The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

13.7 Adjoints for exact functors

Results on adjoint functors between triangulated categories.

Lemma 13.7.1. Let $F : \mathcal{D} \to \mathcal{D}'$ be an exact functor between triangulated categories. If $F$ has a right adjoint, then it is an exact functor.

Proof. Let $G$ be a right adjoint. 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{Mor}\nolimits _\mathcal {D}(X, G(A[1]) & = \mathop{Mor}\nolimits _{\mathcal{D}'}(F(X), A[1]) \\ & = \mathop{Mor}\nolimits _{\mathcal{D}'}(F(X)[-1], A) \\ & = \mathop{Mor}\nolimits _{\mathcal{D}'}(F(X[-1]), A) \\ & = \mathop{Mor}\nolimits _\mathcal {D}(X[-1], G(A)) \\ & = \mathop{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

  1. $F$ and $G$ are exact functors,

  2. $F$ is fully faithful,

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

  4. 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. $\square$


Comments (2)

Comment #2038 by luke on

In the proof of lemma 13.7.1, functor should be homological instead of cohomological.


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