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

1. the functors $S$ and $N$ are exact,

2. there is an isomorphism $g : N \circ S \to \text{id}_\mathcal {B}$ to the identity functor of $\mathcal{B}$,

3. $N$ is faithful, and

4. $S$ 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

$f^{-1} \circ g_{S(B)} \circ S(N(f)) : S(N(A)) \to S(N(S(B))) \to S(B) \to A.$

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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