K(Ab). for this reason this functor has a correct derived functor. it truly is simply noticeable to be functorial in X', and so we have now a bi-~-functor RiiHom" : K(A) O • D+(A) >D(Ab). Now utilizing the lemma, half b), we see that this functor is "exact" within the first variable, complexes, i. e. , takes acyclic complexes and as a result passes to the quotient, into acyclic giving a trivial correct derived functor ~I~ Hom': D(A) ~ • D+(A) (We will denote this functor through > D(Ab).
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