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

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

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

Since $G$ is the adjoint the new morphism determines a morphism $X \to G(C)$ such that the diagram

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

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. $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: