Corollary 4

Let $0 \to E \xrightarrow{f} F \xrightarrow{g} G \to 0$ be an exact sequence of A-modules, S a generating system of E, T a generating system of G. If T' is a subset of F such that $g(T') = T$, T' \cup f(S) is a generating system of F.

The submodule F' of F generated by T' \cup f(S) contains $f(E)$ and, as $g(F')$ contains T, $g(F') = G$; hence $F' = F$.