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\newtheorem{prop}[thm]{����}
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\begin{document}
\parindent=0cm
\section*{IV.3 Cup and Cap Product}

\textbf{Cup Product}\\
coefficient: $R$, a commutative ring with 1.
\begin{defn} $a\in S^p(X), b\in S^q(X), $ define a cup product of $a$
and $b$, $a\cup b \in S^{p+q}(X)$: \\
Let $\lambda_p : \triangle_p \rightarrow \triangle_{p+q}$ be a
linear map determined by\\
$(e_0, \cdots , e_p)\mapsto (e_0, \cdots, e_p ) <
\triangle_{p+q}\hspace{0.5em}(\lambda_p$ is called "front
$p$-face")\\
$\rho_q : \triangle_q \rightarrow \triangle_{p+q}$ \\
$(e_0, \cdots , e_q)\mapsto (e_p, \cdots, e_{p+q} ) <
\triangle_{p+q}\hspace{0.5em}(\rho_q$ is called "back
$q$-face")\\

{\bf ��}\\
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Notation: $a\in S^p, x\in S_p \Rightarrow a(x) =: (x, a) :$
pairing.\\
This induces a pairing on homology - cohomology level.
\footnote{($\xi, \alp ) = ([x], [a]) = (x,a) \\
(i)(x+\bd y, a) =
 (x,a) + (\bd y, a) = (x, a)(\because (\bd y ,a ) = (y , \delta a) = 0)\\
 (ii)(x, a+\delta b ) = (x, a) + (x, \delta b) = (x,a) (\because (x, \delta b)
 = (\bd x, b)=0$)} \\

Define ($\sigma , a\cup b) := (\sigma\lambda_p, a)(\sigma\rho_q ,
b)$ and extend linearly on $S_{p+q},$ so that $a\cup b \in
S^{p+q}$, where $\sigma$ is a singular $(p+q)$-simplex.\\
(1) $\cup$ is bilinear : clear\\

(2) $\cup$ is associative : $(a\cup b)\cup c = a\cup (b\cup c),
\hspace{1em}a\in S^p, b\in S^q, c\in S^r :\\
(\sigma , (a\cup b)\cup c) = (\sigma\lambda_{p+q}, a\cup
b)(\sigma\rho_r, c) = (\sigma\lambda_{p+q}\lambda_p ,
a)(\sigma\lambda_{p+q}\rho_q , b)(\sigma\rho_r , c)\\
(\sigma , a\cup( b\cup c)) = (\sigma\lambda_p,
a)(\sigma\rho_{q+r},b\cup c) = (\sigma\lambda_p ,
a))(\sigma\rho_{p+q}\lambda_q , b)(\sigma\rho_{q+r}\rho_r , c)$\\
and it is easy to check $\sigma \lambda_{p+q}\rho_q = \sigma
\rho_{q+r}\lambda_q$ and $\sigma\rho_r = \sigma\rho_{q+r}\rho_r$.\\

(3) $1\cup a = a\cup 1 = a,\hspace{1em} 1\in S^0(X)$ with ($x$,1)
=
1, $\forall x\in S_0(X)$\\

$\therefore$ Let $S^* = \underset{p}{\oplus}S^p . $ Then $\cup$ is
defined on $S^*$ by extending linearly, i.e., $a =
\underset{p}{\Sigma}a^p \in S^*, b = \underset{q}{\Sigma}b^q \in
S^* \Rightarrow a\cup b = \underset{p,q}{\Sigma}(a^p \cup b^q ) $
and $\cup$ is an
associative product with 1 on $S^*$ giving a ring structure.(or $R$-algebra structure)\\
\end{defn}

\textbf{2. Derivation property :} $\delta(a\cup b) = \delta a\cup
b +
(-1)^p a\cup \delta b$.\\
\begin{pf} $(\sigma_{p+q+1} , \delta(a\cup b)) = (\bd \sigma ,
a\cup b) = \underset{0}{\overset{p+q+1}{\Sigma}}(-1)^i(\sigma f^i
_{p+q+1}, a\cup b)\hspace{0.5em}$\footnote{$f^i_p :
\triangle_{p-1}\longrightarrow
\triangle_{p} \\
\hspace*{3em}(e_0, \cdots , e_{p-1}) \mapsto (e_0, \cdots ,
\hat{e_i} , \cdots, e_p )$}
\\ $=\underset{0}{\overset{p}{\Sigma}}(-1)^i (\sigma
f^i_{p+q+1}\lambda_p , a)(\sigma f^i_{p+q+1}\rho_q, b) +
\underset{p+1}{\overset{p+q+1}{\Sigma}}(-1)^i(\sigma
f^i_{p+q+1}\lambda_p , a)(\sigma f^i_{p+q+1}\rho_q, b)$\\

Note. $0\leq i \leq p+1 \Rightarrow \sigma f^i_{p+q+1}\lambda_p =
\sigma \lambda_{p+1}f^i_{p+1}\\
\hspace*{3em}0\leq i \leq p \Rightarrow \sigma f^i_{p+q+1}\rho_q =
\sigma
\rho_q$\\

$\Rightarrow \underset{0}{\overset{p}{\Sigma}} = (
\underline{\underset{0}{\overset{p+1}{\Sigma}} (-1)^i \sigma
\lambda_{p+1} f^i_{p+1}} , a) (\sigma \rho_q , b) -
(-1)^{p+1}(\sigma
f^{p+1}_{p+q+1}\lambda_p , a)(\sigma\rho_q , b) \\\hspace*{5em} =\bd(\sigma \lambda_{p+1})\\
\hspace*{2.5em}= (\sigma , \del a\cup b ) + (-1)^p(\sigma\lambda_p
,  a) (\sigma\rho_q, b)$\\

Similarly, $p\leq i \leq p+q+1 \Rightarrow f^i_{p+q+1}\rho_q =
\rho_{q+1}f^{i-p}_{q+1} \\ \hspace*{4.5em} p+1\leq i \leq p+q+1
\Rightarrow
f^i_{p+q+i}\lambda_p = \lambda_p $\\

$\therefore \underset{p+1}{\overset{p+q+1}{\Sigma}}
 = (\sigma \lambda_p , a)
 (\underline{\underset{p}{\overset{p+q+1}{\Sigma}}(-1)^i
 \sigma\rho_{q+1}f^{i-p}_{q+1}} , b) - (-1)^p (\sigma\lambda_p ,
 a)(\sigma\rho_q , b)\\ \hspace*{9em}=
 \underset{0}{\overset{q+1}{\Sigma}} (-1)^{j+p}\sigma
 \rho_{q+1}f^j_{q+1} = (-1)^p\bd(\sigma \rho_{q+1})
  \\ \hspace*{3em} = (-1)^p(\sigma , a\cup \delta b)- (-1)^p
 (\sigma\lambda_p , a)(\sigma\rho_q , b)$\end{pf}\\

 $S^*$ : a ring with respect to $\cup\\
 Z^* = \oplus Z^p $ : subring by derivation property \\
 $B^* = \oplus B^p$ : ideal in $Z^*\\
 \Rightarrow \cup$ is induced on $Z^* / B^* = H^* = \oplus H^p$ :
 a "graded ring"(algebra) with 1.\\
 (Note. 1 is a cocycle since $\delta 1(e) = 1(\bd e )= 1(e_1 -e_0 )
 = 1-1 =0)$\\

\textbf{ 3.} Let $f : X\rightarrow Y \\ \Rightarrow
f^{\sharp}(a\cup b) =
 f^{\sharp}(a)\cup f^{\sharp}(b)$ and hence $f^* (\alp \cup \bet )
 = f^*(\alp)\cup f^*(\bet)\\(f^{\sharp} = \tilde{f_{\sharp}} :
 S^*(X)\leftarrow S^*(Y) $ and $f^* :  H^*(X)\leftarrow H^*(Y))$\\

\begin{pf}$(\sigma , f^{\sharp}(a\cup
 b)) = (f_{\sharp}(\sigma), a\cup b) =
 (f_{\sharp}(\sigma)\lambda_p , a)(f_{\sharp}(\sigma)\rho_q , b) \\
 \hspace*{7em}= (f\cdot\sigma\cdot\lambda_p, a)(f\cdot\sigma\cdot\rho_q, b)=
  (f_{\sharp}(\sigma\cdot\lambda_p), a)(f_{\sharp}(\sigma\cdot\rho_q),b)\\
\hspace*{7em} =
 (\sigma\cdot \lambda_p , f^{\sharp}a)(\sigma\rho_q ,
 f^{\sharp}b)= (\sigma, f^{\sharp}a\cup f^{\sharp}b )$\end{pf}\\

 $\therefore S^* , H^*$ : contravariant functor
 $\mathcal{T}opology
 \rightarrow \mathcal{R}ing (\mathcal{R} - algebra$)\\

\textbf{ 4. (anti-) commutativity}\\
 $\alp \in H^p , \bet \in H^q \Rightarrow \alp \cup \bet =
 (-1)^{pq}\bet\cup \alp$\\ This formula is not true on cochain
 level.( but almost true)\\
 But true in simplicial case : \\
  Let $\theta _p
 :\triangle_p\rightarrow \triangle_p \\ \hspace*{3em}(e_0,
 \cdots, e_p ) \mapsto (e_p , \cdots , e_0 )$\\
 To get the same oriented simplex, we have to multiply $\epsilon_p
 = (-1)^{\frac{p(p+1)}{2}}$ so that $\epsilon_p\theta_p = $ id on
 the simplicial chain complex.\\

 Let $\theta(\sigma ) = \epsilon_p \sigma\cdot \theta_p$ for both simplicial and singular case.\\
 Then $\theta $ is a natural chain map : \\
 (1) Show $\theta \bd =\bd\theta : \\
 \bd\theta(\sigma ) = \epsilon_p
 \bd(\sigma\cdot \theta_p ) = \epsilon_p
 \underset{0}{\overset{p}{\Sigma}}(-1)^i\sigma\theta_p \cdot f^i_p
 = \epsilon_p\Sigma (-1)^i \sigma f^{p-i}_{p}\theta_{p-1} =
 \epsilon_p \underset{0}{\overset{p}{\Sigma}}(-1)^{p-j}\sigma
 f^j_p \theta_{p-1} = (-1)^p\epsilon_p
 \underset{0}{\overset{p}{\Sigma}}(-1)^j\sigma f^j_p\theta_{p-1} =
 (-1)^p\epsilon_p\epsilon_{p-1}\theta(\bd\sigma)=\theta(\bd\sigma).$\\

 (2) Naturality : \\
\[\xymatrix  @C=4em @*[l]{%
 S(X) \ar[r]^{f_{\sharp}} \ar[d]^{\theta_{X}}& S(Y) \ar[d]^{\theta_{Y}}
 & &\theta(f_{\sharp}\sigma)=\epsilon_p f_{\sharp}\sigma\cdot \theta_p=
 \epsilon_p f\cdot\sigma \cdot \theta_p\\
 S(X) \ar[r]^{f_{\sharp}}& S(Y)&&  =\epsilon_p
 f_{\sharp}(\sigma\cdot \theta_p )=
 f_{\sharp}(\epsilon_p\sigma\theta_p)= f_{\sharp}(\theta(\sigma)).
 } \]\\

 Now, $(\sigma , \tilde{\theta} (a\cup b)) := (\theta(\sigma) ,
 a\cup b) = \epsilon_{p+q}(\sigma\theta_{p+q}, a\cup b) \\ \hspace*{8em}=
 \epsilon_{p+q}(\sigma\cdot\theta_{p+q}\lambda_p ,
 a)(\sigma\cdot\theta_{p+q}\rho_q , b) \\ \hspace*{8em}= \epsilon_{p+q}(\sigma\cdot\rho_p\cdot
 \theta_p, a)(\sigma\cdot\lambda_q\cdot\theta_q, b)\\ \hspace*{8em}=\epsilon_{p+q}\epsilon_p\epsilon_q
 (\theta(\sigma\rho_p ), a)(\theta(\sigma\lambda_q), b) \\ \hspace*{8em}=
 (-1)^{pq}(\sigma\rho_q , \tilde{\theta}(a))(\sigma\lambda_p,
 \tilde{\theta}(b))\\ \hspace*{8em} = (-1)^{pq}(\sigma , \tilde{\theta}(b)\cup
 \tilde{\theta}(a) )$\\

 $\therefore \tilde{\theta}(a\cup b) =
 (-1)^{pq}\tilde{\theta}(b)\cup \tilde{\theta}(a)$\\

 In simplicial case, $\theta =$ id $\Rightarrow \tilde{\theta} =$
 id $ \Rightarrow a\cup b = (-1)^{p+q}b\cup a$\\
 In singular case, $\theta$ is not id but $\theta^* =$ id on
 $H^*$, since $\theta : S(X)\rightarrow S(X)$ is a natural chain
 map and hence $\theta \simeq$ id by Acyclic model theorem.\\

 $\therefore \alp\cup \bet = (-1)^{pq}\bet\cup\alp.$\\

 Recall Acyclic model theorem.\\

 \textbf{Acyclic Model Theorem}\\

 Let $\mathcal{T}$ be a category and $\mathcal{C}$ be a category
 of chain complexes and chain maps. Let $\mathcal{S}$ and
 $\mathcal{S'} : \mathcal{T}\rightarrow\mathcal{C}$ be functors
 and $\mathcal{M}\subset Ob(\mathcal{T}).\\
 (1) \mathcal{S'}$ is acyclic relative to $\mathcal{M}$, i.e.,
 $\mathcal{S'}(M)$ is acyclic for $\forall M\in \mathcal{M}.\\
 (2) \mathcal{S}$ is free relative to $\mathcal{M}$, i.e.,
 $\forall p, \exists$ an indexed family $\{M_{\alp}\}_{\alp\in
 J_p}$ and $\{i_{\alp}\}_{\alp\in J_p},\hspace{0.5em}
 M_\alp\in\mathcal{M}, i_\alp\in S_p(M_\alp)$ such that the
 indexed family $\{S(\sigma)i_\alp\}_{\alp\in J_p}, \hspace{0.5em}
 \sigma\in hom(M_\alp, X)$ is a basis for $S_p(X)$.\\

 Then (i) $\exists$ a natural transformation $\tau :
 \mathcal{S}\rightarrow\mathcal{S'}$ which induces a given natural
 transformation $\tau_0 : H_0(\mathcal{S})\rightarrow
 H_0(\mathcal{S'}).$\\
 (ii) Given two each natural transformation $\tau$ and $\tau ',
 \hspace{0.5em} \tau\simeq\tau' . $\\

\psset{unit=2cm}
{\bf 5. Examples}\\
(1) $T= S^1\times S^1$ $\Rightarrow H^2=\zb, H^1=\zb\oplus\zb,
H^o=\zb$\\
\begin{center}
\begin{pspicture}(0,0.3)(4,1.6)%
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% Torus%
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     \rput(1.15,.5){$y$}
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     \rput(.3,.7){$\sig_2$}
    \rput(.7,.3){$\sig_1$}

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\end{center}

$\sig_1$�� $\sig_2$�� �� �׸��� ���� 2-simplex�� ����.(standard
3-simplex�� vertex 0,1,2�� ȭ��ǥ ������ ������ simplex�̴�.)\\
 $\bd(\sig_1-\sig_2)=0$�̹Ƿ�
$\zeta=\{\sig_1-\sig_2\}\in H_2$��
�θ� �̴� $H_2$�� generator�̹Ƿ� $T^2$�� fundamental orientation class�� �ȴ�.\\
$H^1\cong \Hom(H_1,\zb)$�̹Ƿ� $H_1=\zb\oplus\zb= \langle \{ x \},
\{y\} \rangle$�� �ϸ� $( \{x\},\{y\})$�� dual basis $\alp=\{a\},
\bet=\{b\}$�� $H^1$�� basis�� �ȴ�.\\
����,
$$\begin{array}{rcl}
(\zeta,\alp\cup\bet)&=& (\sig_1-\sig_2,a\cup b)\\
&=& (\sig_1,a\cup b)-(\sig_2,a\cup b)\\
&=& (\sig_1\lam_1,a)(\sig_1\rho_1,b)-
(\sig_2\lam_1,a)(\sig_2\rho_1,b) \\
&=& (x,a)(y,b)-(y,a)(x,b) \\
&=& (\{x\},\alp)(\{y\},\bet)-(\{y\},\alp)(\{x\},\bet) \\
&=& 1
\end{array}
$$
�̹Ƿ�, $\alp\cup\bet$�� $H^2$�� generator�̴�.\\

Ring structure of $H^*(T^2)$:\\
This is generated by $\alp$ and $\bet$ with the relation
$\alp\cup\alp=0$, $\bet\cup\bet=0$,
$\alp\cup\bet=-\bet\cup\alp$.\\
���� ������ �� �� �ִ�.
$$H^*(T,\rb)\cong \Lam(\rb^2)$$
(�� ��, cup product�� wedge product�� �����ȴ�.)\\

{\bf ���� 20.} Compute the cohomology ring of $\Sig_2=T\sharp
T$.\\


(2) K=Klein bottle. $\Rightarrow H^2=0, H^1=\zb\oplus\zb/2,
H^o=\zb$\\
Universal coefficient theorem���κ��� $H^*(K;\zb/2): H^2=\zb/2$,
$H^1=\zb/2 \oplus \zb/2$, $H^0=\zb/2$���� �� �� �ִ�.\\

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���� �׸��� ���� $\sig_1$�� $\sig_2$�� �����ϸ� $$\bd
(\sig_1+\sig_2)=2y\equiv0 \ (\textrm{mod } 2)$$ ���� �� �� �ִ�.
���� $\zeta=\{\sig_1+\sig_2\}$�� �ϸ� $H_2$�� generator��
�ȴ�.\\

�ռ��� ���������� $\alp, \bet$�� $\{x\}$, $\{y\}$�� dual�̶� �ϰ�
����� ���� ���� ����� �� �� �ִ�.\\

Ring structure of $H^*(K;\zb/2)$:\\
$H^*(K;\zb/2)$ is a ring generated by $\alp, \bet$ with the
relations : $\alp^2=0, \alp\bet=\bet\alp=\bet^2$.\\

{\bf ���� 21.} Compute the ring structure of $H^*(P^2;\zb/2)$.\\

{\bf 6. Relative cup product}\\
(1) $H^p(X,A)\times H^q(X) \overset{\cup}{\longrightarrow}
H^{p+q}(X,A)$ is well defined:  \\
Note. An element of $H^p(X,A)$ can be represented by a cocyle in
$S^p(X)$ which vanishes on $S_p(A)$ :\\
\[
\xymatrix @=2em @*[c] { %
0&S^{p+1}(A) \ar[l] &S^{p+1}(X)\ar[l]&S^{p+1}(X,A)\ar[l]&0\ar[l]\\
0&S^{p}(A)\ar[l] \ar[u]&S^{p}(X)\ar[l]_{\widetilde{i}}\ar[u]&S^{p}(X,A)\ar[l]_{\widetilde{p}}\ar[u]_{\del}&0\ar[l]\\
}\] %
�� diagram���� $\widetilde{i}$�� restriction map�̴�. ����
$S^p(X,A)$�� cocycle�� $S^p(X)$�� cocycle �� $S_p(A)$��
restrict�ϸ� 0�� �Ǵ� �͵��̴�.\\

Suppose $a\in S^p(X,A)\subset S^p(X)$ (vanishing on $A$). Then
$a\cup b $ also vanishes on $A$ since front $p$-face of $\sig \in
S_{p+q}(A) $ also lies in $S_p(A)$.\\

(2) $H^p(X,A)\times H^q(X,A) \overset{\cup}{\longrightarrow}
H^{p+q}(X,A)$ is a restriction of (1).  \\

{\bf \large Cap product}\\

{\bf 7.} The cap product is a bilinear pairing,
$$ \cap : S_{p+q}(X)\times S^p(X) \longrightarrow S_q(X)$$
given by $(\sig \cap a, b)=(\sig, a \cup b)$, i.e., $\cap a $ is
the adjoint of $a\cup$, or $\sig\cap a =(\sig\lam_p, a)\sig
\rho_q$. \\
($\Rightarrow \cap : S_*(X)\times S^*(X)\to S_*(X) $ : extend
linearly)\\

{\bf 8.} (0) $\sig\cap 1=\sig$\\
(1) $\sig\cap (a \cup b) =(\sig \cap a)\cap b$ \\
(2) $\bd (\sig \cap a)= (-1)^p (\bd\sig\cap a -\sig \cap \del
a)$, $\sig\in S_{p+q}$, $a\in S^p$\\
\begin{pf}
(0) clear from $1\cup b=b$.\\
(1) $((\sig\cap a)\cap b , c)=(\sig\cap a, b \cup c)=(\sig, a\cup
(b\cup c))=(\sig,( a\cup b)\cup c)=(\sig\cap ( a\cup b), c)$\\
(2) $(\bd (\sig\cap a) ,b)=(\sig\cap a , \del b)=(\sig, a\cup \del
b )$\\ $=(-1)^p((\sig, \del(a\cup b))-(\sig,\del a \cup b
))=(-1)^p((\bd\sig\cap a, b)-(\sig\cap\del a , b))$
\end{pf}\\


���� (2)�� ���Ͽ� ������ �� �� �ִ�.\\
Cap product induces a bilinear pairing,
$$ \cap : H_{p+q}(X)\times H^p(X) \longrightarrow H_q(X)$$
($\because$ $\{\sig\} \in H_{p+q}(X)$, $\{a\} \in H^p(X)$�� �ϸ�
(2)�� ���Ͽ� $$\bd (\sig \cap a)=0$$�̹Ƿ� $\sig \cap a$�� cycle��
�ȴ�. ���� (2)���� $$(\bd \tau)\cap a= \pm \bd(\tau\cap a)\pm
\tau\cap\del a=\pm \bd(\tau\cap a)$$�̰� $$\sig \cap\del
b=\pm\bd(\sig\cap b)\pm\bd \sig\cap b=\pm\bd(\sig\cap b)$$�̹Ƿ�
homology������ map�� �� ���ǵȴ�.)
\\


{\bf 9.} $f: X\to Y \Rightarrow f_\sharp(\sig\cap f^\sharp
a)=f_\sharp \sig\cap a$. (Same for $f_*$ and $f^*$)

\begin{pf}
$$\begin{array}{rcl}
(f_\sharp(\sig\cap f^\sharp a),b)&=& (\sig\cap f^\sharp a,f^\sharp b)\\
&=& (\sig, f^\sharp a \cup f^\sharp b)\\
&=& (\sig, f^\sharp (a \cup b)) \\
&=& (f_\sharp \sig, a \cup b)\\
&=& (f_\sharp \sig\cap a , b) \\
\end{array}
$$
\end{pf}

{\bf 10. Relative cap product }\\
$$ \cap : H_{p+q}(X,A)\times H^p(X,A) \longrightarrow H_q(X)(\longrightarrow H_q(X,A)) $$
and \hspace{3em}$ \cap : H_{p+q}(X,A)\times H^p(X)
\longrightarrow H_q(X,A) $\\
are well-defined.\\
\begin{pf}
%�� map���� 8���� ������ map�� restriction���̹Ƿ� �� image��
%������ set�� ��������� ���̸� ������ ������.\\
(1) $z\in Z_{p+q}(X,A)$, $c\in Z^p(X,A)$�� �θ� $$\bd z \in S(A)$$
�̰� \begin{center}$c$ is a cocycle in $Z^p(X)$ which vanishes on
$S_p(A)$.\end{center} ���� �ȴ�. ����
$$\bd(z\cap c)=\pm (\bd z)\cap c\pm z\cap (\del c)=0\footnote{$\because \del c=0$ and $\bd z\in A$}$$
�̹Ƿ� $z\cap c$�� $X$�� cocycle�� �Ǿ� ������ ������.\\

(2) $z\in Z_{p+q}(X,A)$, $c\in Z^p(X)$�� �θ�
$$\bd(z\cap c)=\pm (\bd z)\cap c\pm z\cap (\del c)=\pm (\bd z)\cap c$$
���� $\bd(z\cap c)$�� $A$�� chain�� �Ǿ� $z\cap c$�� relative
cycle�� �ȴ�.
\end{pf}
\end{document}
