Proposition 8

a) If the A-module E is of finite type (resp. of finite presentation), the canonical homomorphism (16) is injective (resp. bijective).

b) Suppose that the B-module F is flat; if the A-module E is of finite type (resp. of finite presentation), the canonical homomorphism (17) is injective (resp. bijective).

Let us prove for example b), the proof of a) being analogous. Consider A, B, F, G as fixed, and, for every right A-module E, put

and denote by $\nu_E$ the homomorphism (17); for every homomorphism $v:E\to E'$ of right A-modules, put $T(v)=\operatorname{Hom}(v,1_G)\otimes 1_F$ and $T'(v)=\operatorname{Hom}(v,1_G\otimes 1_F)$.

Let $L_1 \xrightarrow{v} L_0 \xrightarrow{w} E \to 0$ be a presentation of $E$; we suppose the free module $L_0$ (resp. the free modules $L_0$ and $L_1$) of finite type. The diagram

is commutative, and its second line is exact (II, p. 36, th. 1); moreover, the sequence

is exact (loc. cit.), and since $F$ is flat, the first line of (18) is also an exact sequence (X, p. 8, déf. 1). This being so, we know that $v_{L_0}$ is bijective (resp. that $v_{L_0}$ and $v_{L_1}$ are bijective) (II, p. 75, prop. 2). If we suppose only that $v_{L_0}$ is bijective, it follows from (18) that $v_{L_0} \circ T(w) = T'(w) \circ v_E$ is injective, hence $v_E$ is so also. If we suppose that $v_{L_0}$ and $v_{L_1}$ are both bijective, we deduce from cor. 2 (ii) of X, p. 6 that $v_E$ is surjective, and since we have just seen that $v_E$ is injective, it is bijective.