Exercise 3, § 7

Tag 03OT, permanent. Cite it as [Bourbaki, Tag 03OT].

Theory of Sets, III, 251

Let $(\mathrm{E}_\alpha, f_{\alpha\beta})$ and $(\mathrm{F}_\alpha, g_{\alpha\beta})$ be two inverse systems of sets relative to the same index set I. For each $\alpha \in \mathrm{I}$ , let $u_\alpha$ be a mapping of $\mathrm{E}_\alpha$ into $\mathrm{F}_\alpha$ , such that the $u_\alpha$ form an inverse system of mappings. Let $\mathrm{G}_\alpha \subset \mathrm{E}_\alpha \times \mathrm{F}_\alpha$ be the graph of $u_\alpha$ . Show that $(\mathrm{G}_\alpha)$ is an inverse system of subsets of $\mathrm{E}_\alpha \times \mathrm{F}_\alpha$ and that its inverse limit may be canonically identified with the graph of $u = \varprojlim u_\alpha$ .

Solution

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.

Let $(\mathrm{E}_\alpha, f_{\alpha\beta})$ and $(\mathrm{F}_\alpha, g_{\alpha\beta})$ be two inverse systems of sets relative to the same index set $\mathrm{I}$ . For each $\alpha \in \mathrm{I}$ , let $u_\alpha : \mathrm{E}_\alpha \to \mathrm{F}_\alpha$ be a mapping such that the family $(u_\alpha)$ is an inverse system of mappings, i.e. $g_{\alpha\beta} \circ u_\beta = u_\alpha \circ f_{\alpha\beta}$ whenever $\alpha \leqslant \beta$ . Let $\mathrm{G}_\alpha \subset \mathrm{E}_\alpha \times \mathrm{F}_\alpha$ be the graph of $u_\alpha$ , so $\mathrm{G}_\alpha = \{ (x, u_\alpha(x)) \mid x \in \mathrm{E}_\alpha \}$ .

For $\alpha \leqslant \beta$ , the mapping $f_{\alpha\beta} \times g_{\alpha\beta} : \mathrm{E}_\beta \times \mathrm{F}_\beta \to \mathrm{E}_\alpha \times \mathrm{F}_\alpha$ sends $(x,y)$ to $(f_{\alpha\beta}(x), g_{\alpha\beta}(y))$ . If $(x,y) \in \mathrm{G}_\beta$ , then $y = u_\beta(x)$ , and

(f_{\alpha\beta}(x), g_{\alpha\beta}(y)) = (f_{\alpha\beta}(x), g_{\alpha\beta}(u_\beta(x))) = (f_{\alpha\beta}(x), u_\alpha(f_{\alpha\beta}(x))) \in \mathrm{G}_\alpha,

where the second equality uses the hypothesis

g_{\alpha\beta} \circ u_\beta = u_\alpha \circ f_{\alpha\beta}

. Hence the restriction

h_{\alpha\beta} = (f_{\alpha\beta} \times g_{\alpha\beta})|_{\mathrm{G}_\beta}

maps

\mathrm{G}_\beta

into

\mathrm{G}_\alpha

. We verify that

(\mathrm{G}_\alpha, h_{\alpha\beta})

satisfies the inverse system axioms. - For each

\alpha

f_{\alpha\alpha} = \operatorname{id}_{\mathrm{E}_\alpha}

and

g_{\alpha\alpha} = \operatorname{id}_{\mathrm{F}_\alpha}

, so

f_{\alpha\alpha} \times g_{\alpha\alpha} = \operatorname{id}_{\mathrm{E}_\alpha \times \mathrm{F}_\alpha}

. Restricting to

\mathrm{G}_\alpha

gives

h_{\alpha\alpha} = \operatorname{id}_{\mathrm{G}_\alpha}

. - If

\alpha \leqslant \beta \leqslant \gamma

, then

f_{\alpha\gamma} = f_{\alpha\beta} \circ f_{\beta\gamma}

and

g_{\alpha\gamma} = g_{\alpha\beta} \circ g_{\beta\gamma}

, hence

f_{\alpha\gamma} \times g_{\alpha\gamma} = (f_{\alpha\beta} \times g_{\alpha\beta}) \circ (f_{\beta\gamma} \times g_{\beta\gamma})

. Since

h_{\beta\gamma}

maps

\mathrm{G}_\gamma

into

\mathrm{G}_\beta

, the restriction of the composition to

\mathrm{G}_\gamma

equals

h_{\alpha\beta} \circ h_{\beta\gamma}

, while it is also

h_{\alpha\gamma}

. Thus

h_{\alpha\gamma} = h_{\alpha\beta} \circ h_{\beta\gamma}

. Therefore

(\mathrm{G}_\alpha, h_{\alpha\beta})

is an inverse system of subsets of

\mathrm{E}_\alpha \times \mathrm{F}_\alpha

Let $\mathrm{E} = \varprojlim \mathrm{E}_\alpha$ and $\mathrm{F} = \varprojlim \mathrm{F}_\alpha$ , with canonical projections $f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha$ and $g_\alpha : \mathrm{F} \to \mathrm{F}_\alpha$ . By Proposition 1 of § 1 (tag 03OC) applied to the inverse system $(\mathrm{F}_\alpha, g_{\alpha\beta})$ and the family of mappings $v_\alpha = u_\alpha \circ f_\alpha : \mathrm{E} \to \mathrm{F}_\alpha$ , which satisfies $g_{\alpha\beta} \circ v_\beta = v_\alpha$ for $\alpha \leqslant \beta$ , there exists a unique mapping $u : \mathrm{E} \to \mathrm{F}$ such that $v_\alpha = g_\alpha \circ u$ for all $\alpha$ , i.e.

u_\alpha \circ f_\alpha = g_\alpha \circ u \qquad (\alpha \in \mathrm{I}).

(Although Proposition 1 is stated for an ordered index set, its proof only uses the preorder relations — transitivity and reflexivity — and does not require antisymmetry; hence it remains valid for preordered sets.) This

u

is the inverse limit

\varprojlim u_\alpha

The product inverse system $(\mathrm{E}_\alpha \times \mathrm{F}_\alpha, f_{\alpha\beta} \times g_{\alpha\beta})$ has inverse limit canonically isomorphic to $\mathrm{E} \times \mathrm{F}$ by Corollary 2 of Proposition 4 (tag 03OH). Under this isomorphism, an element $(z_\alpha) \in \varprojlim (\mathrm{E}_\alpha \times \mathrm{F}_\alpha)$ corresponds to $(x,y) \in \mathrm{E} \times \mathrm{F}$ with $x_\alpha = \operatorname{pr}_1(z_\alpha)$ , $y_\alpha = \operatorname{pr}_2(z_\alpha)$ .

The inverse limit $\varprojlim \mathrm{G}_\alpha$ is the subset of $\varprojlim (\mathrm{E}_\alpha \times \mathrm{F}_\alpha)$ consisting of families $(z_\alpha)$ with $z_\alpha \in \mathrm{G}_\alpha$ for all $\alpha$ . Under the canonical isomorphism, this corresponds to the set of $(x,y) \in \mathrm{E} \times \mathrm{F}$ such that for every $\alpha$ ,

(x_\alpha, y_\alpha) \in \mathrm{G}_\alpha \iff y_\alpha = u_\alpha(x_\alpha).

(x,y)

satisfies this condition, then

y_\alpha = u_\alpha(x_\alpha) = u_\alpha(f_\alpha(x)) = g_\alpha(u(x))

for all

\alpha

, so

y = u(x)

because the canonical map

\mathrm{F} \to \prod_\alpha \mathrm{F}_\alpha

is injective. Conversely, if

y = u(x)

, then

g_\alpha(y) = g_\alpha(u(x)) = u_\alpha(f_\alpha(x))

, so

y_\alpha = u_\alpha(x_\alpha)

for all

\alpha

. Hence the image of

\varprojlim \mathrm{G}_\alpha

is exactly the graph of

u

. This identification is canonical because it is induced by the canonical projections.

Cited by

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Cites

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