Lemma 14.24.2. Let $\mathcal{A}$ and $\mathcal{B}$ be abelian categories. Let $N : \mathcal{A} \to \mathcal{B}$, and $S : \mathcal{B} \to \mathcal{A}$ be functors. Suppose that
the functors $S$ and $N$ are exact,
there is an isomorphism $g : N \circ S \to \text{id}_\mathcal {B}$ to the identity functor of $\mathcal{B}$,
$N$ is faithful, and
$S$ is essentially surjective.
Then $S$ and $N$ are quasi-inverse equivalences of categories.
Proof. It suffices to construct a functorial isomorphism $S(N(A)) \cong A$. To do this choose $B$ and an isomorphism $f : A \to S(B)$. Consider the map
It is easy to show this does not depend on the choice of $f, B$ and gives the desired isomorphism $S \circ N \to \text{id}_\mathcal {A}$. $\square$
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