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The cateogory $ D(\mathcal{C})$

We assume $ \mathcal{C}$ is an abelian category. We then add some inverses of quasi isomorphisms in $ K(\mathcal{C})$ to define $ D(\mathcal{C})$ . $ D(\mathcal{C})$ again is not necessarily be an abelian category, but it is a triangulated category which has distinguished triangles which satisfy certain axioms.

By considering only complexes which are bounded below, we may define $ C^+(\mathcal{C}), K^+(\mathcal{C}), D^+(\mathcal{C})$ etc.

PROPOSITION 11.4   [1, 4.8] If $ \mathcal{C}$ has enoufh injectives then $ D^+(\mathcal{C})$ is equivalent to $ K^+(I(\mathcal{C}))$ , where $ I(\mathcal{C})$ is the category of injective objects in $ \mathcal{C}$ .

So, in a sence, to consider an object $ X^\bullet$ of $ D^+(\mathcal{C})$ is to consider an injective resolution $ I^\bullet$ of $ X^\bullet$ and treat it up to homotopy.

For left-exact functor $ \mathcal{C}_1\to \mathcal{C}_2$ , we may ``define'' (the actual definiton should be done more carefully. See [1])

$\displaystyle \mathbb{R}F: D^+(\mathcal{C}_1)\to D^+(\mathcal{C}_2)
$

by

$\displaystyle \mathbb{R}F(X^\bullet)=F(I^\bullet)
$

where $ I^\bullet$ is an injective resolution of $ X^\bullet$ .

A good thing about treating derived functors in this way is that we may easily treat derived functors of compositions:

$\displaystyle \mathbb{R}(F\circ G)
\cong (\mathbb{R}F)\circ (\mathbb{R}G).
$


next up previous
Next: Bibliography Up: Categories, abelian categories and Previous: The category
2009-07-31