FALSE: derived functors in two variables are automatically balanced

The functor $A\mapsto A^{\ast }$ is exact on finite-dimensional vector spaces, $C\mapsto {\mathrm{Hom}}_R(k,C)$ is left exact, and tensoring over $k$ is exact. Hence $B$ is additive and left exact in each variable in the sense required by [L1]. Give $k\in \mathcal{A}$ its length-zero projective resolution. The first-variable right-derived object at $(k,k)$ is then zero in every positive degree.

The $R$-module $R$ is injective: the coefficient-of-$\epsilon$ functional identifies $R$ with ${\mathrm{Hom}}_k(R,k)$ as an $R$-module, and ${\mathrm{Hom}}_R(-,{\mathrm{Hom}}_k(R,k))\cong {\mathrm{Hom}}_k(-,k)$ is exact. Thus $0\to k\stackrel{1\mapsto \epsilon }{\to }R\stackrel{\epsilon }{\to }R\stackrel{\epsilon }{\to }R\to \cdots$ is an injective resolution of $k$: at every copy of $R$, both the image and kernel of multiplication by $\epsilon$ are the ideal $(\epsilon )$.

Applying $B(k,-)={\mathrm{Hom}}_R(k,-)$ to the deleted resolution in step 1.2 gives a cochain complex with one copy of $k$ in every degree and zero differentials, since multiplication by $\epsilon$ annihilates ${\mathrm{Hom}}_R(k,R)\cong (\epsilon )$. Consequently the second-variable right-derived object in degree $1$ is $k$, whereas the first-variable object from step 1.1 is $0$. They cannot be isomorphic, so the balance data required by [L2] do not exist and the displayed automatic-balance claim is false.