Proposition 5

With the hypotheses and notation of Proposition 4, suppose that for all $\alpha \in I$, $E_\alpha$ is the direct sum of a family $(M^\lambda_\alpha)_{\alpha \in L}$ of submodules (the indexing set $L$ being independent of $\alpha$) and that $f_{\beta \alpha}(M^\lambda_\alpha) \subset M^\lambda_\alpha$ for $\alpha \leq \beta$ and for all $\lambda \in L$. Then $E$ is the direct sum of the submodules $M^\lambda = \lim \to M^\lambda_\alpha$ $(\lambda \in L)$.

It follows from Proposition 4 that $E$ is the sum of the $M^\lambda$. Let $(y^\lambda)_{\lambda \in L}$ be a family such that $y^\lambda \in M^\lambda$ for all $\lambda \in L$ and whose support is finite and suppose that $\sum_\lambda y^\lambda = 0$. By virtue of Set Theory, III, § 7, no. 5, Lemma 1, there exist an $\alpha \in I$ and a family $(x^\lambda_\alpha)_{\alpha \in L}$ of finite support consisting of elements of $E_\alpha$ such that $x^\lambda_\alpha \in M^\lambda_\alpha$ and $y^\lambda = f_\alpha(x^\lambda_\alpha)$ for all $\lambda \in L$. The relation $f_\alpha \left( \sum_{\lambda \in L} x^\lambda_\alpha \right) = 0$ implies the existence of a $\beta \geq \alpha$ such that $f_{\beta \alpha} \left( \sum_{\lambda \in L} x^\lambda_\alpha \right) = 0$ *(Set Theory, III, § 7, no. 5, Lemma 1), which may be written as $\sum_{\lambda \in L} x_\beta^\lambda = 0$, where $x_\beta^\lambda = f_{\beta \alpha}(x_\alpha^\lambda) \in M_\beta^\lambda$ by hypothesis; hence $x_\beta^\lambda = 0$ for all $\lambda \in L$ and therefore $y^\lambda = f_\beta(x_\beta^\lambda) = 0$ for all $\lambda \in L$, which proves that the sum of the $M^\lambda$ is direct.