be the canonical mapping. For each $\alpha \in I$, let $u_\alpha$ be a mapping of a set F into $E_\alpha$ such that

Then :

(a) there exists a unique mapping $u$ of F into E such that

(b) the mapping $u$ is injective if and only if, for each pair of distinct elements $y$, $z$ of F, there exists $\alpha \in I$ such that $u_\alpha(y) \neq u_\alpha(z)$.

For the relation $u_\alpha = f_\alpha \circ u$ means that for each $y \in F$ we have

the element $u(y) \in \prod_{\alpha \in I} E_\alpha$ is uniquely determined by $u(y) = (u_\alpha(y))_{\alpha \in I}$. It remains to be shown that $u(y) \in E$ for all $y \in F$, in other words, that

whenever $\alpha \leqslant \beta$. But this can be written in the form

and therefore follows from (5). The second part of the Proposition follows immediately from the definitions.

is commutative [^1] whenever $\alpha \leqslant \beta$. Then there exists a unique mapping $u : E \to F$ such that for each $\alpha \in I$ the diagram

is commutative.

Put $v_\alpha = u_\alpha \circ f_\alpha$. If $\alpha \leqslant \beta$, we have, by (2),

and we may therefore apply Proposition 1 to the mappings $v_\alpha$; hence the existence and uniqueness of a mapping $u:E\to F$ such that

for each $\alpha\in I$.

¶ A family of mappings $u_\alpha:E_\alpha\to F_\alpha$ which satisfies the conditions of Corollary 1 is called an inverse system of mappings of $(E_\alpha,f_{\alpha\beta})$ into $(F_\alpha,g_{\alpha\beta})$. The mapping $u$ defined in Corollary 1 is called the inverse limit of the family $(u_\alpha)$ and is written $u=\underset{\leftarrow}{\lim}\,u_\alpha$ when there is no risk of confusion.

(7)

For if we put $w_\alpha=v_\alpha\circ u_\alpha$, then if $\alpha\leq\beta$ we have

which shows that $(w_\alpha)$ is an inverse system of mappings. Furthermore, if $u=\underset{\leftarrow}{\lim}\,u_\alpha$ and $v=\underset{\leftarrow}{\lim}\,v_\alpha$, then $h_\alpha\circ(v\circ u)=(v_\alpha\circ g_\alpha)\circ u=(v_\alpha\circ u_\alpha)\circ f_\alpha$ for each $\alpha\in I$, and therefore, by the uniqueness of the inverse limit, we have $v\circ u=\underset{\leftarrow}{\lim}\,w_\alpha$.

¶ Let $(E_\alpha,f_{\alpha\beta})$ be an inverse system of sets, and for each $\alpha\in I$ let $M_\alpha$ be a subset of $E_\alpha$. If $f_{\alpha\beta}(M_\beta)\subset M_\alpha$ whenever $\alpha\leq\beta$, the $M_\alpha$ are said to form an inverse system of subsets of the $E_\alpha$. Let $g_{\alpha\beta}$ be the mapping of $M_\beta$ into $M_\alpha$ (where $\alpha\leq\beta$) whose graph is the same as that of the restriction of $f_{\alpha\beta}$ to $M_\beta$. Then it is clear that $(M_\alpha,g_{\alpha\beta})$ is an inverse system of sets and that

(8)

For if $\alpha \leqslant \beta$ and $x_\beta \in \overset{-1}{u_\beta}(x'_\beta)$, we have

from which the first assertion follows; and to say that $x = (x_\alpha) \in \mathrm{E} = \varprojlim \mathrm{E}_\alpha$ is such that $u(x) = x'$ means, by definition, that $u_\alpha(x_\alpha) = x'_\alpha$ for each $\alpha \in \mathrm{I}$.

With the notation of Proposition 2, the images $u_\alpha(\mathrm{E}_\alpha)$ also form an inverse system of subsets of the $\mathrm{E}'_\alpha$, and we have

(9)

but the two sides of this relation are not necessarily equal (Exercise 4).

The verification is similar to that of Proposition 4.

Consider the double inverse system $(\mathrm{E}^\lambda_\alpha, g^{\lambda\mu}_{\alpha\beta})$ relative to $\mathrm{I} \times \mathrm{L}$, where the order relation on L is equality (no. 1, Example 1), and apply Proposition 4.

Let $(\alpha_n)$ be a sequence of elements of I which form a cofinal subset of I. Since I is directed, we can define a sequence $(\beta_n)$ of elements of I inductively by the conditions $\beta_0 = \alpha_0, \beta_n \geqslant \beta_i$ for $i < n$ and $\beta_n \geqslant \alpha_n$. Clearly the sequence $(\beta_n)$ is increasing and forms a cofinal subset of I. In view of Proposition 1 of no. 1 and the relations $f_\alpha = f_{\alpha\beta_n} \circ f_{\beta_n}$ for $\alpha \leqslant \beta_n$, we need only prove the Proposition for the case $\mathrm{I} = \mathbf{N}$. Moreover, it is clear that it suffices to prove that $f_0$ is surjective. Let $x_0 \in \mathrm{E}_0$. Define $x_n \in \mathrm{E}_n$ (for $n \geqslant 1$) inductively to be an element of the set $\overset{-1}{f}_{n-1,n}(x_{n-1})$, which is possible because the latter set is non-empty by hypothesis. We then show by induction on $n - m$ that $x_m = f_{mn}(x_n)$ for $m \leqslant n$, and it follows that $x = (x_n)$ belongs to E.

The second criterion concerns inverse systems $(\mathrm{E}_\alpha, f_{\alpha\beta})$ relative to an index set I such that for each $\alpha \in \mathrm{I}$ we are given a set $\mathfrak{S}_\alpha$ of subsets of $\mathrm{E}_\alpha$ which satisfy the following conditions :

(i) Every intersection of sets belonging to $\mathfrak{S}_\alpha$ also belongs to $\mathfrak{S}_\alpha$.

It follows in particular (by considering the intersection of the empty family) that $\mathrm{E}_\alpha \in \mathfrak{S}_\alpha$.

(ii) If a set of subsets $\mathfrak{F} \subset \mathfrak{S}_\alpha$ is such that every finite intersection of sets belonging to $\mathfrak{F}$ is non-empty, then $\bigcap\limits_{\mathbf{M} \in \mathfrak{F}} \mathrm{M}$ is non-empty.

In view of (i), it is clear that (ii) is equivalent to the following condition :

(ii)′ If $\mathfrak{G} \subset \mathfrak{S}_\alpha$ is left directed (with respect to inclusion) and does not contain the empty set, then $\bigcap\limits_{\mathbf{M} \in \mathfrak{G}} \mathrm{M}$ is non-empty.

(iii) For each pair of indices $\alpha$, $\beta$ such that $\alpha \leqslant \beta$, and each $x_\alpha \in \mathrm{E}_\alpha$, we have $\overset{-1}{f}_{\alpha\beta}(x_\alpha) \in \mathfrak{S}_\beta$.

(iv) For each pair of indices $\alpha$, $\beta$ such that $\alpha \leqslant \beta$, and each $\mathrm{M}_\beta \in \mathfrak{S}_\beta$, we have $f_{\alpha\beta}(\mathrm{M}_\beta) \in \mathfrak{S}_\alpha$.

Let $\mathrm{E} = \varprojlim \mathrm{E}_\alpha$ and let $f_\alpha : \mathrm{E} \to \mathrm{E}_\alpha$ be the canonical mapping for each $\alpha \in \mathrm{I}$. Then:

(a) For each $\alpha \in \mathrm{I}$ we have

(b) If $\mathrm{E}_\alpha$ is non-empty for each $\alpha \in \mathrm{I}$, then E is non-empty.

Let $\Sigma$ be the set of families $\mathfrak{A} = (\mathrm{A}_\alpha)_{\alpha \in \mathrm{I}}$ which satisfy the following conditions :

If $\mathfrak{A} = (A_\alpha)$ and $\mathfrak{A}' = (A'_\alpha)$ are any two elements of $\Sigma$, let the relation $\mathfrak{A} \leqslant \mathfrak{A}'$ mean that $A_\alpha \supset A'_\alpha$ for all $\alpha$. Clearly $\Sigma$ is ordered by this relation.

(1) Let us first show that the ordered set $\Sigma$ is inductive. Let L be a totally ordered set and let $\lambda \to \mathfrak{A}^\lambda = (A^\lambda_\alpha)_{\alpha \in I}$ be a strictly increasing mapping of L into $\Sigma$. For each $\alpha \in I$, put $B_\alpha = \bigcap_{\lambda \in L} A^\lambda_\alpha$. Then it is immediately seen that the family $\mathfrak{B} = (B_\alpha)_{\alpha \in I}$ satisfies (21); by reason of (i) and (ii), it also satisfies (20), hence belongs to $\Sigma$; and it is clear that $\mathfrak{B}$ is an upper bound of the set of the $\mathfrak{A}^\lambda$.

(2) Let $\mathfrak{A} = (A_\alpha)$ be a maximal element of $\Sigma$. We shall show that $A_\alpha = f_{\alpha\beta}(A_\beta)$ whenever $\alpha \leqslant \beta$. Let $A'_\alpha = \bigcap_{\beta \geqslant \alpha} f_{\alpha\beta}(A_\beta)$ for all $\alpha \in I$, and let us show that $\mathfrak{A}' = (A'_\alpha)$ belongs to $\Sigma$. Note first that if $\alpha \leqslant \beta \leqslant \gamma$, we have $f_{\alpha\gamma}(A_\gamma) = f_{\alpha\beta}(f_{\beta\gamma}(A_\gamma)) \subset f_{\alpha\beta}(A_\beta)$ by (21); moreover, $f_{\alpha\beta}(A_\beta) \in \mathfrak{S}_\alpha$ by (iv), and $f_{\alpha\beta}(A_\beta) \neq \emptyset$ by (20). Hence conditions (i) and (ii) show that $\mathfrak{A}'$ satisfies (20). Also, if $\alpha \leqslant \beta$, we have

and for each $\delta \geqslant \alpha$ there exists $\gamma \in I$ such that $\gamma \geqslant \delta$ and $\gamma \geqslant \beta$, so that $f_{\alpha\gamma}(A_\gamma) \subset f_{\alpha\delta}(A_\delta)$ and consequently

Hence $\mathfrak{A}'$ satisfies (21) and therefore belongs to $\Sigma$. Since $A'_\alpha \subset A_\alpha$ for all $\alpha$, the maximality of $\mathfrak{A}$ in $\Sigma$ implies $\mathfrak{A}' = \mathfrak{A}$, and our assertion is proved.

(3) We shall establish next that if $\mathfrak{A} = (A_\alpha)$ is a maximal element of $\Sigma$, then each of the $A_\alpha$ consists of a single element. Let $x_\alpha \in A_\alpha$. For each $\beta \geqslant \alpha$, put $B_\beta = A_\beta \cap \overset{-1}{f}_{\alpha\beta}(x_\alpha)$; if $\beta$ is not $\geqslant \alpha$, put $B_\beta = A_\beta$; then we shall see that $\mathfrak{B} = (B_\beta)$ belongs to $\Sigma$. If $\beta$ is not $\geqslant \alpha$, the relation $\beta \leqslant \gamma$ implies $f_{\beta\gamma}(B_\gamma) \subset f_{\beta\gamma}(A_\gamma) \subset A_\beta = B_\beta$. If, on the other hand, $\alpha \leqslant \beta \leqslant \gamma$, then since

we have

and since $f_{\beta\gamma}(A_\gamma) \subset A_\beta$, we again have $f_{\beta\gamma}(B_\gamma) \subset B_\beta$, so that the family $\mathfrak{B}$ satisfies (21). Since $A_\alpha = f_{\alpha\beta}(A_\beta)$ whenever $\alpha \leqslant \beta$ by part (2) of the proof, it is clear that $\mathrm{B}_\beta \neq \emptyset$ for all $\beta \in \mathrm{I}$. Finally, by virtue of (i) and (iii), we have $\mathrm{B}_\beta \in \mathfrak{S}_\beta$ for all $\beta \in \mathrm{I}$, and hence $\mathfrak{B} \in \Sigma$. Since $\mathrm{B}_\beta \subset \mathrm{A}_\beta$ for all $\beta \in \mathrm{I}$, the maximality of $\mathfrak{A}$ implies that $\mathrm{B}_\beta = \mathrm{A}_\beta$ for all $\beta$, and in particular $\mathrm{A}_\alpha = \{x_\alpha\}$.

(4) We are now in a position to prove Theorem 1. Let us start with (a). We have

Conversely, let

and put

if $\beta \geqslant \alpha$, and $\mathrm{B}_\beta = \mathrm{E}_\beta$ otherwise. By the definition of $x_\alpha$, the $\mathrm{B}_\beta$ are non-empty, and we have $\mathrm{B}_\beta \in \mathfrak{S}_\beta$ for all $\beta \in \mathrm{I}$ by virtue of (iii) and (i); moreover, it is evident that $f_{\beta\gamma}(\mathrm{B}_\gamma) \subset \mathrm{B}_\beta$ whenever $\beta \leqslant \gamma$. Hence $\mathfrak{B} = (\mathrm{B}_\beta) \in \Sigma$. Let $\mathfrak{A} = (\mathrm{A}_\alpha)$ be a maximal element of $\Sigma$ such that $\mathfrak{A} \geqslant \mathfrak{B}$ (the existence of $\mathfrak{A}$ follows from (1) and §2, no. 4, Theorem 2, Corollary 1). Since, by (3), $\mathrm{A}_\beta$ is of the form $\{y_\beta\}$ for all $\beta \in \mathrm{I}$, it follows that $y = (y_\beta)$ belongs to E, and $f_\alpha(y) = y_\alpha = x_\alpha$ by definition.

Finally, (a) implies (b). We may as well assume that I is not empty (otherwise there is nothing to prove). The hypothesis that the $\mathrm{E}_\alpha$ are non-empty implies that $f_{\alpha\beta}(\mathrm{E}_\beta) \neq \emptyset$ for all $\beta \geqslant \alpha$. Since the $f_{\alpha\beta}(\mathrm{E}_\beta)$, for fixed $\alpha$ and variable $\beta \geqslant \alpha$, form a left directed set of subsets of $\mathrm{E}_\alpha$ belonging to $\mathfrak{S}_\alpha$, condition (ii)$'$ proves that

Hence $f_\alpha(\mathrm{E}) \neq \emptyset$ by (a), and a fortiori $\mathrm{E} \neq \emptyset$. Q.E.D.