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\begin{document}
\parindent=0cm
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\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.\\

���� ��� $M=\rb^1$�� ��쿡 ������ ���� �־��� 1-cochain $Dx$,
$x\in \rb$�� ������ ����. $$D(x)(e):=H(e(1))-H(e(0))$$ ���⼭
$H\in S^0(\rb)$�� �Լ�
 $H(y)=H_x(y)=\left\{\begin{array}{cc}
  1 & y\geq x \\
  0 & y<x
\end{array}
\right.$�� ���� ������ $0$-cochain�̴�. �׷��� $Dx=\del H$�̴�.
���� $Dx$�� cocycle�̴�. \\
���� $Dx=\del H$�̹Ƿ� $H^1(\rb)$������ $\{Dx\}=0$�̴�. �׷���,
$H$�� compact support�� ���� �����Ƿ� $H_c^1(\rb)$������
$\{Dx\}\neq 0$�̴�. ��ǻ� $D$�� $H_0(\rb)$�� $H^1_c(\rb)$
������ duality isomorphism�� induce�Ѵ�. \\

% ���� Poincar\'{e} duality�� �����ϱ� ���Ͽ�
1. $\hqc(X)$\\
Let $X$ be a space and $J=\{ K^\textrm{compact}\subset X\}$ with
$K\leq K'$ (if $K\subset K')$ be a directed set.\\
Then $\{H^q(X,X-K), i_{K'K}= i^*: H^q(X,X-K)\to H^q(X,X-K')$ where
$i:(X,X-K)\hookrightarrow (X,X-K')\}$ is a direct system.\\

Define $$ \hqc(X)=\dlim \hq(X,X-K)$$

Note. Since homology commutes with direct limit\footnote{������
8.Cor 1.}, if we let $$ S^*_c(X)=\dlim S^*(X,X-K),$$ then
$\hqc(X)$ is
$q$-th homology of $S^*_c(X)$.\\
\underline{Recall} $S^q(X,X-K)$ is a collection of cochains in
$S^q(X)$
which vanish on $S_q(X-K)$. \\



2. $f: X\to Y$�� �־�������, \\ Note that $f(X-K)\nsubseteq Y-f(K)$.\\
If $f$ is \emph{proper}, i.e. $f^{-1}($compact$)=$compact. Then
$\forall L^\textrm{compact}\subset Y$ $$ f: X-f^{-1}(L)\to Y-L.$$
So $f$ induces
$$f^*:\hq(Y-L) \to \hq(X-f^{-1}(L))\to \hqc(X)$$
And $f^*$ induces
$$ "f^*": \hqc(Y)=\dlim[L] \hq(Y,Y-L)\to \hqc(X)$$\\



3. Let $K$ be a compact subset of an $R$-orientable manifold $M$. \\
Recall $H_n(M,M-K)\cong \Gam K (\cong R^k, k=$number of components
of $K$.)\\
Let $\zeta_K\in\Gam K$ be a restriction of an orientation of
$M$(as a section) on $K$.
\begin{center}
$\zeta_K $: "fundamental class of $H_n(M,M-K)$"
\end{center} (If
$M$ is compact $\zeta_K=i^*(\zeta_M), \zeta_M$: fundamental
class of $M$)\\
Consider
\begin{center}
$\zeta_K \cap $: \raisebox{-1.4ex}{\parbox{6cm}{$\hq(M,M-K)\to
\hnq(M)\\ \hspace*{9ex}a \hspace{5ex} \mapsto\ \ \zeta_K\cap a$}}
\end{center}
If $K\subset K'$,
\[
\xymatrix @=2em @C=3.5em @*[c] { %
\hq(M,M-K)\ar[d] \ar[dr]^{\zeta_K \cap}& \\
\hq(M,M-K') \ar@{.>}[d] \ar[r]^{\zeta_{K'} \cap}&\hnq(M)\\
\hqc(M)\ar@{.>}[ur]_{\exists! D }&
 }\]
IV.3 cup and cap product�� 9���� ���Ͽ� �� �ﰢ���� commute�ϰ�
���� $D$�� induce�ȴ�. �� �� $D$�� �ٷ� duality
homomorphism�̴�.\\

4. {\bf ���� [Poincar\'{e} duality]} Let $M$ be an $R$-orientable
$n$-manifold. Then
$$ D: \hqc(M) \to \hnq(M)$$ is an isomorphism.\\

������ �����ϱ� ���Ͽ� ���� ������ Lemma�� �����Ѵ�.\\

{\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 $(U,V)$
\[
\xymatrix @=1em @R=2ex @*[c] { %
H(B) \ar[r]& H(U)\oplus H(V) \ar[r] & H(Y)
 }\]
Lemma�� ������ key diagram���κ��� �������.\\
\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}
%��� label�ֱ�
\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.} �� diagram���� (1),(2)�� commute�ϰ� (3)�� up to
sign���� commute�Ѵ�.\\

 % ���� Claim�� �����ϸ� �� Lemma�� ������ �����µ�,
�� Claim�� �����ϰ� key diagram�� direct limit�� ���ϸ� ������
����.\footnote{���⼭ $Y$�� ������ compact set�� $K\cup L$�� �÷�
��Ÿ���� ������ �� �ִ�. (Exercise)}
\[
\xymatrix @=1em @R=2.5em @*[c] { %
H^{q+1}_c(B) \ar[d]^{D}_{\cong}
\ar@{}[dr]|{\pm\displaystyle{\circlearrowleft}} & \hqc(Y)\ar[l]
\ar[d]^{D} \ar@{}[dr]|{\displaystyle{\circlearrowleft}}&
\hqc(U)\oplus\hqc(V) \ar[l]\ar[d]^{D\oplus D}_{\cong}
\ar@{}[dr]|{\displaystyle{\circlearrowleft}}
 &
 \hqc(B) \ar[l]\ar[d]^{D}_{\cong}   \\
H_{n-q-1}(B)  & \hnq(Y)  \ar[l]& \hnq(U)\oplus\hnq(V) \ar[l] &
\hnq(B)\ar[l]
 }\]
�� diagram���� commute up to sign�̶� 5-lemma�� ������
�����ϹǷ� 5-lemma�� ���Ͽ� ���ϴ� ����� ��´�. \\

���� Claim�� �����ϸ� �ȴ�.\\
$V$-part of (1) Commutes ($U$-part�� ����ϰ�):\\
\[
\xymatrix @=1em @R=1em @C=-.5em @*[c] { %
&\hq(V,V-L)
\ar[dd]\ar@{}[dddr]|{\circlearrowleft\textrm{(Want)}}&&
\hq(Y,Y-K\cap L) \ar[ll]
\ar[dl]_{\cong}^<<{\textrm{excision}}%
\\ \hq(V,V-K\cap L)\ar[dd]\ar[ur] \ar[rr]^<<<<{\cong}_<<<<<{\textrm{excision}}&&
 \hq(B,B-K\cap L) \ar[ul]\ar[dd]&\\
&\hnq(V)\ar@{=}[dl]&&\\
\hnq(V)&&\hnq(B)\ar[ll]\ar[ul]&
 }\]
�� diagram���� ������ ����� ��� commute���� ���� Ȯ���� �� �ִ�.
(����� �Ʒ����� ���ǿ� ���Ͽ�, ������� inclusion�� cap product��
commute�Ѵٴ� ��ǿ� ���Ͽ�
commute�Ѵ�.) ���� (1)�� commute�Ѵ�.\\

$U$-part of (2) commutes ($V$-part�� ��������):\\
\[
\xymatrix @=1em @R=1em @C=-.5em @*[c] { %
\hq(Y,Y-K\cup L) \ar[dd]
\ar@{}[dddr]|{\circlearrowleft\textrm{(Want)}} && \hq(Y,Y-K)
\ar[ll]
\ar[dl]_{\cong}^<<{\textrm{excision}}\ar[dd] \\
 & \hq(U,U-K) \ar[ul]\ar[dd]&\\
\hnq(Y)\ar@{=}[rr]&&\hnq(Y)\\
&\hnq(U)\ar[ur]\ar[ul]&
 }\]
���������� ������ ����� ��� commute���� �� �� �ְ� ���� (2)��
commute�Ѵ�.\\

(3) commutes up to sign :\\
���� diagram�� 2������ ���� $\del$, $\bd$�� �����ϸ� ���ϴ�
����� ��´�. ��ȣ�� $(-1)^q$��ŭ ���̳��� Ȯ���� �� �ִ�.(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(���� 9),
$H_{c}^{q}(U) = \underset{\longrightarrow}{\lim} H_{c}^{q}(U_{i})$
and the corollary of 8(����) implies $H_{n-q}(U) =
\underset{\longrightarrow}{\lim} H_{n-q}(U_{i})$. Then by using
the fact 5(����), 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 ��������} (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.

�켱 poincar\'{e} duality�� ���ؼ� $H^{q}(M) \cong H_{n-q}(M)$
���� �ȴ�. �׸��� the universal coefficient theorem�� ���ؼ�
$H^{q}(M) = \htx(H_{q}(M),\zb) \bigoplus \etx(H_{q-1};\zb)$. ����
$\htx(H_{q}(M),\zb)$�� �����غ���. $H_{q}(M)$�� Hom-functor��
���ϸ� $H_{q}(M)$�� free part�� ���� torsion part�� ��������.
�̰��� $\forall d(\neq 0)$�� ���ؼ� $\zb/d \to \zb$�� 0�ۿ� ����
�����̴�. �ݴ�� $H_{q-1}(M)$�� Ext�� ���ϸ� $H_{q-1}(M)$�� free
part�� �������� torsion part���� ���´�. ����, $\bet_{n-q} =
\bet_{q}$�̰� $T_{n-q} = \etx(H_{q-1}(M),\zb) = T_{q-1}$��
�����Ѵ�.

{\bf Remark} If $M$ is compact, $H_{*}(M)$ is finitely
generated.(���߿� ����)


(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 ��������  }  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 :
\[\xymatrix @*[l]{%
H^2(\mathbb{C}P^2)\ar[r]^<<{i^* :\cong}\ar@{=}[d] &
H^2(\mathbb{C}P^1)\ar@{=}[d]^{let} \\
<"a"> \ar[r]& <a>\ar[l] \\
}\] Poincar\'{e} Duality :$<a>
=H^2(\mathbb{C}P^2)\overset{\zeta\cap :\cong}
{\longrightarrow}H_2(\mathbb{C}P^2)\\
 \Rightarrow \zeta {\cap}a$ 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 $H^2(\mathbb{C}P^3)=<a>$ and
$H^4(\mathbb{C}P^3)\cong H^4(\mathbb{C}P^2)=<a^2>=\mathbb{Z}$\\

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 ����.  } $H^*(\mathbb{C}P^n)$ 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{���� 24.} Show the same thing for
$H^*(\mathbb{H}P^n,\mathbb{Z})$ and
$H^*(\mathbb{R}P^n,\mathbb{Z}/2). (n=\infty$ �� ���� ��������)\\

\textbf{8. �������� } 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 $H^*(P^m)=<a|a^{m+1}=0>$ and $H^*(P^n)=<b|b^{n+1}=0>.\\
\Rightarrow 0 = f^*(a^{m+1})=(f^*(a))^{m+1}\\
\Rightarrow f^*(a) =rb =0\\
\Rightarrow f^*\equiv0$ except at 0-dimension.\end{pf}\\


{\bf ��������  }  $P^m$ 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 ��������  } [\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: å 26��  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 ��������  } $M$: 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}
