Proposition 3

Let $\mathrm{I}$ be a preordered set, let $(\mathrm{E}_\alpha, f_{\alpha\beta})$ be an inverse system of sets relative to $\mathrm{I}$, and let $\mathrm{E} = \varprojlim \mathrm{E}_\alpha$. Let $\mathrm{J}$ be a cofinal subset of $\mathrm{I}$ such that $\mathrm{J}$ is right directed, and let $\mathrm{E}'$ be the inverse limit of the inverse system of sets obtained from $(\mathrm{E}_\alpha, f_{\alpha\beta})$ by restricting the index set to $\mathrm{J}$. Then the canonical mapping $g$ of $\mathrm{E}$ into $\mathrm{E}'$ (no. 1, formula (3)) is bijective.

For each $\alpha \in \mathrm{J}$, let $f'_\alpha$ be the canonical mapping $\mathrm{E}' \to \mathrm{E}_\alpha$. Then, by (2) and (5), $g$ is the unique mapping of $\mathrm{E}$ into $\mathrm{E}'$ such that $f_\alpha = f'_\alpha \circ g$ for all $\alpha \in \mathrm{J}$ (Proposition 1). We shall show that $g$ is injective by using the criterion of Proposition 1. If $x$, $y$ are distinct elements of $\mathrm{E}$, then by definition there exists $\alpha \in \mathrm{I}$ such that $f_\alpha(x) \neq f_\alpha(y)$; since $\mathrm{J}$ is cofinal in $\mathrm{I}$, there exists $\lambda \in \mathrm{J}$ such that $\alpha \leqslant \lambda$; since $f_{\alpha\lambda}(f_\lambda(x)) \neq f_{\alpha\lambda}(f_\lambda(y))$, we have $f_\lambda(x) \neq f_\lambda(y)$. It remains to be shown that $g$ is surjective. Let $x' = (x'_\lambda)_{\lambda \in \mathrm{J}}$ be an element of $\mathrm{E}'$. For each $\alpha \in \mathrm{I}$ there exists $\lambda \in \mathrm{J}$ such that $\alpha \leqslant \lambda$, and the element $f_{\alpha\lambda}(x'_\lambda)$ does not depend on the index $\lambda \in \mathrm{J}$ such that $\alpha \leqslant \lambda$; for if $\mu \in \mathrm{J}$ is such that $\alpha \leqslant \mu$, there exists $\nu \in \mathrm{J}$ such that $\lambda \leqslant \nu$ and $\mu \leqslant \nu$, hence $f_{\alpha\lambda}(x'_\lambda) = f_{\alpha\lambda}(f_{\lambda\nu}(x'_\nu)) = f_{\alpha\nu}(x'_\nu)$, and similarly $f_{\alpha\mu}(x'_\mu) = f_{\alpha\nu}(x'_\nu)$. Let $x_\alpha$ be the common value of the $f_{\alpha\lambda}(x'_\lambda)$ for the $\lambda \in \mathrm{J}$ such that $\alpha \leqslant \lambda$, and let $x = (x_\alpha)_{\alpha \in \mathrm{I}}$. Then $x \in \mathrm{E}$, because if $\alpha \leqslant \beta$ and if $\lambda \in \mathrm{J}$ is such that $\beta \leqslant \lambda$, we have

Finally, we have $x'_\lambda = f_{\lambda\lambda}(x'_\lambda)$ for all $\lambda \in \mathrm{J}$, hence $x_\lambda = x'_\lambda$ for all $\lambda \in \mathrm{J}$; in other words, $f(x_\lambda) = x'_\lambda$, so that $g(x) = x'$. Therefore $g$ is surjective and the proof is complete.

In particular, if $\mathrm{I}$ has a greatest element $\omega$, we may take $\mathrm{J} = \{\omega\}$, so that $\varprojlim \mathrm{E}_\alpha$ is canonically identified with $\mathrm{E}_\omega$.

Remarks