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Let $P \in {}_B\text{Mod}_A$ a $B$-$A$ bimodule (for $A, B$ rings). We can consider the induction functor $$P \otimes_A -:{}_A\text{Mod} \to {}_B\text{Mod}:M \mapsto P\otimes_A M$$ and by the hom-tensor relation we get $${}_B\text{Hom}(P\otimes_A M, N) \cong {}_A\text{Hom}(N, {}_B\text{Hom}(P, N)).$$ Therefore, $${}_B\text{Hom}(P, -):{}_B\text{Mod} \to {}_A\text{Mod}:N \mapsto {}_B\text{Hom}(P, N)$$ is the right adjoint of the induction functor. However, in my representation theory course, we are studying Morita context and we saw that ${}_A\text{Mod},{}_B\text{Mod}$ are equivalent additive categories if and only if there is a strict Morita context $(B, A, P, Q, \mu, \tau)$ where $P, Q$ are in ${}_B\text{Mod}_A$, ${}_A\text{Mod}_B$ respectively. In the case that we have an equivalence of additive categories, the functors of the adjunction are exactly given by $$P \otimes_A -:{}_A\text{Mod} \to {}_B\text{Mod} \quad \text{and} \quad Q \otimes_B -:{}_B\text{Mod} \to {}_A\text{Mod}.$$ My question is the following: In the circumstance of an equivalence of additive categories, do we then have $Q \otimes_B - \cong {}_B\text{Hom}(P, -)$ in a natural way? If the answer is yes, is there any other case where the hom-tensor can be writen as an induction functor ?

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If $P\otimes_A -$ and $Q\otimes_B-$ are quasi-inverse, then this equivalence can be promoted to an adjoint equivalence (this doesn't change the functors, at most the isomorphisms to the identity functor)

In particular, $Q\otimes_B-$ is right adjoint to $P\otimes_A-$, and since right adjoints are unique, it follows that $Q\otimes_B - \cong \hom_B(P-,)$, and in fact uniquely so if you fixed the adjoint equivalence mentioned above and require that the natural isomorphism preserve all the available structure (i.e. if you require that it be an isomorphism "of right adjoints").

To answer your second question, one needs to talk about dualizable objects. Indeed, let me make the context extremely simple for a second: suppose $A=B$ is commutative and $P$ is just an $A$-module (so with the bimodule structure that comes with it). Then if $P$ is a dualizable $A$-module, one has a natural isomorphism $\hom_A(P,M)\cong \hom_A(P,A)\otimes_A M$, in which case you can write everything with tensors ($\hom_A(P,A)$ plays the role of $Q$ in that context).

More generally, Morita theory builds on the following general fact, which isn't restricted to equivalences:

Suppose $F: \mathrm{Mod}_A\to \mathrm{Mod}_B$ is a colimit-preserving functor. Then there is a unique $(B,A)$-bimodule $P$ such that $F\cong P\otimes_A -$

In particular, the right adjoint $\hom_B(P,-) : \mathrm{Mod}_B\to \mathrm{Mod}_A$ is of the form $Q\otimes_B-$ if and only if it preserves all colimits. For this, a number of things are required. Since colimits in $\mathrm{Mod}_A$ are detected in abelian groups, this is equivalent to $\hom_B(P,-) : \mathrm{Mod}_B\to \mathrm{Ab}$ preserving all colimits. In particular, this only depends on the $B$-module structure of $P$; and is in fact equivalent to $P$ being finitely presented and projective as a $B$-module.

Indeed, finitely presented is equivalent to preserving filtered colimits, so it is definitely required, and projective is equivalent to preserving epimorphisms, so it is also required. Conversrly, if you're finitely presented and projective as a $B$-module, you preserve filtered colimits, and finite colimits (finite coproducts and equalizers), so you preserve all colimits.

As a conclusion :

$\hom_B(P,-) \cong Q\otimes_B-$ for some $(A,B)$-bimodule $Q$ if and only if $P$ is finitely generated and projective as a $B$-module.

[note that for projective modules, finitely generated and finitely presented are equivalent]

When $A=B$ is commutative, this is equivalent to dualizability so we recover the example I gave earlier.

Side note: I'm not sure whether this is in fact usually called an induction functor, in my experience this name is reserved for when it's of the form $B\otimes_A -$, given a ring morphism $A\to B$.

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    $\begingroup$ Hi Maxime :) at some point you say "being finitely presented and free as a $B$-module" I think you wanted to say projective, and it is indeed what you prove later in the "indeed" part ! $\endgroup$ Jan 11 at 14:56
  • $\begingroup$ @jeanmfischer : yes of course ! Merci ! :) $\endgroup$ Jan 11 at 15:04

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