Corollary

Let $(P_\alpha)$ be a direct system of subsets of $E_\alpha$ and let $P = \lim \overrightarrow{P_\alpha}$. If, for all $\alpha \in I$, $P_\alpha$ is a free subset (resp. basis) of $E_\alpha$, then $P$ is a free subset (resp. basis) of $E$.

The second assertion follows immediately from the first and Proposition 4. It is therefore sufficient to prove that if the $P_\alpha$ are free every subset $\{ y^{(i)} \}_{1 \leq i \leq n}$ consisting of distinct elements of $P$, is free. There exists an $\alpha \in I$ and elements $x_\alpha^{(i)} \in P_\alpha$ such that $y^{(i)} = f_\alpha(x_\alpha^{(i)})$ for $1 \leq i \leq n$ (Set Theory, III, § 7, no. 5, Lemma 1); if $\sum_i \lambda^{(i)} y^{(i)} = 0$, it may be assumed that $\lambda^{(i)} = \phi_\alpha(\lambda_\alpha^{(i)})$ for $1 \leq i \leq n$ and hence $f_\alpha \left( \sum_i \lambda_\beta^{(i)} x_\beta^{(i)} \right) = 0$; this implies $\sum_i \lambda_\beta^{(i)} x_\beta^{(i)} = 0$ for some $\beta \geq \alpha$, where $\lambda_\beta^{(i)} = \phi_{\beta \alpha}(\lambda_\alpha^{(i)})$, $x_\beta^{(i)} = f_{\beta \alpha}(x_\alpha^{(i)})$ and the $x_\beta^{(i)}$ belong to $P_\beta$ and are distinct since $y^{(i)} = f_\beta(x_\beta^{(i)})$; then $\lambda_\beta^{(i)} = 0$ for $1 \leq i \leq n$, whence

for $1 \leq i \leq n$.

Suppose now that all the rings $A_\alpha$ are equal to the same ring $A$ and the $\phi_{\beta \alpha}$ to $1_A$; then, for every direct system $(E_\alpha, f_{\beta \alpha})$ of $A$-modules, $E = \lim \overrightarrow{E_\alpha}$ is an $A$-module. Let $F$ be an $A$-module and for all $\alpha$ let $u_\alpha : E_\alpha \to F$ be an $A$-linear mapping such that $(u_\alpha)$ is a direct system of mappings; then $u = \lim \overrightarrow{u_\alpha}$ is an $A$-linear mapping of $E$ into $F$. Conversely, for every $A$-linear mapping $v : \lim \overrightarrow{E_\alpha} \to F$, the family of $v_\alpha = v \circ f_\alpha$ is a direct system of $A$-linear mappings such that $v = \lim \overrightarrow{v_\alpha}$. On the other hand we note that for $\alpha \leq \beta$ the mapping

is a $\mathbf{Z}$-module homomorphism such that $(\operatorname{Hom}_A(E_\alpha, F), \bar{f}_{\alpha \beta})$ is an inverse system of $\mathbf{Z}$*-modules*; as $\bar{f}_{\alpha \beta}(v_\beta) = v_\beta \circ f_{\beta \alpha}$, the above remarks can be expressed as follows: