Duality between Tor and Ext? Let $A$ be a $\mathbb{N}$-graded, locally finite $\mathbb{k}$-algebra, $\mathbb{k}$ being a field, $A = \bigoplus_{n \geq 0} A^n$, each $A^i$ being finitely-dimensional as a $\mathbb{k}$ vector space. Assume also that $A$ is connected, i.e., $A^0 = \mathbb{k}$.
Let $A^\ast$ denote its graded dual, i.e., $A^\ast = \bigoplus_{n \geq 0} \mathrm{Hom}_{\mathbb{k}}(A^n, \mathbb{k})$, which is a connected, graded $\mathbb{k}$-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 $\mathrm{Tor}_\bullet^A(\mathbb{k},\mathbb{k})$ and $\mathrm{Ext}^\bullet_{A^\ast} (\mathbb{k},\mathbb{k})$?
Computing them using the normalized (co)bar complexes, one obtains, in both cases, bigraded structures, $\mathrm{Ext}^\bullet_{A^\ast} (\mathbb{k}, \mathbb{k})$ being the Yoneda algebra of $A^\ast$ and $\mathrm{Tor}_\bullet^A(\mathbb{k}, \mathbb{k})$ 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.
 A: If $k$ is a field as you say it is, it is not difficult to see using the standard tensor-hom adjunction that $\mathrm{Ext}^*_R(k, k) \cong (\mathrm{Tor}_*^R(k, k))^*$ as $k$-graded modules, where $R$ is a $k$-algebra. I've learned this from these $\mathrm{Tor}$-$\mathrm{Ext}$ 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:

Remember that when $R$ is commutative the $\mathrm{Ext}$ groups and all maps in sight between them take values in the category of $R$-modules.
More generally, if $R$ is an algebra over a commutative ring $k$, then the $\mathrm{Ext}$ groups and all maps in sight between them take values in the category of $k$-modules.
When $R$ is commutative and augmented over $k$, we have seen that $\mathrm{Tor}^R_*(k, k)$ is a graded $k$-algebra.
We now see that $\mathrm{Ext}_R^*(k, k)$ is also a graded $k$-algebra, via the Yoneda product.
If $k$ is a field and $X$ is an $R$-free resolution of $k$, then we have
$$
  \mathrm{Hom}_R(X, k) ≅ \mathrm{Hom}_k(k ⊗_R X, k)
$$
as $k$-chain complexes.
Writing $M^* = \mathrm{Hom}_k(M, k)$ for the vector space dual of $M$, this implies that
$$
  \mathrm{Ext}_R^*(k, k) ≅ (\mathrm{Tor}^R_*(k, k))^* \,.
$$
Therefore $\mathrm{Ext}_R^*(k, k)$ has both an algebra structure and the dual of an algebra structure, which is called a coalgebra structure.
We shall return to consideration of such structures later, when we shall talk about bialgebras and Hopf algebras.
For now, we sum up by saying that for any $k$-algebra $R$, $\mathrm{Ext}_R^*(k, k)$ is a $k$-algebra under the Yoneda product.
It is not necessarily commutative even when $R$ is commutative, but then $\mathrm{Ext}_R^*(k, k)$ is a Hopf algebra.
We shall later return to this point and show that if $R$ is a Hopf algebra over $k$, then $\mathrm{Ext}_R^*(k, k)$ is a commutative $k$-algebra, but not necessarily a Hopf algebra.
As we shall see, when specialized to group algebras this is closely related to the fact that the cohomology of a space with coefficients in $k$ is a commutative algebra.
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