\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