Again, the Nakayama functor The question has already been asked here, but without an answer, so I want to ask it again.
Let $A$ be a fd. $k$-algebra, $\mathcal{I}$ and $\mathcal{P}$ be the categories of fd. injective and projective $A$-modules, resp. Let $\nu = D\hom_A(-, A)$ and $\nu^{-1} = \hom_A(DA, -)$, where $D = \hom_k(-, k)$. I want to show that $\nu\colon \mathcal{P} \to \mathcal{I}$ and $\nu^{-1}\colon \mathcal{I} \to \mathcal{P}$, and that both form a pair of mutually quasi-inverse quasi-equivalences of $A$-modules.

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*Let $P \in \mathcal{P}$, $M$ and $N$ be modules, such that $$\nu P \xleftarrow{f} M \stackrel{i}\hookrightarrow N$$. I need to find a morphism $N \xrightarrow{g} \nu P$ such that $f = gi$.  Since everything is fd., I can dualize: $$\hom_A(P, A) \xrightarrow{Df} DM \stackrel{Di}\twoheadleftarrow DN$$ and I need to find $DN \xleftarrow{Dg} \hom_A(P, A)$.
The only way to use projectivity of $P$ I see is to pass to $$\hom_A(P, A) \xrightarrow{\hat{f}} \hom(P, DM) \stackrel{\hom(P, Di)}\twoheadleftarrow \hom(P, DN)$$ with the constant map $\hat{f}: \alpha \mapsto (p \mapsto Df(\alpha))$. That looks wrong, and I have no idea how to proceed.


*I hope once I understand 1., I also understand why $\nu^{-1}\colon \mathcal{I} \to \mathcal{P}$.


*To show quasi-inverseness, I certainly have to use hom-tensor-adjunctions. How about $$\begin{aligned}
P &\to \hom(\hom(P, A), A) = D(DA \otimes \hom(P, A)) = \hom(DA, D\hom(P, A)) = \nu^{-1} \nu P,\\p &\mapsto f \mapsto f(p).\end{aligned}$$
I guess I can argue that this map is injective and by finite dimension an isomorphism. Does that suffice for $id \simeq \nu^{-1}\nu$?
For the other composition $$\begin{aligned} \nu\nu^{-1}I = D\hom(\hom(DA, I), A) = \hom(D\hom(DA, I)\otimes A, k) = \hom(A \otimes DA, I) &\to I\\f &\mapsto {?}\end{aligned}$$
I'm not sure. Where do I have to map $f$? I guess $f(\sum_a a \otimes a^*)$, where the $a$ form a basis of $A$? How do I see this gives a quasi-equivalence?
 A: The functor $\nu$ is additive and sends $A$ to the injective module $DA$, so sends finite dimensional projectives to finite dimensional injectives.
Similarly the functor $\nu^-$ is additive and sends the injective cogenerator $DA$ to
$$ \mathrm{Hom}_A(DA,DA) \cong \mathrm{Hom}_k(A\otimes_ADA,k) \cong D^2A \cong A, $$
so send finite dimensional injectives to finite dimensional projectives.
To see that the functors form an adjoint pair on these subcategories, we can show that there is a natural isomorphism of bifunctors
$$ \mathrm{Hom}(\nu^-I,P) \cong \mathrm{Hom}(I,\nu P) $$
with $P$ finite dimensional projective and $I$ finite dimensional injective.
We have natural isomorphisms
$$ \mathrm{Hom}(I,D\mathrm{Hom}(P,A)) \cong
D(\mathrm{Hom}(P,A)\otimes_AI) \cong
D\mathrm{Hom}(P,I) $$
where the last follows since $P$ is finite dimensional projective (dual basis Lemma).
Next, for any two finite dimensional (right and left) modules $X$ and $Y$ we have
$$ \mathrm{Hom}_A(X,DY) \cong D(X\otimes_AY) \cong
\mathrm{Hom}_A(Y,DX). $$
We now get natural isomorphisms
$$ \mathrm{Hom}(\mathrm{Hom}(DA,I),P) \cong
\mathrm{Hom}(\mathrm{Hom}(DI,A),P) \cong
D(DP\otimes_A\mathrm{Hom}(DI,A)) \cong
D\mathrm{Hom}(DI,DP). $$
The last uses the dual basis Lemma for the finite dimensional projective right module $DI$. Applying the observation above once more, this is isomorphic to $D\mathrm{Hom}(P,I)$.
Now that we have adjointness, we have the unit and count, and the computations we did at the beginning show that these are isomorphisms for $DA$ and $A$ respectively. Since they are additive, they are isomorphisms for all (finite dimensional) injectives and projectives, and so we have the desired (adjoint) equivalence between these subcategories.
