Proposition 8

Let M and $M'$ be A-modules, $(P, u)$ and $(P', u')$ projective covers of M and $M'$, respectively, and $g: M\rightarrow M'$ an A-linear mapping.

a) There exists an A-linear mapping $f: P\rightarrow P'$ such that $u'\circ f=g\circ u$.

b) Let $f$ be such a mapping. If $g$ is surjective (resp. bijective), then $f$ is surjective (resp. bijective). If $g$ is injective and its image is a direct factor of $M'$, then $f$ is injective and its image is a direct factor of $P'$.

By assumption, the A-module P is projective and the mapping $u'$ is surjective. Consequently, there exists an A-linear mapping $f: P\rightarrow P'$ such that $g\circ u=u'\circ f$ (II, §2, No. 2, p. 231, Proposition 4). Assertion a) follows.

Let $f$ be a mapping as in a). Suppose that $g$ is surjective. Since $u$ is surjective, we have $M'=g(u(P)) =u'(f(P))$. Since $(P', u')$ is a projective cover of M’, we have $f(P) = P'$, so $f$ is surjective. By loc. cit., the kernel of $f$ admits a supplementary submodule $P_1$, so $f(P_1) = P'$. Now suppose that $g$ is bijective. We have $g(u(P_1)) =u'(f(P_1)) =u'(P') = M'$, hence $u(P_1) = M$. Since $(P, u)$ is a projective cover of M, we have $P_1= P$, hence Ker($f$) $= 0$. So $f$ is injective; since we already know that $f$ is surjective, $f$ is bijective.

To conclude, suppose that $g$ is injective and that its image is a direct factor of $M'$. Then there exists an A-linear mapping $g': M'\rightarrow M$ such that $g'\circ g= 1_M$. By a), there exists an A-linear mapping $f': P'\rightarrow P$ such that $u\circ f'=g'\circ u'$. We have $u\circ (f'\circ f) =g'\circ u'\circ f= (g'\circ g)\circ u$; by the previous paragraph, the mapping $f'\circ f$ is bijective. Denote the converse bijection by $h$; we have $(h\circ f')\circ f= 1_P$, so $f$ is injective, and its image is a direct factor of $P'$ (II, §1, No. 9, p. 212, Corollary 2).