Proposition 4

Let $N$ and $P$ be two submodules of an $A$-module $M$ such that their sum $N + P$ is a direct factor in $M$ and their intersection $N \cap P$ is a direct factor in $N$ and in $P$. Then the homomorphisms $T(N) \to T(M)$, $T(P) \to T(M)$ and

canonical extensions of the canonical injections, are injective; if $T(N)$, $T(P)$ and $T(N \cap P)$ are identified with subalgebras of $T(M)$ by means of these homomorphisms, then (5)

By hypothesis, there exist submodules $N' \subset N$ and $P' \subset P$ such that $N = N' \oplus (N \cap P)$, $P = P' \oplus (N \cap P)$; then

and there exists by hypothesis a submodule $M'$ of $M$ such that

In particular, $N + P$, $N$, $P$ and $N \cap P$ are direct factors in $M$, which implies, as has been seen above, that the canonical homomorphisms

are injective. The three algebras $T(N)$, $T(P)$ and $T(N \cap P)$ are thus identified with subalgebras of $T(N + P)$ and the latter with a subalgebra of $T(M)$;

writing $Q = N \cap P$, it remains to show that, if $T(Q)$, $T(N' \oplus Q)$ and $T(P' \oplus Q)$ are identified with subalgebras of $T(N' \oplus P' \oplus Q)$, then

Now, consider the commutative diagram

where the horizontal arrows are the canonical injections and the vertical arrows the canonical projections. We derive a commutative diagram

where $r$ and $s$ are surjective homomorphisms (Proposition 3) and $u$ and $v$ injective homomorphisms. Hence, to prove (6), note that the right hand side is obviously contained in the left; it therefore suffices to verify that if

then $x \in T(Q)$. Now the definition of the homomorphism $s$ shows that its restriction to $T(P' \oplus Q)$ (identified with a subalgebra of $T(N' \oplus P' \oplus Q)$) is the identity mapping; the hypothesis on $x$ therefore implies that $s(u(x)) = x$. But then also $v(r(x)) = x$, in other words $x$ belongs to the image of $T(Q)$ in $T(P' \oplus Q)$, which was to be proved.