Lemma 13.7.1. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor between triangulated categories. If $F$ admits a right adjoint $G: \mathcal{D'} \to \mathcal{D}$, then $G$ is also an exact functor.
13.7 Adjoints for exact functors
Results on adjoint functors between triangulated categories.
Proof. Let $X$ be an object of $\mathcal{D}$ and $A$ an object of $\mathcal{D}'$. Since $F$ is an exact functor we see that
\begin{align*} \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X, G(A[1])) & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(X), A[1]) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(X)[-1], A) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathcal{D}'}(F(X[-1]), A) \\ & = \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[-1], G(A)) \\ & = \mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X, G(A)[1]) \end{align*}
By Yoneda's lemma (Categories, Lemma 4.3.5) we obtain a canonical isomorphism $G(A)[1] = G(A[1])$. Let $A \to B \to C \to A[1]$ be a distinguished triangle in $\mathcal{D}'$. Choose a distinguished triangle
\[ G(A) \to G(B) \to X \to G(A)[1] \]
in $\mathcal{D}$. Then $F(G(A)) \to F(G(B)) \to F(X) \to F(G(A))[1]$ is a distinguished triangle in $\mathcal{D}'$. By TR3 we can choose a morphism of distinguished triangles
\[ \xymatrix{ F(G(A)) \ar[r] \ar[d] & F(G(B)) \ar[r] \ar[d] & F(X) \ar[r] \ar[d] & F(G(A))[1] \ar[d] \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] } \]
Since $G$ is the adjoint the new morphism determines a morphism $X \to G(C)$ such that the diagram
\[ \xymatrix{ G(A) \ar[r] \ar[d] & G(B) \ar[r] \ar[d] & X \ar[r] \ar[d] & G(A)[1] \ar[d] \\ G(A) \ar[r] & G(B) \ar[r] & G(C) \ar[r] & G(A)[1] } \]
commutes. Applying the homological functor $\mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(W, -)$ for an object $W$ of $\mathcal{D}'$ we deduce from the $5$ lemma that
\[ \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(W, X) \to \mathop{\mathrm{Hom}}\nolimits _{\mathcal{D}'}(W, G(C)) \]
is a bijection and using the Yoneda lemma once more we conclude that $X \to G(C)$ is an isomorphism. Hence we conclude that $G(A) \to G(B) \to G(C) \to G(A)[1]$ is a distinguished triangle which is what we wanted to show. $\square$
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 4.24.4). 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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