Exercise 1, § 7

This solution was written by nemotron-3-ultra-free and judged by a machine. It is not Bourbaki's and it has not been checked by a person.

We first show that $(\mathrm{F}_\lambda, g_{\lambda\mu})$ is an inverse system relative to $L$. For each $\lambda \in L$ the set $\mathrm{F}_\lambda$ is the inverse limit of the system restricted to $\mathrm{J}_\lambda$. If $\lambda \leqslant \mu$ then $\mathrm{J}_\lambda \subset \mathrm{J}_\mu$ by (ii); the canonical mapping $g_{\lambda\mu} : \mathrm{F}_\mu \to \mathrm{F}_\lambda$ is defined by restricting a family indexed by $\mathrm{J}_\mu$ to $\mathrm{J}_\lambda$ (no. 1). By property (4) of no. 1, for $\lambda \leqslant \mu \leqslant \nu$ we have $g_{\lambda\nu} = g_{\lambda\mu} \circ g_{\mu\nu}$, and $g_{\lambda\lambda}$ is the identity on $\mathrm{F}_\lambda$. Hence the conditions $(\mathrm{LP}_\mathrm{I})$ and $(\mathrm{LP}_\mathrm{II})$ hold for $(\mathrm{F}_\lambda, g_{\lambda\mu})$ on $L$; it is an inverse system.

Let $\mathrm{F} = \varprojlim_{\lambda \in L} \mathrm{F}_\lambda$; denote by $p_\lambda : \mathrm{F} \to \mathrm{F}_\lambda$ the canonical projections.

We now construct a canonical bijection $\mathrm{F} \to \mathrm{E}$ using the universal property of inverse limits (Proposition 1 of no. 2, tag 03OC).

Definition of $u : \mathrm{F} \to \mathrm{E}$. For each $\alpha \in I$ we define a mapping $u_\alpha : \mathrm{F} \to \mathrm{E}_\alpha$. Let $z = (x_\lambda)_{\lambda \in L} \in \mathrm{F}$. By (iii) there exists $\lambda$ with $\alpha \in \mathrm{J}_\lambda$; we set $u_\alpha(z) = x_\lambda(\alpha)$. If also $\alpha \in \mathrm{J}_\mu$, choose $\nu \in L$ with $\nu \geqslant \lambda,\mu$ (possible because $L$ is directed). Then $\mathrm{J}_\lambda, \mathrm{J}_\mu \subset \mathrm{J}_\nu$ by (ii), and the compatibility $g_{\lambda\nu}(x_\nu) = x_\lambda$, $g_{\mu\nu}(x_\nu) = x_\mu$ gives $x_\lambda(\alpha) = x_\nu(\alpha) = x_\mu(\alpha)$. Hence $u_\alpha$ is well defined.

Let $\alpha \leqslant \beta$ in $I$. Choose $\lambda$ with $\alpha \in \mathrm{J}_\lambda$ and $\mu$ with $\beta \in \mathrm{J}_\mu$. Pick $\nu \geqslant \lambda,\mu$ in $L$. Then $\alpha,\beta \in \mathrm{J}_\nu$. Since $x_\nu \in \mathrm{F}_\nu = \varprojlim_{\gamma \in \mathrm{J}_\nu} \mathrm{E}_\gamma$, we have $f_{\alpha\beta}(x_\nu(\beta)) = x_\nu(\alpha)$. By definition $u_\alpha(z) = x_\nu(\alpha)$ and $u_\beta(z) = x_\nu(\beta)$, so $f_{\alpha\beta}(u_\beta(z)) = u_\alpha(z)$. Thus $f_{\alpha\beta} \circ u_\beta = u_\alpha$ for all $\alpha \leqslant \beta$.

By Proposition 1 (no. 2) there exists a unique mapping $u : \mathrm{F} \to \mathrm{E}$ such that $f_\alpha \circ u = u_\alpha$ for every $\alpha \in I$, where $f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha$ are the canonical projections.

Definition of $v : \mathrm{E} \to \mathrm{F}$. For each $\lambda \in L$ let $v_\lambda : \mathrm{E} \to \mathrm{F}_\lambda$ be the canonical mapping obtained by restricting the index set from $I$ to $\mathrm{J}_\lambda$ (no. 1): $v_\lambda(y) = (f_\alpha(y))_{\alpha \in \mathrm{J}_\lambda}$. If $\lambda \leqslant \mu$ then $\mathrm{J}_\lambda \subset \mathrm{J}_\mu$, and by (4) of no. 1 we have $v_\lambda = g_{\lambda\mu} \circ v_\mu$. Applying Proposition 1 to the inverse system $(\mathrm{F}_\lambda, g_{\lambda\mu})$ on $L$, there exists a unique mapping $v : \mathrm{E} \to \mathrm{F}$ such that $p_\lambda \circ v = v_\lambda$ for all $\lambda \in L$.

$u$ and $v$ are inverses. Take $y \in \mathrm{E}$. For any $\alpha \in I$ choose $\lambda$ with $\alpha \in \mathrm{J}_\lambda$. Then $f_\alpha(u(v(y))) = u_\alpha(v(y)) = (v_\lambda(y))_\alpha = f_\alpha(y)$. Since the projections $f_\alpha$ separate points in the product $\prod_{\alpha \in I} \mathrm{E}_\alpha$, we obtain $u(v(y)) = y$.

Take $z = (x_\lambda) \in \mathrm{F}$. For any $\lambda \in L$ and any $\alpha \in \mathrm{J}_\lambda$, $(v_\lambda(u(z)))_\alpha = f_\alpha(u(z)) = u_\alpha(z) = x_\lambda(\alpha)$ (using the same $\lambda$ in the definition of $u_\alpha$). Hence $v_\lambda(u(z)) = x_\lambda = p_\lambda(z)$. Because the projections $p_\lambda$ separate points in $\prod_{\lambda \in L} \mathrm{F}_\lambda$, we get $v(u(z)) = z$.

Therefore $u$ is a bijection with inverse $v$. The construction uses only the given data and the canonical mappings, so $u$ is a canonical bijection of $\mathrm{F}$ onto $\mathrm{E}$.