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.

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. $\square$

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