] { % \hq(M) \ar[r]^{\textrm{P.D.}}_{\cong} & \hnq(M) }\] If $M$ is not compact, the above duality as stated is not true. For instance, $$ M=\rb \Rightarrow H^1(\rb)=0 , H_0(\rb)=\zb$$ In this case, the correct duality is $$ \hcq(M)\cong \hnq(M)$$ where $\hcq(M)$ is cohomology with compact support.\\ 예를 들어 $M=\rb^1$인 경우에 다음과 같이 주어진 1-cochain $Dx$, $x\in \rb$를 생각해 보자. $$D(x)(e):=H(e(1))-H(e(0))$$ 여기서 $H\in S^0(\rb)$는 함수 $H(y)=H_x(y)=\left\{\begin{array}{cc} 1 & y\geq x \\ 0 & y}[d] \ar[r]^{\zeta_{K'} \cap}&\hnq(M)\\ \hqc(M)\ar@{.>}[ur]_{\exists! D }& }\] IV.3 cup and cap product의 9번에 의하여 위 삼각형이 commute하고 따라서 $D$가 induce된다. 이 때 $D$가 바로 duality homomorphism이다.\\ 4. {\bf 정리 [Poincar\'{e} duality]} Let $M$ be an $R$-orientable $n$-manifold. Then $$ D: \hqc(M) \to \hnq(M)$$ is an isomorphism.\\ 정리를 증명하기 위하여 먼저 다음의 Lemma를 증명한다.\\ {\bf Lemma 1. }(MV) If theorem holds for open sets $U$, $V$ and $U\cap V$, then theorem holds for $U\cup V$.\\ \begin{pf} Let $K^\textrm{compact}\subset U$, $L^\textrm{compact}\subset V$, $B=U\cap V$, $Y=U\cup V$. \begin{center} \begin{pspicture}(2.2,1.6) \pscircle[linestyle=dashed](.7,.7){.7} \pscircle[linestyle=dashed](1.4,.7){.7} \psellipse(.7,.7)(.5,.3) \psellipse(1.4,.7)(.5,.3) \rput(.6,1.5){$U$} \rput(1.5,1.5){$V$} \rput(.5,1.1){$K$} \rput(1.7,1.1){$L$} \end{pspicture} \end{center} Consider cohomology MV-sequence for triple $(Y, Y-K,Y-L)$: \[ \xymatrix @=1em @R=2ex @*[c] { % H(Y,Y-K\cup L) & H(Y,Y-K)\oplus H(Y,Y-L) \ar[l] \ar@{}[d]_{\wr\|}^{\textrm{excision}} & H(Y,Y-K\cap L)\ar@{}[d]_{\wr\|}^{\textrm{excision}}\ar[l]\\ & H(U,U-K)\oplus H(V,V-L) & H(B,B-K\cap L)\\ }\] and homology sequence for a pair $(U,V)$ \[ \xymatrix @=1em @R=2ex @*[c] { % H(B) \ar[r]& H(U)\oplus H(V) \ar[r] & H(Y) }\] Lemma는 다음의 key diagram으로부터 얻어진다.\\ \begin{list}{}{ \setlength{\leftmargin}{-2cm}} \item $ \xymatrix @=1em @R=2.5em @*[c] { % H^{q+1}(B,B-K\cap L) \ar[d]^{\zeta_{K\cap L}\cap} \ar@{}[dr]|{\textrm{\large (3)}} & \hq(Y,Y-K\cup L)\ar[l]_<<{\del^*} \ar[d]^{\zeta_{K\cup L}\cap} %가운데 label넣기 \ar@{}[dr]|{\textrm{\large (2)}}& %%%%%%%% \hq(U,U-K)\oplus\hq(V,V-L) \ar[l]\ar[d]^{\zeta_{K}\cap\oplus\zeta_L\cap} \ar@{}[dr]|{\textrm{\large (1)}} & \hq(B,B-K\cap L) \ar[l]\ar[d]^{\zeta_{K\cap L}\cap}\\ H_{n-q-1}(B) & \hnq(Y) \ar[l]_{\bd_*}& \hnq(U)\oplus\hnq(V) \ar[l] & \hnq(B)\ar[l] }$ \end{list} \bigskip {\bf Claim.} 위 diagram에서 (1),(2)은 commute하고 (3)은 up to sign으로 commute한다.\\ % 위의 Claim을 증명하면 이 Lemma의 증명이 끝나는데, 위 Claim을 증명하고 key diagram에 direct limit을 취하면 다음과 같다.\footnote{여기서 $Y$의 임의의 compact set이 $K\cup L$의 꼴로 나타남을 증명할 수 있다. (Exercise)} \[ \xymatrix @=1em @R=2.5em @*[c] { % H^{q+1}_c(B) \ar[d]^{D}_{\cong} \ar@{}[dr]|{\pm\displaystyle{\circlearrowleft}} & \hqc(Y)\ar[l] \ar[d]^{D} \ar@{}[dr]|{\displaystyle{\circlearrowleft}}& \hqc(U)\oplus\hqc(V) \ar[l]\ar[d]^{D\oplus D}_{\cong} \ar@{}[dr]|{\displaystyle{\circlearrowleft}} & \hqc(B) \ar[l]\ar[d]^{D}_{\cong} \\ H_{n-q-1}(B) & \hnq(Y) \ar[l]& \hnq(U)\oplus\hnq(V) \ar[l] & \hnq(B)\ar[l] }\] 이 diagram에서 commute up to sign이라도 5-lemma는 여전히 성립하므로 5-lemma에 의하여 원하는 결과를 얻는다. \\ 따라서 Claim만 증명하면 된다.\\ $V$-part of (1) Commutes ($U$-part도 비슷하게):\\ \[ \xymatrix @=1em @R=1em @C=-.5em @*[c] { % &\hq(V,V-L) \ar[dd]\ar@{}[dddr]|{\circlearrowleft\textrm{(Want)}}&& \hq(Y,Y-K\cap L) \ar[ll] \ar[dl]_{\cong}^<<{\textrm{excision}}% \\ \hq(V,V-K\cap L)\ar[dd]\ar[ur] \ar[rr]^<<<<{\cong}_<<<<<{\textrm{excision}}&& \hq(B,B-K\cap L) \ar[ul]\ar[dd]&\\ &\hnq(V)\ar@{=}[dl]&&\\ \hnq(V)&&\hnq(B)\ar[ll]\ar[ul]& }\] 위 diagram에서 나머지 면들이 모두 commute함은 쉽게 확인할 수 있다. (윗면과 아랫면은 정의에 의하여, 옆면들은 inclusion과 cap product가 commute한다는 사실에 의하여 commute한다.) 따라서 (1)은 commute한다.\\ $U$-part of (2) commutes ($V$-part도 마찬가지):\\ \[ \xymatrix @=1em @R=1em @C=-.5em @*[c] { % \hq(Y,Y-K\cup L) \ar[dd] \ar@{}[dddr]|{\circlearrowleft\textrm{(Want)}} && \hq(Y,Y-K) \ar[ll] \ar[dl]_{\cong}^<<{\textrm{excision}}\ar[dd] \\ & \hq(U,U-K) \ar[ul]\ar[dd]&\\ \hnq(Y)\ar@{=}[rr]&&\hnq(Y)\\ &\hnq(U)\ar[ur]\ar[ul]& }\] 마찬가지로 나머지 면들이 모두 commute함을 알 수 있고 따라서 (2)가 commute한다.\\ (3) commutes up to sign :\\ 다음 diagram을 2층으로 놓고 $\del$, $\bd$을 추적하면 원하는 결과를 얻는다. 부호는 $(-1)^q$만큼 차이남을 확인할 수 있다.(Exercise) \\ \begin{list}{}{ \setlength{\leftmargin}{-2cm}} \item $ \xymatrix @=1em @R=2.5em @*[c] { % 0 &S^q(Y,Y-K\cup L)\ar[l]\ar[dd]^{\zeta\cap} &S^q(Y,Y-K)\oplus S^q(Y,Y-L) \ar[d]^{i^\sharp\oplus i^\sharp} \ar[l]& S^q(Y,Y-K\cap L)\ar[l]\ar[dd]^{\zeta\cap}\ar[dr]^{\textrm{excision}}_\cong &0\ar[l]\\ && S^q(U,U-K)\oplus S^q(V,V-L) \ar[d]^{\zeta\cap\oplus\zeta\cap}&& S^q(B,B-K\cap L)\ar[d]^{\zeta\cap}\\ &S_{n-q}(Y) &S_{n-q}(U)\oplus S_{n-q}(V)\ar[r]^-{i_\sharp\oplus i_\sharp} \ar[l]_-{i_\sharp\oplus i_\sharp}&S_{n-q}(Y) &S_{n-q}(B)\ar[l]_{i_\sharp} }$ \end{list} \end{pf} {\bf Lemma 2. } Let $\{U_{i}\}$ be a system of open sets totally ordered by inclusion and let $U=\cup U_{i}$. If theorem is true for each $U_{i}$, then true for $U$.\\ \begin{pf} \,\, By the iterated limit argument proved earlier(앞절 9), $H_{c}^{q}(U) = \underset{\longrightarrow}{\lim} H_{c}^{q}(U_{i})$ and the corollary of 8(앞절) implies $H_{n-q}(U) = \underset{\longrightarrow}{\lim} H_{n-q}(U_{i})$. Then by using the fact 5(앞절), we can easily check the lemma. See the following diagram. \[ \xymatrix @M=1ex @C=3em @R=1em @*[c]{% H_{c}^{q}(U_{i}) \ar[r]^{D_{i}}_{\cong} \ar[d] & H_{n-q}(U_{i}) \ar[d] \\ \vdots \ar[d] & \vdots \ar[d] \\ H_{c}^{q}(U) \ar@{.>}[r]^{D}_{\cong} & H_{n-q}(U) } \] \end{pf} {\bf Lemma 3. } Theorem holds for $U^{\scriptsize{\textrm{open}}} \subset \rb^{n}$.\\ \begin{pf} {\bf Step 1} $U$ is convex (so that $U \cong \rb^{n}$) : Let $B^{n} = B(0,r), \, n$-ball $\subset \rb^{n}$. Then \[ \xymatrix @M=1ex @C=3em @R=1em @*[c]{% H^{q}(\rb^{n}, \rb^{n} - B) \ar[r]^{\zeta_{B} \cap}_{\cong} \ar[d] & H_{n-q}(\rb^{n}) \\ \vdots \ar[d] & \\ H_{c}^{q}(\rb^{n}) \ar@{.>}[uur]^{\exists ! D}_{\cong}} \] Show $ \zeta_{B} \cap : H^{q}(\rb^{n}, \rb^{n}-B) \cong H_{n-q}(\rb^{n})$ : If $q \neq n$, then both are 0 and if $q=n$, then consider as follows. Note that $H^{n}(\rb^{n}, \rb^{n}-B) \cong \htx(H_{n}(\rb^{n},\rb^{n}-B),R)$. Let $\bar{\zeta}$ be a dual of $\zeta_{B}$. Then $H^{n}(\rb^{n},\rb^{n}-B) = <\bar{\zeta}>$. Furthermore, $(\zeta_{B} \cap \bar{\zeta},1) = (\zeta_{B}, \bar{\zeta} \cup 1) = (\zeta_{B},\bar{\zeta}) = 1 \Rightarrow \zeta_{B} \cap$ is an isomorphism. Combining the fact $\{B(0,r)| r \in \rb\}$ is cofinal in $J=\{K^{\textrm{cpt.}} \subset \rb^{n}\}$, it is clear $ D : H_{c}^{q}(\rb^{n}) = \underset{\underset{B}{\longrightarrow}}{\textrm{lim}} H^{q}(\rb^{n}, \rb^{n}-B) \cong H_{n-q}(\rb^{n})$.\\ {\bf Step 2} $U$ is a finite union of convex open sets. Induction on the number of convex open sets and apply lemma 2. \hspace*{2.0em} \psset{unit=1cm} \begin{pspicture}(-0.5,-0.5)(3.0,2.0)% %\psgrid[gridwidth=0.1pt,subgridwidth=0.1pt,gridcolor=red,subgridcolor=green] \pscircle(0.5,0.5){0.7} \pscircle(1.5,0.5){0.7} \pscircle(1.0,1.0){0.7} \rput(0,0.3){$U_{i}$} \rput(2.0,0.8){$U_{i+1}$} \rput(1.0,1.5){$U_{1}$} \end{pspicture} \raisebox{3.3em}{ \hspace*{3.0em} \parbox{8cm}{$U_{i+1} \cap (U_{1} \cup \cdots \cup U_{i}) = (U_{i+1} \cap U_{1}) \cup \cdots \cup (U_{i+1} \cap U_{i})$ is a union of $i$ convex sets and holds by the induction hypothesis.}}\\ {\bf Step 3} $U$ : arbitrary open set in $\rb^{n}$. Let $\mathcal{W} =\{\overset{\circ}{B}(x,r) | x \in U \,\, \textrm{with rational coordinate and} \,\, \overset{\circ}{B} \subset U, r \in \mathbb{Q}\} = \{W_{1}, W_{2}, \cdots \}$ and let $U_{1} = W_{1}, \cdots , U_{n}= W_{1} \cup \cdots \cup W_{n}, \cdots$ and clearly $U = \underset{n=1}{\overset{\infty}{\cup}} U_{n}.$ Now apply lemma 2. \end{pf} {\bf Proof of Thoerem} $\mathcal{U} = \{U^{\textrm{open}} \subset M | \textrm{Theorem holds for} U\}$ : partially ordered set with inclusion. By lemma 2 and lemma 3, $\mathcal{U}$ satisfies the hypothesis of Zorn's lemma. Then there exists a maximal element $U \in \mathcal{U}$. \,\, For any coordinate neighborhood $V$, theorem holds for $V$ and $V \cap U$ by lemma 3. Then theorem holds for $V \cup U$ by lemma 1 and $V \subset U$ by maximality. Hence $U=M$. \newpage {\bf \large{Consequences and applications}}\\ 5. {\bf 따름정리} (1) Let $M^{n}$ be connected and $R$-orientable. Then $H_{c}^{n}(M) \cong \rb$. (2) Let $M^{n}$ be closed and orientable. $\Rightarrow \begin{cases} \bet_{q} = \bet_{n-q} & \bet_{i}=i-\textrm{th Betti Number} \\ T_{q} \cong T_{n-q-1} & T_{i} = \textrm{torsion part of}\,\, H_{i}(M,\zb)\end{cases}$ (3) Let $M^{n}$ be closed and orientable. Then $H_{n-1}(M,\zb)$ is free abelian. (4) Let $M^{2k+1}$ be closed (and orientable). Then $\chi(M)=0$. \begin{pf} (1) $H_{c}^{n}(M) \cong H_{0}(M) \cong \rb$. (2) Note that if $M$ is compact, then $H_{*}(M)$ is finitely generated. 우선 poincar\'{e} duality에 의해서 $H^{q}(M) \cong H_{n-q}(M)$ 임을 안다. 그리고 the universal coefficient theorem에 의해서 $H^{q}(M) = \htx(H_{q}(M),\zb) \bigoplus \etx(H_{q-1};\zb)$. 이제 $\htx(H_{q}(M),\zb)$를 생각해보자. $H_{q}(M)$에 Hom-functor를 취하면 $H_{q}(M)$의 free part만 남고 torsion part는 없어진다. 이것은 $\forall d(\neq 0)$에 대해서 $\zb/d \to \zb$는 0밖에 없기 때문이다. 반대로 $H_{q-1}(M)$에 Ext를 취하면 $H_{q-1}(M)$의 free part는 없어지고 torsion part만이 남는다. 따라서, $\bet_{n-q} = \bet_{q}$이고 $T_{n-q} = \etx(H_{q-1}(M),\zb) = T_{q-1}$이 성립한다. {\bf Remark} If $M$ is compact, $H_{*}(M)$ is finitely generated.(나중에 증명) (3) Clear from (2) (4) $\chi(M) = \sum (-1)^{q}\bet_{q} = 0$ \end{pf} {\bf 6. Intersection pairing} Let $M$ be a closed, connected and orientable manifold with $R$: P.I.D. and fundamental class $\zeta$. Consider \begin{center} $I : H^{n-q}(M) \bigotimes H^{q}(M) \overset{\cup}{\rightarrow} H^{n}(M) \overset{<\zeta,\,\,>}{\underset{\cong}{\rightarrow}} \zb (\textrm{or} R)$ \hspace*{1.0em} $a \otimes b \mapsto a \cup b$ \end{center} If $a$ is a torsion element, i.e. $ra=0$ for some $r \neq 0$, $0 = ra \cup b = r(a \cup b) \Rightarrow <\zeta ,a \cup b> = 0$. Similarly for $b$. Hence $I$ induces $"I" : H^{n-q}(M)/T^{n-q} \bigotimes H^{q}(M)/T^{q} \overset{"\cup"}{\rightarrow} \zb (\textrm{or} R)$.\\ This pairing is non-degenerate,i.e.,\, $I(\forall \alp , \bet) = 0 \Rightarrow \bet = 0 $ and $I(\alp , \forall \bet) = 0 \Rightarrow \alp = 0$. \begin{pf} Consider \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% H^{n-q}(M) \bigotimes H^{q}(M) \ar[r]^>>>>{\cup} \ar[d]_{\zeta_{\cap} \otimes id.}^{\cong} & H^{n}(M) \ar[r]^{<\zeta, \,\,>}_{\cong} \ar[d]_{\zeta_{\cap}}^{\cong} & \zb (\textrm{or} R) \ar[d]^{=} \\ H_{q}(M) \bigotimes H^{q}(M) \ar[r]^>>>>>{\cap} \ar@(dr,dl)[rr]^{<\, , \,>} & H_{0}(M) \ar[r]^{\cong}_{< \, , 1>} & \zb (\textrm{or} R) } \] \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% a \otimes b \ar@{|->}[r] \ar@{|->}[d] & a \cup b \ar@{|->}[r] \ar@{|->}[d] & <\zeta, a \cup b> \ar@{|->}[d]^{=} \\ ( \zeta \cap a) \otimes b \ar@{|->}[r] \ar@{|->}@(dr,dl)[rr]^{<\, , \,>} & (\zeta_{\cap} a) \cap b \ar@{|->}[r] & <(\zeta \cap a) \cap b , 1> (= <\zeta \cap a, b>) } \] This induces \[ \xymatrix @M=1ex @C=1em @R=1em @*[c]{% H^{n-q}(M)/T \bigotimes H^{q}(M)/T \ar[r] \ar[d] & \zb (\textrm{or} R) \ar[d]^{=} \\ H_{q}(M)/T \bigotimes H^{q}(M)/T \ar[r] & \zb (\textrm{or} R) } \] Let $\alp = \{a\}, \bet = \{b\}, a \in H^{n-q}(M) , b \in H^{q}(M)$ and $b$ torsion free. And suppose \begin{center} $ 0 = I(\forall \alp, \bet) = <\zeta, \forall a \cup b> = <\zeta \cap \forall a , b>$ \end{center} Since, $H^{q}(M) \cong \htx(H_{q}(M),\zb) \bigoplus \etx(H_{q-1}(M),\zb)$= Torsion free part $\bigoplus$ Torsion part, $b =b^{'} \oplus b^{''} \Rightarrow b = b^{'}$. Hence $0 = <\zeta \cap \forall a, b> = <\zeta \cap \forall a, b^{'}> \Rightarrow b^{'}=0 \Rightarrow b=0$, since $\zeta \cap \forall a$ represents every elements in $H_{q}(M)$ by Poincar\'{e} duality and $b^{'} \in \htx(H_{q}(M),\zb)$. Hence $b=b^{'}=0$ and $\bet=0$. \end{pf} {\bf Remark} If we consider with a field coefficient $R$, Tor =0 and have a non-degenerate pairing for a $R$-orientable manifold $M$, \begin{center} $I : H^{n-q}(M;R) \bigotimes H^{q}(M;R) \rightarrow R$ \hspace*{13.0em}$a \otimes b \mapsto <\zeta, a \cup b>$ \end{center} {\bf Remark} If $R = \rb$, then $I(a,b) = \int_{M} a \wedge b$.\\ {\bf 따름정리 } Let $M^{4k+2}$ be closed and orientable. Then (1) dim $H^{2k+1}(M;\rb)$ = even. (2) $\chi(M; \rb)$ = even. \begin{pf} $I : H^{2k+1}(M;\rb) \bigotimes H^{2k+1}(M;\rb) \rightarrow \rb$ is non-degenerate and note that $I(a,b) = -I(b,a)$. Now (1) follows from the following fact : {\bf Fact} If $I$ is a skew-symmetric bilinear form (2-form), then there exists a basis $e= \{e_{1}, \cdots , e_{n}\}$ such that $I$ can be represented as \[ \begin{pmatrix} \begin{tabular}{l|cr} $\begin{matrix} \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} & & \textrm{\Large{0}} \\ & \ddots & \\ \textrm{\Large{0}}& & \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \end{matrix}$ & \Large{0} \\ \hline\\ \hspace*{5.0em} \Large{0} &\Large{0} \end{tabular} \end{pmatrix} \] where $I(e_{i},e_{j}) =a_{ij}$. {\bf Proof of fact} Choose $x_{1} \in \rb^{n}$ such that $I(x_{1},y) \neq 0$ for some $y$. Let $y_{1}$ be such that $I(x_{1},y_{1})=1$. Consider $\theta : \rb^{n} \to \rb^{2}$ given by $\theta(x) = (I(x,x_{1}),I(x,y_{1})$. Then $\theta(x_{1}) = (0,1)$ and $\theta(y_{1}) = (-1,0)$. Clearly $\ker \theta = \rb^{n-2}$ and repeat this process until $I$ becomes trivial. (2) follows from (1) immediately \end{pf}\\ {\bf Remark} We will show later that $\chi(M) = \chi(M;\rb)$.\\ \textbf{7. (Cohomology ring of projective spaces)}\\ $M = \mathbb{C}P^n$\\ Recall : $H_q(Z,Y)\cong H_q(X,A),\hspace{1em}Z=X\underset{f}{\cup}Y\\ \Rightarrow H^q(\mathbb{C}P^n,\mathbb{C}P^{n-1})\cong H^q(D^{2n},S^{2n-1})= \left\{\begin {array}{cc} \mathbb{Z}&,if \hspace{0.5em}q=2n\\ 0&, otherwise\end{array}\right\}$\\ Long exact sequence of pair $(\mathbb{C}P^n,\mathbb{C}P^{n-1})\\ \Rightarrow H^q(\mathbb{C}P^n)\overset{i^*}{\rightarrow}H^q(\mathbb{C}P^{n-1})$ is $\cong$ for $q\leq 2n-2$. \\ Consider $\mathbb{C}P^2$ first : \[\xymatrix @*[l]{% H^2(\mathbb{C}P^2)\ar[r]^<<{i^* :\cong}\ar@{=}[d] & H^2(\mathbb{C}P^1)\ar@{=}[d]^{let} \\ <"a"> \ar[r]& \ar[l] \\ }\] Poincar\'{e} Duality :$ =H^2(\mathbb{C}P^2)\overset{\zeta\cap :\cong} {\longrightarrow}H_2(\mathbb{C}P^2)\\ \Rightarrow \zeta {\cap}a$ genererates $H_2(\mathbb{C}P^2)\\ \Rightarrow \pm 1=<\zeta {\cap}a ,a>=<\zeta,a\cup a>=<\zeta,a^2>\\ \Rightarrow a^2$ generates $H^4(\mathbb{C}P^2)$\\ Similarly let $M=\mathbb{C}P^3$, \\ then $H^2(\mathbb{C}P^3)=$ and $H^4(\mathbb{C}P^3)\cong H^4(\mathbb{C}P^2)==\mathbb{Z}$\\ Poincar\'{e} Duality : $H^4(\mathbb{C}P^3) \overset{\zeta\cap:\cong}{\longrightarrow}H_2(\mathbb{C}P^3)\\ \Rightarrow \zeta\cap a^2$ generates $H_2(\mathbb{C}P^3)\\ \Rightarrow \pm 1=<\zeta\cap a^2,a>=<\zeta,a^3>\\ \Rightarrow a^3$ is a generator of $H^6.\\ \hspace*{5em}\vdots\\ $Inductively we can conclude as follows.\\ {\bf 정리. } $H^*(\mathbb{C}P^n)$ is a ring generated by $a\in H^2(\mathbb{C}P^n)$ with $a^{n+1}=0,$ i.e., truncated polynomial ring( or algebra) generated by $a$ of degree 2 with height $n+1$.\\ \textbf{숙제 24.} Show the same thing for $H^*(\mathbb{H}P^n,\mathbb{Z})$ and $H^*(\mathbb{R}P^n,\mathbb{Z}/2). (n=\infty$ 일 때도 마찬가지)\\ \textbf{8. 따름정리 } Suppose $f :P^n\rightarrow P^m\hspace{0.5em},n>m.$ Then $f^*_q =0$ for $q>0$.\\ \begin{pf} $f^* :H^*(P^m)\rightarrow H^*(P^n)$ (Use $\mathbb{Z}/2$-coefficient for $\mathbb{R}P^n)$ is a ring homomorphism, where $H^*(P^m)=$ and $H^*(P^n)=.\\ \Rightarrow 0 = f^*(a^{m+1})=(f^*(a))^{m+1}\\ \Rightarrow f^*(a) =rb =0\\ \Rightarrow f^*\equiv0$ except at 0-dimension.\end{pf}\\ {\bf 따름정리 } $P^m$ is not a retract of $P^n$ if $n>m.$\\ \begin{pf} If $P^m \overset{i}{\underset{r}{\rightleftarrows}}P^n$ with $r\cdot i=id ,$ then $i^*\cdot r^*=id$ and $r^*$ is 1-1. Then this is a contradiction.($\because r^*=0)$\end{pf}\\ \textbf{9. (Borsuk-Ulam)}\\ (1) $n>m\geq 0, \nexists $ anti-pode preserving map $g$ s.t. $g(-x)=-g(x),\hspace{0.5em}\forall x\in S^n.\\ (2)n\geq k , \hspace{0.5em} f:S^n\rightarrow\mathbb{R}^k \Rightarrow \exists x\in S^n$ s.t. $f(x)=f(-x)$\\ \begin{pf} (1) : Suppose $g$ is anti-pode preserving. $\Rightarrow$ \[\xymatrix @*[l]{% S^n \ar[r]^{g}\ar[d]^{p}& S^m \ar[d]^{p}\\ P^n\ar@{.>}[ur]^{\tilde{f}} \ar@{.>}[r]^{f}&P^m}\] Claim. $f_* :\pi_1(P^n)\rightarrow \pi_1(P^m)$ is 0 : $m=1 : f_* :\mathbb{Z}_2\rightarrow \mathbb{Z}$ : clear\\ $\hspace*{15.7em}m>1 : f^*=0 (*\neq 0)$\\ \[\xymatrix @*[l]{% \mathbb{Z}/2 \ar@{=}[r]&\pi_1(P^n) \ar[rrr]^{f_*}\ar[d]^{\chi:\cong} & & &\pi_1(P^m)\ar[d]^{\chi:\cong}\ar@{=}[r]&\mathbb{Z}/2\\ \mathbb{Z}/2\ar@{=}[r] & H_1(P^n)\ar[rrr]^{f_*=H_1(f)} &&& H_1(P^m)\ar@{=}[r]&\mathbb{Z}/2}\] \hspace{0em}Since $f^*=0$ on $H^1 \Rightarrow f_*=0$ on $H_1$ and hence on $\pi_1$.\\ $\therefore \exists \widetilde{f}$: a lifting of $f \Rightarrow \widetilde{f}\cdot p$ and $ g $: two liftings of $f\cdot p\\ \Rightarrow \widetilde{f}\cdot p =g$. But $\widetilde{f}\cdot p\neq g$ since $g$ is 1 to 1 while $\widetilde{f}\cdot p$ is 2 to 1 on each fiber $p^{-1}(x)$.\\ (2) Suppose not. Then let $g(x) = \frac{f(x)-f(-x)}{|f(x)-f(-x)|}$ and apply (1).\end{pf}\\ {\bf 따름정리 } [\textbf{Ham sandwich Theorem}]\\ Let $A_1 , \cdots , A_n$ be bounded measurable subsets of $\mathbb{R}^n.$ Then $\exists$ a hyperplane that bisects each of $A_i$.\\[2mm] \begin{pf} Let $N$ be the north pole of $S^n\subset\mathbb{R}^{n+1}$. Then each $x\in S^n$ determines a unique hyperplane in $\mathbb{R}^{n+1}$ passing through $N$ and perpendicular to $x$. This hyperplane divides $S^n$ into two parts and consider the part containing $x$. The image of this part under stereographic projection is a half space of $\mathbb{R}^n$ and denote it by $H_x$. \\ Let $f_i(x)$ be the measure of $A_i\cap H_x$. Then note that $f_i(-x)$ is the measure of $A_i\cap H_x^c$. Now let $f : S^n\rightarrow \mathbb{R}^n$ be given by $x\mapsto (f_1(x),\cdots, f_n(x))$.\\ Apply Bolsuk-Ulam theorem to get $x\in S^n$ s.t. $f(x)=f(-x).$\end{pf}\\ \textbf{10. } ANR (Absolute Neighborhood Retract) and AR (Absolute Retract)\\ $X^{normal}$ is AR if \[\xymatrix @*[l]{% \forall Y^{normal}\ar@{.>}[rr]^{\exists \bar{f} :extension \hspace{0.2em} of \hspace{0.2em} f} && X\\ B^{closed}\ar[u]^{\cup}\ar[urr]^{f}}\]e.g. Tietz extension theorem$\Rightarrow I^n$: AR.\\ $X^{normal}$ is ANR if \[\xymatrix @*[l]{% \forall Y^{normal} &\ar[l]^{\supset}\exists U^{open}\ar@{.>}[rr]^{\exists \bar{f} :extension \hspace{0.2em} of \hspace{0.2em} f}&& X\\ B^{closed}\ar[u]^{\cup}\ar[urrr]^{f}\ar[ur]^{\subset}}\] e.g. $S^n$ : ANR. (Consider $S^n\subset D^{n+1}$: AR and let $U= \bar{f}^{-1}(D-0)$)\\ \textbf{Theorem A.} Every paracompact manifold is an ANR.\\ Reference: 책 26장 17.6 for compact case.\\ \hspace*{5em} 2.7 of Munkres, Elementary Differential Topology.\\ \textbf{Theorem B.} Every paracompact $n$-manifold can be embedded in $\mathbb{R}^{2n+1}$ as a closed subset.\\ Reference: p.315 of Munkres, a first course Topology or also the above book of Munkres.\\ {\bf 따름정리 } $M$: a compact manifold $\Rightarrow H_*(M)$ is finitely generated.\\[2mm] \begin{pf} Thm B. $\Rightarrow M\subset \mathbb{R}^N$\\ Thm A. $\Rightarrow \exists U$ a neighborhood of $M$ in $\mathbb{R}^N$ and $\exists$ a retraction $r:U\rightarrow M$.\\ Choose a finite simplicial complex $K$ s.t. $M\subset |K|\subset U$ so that $r| : |K|\rightarrow M$ gives a retraction.\\ Know $H_*(|K|)$ is finitely generated (using simplicial homology or CW-homology) and $H_*(M)\overset{i_*}{\rightarrow}H_*(|K|)\overset{r_*}{\rightarrow}H_*(M)$ and $r_*\cdot i_* = id$\\ $\Rightarrow r_*$ is onto. $\Rightarrow H_*(M)$ is finitely generated.\end{pf} \end{document}