We know already (§ 2, no. 1, Proposition 1) that $\overline{H}$ is a subgroup of $G$. Also if $\alpha$ is any operator on $G$, the image of $H$ under the continuous mapping $x \to x^\alpha$ is contained in $H$, and therefore the image of $\overline{H}$ is contained in $\overline{H}$ (Chapter I, § 2, no. 1, Theorem 1).

Let $H$ be a stable normal subgroup of a topological group $G$ with operators. Then for each operator $\alpha$ on $G$, the mapping of $G/H$ into itself induced by $x \to x^\alpha$ is continuous (§ 2, no. 8, Remark 3), and the structure of group with operators on $G/H$ is therefore compatible with the quotient topology on $G/H$.

Let $(G_i)_{i \in I}$ be a family of topological groups with operators, where each $G_i$ is assumed to have the same operator set $\Omega$. For each $\alpha \in \Omega$, the mapping $x \to ((\mathrm{pr}_i x)^\alpha)$ of $G = \prod_{i \in I} G_i$ into itself is continuous (Chapter I, § 4, no. 1, Proposition 1), and the structure of groups with operators on $G$ is therefore compatible with the product topology on $G$.

If $G$ is a Hausdorff topological group with operators and if $G$ has a completion $\hat{G}$ (§ 3, no. 4), then every endomorphism $x \to x^\alpha$ of $G$ defined by an operator on $G$ can be extended by continuity to an endomorphism of $\hat{G}$ (§ 3, no. 3, Proposition 5). Hence $\hat{G}$ has the structure of a topological group with operators, and the operator set is the same for $\hat{G}$ as for $G$.

For the mapping $x \to (p_i(x))$ is the inverse of the mapping $(x_i) \to \sum_{i=1}^n x_i$.

Since $1 = \sum_{i=1}^n P_i$ (where $1$ denotes the identity mapping of $E$) it is sufficient for $n - 1$ of the projectors $p_i$ to be continuous in order that the $n$ th should also be continuous.

If $E$ is the topological direct sum of two stable subgroups $M, N$, then $N$ is said to be a topological complement of $M$ in $E$; this is the case if and only if the canonical mapping of $E/M$ onto is an isomorphism of topological groups with operators.

a) $M$ has a topological complement in $E$.

b) There is a continuous projector $p$ of $E$ into $E$ such that $p(E) = M$.

c) The identity mapping of $M$ can be extended to a continuous linear mapping of $E$ onto $M$.

It follows from Proposition 2 that a) implies b), and it is clear that b) implies c). Finally, if $p$ is a continuous linear mapping of $E$ onto $M$ which extends the identity mapping of $M$, then $p$ is a continuous projector, and the projectors $p$ and $1 - p$ are associated with the direct sum decomposition $E = M + N$, where $N = \overline{p}(0)$.

The conditions are necessary. For we have then $u(v(F)) = F$ and a fortiori $u(E) = F$; furthermore, if $p = v \circ u$, then $p$ is a continuous linear mapping of $E$ into itself such that $p^2 = p$; therefore (Corollary to Proposition 2) $p(E) = v(u(E)) = v(F)$ has a topological complement $\overline{p}(0)$ in $E$; but since $u(p(x))) = u(x)$ by hypothesis, we have $\overline{u}(0) = \overline{p}(0)$. Finally, the bijective mapping of $E/\overline{u}(0)$ onto $F$, associated with $u$, is the composition of the bijective mapping of $E/\overline{u}(0)$ onto $v(F)$, associated with $p$, and the restriction of $u$ to $v(F)$; since $v$ is continuous, both these mappings are isomorphisms, and therefore $u$ is a strict morphism of $E$ onto $F$.

The conditions are sufficient. For if $\varphi$ is the canonical homomorphism of E onto $E/\overline{u}^1(o)$, to say that $\overline{u}^1(o)$ has a topological complement M in E is to say that the restriction of $\varphi$ to M is an isomorphism of M onto $E/\overline{u}^1(o)$. Since on the other hand $u = w \circ \varphi$, where w is an isomorphism of $E/\overline{u}^1(o)$ onto F, we see that the restriction of u to M is an isomorphism of M onto F, and the inverse isomorphism v is therefore such that $u \circ v$ is the identity mapping of F onto itself.

The conditions are sufficient, for if they are satisfied we obtain a left inverse v of u by taking the composition of the isomorphism of u(E) onto E which is the inverse of u, with a continuous projector of F onto u(E).

The conditions are necessary. For the relation $v(u(x)) = x$ shows that $\overline{u}^1(o) = \{ o \}$; u is therefore a bijection of E onto u(E), and since the restriction of v to u(E) is continuous, it follows that u is an isomorphism of E onto u(E). On the other hand, if we put $q = u \circ v$, then q is a continuous linear mapping of F onto u(E) such that $q^2 = q$, which proves (Corollary to Proposition 2) that u(E) has a topological complement in F.

The proof is the same as for Proposition 8 of § 2, n. 3 : if for example $K$ is a left ideal in $H$, then the mapping $(z, x) \to zx$ is continuous on $A \times A$ and maps $H \times K$ into $K$; hence it maps $A \times K = \overline{H} \times K$ into $\overline{H}$.

Let $H$ be a two-sided ideal in a topological ring $A$. By the same argument as for quotient groups, we see that the quotient of the topology of $A$ by the relation $x - y \in H$ is compatible with the ring structure of $A/H$. In particular, if $A$ is not Hausdorff, then the closure $N$ of $\{0\}$ in $A$ is a closed two-sided ideal, by Proposition 5; the quotient ring, which is Hausdorff ($\S$ 2, no. 5, Proposition 13) is called the Hausdorff ring associated with $A$.

Let $(A_i)_{i \in I}$ be a family of topological rings. On the set $A = \prod_{i \in I} A_i$, the product of the topologies of the $A_i$ is compatible with the product of the ring structures of the $A_i$ (same proof as for product groups); the topological ring $A$ thus defined is called the product of the topological rings $A_i$.

Let $(x_0, y_0)$ be an arbitrary point of $E \times F$, and let $\mathfrak{U}, \mathfrak{B}$ be the traces on A, B respectively of the neighbourhood filters of $x_0, y_0$ ($\mathfrak{U}, \mathfrak{B}$ are filters, by hypothesis). To show that f can be extended by continuity, it is enough to show that $f(\mathfrak{U} \times \mathfrak{B})$ is a Cauchy filter-base on G (Chapter II, § 3, no. 6, Proposition 11). Consider the identity

We shall show that by taking $(x, y)$ and $(x', y')$ in a sufficiently small set of $\mathfrak{U} \times \mathfrak{B}$, and by choosing $x_1$ and $y_1$ suitably, we can make each of the terms on the right-hand side very small. Let W be any neighbourhood of 0 in G; since f is continuous at $(0, 0) \in A \times B$, there is a set $U \in \mathfrak{B}$ and a set $V \in \mathfrak{B}$ such that $f(x' - x, y' - y) \in W$ whenever $x \in U, x' \in U, y \in V, y' \in V$. Take a point $x_1 \in U$ and a point $y_1 \in V$; then for all $x, x'$ in U and all $y, y'$ in V we shall have

On the other hand, the partial mapping $x \to f(x, y_1)$ is continuous on A; hence there is a set $U' \subset U$, belonging to $\mathfrak{U}$, and such that, whenever $x$ and $x'$ belong to $U'$, we have $f(x' - x, y_1) \in W$. Likewise there exists $V' \subset V$ belonging to $\mathfrak{B}$ such that, whenever $y$ and $y'$ belong to $V'$, we have $f(x_1, y' - y) \in W$. Consequently, if $(x, y)$ and $(x', y')$ are any two points of $U' \times V'$, then

this proves the existence of the extension $\overline{f}$ of f. The fact that $\overline{f}$ is $\mathbf{Z}$-bilinear is an immediate consequence of the principle of extension of identities (Chapter I, § 8, no. 1, Corollary 1 to Proposition 2).

Q.E.D.

In the application of this theorem to a Hausdorff topological ring A, we take E, F and G to be $\hat{A}$, A and B to be the ring A, and f to be

(*) $f$ is said to be $\mathbf{Z}$-bilinear if, for all elements $x, x'$ of A and all elements $y, y' \in B$ we have

the $\mathbf{Z}$-bilinear mapping $(x, y) \to xy$, which by hypothesis is continuous. We denote again by $xy$ the value of the extended function on $\hat{A} \times \hat{A}$; this function is a law of composition on $\hat{A}$, and to say that it is $\mathbf{Z}$-bilinear means that it is distributive on both sides with respect to addition; and it is also associative, by the principle of extension of identities. Consequently:

If $A$ is commutative (resp. has an identity element) then the same is true of $\hat{A}$ (principle of extension of identities).

Let $A$ be a topological ring, not necessarily Hausdorff; let $N$ be the closure of $\{0\}$ in $A$, and let $A' = A/N$ be the Hausdorff ring associated with $A$. Then $\hat{A}'$, the completion of $A'$, is called the Hausdorff completion of $A$ and is also denoted by $\hat{A}$. One shows as in § 3, no. 4, Proposition 8 that every continuous ring homomorphism $u$ of $A$ with a complete Hausdorff topological ring $C$ can be uniquely factorized as $u = v \circ \varphi$, where $v$ is a continuous homomorphism of $\hat{A}$ into $C$ and $\varphi$ is the canonical mapping of $A$ into $\hat{A}$. If $A, B$ are two topological rings, and $u : A \to B$ is a continuous homomorphism, there is therefore a unique continuous homomorphism $\hat{u} : \hat{A} \to \hat{B}$ such that the diagram

is commutative ($\varphi$ and $\psi$ being the canonical mappings); we have only to apply the preceding result to $\psi \circ u$.

There are topological division rings in which this condition is not satisfied and in which the ring $\hat{K}$ has zero divisors (see Exercise 26). Moreover, even if the completion $\hat{K}$ is a topological division ring, there is no a priori reason to assume that the multiplicative structures of $\hat{K}$ are structures of a complete space. However, this will be the case for division rings K such that $\hat{K}$ is locally compact (see Chapter I, § 9, no. 7, Proposition 13, and Chapter III, § 3, no. 3, Proposition 4) and for topological fields; for the latter, we have the following proposition:

We shall show that if $\mathfrak{F}$ is a Cauchy filter with respect to the multiplicative structure on K*, then $\mathfrak{F}$ is a Cauchy filter with respect to the additive structure on K, and does not converge to 0; this will establish the result. Let U be any neighbourhood of 0 in K, let V be a closed neighbourhood of 0 such that $V \subset U, VV \subset U$ [axiom (AV$_\text{II}$)] and $-1 \notin V$; then by hypothesis there is a set $A \in \mathfrak{F}$ such that, for all $x \in A$ and $y \in A$, we have $x^{-1}y \in 1 + V$. Let $a \in A$, then $A \subset a + aV$, and $a + aV$ is a closed set which does not contain 0; hence 0 is not in the closure of A and is therefore not a cluster point of $\mathfrak{F}$. Let W be a neighbourhood of 0 such that $aW \subset V$ [axiom (AV$_\text{I}$)]; then by hypothesis there is a set $B \in \mathfrak{F}$ such that $B \subset A$ and, for all $x \in B$ and $y \in B$, we have $x^{-1}y \in 1 + W$; hence $y - x \in xW \subset AW \subset aW + aVW$; but K is commutative, and therefore $aVW = aWV \subset VW \subset U$; consequently $y - x \in U + U$ and the Proposition is proved.

The same proof shows that Proposition 8 can be extended to the case in which every Cauchy filter with respect to one of the multiplicative structures of K is also a Cauchy filter with respect to the other multiplicative structure.