Why is the Kunneth formula an isomorphism of rings? Let's assume we have two topological spaces $X,Y$ such that their cohomology modules over ring $R$ are finitely generated and thus the Kunneth formula holds. Let's assume additionally that monomorphism $\bigoplus H^{*}(X,R) \otimes H^{*}(Y,R) \rightarrow H^{*}(X \times Y,R) $ is an isomorphism (for instance product of spheres over $\mathbb{Z}$).  I've heard that this map is also an isomorphism of rings (the right hand side with cup product structure, the left hand side as a tensor product of graded rings, where $H^{*}(X,R),H^{*}(Y,R)$ have cup product structure). How can one see that this is indeed the case? Any referrals greatly welcomed.
 A: The map $H^*(X,R)\otimes H^*(Y,R)\to H^*(X\times Y,R)$ is always a ring-homomorphism.  Indeed, this is true much more generally: if $R$ is a commutative ring, $A_*$, $B_*$, and $C_*$ are graded commutative $R$-algebras, and $f:A_*\to C_*$ and $g:B_*\to C_*$ are graded $R$-algebra homomorphisms, then the linear map $h:A_*\otimes_R B_*\to C_*$ given by $h(a\otimes b)=f(a)g(b)$ is a $R$-algebra homomorphism as well.
Since $h$ is linear and every element of $A_*\otimes_R B_*$ is a linear combination of elements of the form $a\otimes b$ where $a$ and $b$ are homogeneous, it suffices to check that $h((a\otimes b)(c\otimes d))=h(a\otimes b)h(c\otimes d)$ if $a,b,c,$ and $d$ are homogeneous.  Now just note that 
\begin{align*}
h((a\otimes b)(c\otimes d))&=h((-1)^{|b||c|}ac\otimes bd) \\
&=(-1)^{|b||c|}f(ac)g(bd) \\
&=(-1)^{|b||c|}f(a)f(c)g(b)g(d) \\
&=f(a)g(b)f(c)g(d) \\
&=h(a\otimes b)h(c\otimes d).
\end{align*}
In your case, $A_*=H^*(X,R)$, $B_*=H^*(Y,R)$, $C_*=H^*(X\times Y,R)$, $f=p_1^*$, and $g=p_2^*$, where $p_1:X\times Y\to X$ and $p_2:X\times Y\to Y$ are the projections.
