# 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
@MarianoSuárez-Alvarez: more than two years later I would like to ask: is there a more modern version of cartan-eilenberg? I've learned homological algebra using gelfand-manin, but there are many topics in cartan-eilenberg which I think are treated in greater detail. Is there a book that does that but using more modern language (perhaps even using derived categories?). – bananastack Dec 15 '14 at 20:11
The cup product is just the composition of maps in the derived category. If you want to actually compute, that's not very helpful, though. – Mariano Suárez-Alvarez Dec 16 '14 at 3:19