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Thus A is a complex. D. , A is exact. We are also interested in the opposite direction. 18. 18': Suppose . . + P, 4 P,- --* * * * Po 3 P - -+0 is an exact sequence of projective R-modules. Then there are maps si:pi -+ P, + such that dos-, = 1 , - , and di+,si+ si- ,di = l p i for all i 2 0. -+ , Proof: Since d o is split epic we can find spl with dos-, = l p - , . 0(8) to the diagram to obtain si:&-+ pi+, satisfying di+,si = 1 - si-,di, or di+,si + si-,di = 1. D. The same idea applies to homotopy.

15 to the diagram 7 0 ---+Mi- - - -,- , Mi MJM, 0 j (iii) By (ii) we have an exact sequence 0 + N -+ F s P 0 where F is filtered-free and f , g are strictiy filtered. 6. e. gp = 1, for some p: P + F. Note is a summand of G ( P )and thus is projective; hence the epic fi -+ Pi/e- splits. , piecing to get pi c, fi). D. 39: For any filtered module M we have pd, M Ipd,(,, G ( M ) . - Let n = pdGo, G(M). There is nothing to prove unless n is finite, so we proceed by induction on n. 38(iii) gives the result for n = 0.

Writing u = (a,,. , v is a unimodular vector in R'"- l ) (viewed in R'"))under the corresponding change of base. Hence Rv is a summand of R'"- '), so writing R'"- ') = Rv @ P' we get R(")z (Ru@ P ' ) 0 R = RU@ (P' @ R) = Rv @ R("-'). Factoring out Rv yields P = R'"'/Rv % R("-'). D. 55: Suppose s ( R ) = k. g. stably free module which cannot be generated by k elements then P is free. D. the lemma (since, clearly, n > k). 56: If R is lefr Noetherian with K,(R) = ([R]), and if K-dimR = k, then every f g .