0 Note that, to preserve duality, one should really speak of "direct factor" in (4) and (5), rather than "direct summand". However, the two notions coincide! 1. 4. 2. Let A be a A-module and let G be a divisible abelian group containing A. Show that we may embed A in an injective module by the scheme A = HomA(A, A) ~ Homz(A, A) ~ Homz(A, G). 3. For any A-module A, let A* be the right A-module Homz(A, ~). Show that A is naturally embedded in A**. Use this embedding and a free presentation of A* to embed A in a cofree module.
El Proof. We prove the proposition only for A = PtBQ. The proof in the general case is analogous. First assume P and Q projective. Let s : B- C be surjective and y: P tB Q-+C a homomorphism. Define yp = yip: P-+C and YQ = Y IQ : Q-+ C. Since P, Q are projective there exist f3p, f3 Qsuch that ef3p = YP' ef3Q = YQ' By the universal property of the direct sum there exists f3: PtBQ-+B such that f3lp = f3p and f3IQ = f3Q. It follows that (ef3)/p=ef3p=Yp=Ylp and (ef3)/Q=ef3Q=YQ=YIQ. By the uniqueness part of the universal property we conclude that e f3 = y.