Derived functor

Remarks

Up to a canonical isomorphism, $R^n F(X)$ does not depend on the resolution, since any two injective resolutions are chain homotopy equivalent; the same argument makes $R^n F$ a functor. Left exactness gives $R^0 F \cong F$, and a short exact sequence in $\mathcal{A}$ gives a long exact sequence of the $R^n F$. Example: let $F = \Hom_\Z(\Z/2\Z, -)$ on abelian groups and $X = \Z$, with the injective resolution $0 \to \Z \to \Q \to \Q/\Z \to 0$. Then $F(\Q) = 0$ and $F(\Q/\Z) = \set{0, 1/2} \cong \Z/2\Z$, so $R^0 F(\Z) = 0$ and $R^1 F(\Z) \cong \Z/2\Z$, which is $\operatorname{Ext}^1_\Z(\Z/2\Z, \Z)$. Non-example: the complex $0 \to \Z \xrightarrow{\id} \Z \to 0$ is exact, but $\Z$ is not injective, and using it in place of $I^\bullet$ would give $0$ in degree $1$ instead of $\Z/2\Z$. In general $R^n \Hom_R(M, -) = \operatorname{Ext}^n_R(M, -)$ and $L_n(M \otimes_R -) = \operatorname{Tor}_n^R(M, -)$.