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Lets $A$ and $B$ two algebras, $_AU_B$ and $_B V_A$ two bi-modules such that there exists two isomorphisms $U \otimes_B V \simeq A$ and $V \otimes_A U \simeq B$. Let $F: Mod A \to Mod B$ a functor, if $F \simeq -\otimes_A U$ then $F$ is an equivalence between $Mod A$ and $Mod B$.

There are at least two ways to solve this exercise, the first one is showing that $F$ is faithfull, full and dense. The another way is showing that there exists a functor $G: Mod B \to Mod A$ s.t $FG \simeq 1_{mod B}$ and $1_{Mod A} \simeq GF$.

Using $F \simeq -\otimes_A U$, follows from Watts Theorem that $F$ admit a right adjoint functor $G$, further from that I've proved that there exists a morphism between $FG$ and $1_{Mod B}$, but I'm stuck here.

I need some help...

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1 Answer 1

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Simply define $G$ to be $-\otimes_B V$. Then $GF\simeq -\otimes_A U\otimes_B V$ and $FG\simeq-\otimes_B V\otimes_A U$ are isomorphic to the identity functors since $U \otimes_B V \simeq A$ and $V \otimes_A U \simeq B$.

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  • $\begingroup$ Of course, thank you!! $\endgroup$
    – Joãonani
    Jun 23, 2021 at 3:10

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