The Stacks project

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 $F$ 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. $\square$

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 4.24.4). 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. $\square$

Comments (2)

Comment #2038 by luke on

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

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0A8C. Beware of the difference between the letter 'O' and the digit '0'.