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\begin{document}
\parindent=0cm

\section*{VII.2 \Ak \ and Universal Coefficient theorem for homology}

{\bf {Algebraic K\"{u}nneth formula}}\\

\textbf{9. (Tensor product of chain complexes)}\\
$\cc, \cc'$ : chain complexes $\Rightarrow \cc\otimes\cc'$ :chain
complex :\\

Define $(\cc\otimes\cc')_n := \pon (C_p\otimes C'_{n-p})$ and $\bd
: (\cc\otimes\cc')_n\rightarrow (\cc\otimes\cc')_{n-1}$ by
$\bd(c\otimes c')=\bd c\otimes c' +(-1)^p c\otimes \bd c'$ for
$\forall c\otimes c'\in C_p\otimes C'_{n-p}\\ \Rightarrow \bd^2 =0
$: clear. ($\because (-1)^{p-1}\bd c\otimes \bd c' + (-1)^p\bd
c\otimes \bd c' =0 $)\\

Want to compare $H(\cc\otimes \cc' )$ and $H(\cc)\otimes
H(\cc')$:\\
We have a well-defined canonical homomorphism $i : H(\cc)\otimes
H(\cc') \rightarrow H(\cc\otimes \cc')$\\
given by $H_p(\cc)\otimes H_{n-p}(\cc')\rightarrow H_n(\cc\otimes
\cc').\\
\hspace*{5em}\{z\}\otimes \{z'\}\mapsto\{z\otimes z'\}$\\
Check this is well-defined : (1) $z,z'$: cycles $\Rightarrow
z\otimes z'$ is a cycle. (clear)\\
\hspace*{12em}(2) $(z+\bd c)\otimes z'=z\otimes z' + (\bd c\otimes
z') = z\otimes z' + \bd(c\otimes z')$\\

\textbf{10.} Assume $R$ : PID , $\cc$ : free\\
Show algebraic K\"{u}nneth :\\

Split chain complex $\cc$ into two short
exact sequences as usual :\\
$(1) 0\rightarrow Z \overset{i}{\rightarrow}\cc
\overset{\bd}{\rightarrow}\bar{B}\rightarrow
0\hspace{0.2em}(\bar{B}_p:=B_{p-1})\\ (2) 0\rightarrow
B\overset{j}{\rightarrow} Z\rightarrow H(\cc)\rightarrow 0$\\

Start with tensoring (1) with $\cc'$ to compute $H(\cc\otimes \cc'
).$\\
Note $\bar{B}\subset \cc$ is free and (1) is a splitting s.e.s.\\
(1)$\Rightarrow 0\rightarrow Z\otimes \cc'\overset{i\otimes
1}{\rightarrow}\cc\otimes \cc'\overset{\bd\otimes
1}{\rightarrow}\bar{B}\otimes \cc'\rightarrow 0$ : s.e.s.\\
(i.e., $0\rightarrow Z_p\otimes C'_{n-p}\rightarrow C_p\otimes
C'_{n-p}\rightarrow \bar{B}_p\otimes C'_{n-p}\rightarrow 0$ :
s.e.s.) \\

Here we view $Z$ as a chain complex : $\rightarrow
Z_p\overset{\bd=0}{\rightarrow}Z_{p-1}\overset{\bd=0}{\rightarrow}\cdots$
and similarly for $\bar{B}.$\\

Remark. $f : \cc\rightarrow \cc'$ and $g : \dd\rightarrow \dd'$ :
chain map \\ $\Rightarrow f\otimes g :\cc\otimes\dd \rightarrow
\cc'\otimes
\dd'$ :  chain map\\
\hspace*{4.5em}$c\otimes d\mapsto f(c)\otimes g(d)\\
\hspace*{4em}(\bd c\otimes d+(-1)^p c\otimes \bd d \mapsto
f(\bd c)\otimes g(d) + (-1)^p f(c)\otimes g(\bd d)\\
\hspace*{14em}=\bd f(c)\otimes
g(d)+(-1)^p f(c)\otimes\bd g(d))$\\

Now Snake lemma $\Longrightarrow \\ \cdots\rightarrow H_n(Z\otimes
\cc')\rightarrow H_n(\cc\otimes \cc')\rightarrow
H_n(\bar{B}\otimes \cc')\rightarrow H_{n-1}(Z\otimes
\cc')\rightarrow \cdots$\\

Show $H(Z\otimes \cc')=Z\otimes H(\cc')$ and $H(\bar{B}\otimes
\cc')=\bar{B}\otimes H(\cc')$:\\

$\rightarrow
Z_p\overset{\bd=0}{\rightarrow}Z_{p-1}\overset{\bd=0}{\rightarrow}\cdots
\\
\Rightarrow\rightarrow(Z\otimes \cc')_n\overset{\bd}{\rightarrow}
(Z\otimes \cc')_{n-1}\rightarrow \cdots$ is given by\\ $\cdots
\rightarrow Z_p\otimes C'_{n-p}\overset{"\bd"=(-1)^p
1\otimes\bd}{\longrightarrow}Z_p\otimes C'_{n-p-1}\rightarrow \cdots\\
\hspace*{5em}z\otimes c'\hspace{1em}\mapsto\hspace{1em} (-1)^pz\otimes \bd c'\\
= Z_p\otimes (\cdots \rightarrow C'_{n-p}\overset{\bd '=(-1)^p\bd
}{\longrightarrow}
C'_{n-p-1}\rightarrow \cdots )$\\

$= \left\{\begin{array}{l}Z_p\otimes (1)'_{n-p}\\
Z_p\otimes (2)'_{n-p}\end{array}\right\}$ (Since $Z_p$ is free, $Z_p\otimes
$ preserves s.e.s.)\\

$\Rightarrow H_n(Z\otimes \cc') =
\overset{n}{\underset{p=0}{\oplus}} H(Z_p\otimes \cc'_{n-p}) \cong
\overset{n}{\underset{p=0}{\oplus}} Z_p\otimes H_{n-p}(\cc')=
(Z\otimes H(\cc'))_n\\
\hspace*{3em}\{z\otimes z'\}\hspace{8em}\mapsto
\hspace{8em}z\otimes \{z'\}$\\

Similarly, $H(\bar{B}\otimes \cc') =\bar{B}\otimes H(\cc')\\
\hspace*{4em} \{b\otimes z'\}\leftrightarrow b\otimes \{z'\}\\
\therefore \cdots \overset{\phi_n}{\rightarrow} (Z\otimes
H(\cc'))_n\rightarrow H_n(\cc\otimes \cc')\rightarrow
(\bar{B}\otimes
H(\cc'))_n\overset{\phi_{n-1}}{\rightarrow}\cdots\\
\hspace*{4em}z\otimes\{z'\}\mapsto \{z\otimes
z'\}\hspace{4em}(=(B\otimes H(\cc'))_{n-1})$\\

$\Rightarrow 0\rightarrow cok \phi_n\rightarrow
H_n(\cc\otimes\cc')\rightarrow ker \phi_{n-1}\rightarrow 0$,\\
where
$\phi_n : (B\otimes H(\cc'))_n\rightarrow (Z\otimes H(\cc'))_n$\\

Now, (2)$\Rightarrow 0\rightarrow
B_p\overset{j}{\rightarrow}Z_p\rightarrow H_p(\cc)\rightarrow 0$ :
s.e.s.\\
$\underset{ Z_p:free}{\overset{\otimes
H_{n-p}(\cc')}{\Longrightarrow}}0\rightarrow
Tor(H_p(\cc'),H_{n-p}(\cc'))\rightarrow B_p\otimes
H_{n-p}(\cc')\rightarrow Z_p\otimes H_{n-p}(\cc')\rightarrow
\hspace*{3em}H_p(\cc)\otimes H_{n-p}(\cc')\rightarrow 0$\\

$\therefore cok \phi_n = \pon H_p(\cc)\otimes H_{n-p}(\cc')$ and
$ker \phi_{n-1} =
\overset{n-1}{\underset{p=0}{\oplus}}Tor(H_p(\cc),
H_{n-p-1}(\cc'))$\\

$\therefore 0\rightarrow \pon H_p(\cc)\otimes
H_{n-p}(\cc')\rightarrow H_n(\cc\otimes \cc')\rightarrow
\overset{n-1}{\underset{p=0}{\oplus}}Tor(H_p(\cc),H_{n-p-1}(\cc'))\rightarrow
0\\
\hspace*{5em}\{z\}\otimes\{z'\}\hspace{1em}\mapsto\hspace{1em}\{z\otimes
z'\}$\\
or $0\rightarrow (H(\cc)\otimes H(\cc'))_n\rightarrow
H_n(\cc\otimes \cc')\rightarrow (H(\cc)*H(\cc'))_{n-1}\rightarrow
0$.\\

Sequence splits if $\cc$ and $\cc'$ are free : Since $\bar{B}$ is free,\\
\[\xymatrix @C=2.8em @R=1em@*[l]{%
0 \ar[r]& Z\ar[r]^{i}\ar[d]^{p} & \ar@(ur,ul)[l]^{\exists
\pi}\cc\ar[r]\ar[dl]^{\psi = p\cdot \pi} & \bar{B}\ar[r] & 0 &
\Rightarrow &
\cc\otimes\cc'\ar[r]^<<{\psi\otimes \psi'}&H(\cc)\otimes H(\cc')\\
&H(\cc)&&&&&c\otimes c'\ar@{|->}[r]&\{\pi(c)\}\otimes \{\pi'(c')\}
}\],where again $H(\cc)$ is a trivial chain complex.\\
$\Rightarrow
H(\cc\otimes
\cc')\overset{(\psi\otimes\psi')*}{\longrightarrow}H(H(\cc)\otimes
H(\cc'))= H(\cc)\otimes H(\cc')$\\

Naturality follows as before.\hspace{3em}$\square$\\

\textbf{11. Universal Coefficient Theorem for Homology}\\
In algebraic K\"{u}nneth, let $\cc'_n =\left\{\begin{array}{ll}G&
,n=0\\
0&, otherwise\end{array}\right\}$, where $G
$ is $R$(PID)-mod.\\

Then $H_q(\cc') =\left\{\begin{array}{ll}G& , q=0\\ 0&, q\neq
0\end{array}\right\}$ and $(\cc\otimes \cc')_n = C_n\otimes G$. Hence we have\\

\textbf{ U.C.T.} : If $\cc$ is free, then $0\rightarrow
H_n(\cc)\otimes G\rightarrow H_n(\cc\otimes G)\rightarrow
Tor(H_{n-1}(\cc),G)\rightarrow 0$ which splits (not canonical),
where $H_n(\cc\otimes G)=H_n(\cc;G)$ homology with coefficient
$G$.\\

Note. For a proof for splitting, see the proof of algebraic
K\"{u}nneth :
\[\xymatrix @C=2em @R=1em@*[l]{%
0 \ar[r]& Z'\ar[r]^{i}\ar[d]^{p} & \cc'\ar[r]\ar[dl]^{Want
\hspace{0.2em}\psi = p\cdot
\pi} & \bar{B'}\ar[r] & 0 \\
&H(\cc') }\] but $H_p(\cc')=Z'_p=0$ if $p\neq 0$ and for $p=0
, Z'_0=\cc'_0=H_0(\cc')=G$.\\

\textbf{Note 1.} In particular, if $\cc$ is a chain complex of
free abelian group and $R,\hspace{0.2em}\forall$ commutative ring
with 1 (so abelian group), then $0\rightarrow H_n(\cc)\otimes
R\rightarrow H_n(\cc\otimes R)\rightarrow
H_{n-1}(\cc)*R\rightarrow 0$, where $\otimes =
\underset{\mathbb{Z}}{\otimes}$, i.e., abelian group tensor
product and $* =\underset{\mathbb{Z}}{*}$,i.e., abelian group
torsion product and hence if $\cc=S_n(X)$ (or $S_n(X,A)$), then
$0\rightarrow H_n(X;\mathbb{Z})\otimes R\rightarrow
H_n(X;R)\rightarrow Tor(H_{n-1}(X;\mathbb{Z}),R)\rightarrow 0$\\

\textbf{Note 2.} $A$: abelian group and $R$ : commutative ring
with 1.\\ $\Rightarrow A\otimes R$ has a canonical $R$-module
structure given by $r(a\otimes x)=a\otimes (rx)$\\
$R$-module $S_n(X;R)$ defined earlier is exactly $S_n(X)\otimes
R$.\\

\textbf{���� 28.} Compute $H(P^n;\mathbb{Z}/2)$ using
$H(P^n;\mathbb{Z})$ and compare. (Use U.C.T.)\\

\textbf{���� 29.} $R$ : PID $\Rightarrow \chi(X)=\chi(X;R)$ (Use
U.C.T.)\\

\textbf{Note 3.} If $R$ is a field, then every $R$-module is free.
Hence $H(\cc)\otimes H(\cc')\cong H(\cc\otimes\cc')$ (or
$H_n(\cc\otimes\cc')\cong\pon H_p(\cc)\otimes H_{n-p}(\cc'))$ and
$H_n(\cc)\otimes G\cong H_n(\cc\otimes G)\hspace{1em},\forall$
vector space $G$.\\



{\bf Note}\,\,(4) Let $R$ be a field with ch($R$)=0.

$\Rightarrow H_{n}(X;\zb) \bigotimes R \cong H_{n}(X;R)$, since
$R$ is a torsion free abelian group and hence
$\ttx(H_{n-1}(X;\zb),R)=0$.\\


{\bf Proposition}\,\, Let $B$ be a torsion free abelian group.
Then $B \ast A=0$.

\begin{pf3}

  Note that any $R$-module is a direct limit of its finitely
generated submodule. And finitely generated torsion free is free
if $R$ is a P.I.D.(Structure theorem).

Now tensor product commutes with direct limit(easy exercise).

$\Rightarrow \ast$ commutes with direct limit.

$\Rightarrow \ttx(C,A) = 0$ if $C$ is a torsion free $R$-module with $R$: P.I.D.\\

\end{pf3}

{\bf 12. Eilenberg-Zilber Theorem}

$S(X\times Y)$ and $S(X) \bigotimes S(Y)$ are naturally chain
homotopy equivalent.

Hence $H(X \times Y) \cong H(S(X) \bigotimes S(Y))$.


\begin{pf2}
\,Use Acyclic Model Theorem. Recall AMT.

\vspace*{1.0em} \framebox{\parbox[b]{14cm}{
\textbf{Acyclic Model Theorem}\\
Let $F, F' :\mathcal{T}\rightarrow \mathcal{C}$ be functors and
$\mathcal{M}\subset Ob(\mathcal{T}),$ where $\mathcal{C}$ is a
category of chain complexes.

Suppose

\,\,(1) $F'$ is acyclic relative to $\mathcal{M}$, i.e., $F'(M)$
is acyclic $\forall M \in \mathcal{M}$.

\,\,(2) $F$ is free relative to $\mathcal{M}$, i.e., $\forall p,
\exists \textrm{indexed family} \{M_{\alp}\}_{\alp \in J_{p}}
\textrm{and} \{i_{\alp}\}_{\alp \in J_{p}}, M_{\alp} \in
\mathcal{M}, i_{\alp} \in F_{p}(M_{\alp})$ such that the indexed
family $\{F(\sigma)i_{\alp}\}_{\alp \in J_{p}}, \sigma \in
\textrm{hom}(M_{\alp},X)$ is a basis for $F_{p}(X)$.

Then

\,\,(1) $\exists $ a natural transformation $\tau :F\rightarrow
F'$ which induces a given natural transformation
$\tau_0:H_{0}(F)\rightarrow H_{0}(F')$.

\,\,(2) Given two such natural transformations
$\tau,\tau^{'}:F\rightarrow F'$ with $\tau_{0} = \tau_{0}^{'},
\tau\simeq\tau^{'}$}} \vspace*{1.0em}

{\bf Eilenberg-Zilber Situation}

Let $\mathcal{T}$ be the category of pairs $(X,Y)$ of topological
spaces.

Consider $F : \mathcal{T} \rightarrow \mathcal{C}, F(X,Y) = S(X
\times Y)$ and $F^{'}: \mathcal{T} \rightarrow \mathcal{C},
F^{'}(X,Y)= S(X) \bigotimes S(Y)$.

$\Rightarrow$ These are clearly functors :


\[
\xymatrix @M=1ex @C=1em @R=1em @*[c] {%
(X,Y) \ar[r] \ar[d]_{(f,g)} & S(X \times Y) \ar[d]^{F(f,g)=(f
\times g)_{\sharp}} & \hspace*{1.5em} & S(X) \bigotimes S(Y)
\ar[d]^{F^{'}(f,g) = f_{\sharp} \otimes g_{\sharp} =
\overset{n}{\underset{p=0}{\bigoplus}} f_{\sharp p} \otimes g_{\sharp n-p}}  \\
(X^{'}, Y^{'}) \ar[r] & S(X^{'} \times Y^{'}) & & S(X^{'})
\bigotimes S(Y^{'}) }
\]


Let $\mathcal{M} = \{(\triangle^{p},\triangle^{q}), \, p,q \geq
0\}$, where $\triangle^{p}$ is a standard $p$-simplex.

(1) $F, F^{'}$ are both acyclic relative to $\mathcal{M}$ :


That $F$ is acyclic relative to $\mathcal{M}$ is clear since
$\triangle^{p} \times \triangle^{q}$ is contractible.\\


Consider $F^{'}.$

$H_{n}(S(X)\bigotimes S(Y)) \cong
\overset{n}{\underset{p=0}{\bigoplus}} H_{p}(X) \bigotimes
H_{n-p}(Y) \bigoplus \overset{n-1}{\underset{p=0}{\bigoplus}}
\ttx(H_{p}(X),H_{n-p-1}(Y))$���� $X=\triangle^{p},
Y=\triangle^{q}$��� ����. �׷��� $\widetilde{H}(\triangle^{p}) =
0 = \widetilde{H}(\triangle^{q})$�̰�

$H_{0}(S(X)\bigotimes S(Y)) \cong H_{0}(X) \bigotimes H_{0}(Y)
\cong R \bigotimes R = R$ �̸� $n >0$�� $n$�� ���ؼ��� $H_{n}(S(X) \bigotimes S(Y)) = 0$�̴�.\\


(2) $F$ is free relative to $\mathcal{M}$ :


For each $n$, choose $(\triangle^{n}, \triangle^{n}) \in
\mathcal{M}$ and $d_{n} \in S_{n}(\triangle^{n} \times
\triangle^{n})$, where $d_{n} : \triangle^{n} \rightarrow
\triangle^{n} \times \triangle^{n}$ is the diagonal map $:t
\mapsto (t,t)$.

Now for each $(f,g) \in \textrm{hom}((\triangle^{n},
\triangle^{n}),(X,Y)), \{F(f,g)d_{n}\}$ form a basis for $S_{n}(X
\times Y)$ since $ \xymatrix @=1em @R=2ex @*[c] { %
& Y \\ \forall\sigma : \triangle^{n} \ar[ur]^{\sigma_{Y}} \ar[r]
\ar[dr]_{\sigma_{X}} & X \times Y \ar[u]_{P_{Y}} \ar[d]^{P_{X}} \\
& X }$ can be written uniquely as $\sigma = (\sigma_{X},
\sigma_{Y}) = (\sigma_{X} \times \sigma_{Y})\circ d_{n}$.\\


(3) $F^{'}$ is free relative to $\mathcal{M}$ :

For each $n$, choose $(\triangle^{p}, \triangle^{q}) \in
\mathcal{M}$ with $p+q=n$ and $i_{p} \otimes i_{q} \in
(S(\triangle^{p})\bigotimes S(\triangle^{q}))_{n} =
\underset{i+j=n}{\bigoplus} S_{i}(\triangle^{p}) \bigotimes
S_{j}(\triangle^{q})$, where $i_{p} = id. : \triangle^{p}
\rightarrow \triangle^{p}$.

For each $(\sigma, \tau) \in
\textrm{hom}((\triangle^{p},\triangle^{q}), (X,Y)),$


$\{F^{'}(\sigma,\tau)(i_{p}\otimes i_{q})\}$ form a basis for
$(S(X) \bigotimes S(Y))_{n} (\,\, \textrm{note that}\,\,
F^{'}(\sigma,\tau)(i_{p}\otimes i_{q}) = \sigma_{\sharp} \otimes
\tau_{\sharp}(i_{p}\otimes i_{q}) = \sigma_{\sharp}(i_{p}) \otimes
\tau_{\sharp}(i_{q}) = \sigma \circ i_{p} \otimes \tau \circ i_{q}
= \sigma \otimes \tau$) since $\{\sigma \otimes \tau | \sigma \in
S_{p}(X), \tau \in S_{q}(Y)\}$ form a basis for $S_{p}(X)
\bigotimes S_{q}(Y)$ and hence $\{\sigma \otimes \tau | \sigma \in
S_{p}(X), \tau \in S_{q}(Y),
p+q=n\}$ form a basis for $(S(X) \bigotimes S(Y))_{n}$.\\


{\bf Proof of Eilenberg-Zilber theorem}

Let $\tau_{0} : H_{0}(F) (=H_{0}(X \times Y)) \leftrightarrow
H_{0}(F^{'})(=H_{0}(X) \bigotimes H_{0}(Y))$ be a natural
transformation(isomorphism) determined by path components, i.e.,
if $C, D$ are path-components of $X, Y$, respectively, then $C
\times D$ is a path component of $X \times Y$ and $H_{0}(C \times
D) \overset{\tau_{0}}{\leftrightarrow} H_{0}(C) \bigotimes
H_{0}(D)$.

AMT $\Rightarrow \tau_{0}$ gives rise to a natural chain homotopy
equivalence.

\end{pf2}

{\bf ���� 30} (29.14) Compute $H(\mathbb{P}^{2} \times S^{3}) \neq
H(\mathbb{P}^{3} \times S^{2})$ and show $\pi_{*}(\mathbb{P}^{2}
\times S^{3}) \cong
\pi_{*}(\mathbb{P}^{3} \times S^{2})$.\\


{\bf ���� 31} (29.15) Compare $S^{2} \times S^{4}$ and
$\mathbb{C}P^{3}$. They both have same homology but $\pi_{4}$ are
different.

(If necessary, use the following fibration and the corresponding
long exact sequence of homotopy groups : $S^{1} \to S^{7} \to
\mathbb{C}P^{3} \Rightarrow \cdots \to \pi_{k}(S^{1}) \to
\pi_{k}(S^{7}) \to
\pi_{k}(\mathbb{C}P^{3}) \to \pi_{k-1}(S^{1}) \to \cdots$)\\


{\bf ���� 32} (29.11.2) $\chi(X \times Y) = \chi(X)\chi(Y)$.

(Use the Poincar\'{e} series of $X = f_{X}(t) := \bet_{0} +
\bet_{1}t + \cdots + \bet_{n}t^{n}$ and notice that $\chi(X) = f_{X}(-1)$.)\\


{\bf 13. Homology cross product}

{\scriptsize
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & (H(X) \bigotimes H(Y))_{n} \ar[r]^{i} \ar[dr] & H_{n}(X
\times Y) \ar[r] \ar@2{-}[d] & (H(X)*H(Y))_{n-1} \ar[r] & 0 & &
\{x\}\otimes\{y\} \ar@{|->}[dr]
\ar@{|.>}[r] & \{x\}\times\{y\} \\
&&H_{n}(S(X) \bigotimes S(Y)) \ar[ur] &&&&&\{x\otimes y\}
\ar@{|->}[u] }
\]}


{\bf Note} $ \phi : S(X)\bigotimes S(Y) \rightarrow S(Y)
\bigotimes S(X)$ given by $\phi(x\otimes y) = (-1)^{pq} y \otimes
x$ is a natural chain map.

$\ulcorner \because \phi(\bd(x\otimes y)) = \phi(\bd x \otimes y +
(-1)^{p} x \otimes \bd y) = (-1)^{(p-1)q} y \otimes \bd x +
(-1)^{p}(-1)^{p(q-1)} \bd y \otimes x = \bd ( \phi(x \otimes
y))\lrcorner$\\


$\Rightarrow \xymatrix @M=1ex @C=1em @R=1em @*[c]{%
S(X \times Y) \ar[r]^{T_{\sharp}} \ar[d]^{\simeq}_{f_{1}} & S(Y
\times X) \ar[d]^{\simeq}_{f_{2}} & , & T : X \times Y \rightarrow
Y \times X &
T(x,y) = (y,x) \\
S(X) \bigotimes S(Y) \ar[r]^{\simeq}_{\phi} & S(Y) \bigotimes S(X)
&&&}$

All the maps a natural chain transformations.

By the AMT, $T_{\sharp} \simeq "f_{2}^{-1}" \circ \phi \circ
f_{1}.$
\\ $\Rightarrow T_{*}(\xi \times \eta) = (-1)^{pq}(\eta
\times \xi), \xi \in H_{p}(X), \eta \in H_{q}(Y)$.\\


{\bf 14. Alexander-Whitney diagonal approximation}

We can give a specific chain map( and hence a chain homotopy
equivalence.)\\


$A : S(X \times Y) \rightarrow S(X) \bigotimes S(Y)$ given by $
\forall \ome \in S_{n}(X \times Y), \ome = (\sigma,\tau)$,


$A(\ome) = \overset{n}{\underset{p=0}{\bigoplus}} \sigma\lam_{p}
\otimes \tau \rho_{n-p} \in (S(X)\bigotimes
S(Y))_{n}$.\\


(1) $A$ is natural : only to check the commutativity of the
following diagram.

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
S_{n}(X \times Y) \ar@{}[dr]|{\circlearrowright} \ar[r]^<<{A}
\ar[d]_{(f \times g)_{\sharp}} & (S(X) \bigotimes S(Y))_{n} =
\bigoplus S_{p}(X) \bigotimes S_{n-p}(Y) \ar[d]^{f_{\sharp}
\otimes g_{\sharp} = \oplus
f_{\sharp p} \otimes g_{\sharp n-p}} \\
S_{n}(X' \times Y') \ar[r]^<<{A} & (S(X') \bigotimes S(Y'))_{n} =
\bigoplus S_{p}(X') \bigotimes S_{n-p}(Y') }
\]

Just note $f_{\sharp}(\sigma \lam_{p}) =
(f_{\sharp}\sigma)\lam_{p} \Rightarrow !$.\\


(2) $A$ is a chain map, i.e., $\bd A = A \bd :$

(Proof is essentially same as the proof of derivation property of
cup product.)\\

$\therefore$ Eilenberg-Zilber Theorem $\Rightarrow A_{*} : H_{*}(X
\times Y) \overset{\cong}{\rightarrow} H_{*}(S(X) \bigotimes S(Y))$\\


{\bf 15. Relative K\"{u}nneth formula}

K\"{u}nneth formula for $(X,A) \times (Y,B)$\\


{\bf Lemma \,\,(Relative Eilenberg-Zilber Theorem)}


Suppose $\{X \times B, A \times Y\}$ : excisive couple in $ X
\times Y$. Then

$S(X \times Y) / S(X \times B \cup A \times Y) \simeq S(X)/S(A)
\bigotimes S(Y)/S(B)$ naturally.

Note that $S(X \times Y)/S(X \times B \cup A \times Y) := S((X,A)
\times (Y,B)), S(X)/S(A)=S(X,A)\,\, \textrm{and}\,\,
S(Y)/S(B)=S(Y,B)$.

(Recall $\{A, B\}$ is excisive couple if $S(A) + S(B)
\overset{i}{\underset{\simeq}{\hookrightarrow}} S(A \cup B)$.)\\



\begin{pf2}

Excisive $\Rightarrow S(X \times B) + S(A \times Y)
\overset{\simeq}{\hookrightarrow} S(X \times B \cup A \times Y)$

$\Rightarrow S(X \times Y)/(S(X \times B)+ S(A \times Y))
\overset{\simeq}{\hookrightarrow} S(X \times Y)/S(X \times B \cup
A \times Y)$ naturally.\\


\hspace*{3.0em}$S(A) \bigotimes S(Y)
\underset{\simeq}{\rightarrow} S(A \times Y)$

\hspace*{6.0em} $\cap$ \hspace*{4.0em} $\cap$


EZ $\Rightarrow S(X) \bigotimes S(Y)
\underset{\simeq}{\rightarrow} S(X \times Y)$ naturally.


\hspace*{6.0em} $\cup$ \hspace*{4.0em} $\cup$


\hspace*{3.0em}$S(X) \bigotimes S(B)
\underset{\simeq}{\rightarrow} S(X \times B)$


$\Rightarrow$

\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
S(X \times Y)/(S(X \times B)+S(A \times Y)) \ar[d]_{\simeq} &
\ar[l]^<<<{\simeq} S(X) \bigotimes S(Y) / (S(X)\bigotimes S(B)+
S(A)
\bigotimes S(Y)) \ar[d]^{\cong}_{(*)}\\
S(X \times Y)/S(X \times B \cup A \times Y) & S(X)/S(A) \bigotimes
S(Y)/S(B) }
\]

$(*)$ follows from the general fact that$ (X \bigotimes Y)/(X
\bigotimes B + A \bigotimes Y) \overset{\cong}{\rightarrow} X/A
\bigotimes Y/B$.


Indeed $\cong$ is induced from canonical map $ X \times Y
\rightarrow X/A \bigotimes Y/B$(cf 1.(3)).

\end{pf2}

From the lemma, we have the following relative K\"{u}nneth
formula.

\vspace*{1.0em} \framebox{\parbox[b]{14cm}{ Let $R$ be a P.I.D.
Then

$0 \rightarrow (H(X,A)\bigotimes H(Y,B))_{n} \rightarrow
H_{n}((X,A) \times (Y,B)) \rightarrow (H(X,A) * H(Y,B))_{n-1}
\rightarrow 0$

, a natural short exact sequence which splits.

By definition, $H_{n}((X,A)\times(Y,B)) = H_{n}(X \times Y, X
\times B \cup A\times Y) $}} \vspace*{1.0em}



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