Editing Derived functor (section)

====[[Ext functor]]s====
If <math>R</math> is a [[ring (mathematics)|ring]], then the category of all left [[module (mathematics)|<math>R</math>-modules]] is an abelian category with enough injectives. If <math>A</math> is a fixed left <math>R</math>-module, then the functor <math>\operatorname{Hom}(A,-): R\text{-Mod} \to \mathfrak{Ab}</math> is left exact, and its right derived functors are the [[Ext functor]]s <math>\operatorname{Ext}_R^i(A,-)</math>. Alternatively <math>\operatorname{Ext}_R^i(-,B)</math> can also be obtained as the left derived functor of the right exact functor <math>\operatorname{Hom}_R(-,B): R\text{-Mod} \to \mathfrak{Ab}^{op}</math>.

Various notions of cohomology are special cases of Ext functors and therefore also derived functors.

* '''[[Group cohomology]]''' is the right derived functor of the invariants functor <math>(-)^G : k[G]\text{-Mod}\to k[G]\text{-Mod}</math> which is the same as <math>\operatorname{Hom}_{k[G]}(k,-)</math> (where <math>k</math> is the trivial <math>k[G]</math>-module) and therefore <math>H^i(G,M) = \operatorname{Ext}_{k[G]}^i(k,M)</math>.
* '''[[Lie algebra cohomology]]''' of a [[Lie algebra]] <math>\mathfrak{g}</math> over some commutative ring <math>k</math> is the right derived functor of the invariants functor <math>(-)^{\mathfrak{g}}: \mathfrak{g}\text{-Mod}\to k\text{-Mod}</math> which is the same as <math>\operatorname{Hom}_{U(\mathfrak{g})}(k,-)</math> (where <math>k</math> is again the trivial <math>\mathfrak{g}</math>-module and <math>U(\mathfrak{g})</math> is the [[universal enveloping algebra]] of <math>\mathfrak{g}</math>). Therefore <math>H^i(\mathfrak{g},M) = \operatorname{Ext}_{U(\mathfrak{g})}^i(k,M)</math>.
*'''[[Hochschild cohomology]]''' of some [[associative algebra|<math>k</math>-algebra]] <math>A</math> is the right derived functor of invariants <math>(-)^A: (A,A)\text{-Bimod}\to k\text{-Mod}</math> mapping a [[bimodule]] <math>M</math> to its [[center of a bimodule|center]], also called its set of invariants <math>M^A := Z(M) := \{m\in M \mid \forall a\in A: am=ma\}</math> which is the same as <math>\operatorname{Hom}_{A^e}(A,M)</math> (where <math>A^e:=A\otimes_k A^{op}</math> is the enveloping algebra of <math>A</math> and <math>A</math> is considered an <math>(A,A)</math>-bimodule via the usual left and right multiplication). Therefore <math>HH^i(A,M) = \operatorname{Ext}_{A^e}^i(A,M)</math>: