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\begin{document}
\parindent=0cm
\section*{VII. Products}
\section*{VII.1 The \ku s and torsion product}

Want a formula relating the homology of $X\times Y$ to the
homology of $X$ and $Y$.
\begin{thm}[The \ku] $R$: PID. There
exists a natural s.e.s
$$ 0 \to \bigoplus_{p=0}^{n} H_p(X)\otimes H_{n-p}(Y) \to H_n(\xty)\to \bigoplus_{p=0}^{n-1} \Tor( H_p(X) ,
H_{n-p-1}(Y)) \to 0$$ which splits (but not naturally).
\end{thm}

This result will be proved in 2 stages : \\
(1) The Eilenberg-Zilber theorem : {\it There exists a natural
chain
homotopy equivalence between $S(\xty)$ and $S(X)\otimes S(Y)$.} \\

(2) \Ak : {\it Let $\cc, \cc'$ be chain complexes over PID $R$.
Suppose at least one of $\cc$ and $\cc'$ is a free chain complex.
Then there exists a natural s.e.s
$$ 0 \to \bigoplus_{p=0}^{n} H_p(\cc)\otimes H_{n-p}(\cc') \to H_n(\cc\otimes\cc')\to
 \bigoplus_{p=0}^{n-1} \Tor(H_p(\cc) ,H_{n-p-1}(\cc')) \to 0$$ which splits (but not naturally) if
 both
$\cc$ and $\cc'$ are free.}\\


\underline{\bf Tensor product}\\

Work in the category of $R$-modules($R$: commutative ring with 1)
and write $A\otimes B$ for
$A\otimes_R B$. \\

1. (1) Let $f: A\to B$, $f': A' \to B'$ be homomorphisms.\\
\hspace*{2em} $\Rightarrow \exists !$ homomorphism  $f\otimes f'
:$ \raisebox{-1.4ex}{\parbox{5cm}{$A\otimes
A' \to B\otimes B'$\\ $ a\otimes a' \mapsto f(a)\otimes f'(a')$}}\\

(2) $A \overset{f}{\to} B \overset{g}{\to} C$, $A'
\overset{f'}{\to} B' \overset{g'}{\to} C'$ homomorphisms \\
\hspace*{2em} $\Rightarrow (g\circ f)\otimes (g'\circ
f')=(g\otimes g')\circ (f\otimes f') : A\otimes A'\to B\otimes
B'\to C\otimes C'$\\

(3) $f : A\to B$, $f': A'\to B'$ : surjective\\
\hspace*{2em} $\Rightarrow f\otimes f'$ is surjective and
$ker(f\otimes f')$ is generated by the elements of the form $a\ten
a'$ with $ a\in ker f$ or $a' \in ker f'$.\\
\begin{pf}
$f\ten f' : A\ten A' \to B\ten B'$. For all $b\ten b'\in B\ten B'
$, there exists $a\in A $, $a'\in A'$ such that $f(a)=b$ and $
f'(a')=b'$. Hence, $f\ten f'$ is surjective. \\
Let $K = \langle a\ten a' | a\in ker f\  \textrm{or} \ a'\in ker
f'
\rangle$. Clearly $K\subset ker f\ten f' $. \\
Consider
\[
\xymatrix @=1.5em @*[c] { %
A\ten A' \ar[rr]^-{f\ten f'}\ar[rd]_-p &\ar@{}[d]|-{\displaystyle{\circlearrowleft}}& B\ten B' \ar[ld]^-{\displaystyle{\exists \varphi}}\\
& A\ten A'/K& }\] and define $\vph(b\ten b' )=(a\ten a')$ where
$f(a)=b$ and $f'(a')=b'$. Then  $\vph$ is well-defined since
$$a\ten a' - a_0\ten a_0' = (a-a_0)\ten a' + a_0\ten (a'-a_0')\in
K$$ Therefore $ker (f\ten f')\subset K$.
\end{pf}\\
In particular, $A\overset{f}{\to} B\to 0 \Rightarrow A\ten G
\overset{f\ten id}{\to} B\ten G \to 0$.\\

2. Exactness of $\ten G$\\
(1) \xymatrix{A \ar[r]^f & B \ar[r]^g &C \ar[r] & 0} : exact \\
 \hspace*{3ex}$\Rightarrow $ \xymatrix{A\ten G \ar[r]^{f\ten id} & B\ten G  \ar[r]^{g\ten id } &C\ten G \ar[r] & 0} : exact \\
(2) \xymatrix{0 \ar[r]&A \ar[r]^f & B \ar[r]^g &C \ar[r] & 0} : split exact \\
 \hspace*{3ex}$\Rightarrow $ \xymatrix{0 \ar[r]&A\ten G \ar[r]^{f\ten id} & B\ten G  \ar[r]^{g\ten id } &C\ten G \ar[r] & 0} : split exact \\

\begin{pf}
(1) 1.(3) $\Rightarrow g\ten id$ is onto and $ker(g\ten
id)=\langle b\ten x | b\in ker g \rangle=im(f\ten id)$.\\
(2) There exists $p$ such that $p\circ f=id$. So $$(p\ten id
)\circ(f\ten id)=(p\circ f)\ten id=id.$$ Therefore $f\ten id$ is
1-1 and the sequence is split exact.
\end{pf}\\
\underline{Remark} \raisebox{-1.3em}{\parbox{4cm}{
\begin{center}$A\ten R \cong A \cong R\ten A$\\
$a\ten r \rightarrow ra \leftarrow r\ten a$\\  $a \ten 1
\leftarrow a\rightarrow 1\ten a$ \end{center}}}
\\
e.g. $\zb \ten \zb/2=\zb/2$
\[ (\xymatrix{0\ar[r] &\zb\ar[r]^{\times 2}&\zb\ar[r]&\zb/2\ar[r]&0})\ten \zb/2 \]
\[\Rightarrow  \xymatrix @R=1ex {\zb/2 \ar[r]^{\times 2}_{\textrm{0-map}}&\zb/2 \ar[r]&\zb/2 \ar[r]&0
}\]\\


In general,
\[ \xymatrix{0\ar[r] &\zb\ar[r]^{\times n}&\zb\ar[r]&\zb/n\ar[r]&0}\]
\[\overset{\ten G}{\Rightarrow}  \xymatrix @R=1ex {\zb\ten G \ar@{=}[d]\ar[r]&\zb\ten G \ar@{=}[d]\ar[r]&\zb/n\ten G \ar[r]&0\\
G\ar[r]^{\times n}&G }\] Therefore $\zb/n\ten G\cong G/nG$.\\

Exercise. $\zb/n\ten\zb/m=\zb/d$, $d=(n,m)$\\

Recall : $A\ten B\cong B\ten A$, $(A\ten B)\ten C\cong A\ten
(B\ten
C)$,\\
\hspace*{5ex} $(\oplus A_\alp)\ten B\cong \oplus(A_\alp\ten B)$,
$A\ten (\oplus B_\alp)\cong \oplus (A\ten B_\alp)$\\
Use this to compute $\ten$ of finitely generated abelian groups or
$R$-modules. ($R$: PID)\\

3. If $G$ is free, then $\ten G$ preserves s.e.s.\\
($G$ : free $\Rightarrow G=\oplus R $ and\\[2mm]
\hspace*{5em} \xymatrix @C=2ex {0\ar[r]&A\ar[r]&B} : exact
$\Rightarrow$ \xymatrix @R=1ex @C=2ex {0\ar[r] & A\ten
R\ar[r]\ar@{=}[d] & B\ten R\ar@{=}[d]\\ &
 A&B} : exact \\
\hspace*{14em}$\Rightarrow$  \xymatrix @C=2ex {0\ar[r]&A\ten G
\ar[r]&B\ten G} :
exact) \\


4. (Homology with coefficient $G$) \\
$\cc$ : a chain complex (over $R$) , $G$ : $R$-module.\\
$\Rightarrow \cc\ten G$ : $\cdots \to C_p\ten G \to C_{p-1}\ten
G\to \cdots$ is a chain complex. \\
$\Rightarrow H(C; G):= $ homology of $\cc\ten G$ := homology of
$\cc$ with coefficient $G$.\\

$f: \cc \to \cc'$, a chain map $\Rightarrow f_*: H(\cc;G)\to
H(\cc'; G)$.\\
$\phi: G \to G'$, an $R$-module homomorphism $\Rightarrow \phi_*:
H(\cc;G)\to H(\cc; G')$.\\

If $0\to \cc\to \dd\to\ee\to 0$ is a split s.e.s.(e.g. $\ee$ is
free), by snake lemma there exists a natural l.e.s.,
$$ \cdots \to H_p (\cc;G)\to H_p(\dd;G)\to
H_p(\ee;G)\overset{\bd}{\to} H_{p-1}(\cc;G)\to \cdots.$$\\
If $\cc$ is free and $0\to G'\to G\to G''\to 0$ is a s.e.s., by
snake lemma there exists a natural l.e.s.,
$$ \cdots \to H_p (\cc;G')\to H_p(\cc;G)\to
H_p(\cc;G'')\overset{\bet_*}{\to} H_{p-1}(\cc;G')\to \cdots.$$
In this case, $\beta_*$ is called Bockstein homomorphism.\\

\underline{\bf Torsion product}\\

5. Let $$\cdots \to F_p\overset{\bd}{\to}\cdots\overset{\bd}{\to}
F_1\overset{\bd}{\to}F_0\overset{\eps}{\to}A\to 0$$ be a free
resolution of $A$. \\

Define $\Tor_p(A,G):= H_p(F;G)=H_p(F\ten G)$\\
\hspace*{10ex} = homology of $\cdots \to F_p\ten G {\to}\cdots \to
F_1\ten G{\to} F_0\ten
G{\to}0$.\\

This is well-defined by comparison theorem. ($\ext^p$�� ����
��������)\\

Note.
 (1) $\Tor_0(A,G)\cong A\ten G$ :
\[
\xymatrix @=1.5em @*[c] { %
F_1\ten G \ar[r]^{\bd\ten 1} & F_0\ten G \ar[r]^{\eps\ten 1} &
A\ten G \ar[r] & 0 }\] is exact.
So $\Tor_0(A,G)=F_0\ten G/ im(\bd\ten 1)=F_0\ten G/ ker(\eps\ten 1)=A\ten G$.\\

(2) If $p\geq 1$ and $F$ is free, $\Tor_p(F,G)=0$ and $\Tor_p(G,F)=0$.\\
A free resolution of $F$ is \[
\xymatrix @=1.5em @*[c] { %
\cdots \ar[r] & 0 \ar[r] & F \ar[r]^{id} & F \ar[r] & 0 }\] So if
$p\geq 1$, $C_p=0$ and  $$ \Tor_p(F,G)=0.$$ And since $\ten $free
preserves resolution, $$\Tor_p(G,F)=0$$

(3) If $R$ is a PID, $\Tor_p(A,G)=0$ for $p\geq 2$.\\
A free resolution of $A$ is \[
\xymatrix @=1.5em @R=1ex @*[c] { %
 0 \ar[r] & R\ar[r]\ar@{=}[d] & F \ar[r]^{\eps}
\ar@{=}[d] & A \ar[r] & 0.\\ & ker\eps&free }\] So if $p\geq 2$,
$C_p=0$ and  $$ \Tor_p(A,G)=0.$$
In this case we simply denote $\Tor(A,G)$ or $A*G$ for $\Tor_1(A,G)$.\\

(4) $\Tor_p(A,G)$ is a covariant functor in both variables.\\


6. If $$0\to A\to B\to C\to 0$$ is a short exact sequence, then
there exists a natural long exact sequence,\\[2mm]
\hspace*{1em} $
\begin{array}{l}
  \cdots  \to  \Tor_n(A,G)\to\Tor_n(B,G)\to\Tor_n(C,G) \to \Tor_{n-1}(A,G)\to \\
   \cdots  \to  \Tor_1(C,G)\to A\ten G\to B\ten G\to C\ten G\to 0
\end{array}
$\\

\begin{pf} \hspace{1ex} There exist free resolutions such that
\[
\xymatrix @=1.5em @*[c] { %
0 \ar[r] &X\ar[r] \ar[d]&Y\ar[r]\ar[d]&Z\ar[r]\ar[d]&0\\
0 \ar[r] &A\ar[r]&B\ar[r]&C\ar[r]&0
 }\]
And apply snake lemma. Naturality is same as before.
\end{pf}\\

In particular, if $R$ is a PID, the l.e.s becomes
$$ 0\to A*G\to B*G\to C*G\to A\ten G \to B\ten G\to C\ten G\to 0. $$

\bigskip

7.If $$0\to A\to B\to C\to 0$$ is a short exact sequence, then
there exists a natural long exact sequence,\\[2mm]
\hspace*{1em} $
\begin{array}{l}
  \cdots  \to  \Tor_n(G,A)\to\Tor_n(G,B)\to\Tor_n(G,C) \to \Tor_{n-1}(G,A)\to \\
   \cdots  \to  \Tor_1(G,C)\to G\ten A\to G\ten B\to G\ten C\to 0
\end{array}
$\\

\begin{pf}
$F\to G\to 0 $ : free resolution of $G$. \\
$\Rightarrow 0\to F\ten A \to F\ten B\to F\ten C\to 0$ : s.e.s. of
chain complex. \\
Apply the snake lemma.
\end{pf}\\

\newpage
8. (1) $\Tor_n(A,B)\cong\Tor_n(B,A)$ for all $n$.\\
(2) $\Tor_n(\bigoplus A_\alp,B)\cong \bigoplus \Tor_n(A_\alp,B)$.\\
\hspace*{3ex}($\Tor_n(A, \bigoplus B_\alp)=\bigoplus\Tor_n(A,B_\alp)$)\\
(3) $ R/a* B=\Tor (R/a,B)\cong ker(B\overset{\times
a}{\longrightarrow } B) $
and $R/a \ten B\cong B/aB $.\\
(4) $R$ : PID, $\Tor (R/a,R/b)=R/a*R/b\cong R/d$, $d=(a,b)$.\\

\begin{pf}
(1)
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
\cdots \ar[r] & F \ar[r] & B \ar[r] & 0 & \textrm{: a free
resolution of  }B}
\]
and let $K_p=ker\bd =im\bd\subset F_p$ so that
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & K_p \ar[r] & F_p \ar[r]^-\bd & K_{p-1} \ar[r] & 0 &
\textrm{: s.e.s.}}
\]
By 7.,
\[
\xymatrix @M=1ex @C=1em @R=1ex @*[c]{%
\cdots \ar[r] & \Tor_n(A,K_0) \ar[r] & \Tor_n(A,F_0) \ar[r]
\ar@{=}[d] & \Tor_n(A,B) \ar[r] & \Tor_{n-1}(A,K_0) \ar[r] &
\cdots
\\&&0 }
\]
So, $\Tor_n(A,B)\cong\Tor_{n-1}(A,K_0)\cong \cdots \cong
\Tor_1(A,K_{n-2})$.\\
Now consider
\[
\xymatrix @M=1ex @C=1em @R=1ex @*[c]{%
0 \ar[r] & \Tor_1(A,K_{n-2}) \ar[r] & A\ten K_{n-1} \ar[r]^i &
A\ten F_{n-1} \ar[r] & A\ten K_{n-2} \ar[r] &0 }
\]
Note that
\[
\xymatrix @M=1ex @C=-1ex @R=2ex @*[c]{%
&&&&0&&&&0\\
&&&A\ten K_n\ar[dr]\ar[ur]&&&&A\ten K_{n-2}\ar[ur]\\
\cdots \ar[rr] && A\ten F_{n+1} \ar[rr]^-{\bd_{n+1}}\ar[ur] &&
A\ten F_n \ar[dr]_-{\bd'}
\ar[rr]^{\bd_n} && A\ten F_{n-1} \ar[rr]\ar[ur] && \cdots\\
&&&&&A\ten K_{n-1}\ar[ur]_i \ar[dr]\\
&&&&&&0}
\]
$
\begin{array}{rcl}
\Tor_1(A, K_{n-2}) & = & ker i \cong ker \bd_n/ker \bd' =
ker\bd_n/
im\bd_{n+1}\\
 & = & H_n(A\ten F)=H_n(F\ten A)=\Tor_n(B,A)
\end{array}
$\\

(2) $\Tor_n(A,\bigoplus B_\alp):= H_n(F\ten (\bigoplus
B_\alp))=\bigoplus H_n(F\ten B_\alp)=\bigoplus \Tor
_n(A,B_\alp)$\\
The other follows from (1).\\

(3) Clear from the following.
\[
\xymatrix @M=1ex @C=1em @R=1em @*[c]{%
0 \ar[r] & R \ar[r]^{\times a} & R \ar[r] & R/a \ar[r] & 0 &
\textrm{free resolution of }R/a}
\]
$\Rightarrow$
\[
\xymatrix @M=1ex @C=1em @R=1ex @*[c]{%
0 \ar[r] & R/a*B \ar[r] & R\ten B \ar[r]\ar@{=}[d]& R\ten B
\ar@{=}[d]\ar[r] & R/a\ten B
\ar[r] & 0\\
&&B \ar[r]^{\times a} &B}
\]

(4) {\bf ���� 27.} \\
$R/a*R/b = ker(R/b \overset{\times a}{\longrightarrow } R/b)= R/d$, $d=(a,b)$ \\
$R/a\ten R/b = coker(R/b \overset{\times a}{\longrightarrow }
R/b)= R/d$.
\end{pf}

\end{document}
