Lemma 13.7.2. Let $\mathcal{D}$, $\mathcal{D}'$ be triangulated categories. Let $F : \mathcal{D} \to \mathcal{D}'$ and $G : \mathcal{D}' \to \mathcal{D}$ be functors. Assume that

$F$ and $G$ are exact functors,

$F$ is fully faithful,

$G$ is a right adjoint to $F$, and

the kernel of $G$ is zero.

Then $F$ is an equivalence of categories.

**Proof.**
Since $F$ is fully faithful the adjunction map $\text{id} \to G \circ F$ is an isomorphism (Categories, Lemma 4.24.3). Let $X$ be an object of $\mathcal{D}'$. Choose a distinguished triangle

\[ F(G(X)) \to X \to Y \to F(G(X))[1] \]

in $\mathcal{D}'$. Applying $G$ and using that $G(F(G(X))) = G(X)$ we find a distinguished triangle

\[ G(X) \to G(X) \to G(Y) \to G(X)[1] \]

Hence $G(Y) = 0$. Thus $Y = 0$. Thus $F(G(X)) \to X$ is an isomorphism.
$\square$

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