Proposition 10

Let M, N be two submodules of an A-module E. Then there are two exact sequences

where $i : M \to M + N$, $j : N \to M + N$ are the canonical injections,

the canonical surjections and where the homomorphisms u and v are defined as follow:

if $f : M \cap N \to M \to M \oplus N$ and $g : M \cap N \to N \to M \oplus N$ are the canonical injections, $u = f + g$, and if $r : E/(M \cap N) \to E/M \to (E/M) \oplus (E/N)$ and

are the canonical mappings, $v = r + s$.

We prove the exactness of (29): obviously $i - j$ is surjective and $u$ is injective. On the other hand, to say that $(i - j)(x, y) = 0$, where $x \in M$ and $y \in N$, means that $i(x) - j(y) = 0$, hence $i(x) = j(y) = z \in M \cap N$, whence by definition $x = f(z), y = g(z)$, which proves that $\operatorname{Ker}(i - j) = \operatorname{Im} u$.

We prove the exactness of (30): clearly $p - q$ is surjective. On the other hand, to say that $v(t) = 0$ for $t \in E/(M \cap N)$ means that $r(t) = s(t) = 0$, hence $t$ is the class mod. $(M \cap N)$ of an element $z \in E$ whose classes mod. $M$ and mod. $N$ are zero, which implies $z \in M \cap N$ and $t = 0$. Finally, to say that $(p - q)(x, y) = 0$, where $x \in E/M, y \in E/N$ means that $p(x) = q(y)$, or also that there exist two elements $z', z''$ of $E$ whose classes mod. $M$ and mod. $N$ are respectively $x$ and $y$ and which are such that $z' - z'' \in M + N$. Hence there are $t' \in M, t'' \in N$ such that $z' - z'' = t' - t''$, whence

Let $w$ be the class mod. $(M \cap N)$ of $z; r(w)$ is the class mod. $M$ of $z$ and hence also that of $z'$, that is $x$; similarly $s(w) = y$, which completes the proof that $\operatorname{Ker}(p - q) = \operatorname{Im} v$.