Proposition 6

Given two exact sequences (14) and (16), the homomorphism v \otimes t : E \otimesA F \to E'' \otimes_A F'' is surjective and its kernel is equal to*

Now $v \otimes t = (v \otimes 1_{F''}) \circ (1_E \otimes t)$ (no. 2, formula (5)) and $v \otimes t$ is therefore surjective, being the composition of two surjective homomorphisms by virtue of Proposition 5 and its Corollary. On the other hand, for $z \in E \otimes F$ to be in the kernel of $v \otimes t$, it is necessary and sufficient that $(1_E \otimes t)(z)$ belong to the kernel of $v \otimes 1_{F''}$, that is, by virtue of (15) to the image of

But as the homomorphism $t: F \to F''$ is surjective, so is

by the Corollary to Proposition 5, hence the condition on $z$ reduces to the existence of an $a \in E' \otimes F$ such that

Let $b = z - (u \otimes 1_F)(a)$; then $(l_E \otimes t)(b) = 0$, and by virtue of (17), $b$ belongs to the image of $l_E \otimes s$, which proves the proposition.

In other words: