Corollary

Let $(\mathrm{E}_\alpha, f_{\beta\alpha})$ and $(\mathrm{E}'_\alpha, f'_{\beta\alpha})$ be two direct systems of sets, let $(u_\alpha)$ be a direct system of mappings $u_\alpha : \mathrm{E}_\alpha \to \mathrm{E}'_\alpha$, and let $u = \varinjlim u_\alpha$.

(i) Let $(\mathrm{M}_\alpha)$ be a direct system of subsets of the $\mathrm{E}_\alpha$. Then $(u_\alpha(\mathrm{M}_\alpha))$ is a direct system of subsets of the $\mathrm{E}'_\alpha$, and we have

(26)

(ii) Let $(a'_\alpha)_{\alpha \in \mathbf{I}}$ be a family such that $a'_\alpha \in \mathrm{E}'_\alpha$ for each $\alpha \in \mathrm{I}$ and $f'_{\beta\alpha}(a'_\alpha) = a'_\beta$ whenever $\alpha \leqslant \beta$. Then the sets $\overset{-1}{u}_\alpha(a'_\alpha)$ form a direct system of subsets of the $\mathrm{E}_\alpha$, and we have

(27)

where $a'$ is the unique element of $\varinjlim \mathrm{E}'_\alpha$ which is the canonical image of $a'_\alpha$ for each $\alpha \in \mathrm{I}$.

(i) It is evident that the $u_\alpha(\mathrm{M}_\alpha)$ form a direct system of subsets of the $\mathrm{E}'_\alpha$, and we may write $u_\alpha(\mathrm{M}_\alpha) = v_\alpha(\mathrm{M}_\alpha)$, where $v_\alpha$ is the mapping of $\mathrm{M}_\alpha$ onto $u_\alpha(\mathrm{M}_\alpha)$ whose graph is the same as that of the restriction of $u_\alpha$ to $\mathrm{M}_\alpha$. The formula (26) then follows from Proposition 7 because the $v_\alpha$ are surjective.

(ii) Let $\mathrm{N}_\alpha = \overset{-1}{u}_\alpha(a'_\alpha)$. If $\alpha \leqslant \beta$ and $x_\alpha \in \mathrm{N}_\alpha$, then

hence $f_{\beta\alpha}(x_\alpha) \in \mathrm{N}_\beta$, and the $\mathrm{N}_\alpha$ therefore form a direct system of subsets of the $\mathrm{E}_\alpha$. With the notation of the proof of Proposition 7, consider an element $x \in \varinjlim \mathrm{N}_\alpha$. There exists $\alpha \in \mathrm{I}$ and $x_\alpha \in \mathrm{N}_\alpha$ such that $x = f_\alpha(x_\alpha)$, so that $u(x) = u(f_\alpha(x_\alpha)) = f'_\alpha(u_\alpha(x_\alpha)) = f'_\alpha(a'_\alpha) = a'$. Conversely, if $x \in \overset{-1}{u}(a')$ and if $x = f_\alpha(x_\alpha)$ for some $\alpha \in \mathrm{I}$ and some $x_\alpha \in \mathrm{E}_\alpha$, then we have $a' = u(f_\alpha(x_\alpha)) = f'_\alpha(u_\alpha(x_\alpha)) = f'_\alpha(a'_\alpha)$. Hence (no. 5, Lemma 1) there exists $\beta \geqslant \alpha$ such that $f'_{\beta\alpha}(u'_\alpha(x_\alpha)) = f'_{\beta\alpha}(a'_\alpha) = a'_\beta$; i.e., $u_\alpha(f_{\beta\alpha}(x_\alpha)) = a'_\beta$, and therefore $f_{\beta\alpha}(x_\alpha) \in \mathrm{N}_\beta$. Since $x = f_\beta(f_{\beta\alpha}(x_\alpha))$, it follows that $x \in \varinjlim \mathrm{N}_\alpha$.