B-)C>0 to )A'B'C')0 177 17P 77 0 in RM with exact rows. Prove that: a) If rl and 0 are one-to-one, then so is b. b) If rt and ¢ are onto, then so is V). 15. Suppose A E SMR, B E RM, and C E SM. Then Homs(A, C) becomes a left R-module, and A OR B becomes a left S-module.

The proof of C) is left exact. (b) is similar, since If A is flat and n > 1, then A ® Pn+1 -+ A ® Pn -* A ® Pn_1 is exact since A® is an exact functor. Hence, Torn (A, B) = 0 for n > 0. The proof that Extn(B, C) = 0 if C is injective works pretty much the same way. Finally, if B is projective, then is a projective resolution of B. Applying A® and deleting the last term gives just B)0. Hence, Torn (A, B) = 0 for n > 1. Again, Ext works virtually the same way. 0 The reader who has been wondering how two functors could measure three properties (departure from flatness, projectivity, or injectivity) now has an answer: Ext(B, C) simultaneously measures departure of B from projectivity and departure of C from injectivity.

Choose P2, and d2 : P2 - ker d1 onto. Etc. (By the way, there is a reason for using the notation it : Po -+ B as will become evident. Also, our constructions will eventually ignore 7r, so it has been left from the (Pn, dn) notation. ) As noted earlier, there are choices involved here. The extent to which the choices drop out is covered by the following proposition. 1 Suppose B, B' E RM, and co E Hom(B, B'). Suppose (Pn, dn) is a projective resolution of B, and (,nn d') is a projective resolution of V.