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\begin{document}
\parindent=0cm
\section*{VII.3 The \ku \ for cohomology}

16. (\ku \ for cohomology)\\
(1) \Ak\\
If $\cc^*$(or $\dd^*$) is free cochain complex over PID $R$, then
there exists a natural s.e.s \\[2mm]
$ \displaystyle{0 \to \bigoplus_{p+q=n} H^p(\cc^*)\otimes
H^q(\dd^*) \to H^n(\cc^*\ten\dd^*)\to
\bigoplus_{p+q=n+1} H^p(\cc^*) * H^q(\dd^*) \to 0}$ \\
\hspace*{6em}$\{a\}\ten\{b\}\,\,\ \mapsto \,\,\,\{a\ten b\}$\\[2mm]
 which splits if both are free.\\
(This follows from \ak \ since $\cc^*$ can be viewed as a chain
complex by letting $C^n=C_{-n}$ and homology of this is the
cohomology of $\cc^*$, i.e., $H_{-n}(\cc_{-*})=H^n(\cc^*)$.)\\

(2) Note. \\
(i) $\cc$ free $\nRightarrow \cc^*$ free in general.\\
\hspace*{3ex}($\Hom(\oplus A_\alp, R)=\prod\Hom(A_\alp,R)$)\\
(ii) $S^*(\xty)\underset{EZ}{\simeq} (S(X)\ten S(Y))^* \ncong
S(X)^*\ten S(Y)^*$\\

We get around these difficulties by finite approximation of
$\cc$.\\

(iii) $\del $ in $\cc^*\ten \dd^*$ is defined as before:
$$\del |_{C^p\ten D^{n-p}}=\del \ten 1 +(-1)^p(1\ten \del)$$

(3) $\alp \overset{D}{\simeq} \alp': \cc\to \dd$, chain homotopic
$\Rightarrow \alp\ten 1\overset{D\ten 1}{\simeq} \alp'\ten 1
:\cc\ten\ac\to \dd\ten \ac$ for all chain complex $\ac$\\
(Exercise. Check this)\\

(4) $\cc\overset{\alp}{\simeq}\dd\Rightarrow
\cc^*\overset{\widetilde{\alp}}{\simeq}\dd^*$ and
$\cc\ten\ac\overset{\alp\ten 1}{\simeq} \dd\ten \ac$.\\

(5) $\cc$(or $\dd$) : free and finite type, i.e., each $C_p$ is
finitely generated. Then\\[3mm]
\hspace*{10em}$\cc^*\ten \dd^*\cong (\cc\ten \dd)^*$\\
\hspace*{10em}$\alp\ten \bet\mapsto "\alp\times\bet"$, where
$\alp\times\bet (c\ten d)=\alp(c)\bet(d)$\\


\begin{pf}
Show for $C$, $D$ : $R$-module first. \\
If $D=R$, $ C^*\ten R^*=C^*\ten R=C^*=(C\ten R)^*$.\\
If $D$ is finitely generated and free, $D=\displaystyle{\bigoplus_{\textrm{finite}}} R$\\
$\Rightarrow C^*\ten D^*=C^*\ten (\bigoplus R)^*=C^*\ten
(\bigoplus R^*)=\bigoplus C^*\ten  R^*=\bigoplus (C \ten
R)^*$\\[2mm]
\hspace*{5.3em}$=(\bigoplus C \ten  R)^*= (C \ten \bigoplus
R)^*=(C\ten D)^*$\\

Show for chain complex :\\
$(\cc^*\ten \dd^*)_n=\displaystyle{\bigoplus_{p+q=n}}(C^p\ten
D^q)\cong \displaystyle{\bigoplus_{p+q=n}}(C_p\ten D_q)^*
=(\displaystyle{\bigoplus_{p+q=n}}C_p\ten D_q)^* $\\[2mm]
\hspace*{5em}$=((\cc\ten \dd)_n)^*=(\cc\ten
\dd)^n=((\cc\ten\dd)^*)_n$
\end{pf}\\

{\bf Remark.} (i) In general, \begin{center} $\cc^*\ten \dd^*\to (\cc\ten\dd)^*$\\
$ a\ten b\mapsto a\times b$ \end{center} is a chain map and $\del
(a\times b )=\del a \times b +(-1)^p a\times \del b$, $a\in
C^p$.\\
$\because$\raisebox{-2.8em}{$
\begin{array}{rcl}
\del(a\times b)(x\ten y)&=&a\times b (\bd (x\ten y))\\
&=&a \times b (\bd x\ten y +(-1)^{|x|}x\ten \bd y)\\
&=&a(\bd x)b(y)+(-1)^{|x|}a(x)b(\bd y)\\
&=&\del a(x) b(y) +(-1)^{|x|}a(x)\del b(y)\\
&=&(\del a\times b +(-1)^{|x|}a\times \del b)(x\ten y)
\end{array}$}\\
and $(-1)^{|x|}a\times \del b(x\ten y)$ is nonzero only when
$|x|=p$.\\
Now \[ \xymatrix {a\ten b \ar@{|->}[r] \ar@{|->}[d]^\del \ar@{}[dr]|{\displaystyle{\circlearrowleft}} & a\times b \ar@{|->}[d]^\del\\
\del a \ten b +(-1)^p a\ten \del b \ar@{|->}[r] & \del a \times b
+(-1)^p a\times \del b}\]\\

(ii) $\alp, \bet$ : cocycles $\Rightarrow \alp\times
\del\gam=\pm\del(\alp\times\gam)$ and similarly $\del
\gam\times\bet=\pm\del(\gam\times\bet)$.\\
Therefore the cohomology cross product \[ \xymatrix @R=0ex
{H^*(\cc)\ten H^*(\dd)\ar[r]^-\times
&H^*(\cc\ten\dd)\\
\{a\}\ten\{b\}=\alp\ten \bet  \ar@{|->}[r] &
\alp\times\bet=\{a\times b\} }\] is well-defined.\\


(6) (finite approximation) $\cc$ : free over PID. $H(\cc)$ :
finite type.\\
$\Rightarrow \exists \ \overline{\cc}$ : free and finite type such
that $H(\overline{\cc})\overset{\cong}{\to} H(\cc)$.\\
\begin{pf}
Consider the exact sequences \\
\hspace*{7em} \xymatrix{0 \ar[r] & B \ar[r] &Z \ar[r] & H(\cc)
\ar[r] &0} \\
\hspace*{7em}\xymatrix{0 \ar[r] & Z \ar[r] &C \ar[r] &
\overline{B} \ar[r] &0}, \hspace{3ex}
$\overline{B}_p=B_{p-1}$.\\[2mm]
Since $H(\cc)$ is of finite type, there exist finitely generated
$Z'$ and $B'=ker p$ such that
\[\xymatrix @R=1.3em {0 \ar[r] & B' \com \ar[d] \ar[r] &Z' \com \ar[r]^p \ar[d] &
H(\cc)\ar@{=}[d] \ar[r] &0\\ 0 \ar[r] & B \ar[r] &Z \ar[r]^p &
H(\cc) \ar[r] &0}\]

and let $\overline{C}=Z'\oplus \overline{B}'$ and
$\phi=\begin{pmatrix}*& 0\\0&*\end{pmatrix}$. Then
\[\xymatrix @R=1.3em {0 \ar[r] & Z' \com \ar@{.>}[r] \ar[d] &\overline{C} \com \ar@{.>}[r] \ar@{.>}[d] &
\overline{B}' \ar[r]\ar[d] &0\\ 0 \ar[r] & Z \ar[r]  &C \ar[r]^\bd
& \overline{B} \ar[r]&0}\]

$\Rightarrow \phi : \overline{\cc}\to \cc $ is a chain map and
$\phi_*: H(\overline{\cc})\cong H(\cc)$ from the construction.
\end{pf}\\

(7) Let $\cc$(or $\dd$) be free over PID and $H(\cc)$(or $H(\dd)$)
be of finite type. \\
$\Rightarrow $ There exists a finite approximation
$\overline{\cc}$(or $\overline{\dd}$) : free\\
$\Rightarrow \overline{\cc}^*$(or $\overline{\dd}^*$) : free \\
By (1), $$ \displaystyle{0 \to \bigoplus_{p+q=n}
H^p(\overline{\cc}^*)\otimes H^q(\dd^*) \to
H^n(\overline{\cc}^*\ten\dd^*)\to \bigoplus_{p+q=n+1}
H^p(\overline{\cc}^*) * H^q(\dd^*) \to 0}$$

Since $\overline{\cc}$ and $\cc$ has the same homology,
$\overline{\cc}\simeq\cc$.\footnote{see earlier theorem about
mapping cone construction in the chapter of cohomology(IV.1) or
Munkres, Elements of
algebraic topology, p.279, theorem 46.2}\\
Therefore by (4) and (5), \ak \ for cohomology is \\

{\it If $\cc$(or $\dd$) is free chain complex over PID $R$ and
$H(\cc)$ (or $H(\dd)$) is of finite type, then
there exists a natural s.e.s \\[2mm]
$ \displaystyle{0 \to \bigoplus_{p+q=n} H^p(\cc^*)\otimes
H^q(\dd^*) \to H^n((\cc\ten\dd)^*)\to
\bigoplus_{p+q=n+1} H^p(\cc^*) * H^q(\dd^*) \to 0}$ \\[2mm]
which splits if $\cc$ and $\dd$ are free and $H(\cc)$ and $H(\dd)$ are of finite type.}\\


(8) Let $C=S(X,A)$ and $D=S(Y,B)$.\\
Then $$S(X,A)\ten S(Y,B) \underset{\textrm{rel. EZ}}{\simeq}
S(\xty)/S(X\times B\cup A\times Y) =: S((X,A)\times (Y,B)).$$ \\
By (7), we have relative \ku \ for cohomology over PID:
\begin{list}{}{
\setlength{\leftmargin}{-1.5cm}}
\item
$$ \displaystyle{0 \to \bigoplus_{p+q=n} H^p(X,A)\otimes H^q(Y,B)
\overset{\times}{\to} H^n((X,A)\times(Y,B))\to \bigoplus_{p+q=n+1}
H^p(X,A)* H^q(Y,B) \to 0}$$
\end{list}
natural if $H(X,A)$(or $H(Y,B)$) is of finite type, and splits if
both are of finite type.\footnote{여기서 실제로는 natural에 대한
가정은 반드시 필요한 가정이지만, split에 대한 가정은 생략하여도
무방하다.}\\

17. (Cross and Cup product)\\
(1) \xymatrix{S(\xty)\ar[r]^-\phi_-\simeq & S(X)\ten S(Y)} any
chain
homotopy equivalence.\\
$\Rightarrow$ \xymatrix @R=0ex {S^*(\xty)& (S(X)\ten S(Y))^*
\ar[l]_-{\widetilde{\phi}}^-\simeq & S^*(X)\ten S^*(Y) \ar[l]_-r\\
a\times b & r(a\ten b)="a\times b" \ar@{|->}[l] & a\ten b
\ar@{|->}[l]}\\

Let $$
\begin{array}{cc}
   \zeta=\{x\}, x\in Z_p(X) & \alp=\{a\}, a\in Z^r(X) \\
  \eta=\{y\}, y\in Z_q(Y) & \beta=\{b\}, b\in Z^s(Y).
\end{array}$$
Then $\langle\zeta\times\eta, \alp \times \bet
\rangle=\langle\zeta,\alp \rangle\langle\eta,\bet\rangle$.\\
\begin{pf}
$$ \begin{array}{rcl}
\langle\zeta\times\eta, \alp \times \bet \rangle&=&\langle
\phi_*^{-1}\{x\ten y\}, \phi^*\{r(a\ten b)\} \rangle\\ &=& \langle
\{x \ten y\},\{r(a\ten b)\} \rangle\\&=&\langle x\ten y, r(a\ten b)\rangle\\&=& a(x)b(y)\\
&=& \langle x,a \rangle \langle y,b \rangle\\&=&\langle\zeta,\alp
\rangle\langle\eta,\bet\rangle
\end{array}$$
\end{pf}

(2) Take $\phi=A$ : AW-diagonal approximation\\
\[\xymatrix{X &\xty\ar[r]^-{p_2} \ar[l]_-{p_1} & Y}\]
Then $\alp\times \bet =p_1^*\alp \cup p_2^*\bet$.\\
In fact, this holds on chain level : $a\times b =p_1^\sharp a \cup
p_2^\sharp b$.\\
\begin{pf}
For any $\ome\in S_n(\xty)$, let $\sig=p_{1\sharp}\ome$ and
$\tau=p_{2\sharp}\ome$.
$$
\begin{array}{rcl}
\langle(\sig,\tau), a \times b \rangle&=&\langle(\sig,\tau),
\widetilde{\phi}(r(a
\times b)) \rangle\\
&=&\langle\phi(\sig,\tau),r(a \times b) \rangle\\
&\overset{\phi=A}{=} & \langle \displaystyle{\sum_{k+l=n}}
\sig\lam_k\ten\tau\rho_l, r(a \times b)\rangle\\
&= &\displaystyle{\sum_{k+l=n}}\langle
\sig\lam_k,a\rangle \langle \tau\rho_l, b\rangle\\
&= & \langle
\sig\lam_p,a\rangle \langle \tau\rho_q, b\rangle \hspace{2em} \textrm{if } a\in S^q(X),\ b\in S^q(Y) \\
&=& \langle p_{1\sharp}(\sig,\tau)\lam_p,a\rangle \langle
p_{2\sharp}(\sig,\tau)\rho_q, b\rangle\\
&=& \langle (\sig,\tau)\lam_p,p_1^\sharp a\rangle \langle
(\sig,\tau)\rho_q,p_2^\sharp b\rangle\\
&=& \langle (\sig,\tau), p_1^\sharp a \cup p_2^\sharp b \rangle
\end{array}
$$
\end{pf}

(3) Let $\Del : X\to X\times X$ be the diagonal map.\\
Then $\Del ^\sharp(a \times b)=a\cup b$. So that
$\Del^*(\alp\times\bet)=\alp\cup \bet$.\\
\begin{pf}
$$
\begin{array}{rcl}
\Del^\sharp(a\times b) &=& \Del^\sharp(p_1^\sharp a \cup
p_2^\sharp b)\\&=& \Del^\sharp p_1^\sharp a \cup \Del^\sharp
p_2^\sharp b\\&=& a\cup b \hspace{3em} (\because \Del^\sharp
p_1^\sharp=( p_1\Del)^\sharp=id)
\end{array}
$$
\end{pf}


(4) $(\alp\cup\bet)\times(\gam\cup\del)
=(-1)^{|\bet||\gam|}(\alp\times\gam)\cup(\bet\times\del)$\\
\begin{pf}
$$
\begin{array}{rcl}
(\alp\cup\bet)\times(\gam\cup\del) &=& p_1^*(\alp\cup\bet)\cup p_2^*(\gam\cup\del)\\
&=& p_1^*\alp\cup p_1^*\bet\cup p_2^*\gam\cup p_2^*\del\\
&=&(-1)^{|\bet||\gam|} p_1^*\alp\cup p_2^*\gam\cup p_1^*\bet\cup p_2^*\del\\
&=&(-1)^{|\bet||\gam|}(\alp\times\gam)\cup(\bet\times\del)\\
\end{array}
$$
\end{pf}

(5) Slant product\\
$c\in (S(X)\ten S(Y))^{p+q}$, $x\in S_p(X)$.\\
Define $c/x\in S^q(Y)$ by the formula $$ \langle y, c/x \rangle
:=\langle x\ten y, c \rangle$$ Then $\del (c/x)=(-1)^p(\del c/x
-c/\bd x)$.\\
\begin{pf}
$$
\begin{array}{rcl}
\langle y, \del (c/x) \rangle &=& \langle \bd y, c/x \rangle\\
&=&\langle x\ten \bd y, c \rangle
\\ &=&(-1)^p \{ \langle \bd(x\ten y), c \rangle -\langle \bd x\ten  y, c
\rangle\}
\\
 &=& (-1)^p \{\langle y, \del c/x \rangle -\langle y, c/\bd x \rangle\}
\end{array}
$$
\end{pf}

This implies that we have a well-defined slant product on homology
:
\begin{center}
$H^{p+q}(\xty)\ten H_p(X)\to H^q(Y)$\\
$\gam\ten \xi \hspace{3ex}\mapsto\hspace{3ex} \gam/\xi$
\end{center}

{\bf 숙제 33.} 다음을 증명하라.\\
(6) $(\xi\times\eta)\cap(\alp\times
\bet)=(-1)^{|\bet|(|\xi|-|\alp|)}
(\xi\cap\alp)\times(\eta\cap\bet)$\\
(Use slant product and $\Del$.)\\

(7)
$\{(\alp\times\bet)\cup\gam\}/\xi=(-1)^{|\bet|(|\alp|-|\xi|)}\bet\cup(\gam/\xi\cap\alp)$\\
(Use AMT as in 13. note.)

\end{document}
