Corollary 3

Let (Eλ)_{λ ∈ L} be a family of right A-modules and F a free (resp. finitely generated free) left A-module. Then the canonical mapping (22)

is injective (resp. bijective).

If (b_μ) is a basis of F, every element of $\left( \prod_{\lambda \in L} E_\lambda \right) \otimes_A F$ can be written uniquely as $z = \sum_\mu ((x_\lambda^{(\mu)}) \otimes b_\mu)$ (Corollary 1); to say that its canonical image is zero means that, for all λ ∈ L, $\sum_\mu (x_\lambda^{(\mu)} \otimes b_\mu) = 0$, hence $x_\lambda^{(\mu)} = 0$ for all λ ∈ L and all μ (Corollary 1) and therefore $z = 0$.

Showing that the canonical mapping is bijective when F admits a finite basis is immediately reduced, by virtue of Proposition 7, to the case where

F = A_s; but then the two sides are canonically identified with $\prod_{\lambda \in L} E_\lambda$ (no. 4, Proposition 4) and after these identifications the canonical mapping (22) becomes the identity.