\section{Products in cohomology}

An internal product in cohomology brings a further algebraic
structure. The contravariant functor $H^*$ becomes a cofunctor
into graded rings. It enables us to obtain more information
on topological spaces and homotopy classes of maps.
In this section we will define an internal product -- called
the cup product and a closely related external product
-- called the cross product.

\begin{cislo}\label{PRcup} {\bf Cup product.} Let $R$ be a
commutative ring  with a unite and let $X$ be a space. For two cochains $\p\in C^k(X;R)$
and $\ps\in C^l(X;R)$ we define their \emph{cup
product}\index{cup product} $\p\cup\ps\in C^{k+l}(X;R)$
$$(\p\cup\ps)(\sigma)=\p(\sigma/[v_0,v_1,\dots,v_k])\cdot
\ps(\sigma/[v_k,v_{k+1},\dots,v_{k+l}])$$
for any singular simplex $\s:\Delta^{k+l}\to X$. The notation
$\sigma/[v_0,v_1,\dots,v_k]$ and  $\sigma/[v_k,v_{k+1},$ $\dots,v_{k+l}]$
stands for $\s$  composed with inclusions of the standard simplices
$\Delta^{k}$ and $\Delta^{l}$ into the indicated faces of the
standard simplex $\Delta^{k+l}$, respectively. The coboundary
operator $\delta$ behaves on the cup products of cochains as
graded derivation as shown in the following

\begin{lemma*} $$\delta(\p\cup\ps)=\delta\p\cup \ps +
(-1)^{k}\p\cup\delta\ps.$$
\end{lemma*}

\begin{proof} For $\s\in C_{k+l+1}(X)$ we get
\begin{multline*}
(\delta\p\cup \ps)(\s) +(-1)^{k}(\p\cup\delta\ps)(\s)=
\delta\p(\s/[v_0,v_1,\dots,v_{k+1}]) \ps(\s/[v_{k+1},\dots,
v_{k+l+1}])\\
+(-1)^{k}\p(\s[v_0,v_1,\dots,v_k])\delta\ps
(\s/[v_k,\dots,v_{k+l+1}])\\
=\sum_{i=0}^{k+1}(-1)^i\p(\s/[v_0,\dots,\hat v_i,\dots,v_{k+1}])
(\ps(\s/[v_{k+1},\dots,v_{k+l+1}]))\\
+ (-1)^k \left(
\sum_{j=k}^{k+l+1}(-1)^{j-k}\p(\s/[v_0,\dots,v_{k}])
\ps(\s/[v_{k},\dots,\hat v_j,\dots,v_{k+l+1}])
\right)\\
=\sum_{i=0}^{k+l+1}(-1)^{i}(\p\cup\ps)(\s/[v_{0},\dots,\hat v_i,
\dots,v_{k+l+1}])=\delta(\p\cup\ps)(\s).
\end{multline*}
\end{proof}
Lemma implies that
\begin{enumerate}
\item If $\p$ and $\ps$ are cocycles, then $\p\cup\ps$ is
a cocycle.
\item If one of the cochains $\p$ and $\ps$  is a coboundary,
then $\p\cup\ps$ is a coboundary.
\end{enumerate}
It enables us to define the $\emph{cup product}$
$$\cup: H^k(X;R)\times H^l(X;R)\to H^{k+l}(X;R)
$$
by the prescription
$$[\p]\cup[\ps]=[\p\cup\ps]$$
for cocycles $\p$ and $\ps$. Since $\cup$ is an $R$-bilinear map
on $H^k(X;R)\times H^l(X;R)$, it can be considered as an
$R$-linear map on the tensor product
$H^k(X;R)\otimes_R H^l(X;R)$. Given a pair of spaces $(X,A)$
we can define the cup product as a linear map
\begin{align*}
\cup:& H^k(X,A;R)\otimes_R H^l(X;R)\to H^{k+l}(X,A;R),\\
\cup:& H^k(X;R)\otimes_R H^l(X,A;R)\to H^{k+l}(X,A;R),\\
\cup:& H^k(X,A;R)\otimes_R H^l(X,A;R)\to H^{k+l}(X,A;R).
\end{align*}
Moreover, if $A$ and $B$ are open in $X$ or $A$ and $B$ are
subcomplexes of CW-complex $X$, one can define
$$\cup: H^k(X,A;R)\otimes_R H^l(X,B;R)\to H^{k+l}(X,A\cup B;R).$$

\begin{ex*} Prove that the previous definitions of cup product
for pairs of spaces are correct. For the last case you need
Lemma 3.12. % \ref{HOet}.
\end{ex*}

\begin{remark*} In the same way as the singular cohomology groups and
the cup product
have been defined using the singular chain complexes, we can
introduce simplicial cohomology groups for $\Delta$-complexes
and a cup product in these groups.
\end{remark*}
\end{cislo}

\begin{cislo}\label{PRprop}{\bf Properties of the cup product}
are following:
\begin{enumerate}
\item The cup product is associative.
\item If $X\ne\emptyset$, there is an element $1\in H^0(X;R)$ such that for all
$\alpha\in H^k(X,A;R)$
$$1\cup\alpha=\alpha\cup 1=\alpha.$$
\item For all $\alpha\in H^k(X,A;R)$ and $\beta\in H^l(X,A;R)$
$$\alpha\cup\beta=(-1)^{kl}\beta\cup\alpha,$$
i.~e. the cup product is graded commutative.
\item Naturality of the cup product. For every map $f:(X,A)\to (Y,B)$ and
any $\alpha\in H^k(Y,B;R)$, $\beta\in H^l(Y,B;R)$ we have
$$f^*(\alpha\cup\beta)=f^*(\alpha)\cup f^*(\beta).$$
\end{enumerate}

\begin{remark*} Properties (1) -- (3) mean that
$H^*(X,A;R)=\bigoplus_{i=0}^{\infty}H^i(X,A;R)$ with the cup
product is not only a graded group but also a graded ring and that
$H^*(X;R)$ is even a graded ring with
a unit if $X\ne\emptyset$. Property (4) says that $f:(X,A)\to (Y,B)$
induces a ring homomorphism $f^*:H^*(Y,B;R)\to H^*(X,A;R)$.
\end{remark*}

\begin{proof} To prove properties (1), (2) and (4) is easy and
left to the reader as an exercise. To prove property (3) is more
difficult. We refer to [Hatcher], %\cite{Ha} 
Theorem 3.14, pages 215 -- 217 for geometrically motivated proof. Another approach is outlined later in \ref{PRez}.
\end{proof}
\end{cislo}

\begin{cislo}\label{PRcross} {\bf Cross product.} Consider spaces
$X$ and $Y$ and projections $p_1:X\times Y\to X$ and $p_2:X\times Y\to
Y$. We will define the \emph{cross product}\index{cross product} or
\emph{external product}. \index{external product} The absolute
and relative forms are the linear maps
\begin{align*}
\mu &:H^k(X,R)\otimes H^l(Y;R)\to H^{k+l}(X\times Y;R),\\
\mu &:H^k(X,A;R)\otimes H^l(Y,B;R)\to H^{k+l}(X\times Y, A\times
Y\cup X\times B;R)
\end{align*}
given by
$$\mu(\alpha\otimes\beta)=p_1^*(\alpha)\cup p_2^*(\beta).$$
For the relative form of the cross product we suppose that $A$
and $B$ are open in $X$ and $Y$, or that $A$ and
$B$ are subcomplexes of $X$ and $Y$, respectively. (See the
definition of the cup product.) The name cross product comes from
the notation since  $\mu(\alpha\otimes\beta)$ is often written as
$\alpha\times\beta$.

\begin{ex*} Let $\Delta:X\to X\times X$ be the diagonal
$\Delta(x)=(x,x)$. Show that for $\alpha,\beta\in H^*(X;R)$
$$\alpha\cup\beta=\Delta^*\big(\mu(\alpha\otimes\beta)\big).$$
\end{ex*}
\end{cislo}

\begin{cislo}\label{PRtp}{\bf Tensor product of graded rings.}
Let $A^*=\bigoplus_{n=0}^{\infty} A^n$ and $B^*=\bigoplus_{n=0}^{\infty}
B^n$ be graded rings. Then the \emph{tensor product of graded
rings}\index{tensor product of graded rings} $A^*\otimes B^*$
is  the graded ring $C^*=\bigoplus_{n=0}^{\infty}C^n$ where
$$C^n=\bigoplus_{i+j=n} A^i\otimes B^j$$
with the multiplication given by
$$(a_1\otimes b_1)\cdot (a_2\otimes b_2)=(-1)^{\vert
b_1\vert\cdot\vert a_2\vert}(a_1\cdot a_2)\otimes(b_1\cdot b_2).$$
Here $|b_1|$ is the degree of $b_1\in B^{*}$, i.e. $b_1\in B^{|b_1|}$.
If $A^*$ and $B^*$ are graded commutative, so is $A^*\otimes
B^*$.

\begin{lemma*}  The cross product
$$\mu :H^k(X,A;R)\otimes H^l(Y,B;R)\to H^{k+l}(X\times Y, A\times
Y\cup X\times B;R)$$
is a homomorphism of graded rings.
\end{lemma*}

\begin{proof} Using the definitions of the cup and cross products
and their properties we have
\begin{align*}
\mu\big((a\times b)\cdot(c\times d\big))&= (-1)^{\vert
b\vert\cdot\vert c\vert}\mu\big((a\cup c)\otimes (b\cup d)\big)
=(-1)^{\vert b\vert\cdot\vert c\vert}p_1^*(a\cup c)\cup
p_2^*(b\cup d)\\
&=(-1)^{\vert b\vert\cdot\vert c\vert}p_1^*(a)\cup
p_1^*(c)\cup p_2^*(b)\cup p_2^*(d)\\
&=p_1^*(a)\cup
p_2^*(b)\cup p_1^*(c)\cup p_2^*(d)=\mu(a\otimes b)\cup\mu(c\otimes
d).
\end{align*}
\end{proof}
\end{cislo}

\begin{cislo}\label{PRkf}{\bf K\" unneth formulas} tell us how to
compute the graded $R$-modules  $H_*(X\times Y;R)$ or
$H^*(X\times Y;R)$ out of the graded modules $H_*(X;R)$ and
 $H_*(Y;R)$ or $H^*(X;R)$ and $H^*(Y;R)$, respectively. Under
certain conditions it even determines the ring structure of
$H^*(X\times Y;R)$.

\begin{thm*}[K\" unneth formula] Let $(X,A)$ and $(Y,B)$  be
pairs of CW-complexes. Suppose that $H^k(Y,B;R)$ are free
finitely generated $R$-modules for all $k$. Then
$$\mu:H^*(X,A;R)\otimes H^*(Y,B;R)\to H^*(X\times Y,A\times Y
\cup X\times B;R)$$
is an isomorphism of graded rings.
\end{thm*}

\begin{example*}  $H^*(S^k\times S^l)\cong \Z[\alpha,\beta]/\mathcal I$
where $\mathcal I$ is the ideal generated by elements
$\alpha^2$, $\beta^2$, $\alpha\beta=(-1)^{kl}\beta\alpha$
and $\deg\alpha=k$, $\deg \beta=l$.
\end{example*}

\begin{proof} Consider the diagram
$$
\xymatrix{
H^*(X,A)\otimes_R H^*(Y)\ar[rr]\ar[dd]^-{\mu}&
{}&
H^*(X)\otimes_R H^*(Y)\ar[ld]\ar[dd]^-{\mu}\\
{}& H^*(A)\otimes_RH^*(Y)\ar[lu]_{\delta^*\otimes\id}
\ar[dd]^<<<<<{\mu}& {}\\
H^*(X\times Y,A\times Y)\ar[rr] H^*(X\times Y) &{}&
H^*(X\times Y)\ar [ld]\\
{}&H^*(A\times Y)\ar[lu]_-{\delta^*}&{}
}
$$
where the  upper and the lower triangles come from the long exact
sequences for pairs $(X,A)$ and $(X\times Y,A\times Y)$,
respectively. The right rhomb commutes as a consequence of
the naturality of the cross product. We prove that the left rhomb
also commutes.

Let $\p$ and $\ps$ be  cocycles in $C^*(A)$ and $C^*(Y)$,
respectively. Let $\Phi$ be a cocycle in $C^*(X)$ extending $\p$.
Then $p_1^*\Phi\cup p_2^*\ps\in C^*(X\times Y)$ extends
$p_1^*\p\cup p_2^*\ps\in C^*(A\times Y)$.
Using the definition of the connecting homomorphism in
cohomology (see Remark 5.6 B) %\ref{COles}) 
we get
\begin{align*}
\mu\big((\delta^*\otimes\id)([\p]\otimes[\ps])\big)&=
\mu[\delta\Phi\otimes\ps]=p_1^*[\delta\Phi]\cup p_2^*[\ps],\\
\delta^*\big(\mu([\p]\otimes[\ps])\big)&=\delta^*[p_1^*\p\cup
p_2^*\ps]
=[\delta(p_1^*\Phi\cup p_2^*\ps)]=p_1^*[\delta\Phi]\cup
p_2^*[\ps].
\end{align*}

First, we prove the statement of Theorem for a finetedimensional CW-complex $X$
and $A=B=\emptyset$
using the induction by the dimension of $X$ and Five Lemma. If $\dim X=0$,
$X$ is a finite discrete set and the statement of Theorem
is true. Suppose that Theorem holds for spaces of dimension $n-1$
or less. Let $\dim X=n$. It suffices to show that
$$\mu:H^*(X^n,X^{n-1})\otimes H^*(Y)\to H^*(X^n\times
Y,X^{n-1}\times Y)$$
is an isomorphism and than to use Five Lemma in the diagram above
with $A=X^{n-1}$ to prove the statement for $X=X^n$.
$X^n/X^{n-1}$ is homeomorphic to
$\bigsqcup_i D^n_i/\bigsqcup_i \partial D^n_i$. 
To prove that
$$\mu:H^*\bigl (\bigvee_i S^n_i\bigr )\otimes H^*(Y)\to H^*\bigl (\bigvee_i S^n_i\times
Y\bigr )$$
is an isomorphism, 
we use again the diagram above for $X=\bigsqcup_i D^n_i$ and $A=\bigsqcup_i \partial
D^n_i$ and the induction with respect to $n$.

So we have proved the theorem for $X$ a finite dimensional
CW-complex and $A=B=\emptyset$.
Using once more our diagram and Five Lemma, we can easily prove Theorem
for any pairs $(X,A)$, $(Y,\emptyset)$ with $X$ of finite
dimension.
For $X$ of infinite dimension, we have to prove $H^i(X)=H^i(X^n)$
for $i<n$ which is equivalent to $H^i(X/X^n)=0$. We omit the
details and refer the reader to [Hatcher], %\cite{Ha}, 
pages 220 -- 221.
\end{proof}
\end{cislo}

\begin{cislo}\label{PRappl}{\bf Application of the cup product.}
In this paragraph we show how to use the cup product to prove
that  $S^{2k}$  is not an H-space.
A space $X$ is called an \emph{ H-space}\index{H-space} if there is a
map $m:X\times X\to X$ called a multiplication and an element $e\in X$
called a unit such that
$m(e,x)=m(x,e)=x$ for all $x\in X$.

Suppose that there is a multiplication $m:S^{2k}\times S^{2k}\to S^{2k}$
with a unit $e$. According to  Example after Theorem \ref{PRkf}
$$H^*(S^{2k}\times S^{2k};\Z)=\Z[\alpha,\beta]/
\mathcal I$$
where $\mathcal I$ is the ideal generated by relations
$\alpha^2=0$,
$\beta^2=0$ and $\alpha\beta=\beta\alpha$. The last relation is
due to the fact that the dimension of the sphere is even.
Moreover, $\alpha=\gamma\otimes 1$ and $\beta=1\otimes\gamma$
where $\gamma\in H^{2k}(S^{2k};\Z)$ is a generator. Let us
compute $m^*:H^*(S^{2k};\Z)\to H^*(S^{2k}\times S^{2k};\Z)$.
We have
$$m^*(\gamma)=a\alpha+b\beta,\quad a,b\in\Z.$$
Since the composition
$$S^{2k}\xrightarrow{\id\times e}S^{2k}\times
S^{2k}\xrightarrow{\ m\ }
S^{2k}$$
is the identity, we get that $a=1$. Similarly, $b=1$.
Now compute $m^*(\gamma^2)$:
$$0=m^*(0)=m^*(\gamma^2)=\big(m^*(\gamma)\big)^2=(\alpha+\beta)^2=
2\alpha\beta\ne 0,$$
a contradiction. Does  this proof go through for odd
dimensional spheres?
\end{cislo}

\begin{cislo}\label{PRkfha}{\bf K\" unneth formula in homological
algebra.}
Consider two chain complexes $(C_*,\partial_C)$, $(D_*,\partial_D)$
of $R$-modules. Suppose there is an integer $N$ such that
$C_n=D_n=0$ for all $n<N$. Then their tensor product is the chain complex
$(C_*\otimes D_*,\partial)$ with
$$(C_*\otimes D_*)_n=\bigoplus_{i+j=n}C_i\otimes D_i$$
and the boundary operator on $C_i\otimes D_j$
$$\partial(c\otimes d)=\partial_Cc\otimes
d+(-1)^ic\otimes\partial_Dd.$$
It is easy to make sure that $\partial\partial=0$.

Next we can define the graded $R$-module $C_**D_*$ as
$$(C_**D_*)_n=\bigoplus_{i+j=n}\tor_1^R(C_i,D_j).$$

A ring $R$ is called \emph{hereditary}\index{hereditary ring}
if any submodule of a free $R$-module is free. Examples of
hereditary rings are $\Z$ and all fields.

\begin{thm*}[Algebraic K\"unneth formula] Let $R$ be a hereditary
ring and let $C_*$ and $D_*$ be chain complexes of R-modules.
If $C_*$ is free, then the homology groups of $C_*\otimes D_*$
are determined by the splitting short exact sequence
$$0\to (H_*(C)\otimes H_*(D))_n\xrightarrow{l} H_n(C_*\otimes D_*)
\to (H_*(C)*H_*(D))_{n-1}\to 0$$
where $l([c]\otimes [d])=[c\otimes d]$. This sequence is natural
but the splitting is not.
\end{thm*}

Notice that for the chain complex
$$D_n=
\begin{cases}0\quad&\text{for }n\ne 0,\\
G\quad\text{for }n=0
\end{cases}$$
the K\" uneth formula gives the universal coefficient theorem for
homology groups, see Theorem 6.11 B. %\ref{HAukh}.

The proof of the K\" unneth formula is similar to the proof of
the universal coefficient theorem and we omit it.
\end{cislo}

\begin{cislo}\label{PRez}{\bf Eilenberg-Zilbert Theorem.}
To be able to apply the previous K\" unneth formula in topology
we have to show that the singular chain complex
$C_*(X\times Y)$ of a product $X\times Y$ is chain homotopy
equivalent to the tensor product of the  singular chain complexes
$C_*(X)\otimes C_*(Y)$.

\begin{thm*}[Eilenberg-Zilbert] Up to chain homotopy
there are unique  natural chain homomorphisms
\begin{align*}
\Phi:&C_*(X)\otimes C_*(Y)\to C_*(X\times Y),\\
\Psi:&C_*(X\times Y)\to C_*(X)\otimes C_*(Y)
\end{align*}
such that for  $0$-simplices $\s$ and $\tau$
$$\Phi(\s\otimes\tau)=(\s,\tau),\quad
\Psi(\s,\tau)=\s\otimes\tau.$$


Moreover, such chain homomorphisms are chain homotopy equivances:
there are natural chain homotopies such that
$$\Psi\Phi\sim\id_{C_*X\otimes C_*(Y)},\quad
\Phi\Psi\sim\id_{C_*(X\times Y)}.$$
\end{thm*}

For the proof of this theorem see [Dold], %\cite{Do}, 
IV.12.1. The chain homomorphism $\Psi$ is called the Eilenberg-Zilbert homomorphism
and denoted $EZ$. It enables a different and more abstract
approach to the definitions of the
cross  and cup products. The cross product is
$$\mu([\alpha]\otimes[\beta])=[(\alpha\otimes\beta)\circ EZ]$$
for cocycles $\alpha\in C^*(X;R)$ and $\beta\in C^*(Y;R)$
and the cup product is
$$([\p]\otimes[\ps])=[(\p\otimes\ps)\circ EZ\circ\Delta_*]$$
for cocycles $\p,\ps\in C^*(X;R)$ and the diagonal $\Delta:X\to
X\times X$. In our definition in \ref{PRcup} we have used
for $EZ\circ\Delta_*$ the homomorphism
$$\s\to\s/[v_0,v_1,\dots,v_k]\otimes\s/[v_k,\dots,v_n].$$
The properties of $EZ$ can be used for a different proof of the
graded commutativity of the cup product.
\end{cislo}

\begin{cislo}\label{PRkft}{\bf K\" unneth formulas in topology.}
The following statement is an immediate consequence of the
previous paragraph.

\begin{thma}[K\" unneth formula for homology] Let $R$ be a hereditary
ring. The homology groups
of the product of two spaces $X$ and $Y$ are determined by the
following splitting short exact sequence
$$0\to (H_*(X;R)\otimes H_*(Y;R))_n\xrightarrow{l} H_n(X\times
Y;R)
\to (H_*(X;R)*H_*(Y;R))_{n-1}\to 0$$
where $l([c]\otimes [d])=[c\otimes d]$. This sequence is natural
but the splitting is not.
\end{thma}

For cohomology groups one can prove

\begin{thma}[K\" unneth formula for cohomology groups] Let $R$ be a hereditary
ring. The cohomology groups
of the product of two spaces $X$ and $Y$ are determined by the
following splitting short exact sequence
$$0\to (H^*(X;R)\otimes H^*(Y;R))_n\xrightarrow{\mu} H^n(X\times
Y;R)
\to (H^*(X;R)*H^*(Y;R))_{n+1}\to 0.$$
This sequence is natural but the splitting is not.
\end{thma}

For the proof and other forms of K\" unneth formulas see [Dold], %\cite{Do}, 
Chapter VI, Theorem 12.16 or [Spanier], %\cite{Sp}, 
Chapter 5, Theorems 5.11. and 5.12.
\end{cislo}