2
$\begingroup$

Let $F:\mathcal C\to\mathcal C'$ be an additive functor of abelian categories, and assume that the right derived functor $RF:\text D(\mathcal C)\to\text D(\mathcal C')$ exists (see this question for the definition of $RF$).

On p. 330 of the book Categories and Sheaves by Kashiwara and Schapira it is claimed that $RF$ is triangulated.

I'm unable to prove this.

I'm not even able to show that $RF$ is additive.

Thank you very much in advance for your help.

$\endgroup$

1 Answer 1

2
$\begingroup$

Consider the (non-commutative) diagram $$ \begin{matrix} \text K(\mathcal C)&\overset{\text K(F)}{\to}&\text K(\mathcal C')\\ {\scriptstyle Q}\downarrow&&\downarrow\scriptstyle Q'\\ \text D(\mathcal C)&\underset{RF}{\to}&\text D(\mathcal C'). \end{matrix} $$

We have $$ RF(X)\simeq\operatorname*{colim}_{Y\to X}\ \text K(F)(Y), $$ where $Y\to X$ runs over the morphisms from $Y$ to $X$ in $\text D(\mathcal C)$. Using this expression for $RF(X)$ it is easy to see that $RF$ commutes with finite products. By Proposition 8.2.15 p. 173 of Categories and Sheaves by Kashiwara and Schapira, this implies that $RF$ is additive.

As $RF$ commutes with finite products, it commutes with mapping cones, and is thus triangulated.

(We have implicitly used the fact that $\text K(\mathcal C)$ and $\text D(\mathcal C)$ have the same set of objects, that $Q$ acts on this set as the identity, and that the same holds for $\mathcal C'$.)

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .