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])

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

by

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

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:

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