Proposition 15

Let $u$ be a linear mapping of a module $E$ into a module $F$. If the equation $u(x) = y_0$ has at least one solution, $y_0$ is orthogonal to the kernel of $^t u$.

To say that $u(x) = y_0$ admits a solution means that $y_0 \in u(E)$ and the proposition follows from no. 5, Corollary to Proposition 8.

† This is in fact an abuse of language; from the logical point of view, we are not here defining the word "solution", but simply the sentence "$x_0$ is a solution of the equation $u(x) = y_0$" as equivalent to the relation "$x_0 \in E$ and $u(x_0) = y_0$". Observe that in a mathematical theory $\mathscr{T}$ where the relation "$A$ is a ring, $E$ and $F$ are $A$-modules, $u$ is a homomorphism of $E$ into $F$, $y_0$ an element of $F$" is a theorem, every term $T$ of $\mathscr{T}$ such that the relation "$T \in E$ and $u(T) = y_0$" is true in $\mathscr{T}$ is a solution of the equation $u(x) = y_0$ in the sense of Set Theory, I, § 5, no. 2; this justifies the above abuse of language.

Observe that the necessary criterion for the existence of a solution of $u(x) = y_0$, given by Proposition 15, is sufficient when $A$ is a field (§ 7, no. 6, Proposition 12), but not in general (Exercise 10).