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Let $A$ be a $\mathbb{N}$-graded, locally finite $\Bbbk$-algebra, $\Bbbk$ being a field, $A=\oplus_{n \geq 0} \ A^n$, each $A^i$ being finitely dimensional as a $\Bbbk$ vector space. Assume also that $A$ is connected, i.e. $A^0=\Bbbk$.

Let $A^\ast$ denote its graded dual, i.e. $A^\ast=\oplus_{n \geq 0} \ \mathrm{Hom}_\Bbbk(A^n,\Bbbk)$, which is a connected, graded $\Bbbk$-coalgebra.

Starting from some computations involving the (co)bar complexes for Hochschild (co)homology, I came up with the following question:

What is the connection between $\mathrm{Tor}_\bullet^A(\Bbbk,\Bbbk)$ and $\mathrm{Ext}^\bullet_{A^\ast} (\Bbbk,\Bbbk)$ ?

Computing them using the normalized (co)bar complexes, one obtains, in both cases, bigraded structures, $\mathrm{Ext}^\bullet_{A^\ast} (\Bbbk,\Bbbk)$ being the Yoneda algebra of $A^\ast$ and $\mathrm{Tor}_\bullet^A(\Bbbk,\Bbbk)$ being a coalgebra, obtained in a "dual" way as for the Yoneda algebra.

It would be great to have a duality between these structures, but I don't know how to approach an attempt of proof.

Thank you.

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After reading some more about this subject and trying to simplify the question as much as possible, I'm starting to think that one could use Universal Coefficient Theorem for cohomology. How exactly, I'm still wondering... – AdrianM Feb 4 '14 at 17:36

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up vote 2 down vote accepted

If $k$ is a field as you say it is, it is not difficult to see using the standard tensor-hom adjunction that $Ext^*_R(k,k)\cong (Tor_*^R(k,k))^*$ as $k$-graded modules, where $R$ is a $k$-algebra. I've learned this from these Tor-Ext notes by May.

Here's the relevant passage, which contains other comments which might be useful to you. It's at the very end of the note:

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