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 $\xi _ X : F(X[1]) \to F(X)[1]$ be as in Definition 13.3.3. Let $\epsilon _ A : F(G(A)) \to A$ be the adjunction map. Consider the composition
\[ F(G(A)[1]) \xrightarrow {\xi _{G(A)}} F(G(A))[1] \xrightarrow {\epsilon _ A[1]} A[1] \]
This map is adjoint to a map $G(A)[1] \to G(A[1])$ which we claim to be an isomorphism. To see this, by the Yoneda lemma it suffices to show that we get a bijection on applying $\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[1], -)$ for every object $X$ of $\mathcal{D}$. Now, every morphism $f : X[1] \to G(A)[1]$ is of the form $g[1]$ for a unique $g : X \to G(A)$ and every $g$ is adjoint to a unique $h : F(X) \to A$ and in turn $h[1] \circ \xi _ X$ is adjoint to a unique $i : X[1] \to G(A[1])$. The rule sending $f$ to $i$ gives a bijection between $\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[1], G(A)[1])$ and $\mathop{\mathrm{Mor}}\nolimits _\mathcal {D}(X[1], G(A[1]))$. That this bijection is given by the morphism above follows from the discussion in Categories, Section 4.24 and the following calculation
\begin{align*} \epsilon _ A[1] \circ \xi _{G(A)} \circ F(f) & = \epsilon _ A[1] \circ \xi _{G(A)} \circ F(g[1]) \\ & = \epsilon _ A[1] \circ F(g)[1] \circ \xi _ X \\ & = (\epsilon _ A \circ F(g))[1] \circ \xi _ X \\ & = h[1] \circ \xi _ X \end{align*}
Some details omitted. We will show that $G$ is an exact functor using the inverse of the isomorphisms $G(A)[1] \to G(A[1])$. These isomorphisms are functorial in $A$ (details omitted).
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
\[ \xymatrix{ F(G(A)) \ar[r] \ar[d]^{\epsilon _ A} & F(G(B)) \ar[r] \ar[d]^{\epsilon _ B} & F(X) \ar[r] \ar[d] & F(G(A))[1] \ar[d]^{\epsilon _ A[1]} \\ A \ar[r] & B \ar[r] & C \ar[r] & A[1] } \]
Recall that the arrow $F(X) \to F(G(A))[1]$ is the composition of $F(X) \to F(G(A)[1])$ with $\xi _{G(A)}$. Hence using that $G$ is the right adjoint of $F$ and our discussion above, we conclude the existence of 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 and the right vertical arrow is the morphism constructed above. 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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