Lemma 3

Under the hypotheses of Lemma 2, for $\nu_F$ to be injective (resp. surjective), it is necessary and sufficient that $\nu_M$ and $\nu_N$ be so.

This follows from Lemma 2 and § 1, no. 6, Corollary 1 to Proposition 6.

Then Lemma 3, together with § 2, no. 2, Proposition 4, shows that it suffices to consider the case where $F$ is a free module. But, if $(b_\mu)$ is a basis of $F$, every element of $\mathrm{Hom}_A(E, G) \otimes_B F$ may then be written uniquely as $\sum_\mu u_\mu \otimes b_\mu$, where $u_\mu \in \mathrm{Hom}_A(E, G)$ (§ 3, no. 7, Corollary 1 to Proposition 7); the image of this element under $\nu$ is the $A$-linear mapping $x \mapsto \sum_\mu u_\mu(x) \otimes b_\mu$; it cannot be zero for all $x \in E$ unless $u_\mu(x) = 0$ for all $x \in E$ and all $\mu$, which is equivalent to saying that $u_\mu = 0$ for all $\mu$; hence $\nu$ is injective. When also $F$ admits a finite basis, Lemma 3 shows (by induction on the number of elements in the basis of $F$) that to prove that $\nu$ is surjective, it suffices to do so when $F = B_s$; but in this case the two sides of (7) are canonically identified with $\mathrm{Hom}_A(E, G)$ (§ 3, no. 4, Proposition 4) and $\nu$ becomes the identity.

(ii) To show the proposition when $E$ is projective and finitely generated, this time we fix $F$ and $G$ and write, for every left $A$-module $E$,

and, for every left $A$-module homomorphism $v : E \to E'$,

on the other hand, we write $\nu_E$ instead of $\nu$. Then we have the two lemmas: