Defining the cup product in Ext using a Kunneth formula

2012-08-05T20:25:31Z

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!

Defining the cup product in Ext using a Kunneth formula - MathOverflow

Defining the cup product in Ext using a Kunneth formula