# Defining the cup product in Ext using a Kunneth formula

I want to make a Kunneth product of sorts on Ext. In particular, letting $C_\*$ be a $R$-free resolution for $k$ over a $k$-hopf algebra $R$, elements in $Ext_R(k,k)$ are represented by maps in $Hom_R(C_\*,k)$. Of course, we can consider $Hom_R(C_\* \otimes C_\*,k)$, the cohomology of this being the same since $k\otimes k \cong k$.

What I want then is a sort of Kunneth product:

$\times :Hom_R(C_\*,k)\otimes Hom_R(C_\*,k)\to Hom_R(C\_*\otimes C_\*,k)$

I want it to be the case that this defines the cup product (aka Yoneda product). If we get a magical map

$\Delta : C_\*\to C_\*\otimes C_\*$,

then

$\cup : Ext_R(k,k)\otimes Ext_R(k,k)\to Ext_R(k,k)$

can be given, on the co-chain level,

$(a\cup b)(\sigma) = (a\times b)(\Delta \sigma)$

Now, $\Delta$ is possibly defined up to homotopy by the requirement that it commutes with the augmentation in $R$.

My question is this: is there a way to describe this Kunneth product using any $C_\*$, or do I need to use a Bar resolution of sorts like in group cohomology. Does this method even make sense? Can you help flesh this out?

Thanks!

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This is done in Cartan-Eilenberg. –  Mariano Suárez-Alvarez Aug 5 '12 at 20:57
I can't seem to improve on Mariano's observation... –  Sean Tilson Aug 5 '12 at 22:02
I'm reading it on Google Books preview right now, and it seems to walk through my reasoning and clear up my confusion. Thanks for the reference. I think I'll grab a real copy when the library opens Monday. –  Joseph Victor Aug 5 '12 at 22:04