Editing Derived functor (section)

== Motivation ==

It was noted in various quite different settings that a [[short exact sequence]] often gives rise to a "long exact sequence". The concept of derived functors explains and clarifies many of these observations.

Suppose we are given a covariant [[left exact functor]] ''F'' : '''A''' → '''B''' between two [[abelian category|abelian categories]] '''A''' and '''B'''. If 0 → ''A'' → ''B'' → ''C'' → 0 is a short exact sequence in '''A''', then applying ''F'' yields the exact sequence 0 → ''F''(''A'') → ''F''(''B'') → ''F''(''C'') and one could ask how to continue this sequence to the right to form a long exact sequence. Strictly speaking, this question is ill-posed, since there are always numerous different ways to continue a given exact sequence to the right. But it turns out that (if '''A''' is "nice" enough) there is one [[canonical form|canonical]] way of doing so, given by the right derived functors of ''F''. For every ''i''≥1, there is a functor ''R<sup>i</sup>F'': '''A''' → '''B''', and the above sequence continues like so: 0 → ''F''(''A'') → ''F''(''B'') → ''F''(''C'') → ''R''<sup>1</sup>''F''(''A'') → ''R''<sup>1</sup>''F''(''B'') → ''R''<sup>1</sup>''F''(''C'') → ''R''<sup>2</sup>''F''(''A'') → ''R''<sup>2</sup>''F''(''B'') → ... . From this we see that ''F'' is an exact functor if and only if ''R''<sup>1</sup>''F'' = 0; so in a sense the right derived functors of ''F'' measure "how far" ''F'' is from being exact.

If the object ''A'' in the above short exact sequence is [[injective object|injective]], then the sequence [[Splitting lemma|splits]]. Applying any additive functor to a split sequence results in a split sequence, so in particular ''R''<sup>1</sup>''F''(''A'') = 0. Right derived functors (for ''i>0'') are zero on injectives: this is the motivation for the construction given below.