The Stacks project

Definition 63.6.4. Let $F: \mathcal{A} \to \mathcal{B}$ be a left exact functor and assume that $\mathcal{A}$ has enough injectives. We define the total right derived functor of $F$ as the functor $RF: D^+(\mathcal{A}) \to D^+(\mathcal{B})$ fitting into the diagram

\[ \xymatrix{ D^+(\mathcal{A}) \ar[r]^{RF} & D^+(\mathcal{B}) \\ K^+(\mathcal I) \ar[u] \ar[r]^ F & K^+(\mathcal{B}). \ar[u] } \]

This is possible since the left vertical arrow is invertible by the previous lemma. Similarly, let $G: \mathcal{A} \to \mathcal{B}$ be a right exact functor and assume that $\mathcal{A}$ has enough projectives. We define the total left derived functor of $G$ as the functor $LG: D^-(\mathcal{A}) \to D^-(\mathcal{B})$ fitting into the diagram

\[ \xymatrix{ D^-(\mathcal{A}) \ar[r]^{LG} & D^-(\mathcal{B}) \\ K^-(\mathcal{P}) \ar[u] \ar[r]^ G & K^-(\mathcal{B}). \ar[u] } \]

This is possible since the left vertical arrow is invertible by the previous lemma.


Comments (2)

Comment #3411 by Dongryul Kim on

Typo: the second derived functor should be a left derived functor

There are also:

  • 4 comment(s) on Section 63.6: Derived categories

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Your email address will not be published.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 03T7. Beware of the difference between the letter 'O' and the digit '0'.