By L. D. Olson

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2. (FR1) and sequence of a 36 be given, with s 6 S. (Z,Y,N,s,f,g). Then (u, idN) the first, Complete Complete fu s to a triangle to a triangle (W,X,N,t,fu,h). is a map of two sides of the second triangle so there is a map v: W >Z into g i v i n g a m o r p h i s m of triangles. V Z t/u idN X >Y U Now sv = ut, so it remains = 0 to prove t 6 S. s 6 S, we have H(Ti(N)) for all i 6 ~ sequence of the first triangle. sequence of the second triangle, Indeed, since by the long exact Applying this to the long exact we H(Ti(t)) find is an i s o m o r p h i s m for all i 6 ~.

ZP+I(x ") > I p+l to a map easily that the resulting map isomorphism, as required. fp+l: f: xP+l X" are in IP/im I p-I A' is Extend the natural map > I p+I. 7. let I be the (additive) of A. A be an abelian category, and subcategory of injective objects Then the natural functor K+(I) is fully faithful. > D+(A) (Note that the results of section 3 carry over to additive subcategories of abelian categories. , if every admits an injection into an inJective object) is an equivalence of categories. Proof.

Of Lemma call the h o m o t o p y (k,t): morphism > Y" as above. cone of s. 4, and so is h o m o t o p i c operator T(I') 9 Y" i Then we have the e q u a t i o n V = (idi. , O) = (k,t) Separating the components, we d Z + d I (k,t) find Let Z" is acyclic, satisfies to zero. the Let us 42 id I = dk + kd + ts and dt - td = O . Thus t: Y" homotopic > I" to 1). 6. Lemma Let (i) element of is a m o r p h i s m Let P is a h o m o t o p y A be an a b e l i a n category. A E v e r y object of Ob A X" Assume (ii) and assume admits an i n j e c t i o n I" of o b j e c t s of o > Y >x P that into an X 6 P, then x~ > xI is an exact sequence, Then every X" I" of objects Let A' of A'.