UDC 513.83

MATHEMATICS

P. KENDEROV

ON TOPOLOGICAL GROUPS

(Presented by Academician P. S. Aleksandrov on 12 III 1970)

Let (X_1) be a subset of a topological space (X). A system of sets ({O_\alpha:\alpha\in\Gamma}), where (X_1\subset \operatorname{int} O_\alpha\subset X)*, is called a character of the set (X_1) in (X), if for every open set (U\supset X_1) there exists (O_{\alpha_0}) ((\alpha_0\in\Gamma)) such that (O_{\alpha_0}\subset U). B. A. Pasynkov calls a topological group almost metrizable ((^1)) if it contains a bicompactum of countable character.

Theorem.** Let (G) be a separable topological group and let (H) be its Weil-complete almost metrizable normal subgroup. Then there exists a set (A\subset G) such that (\pi(A)=G/H), where (\pi:G\to G/H) is the natural projection, and the restriction (\pi/A:A\to G/H) is a perfect mapping.

Proof. Let ({O_i}{i=1}^{\infty}) be a character in (H) of a bicompact set (B\subset H); we may assume that (B) contains the identity element (e) of the group (G). Choose a system of neighborhoods ({U_i}=U_n); c) (U_n\cap H\subset O_n). Put}^{\infty}) of the identity so that: a) (U_{n=1}^2\subset U_n); b) (U_n^{-1
[
L=\bigcap_{i=1}^{\infty} U_i;
]
(L) is a closed subgroup of the group (G), and therefore the subgroup
[
K=L\cap H=\left(\bigcap_{i=1}^{\infty} U_i\right)\cap H=
]
[
=\bigcap_{i=1}^{\infty}(U_i\cap H)\subset \bigcap_{i=1}^{\infty} O_i=B
]
is bicompact.

Lemma 1. The system of sets ({V_i}_{i=1}^{\infty}), where (V_i=U_i\cap H), is a character of the set (K) in (H).

The proof of this lemma is left to the reader.

We shall denote the space of left cosets of the group (G) modulo the subgroup (L) by (G/L), and the natural mapping (x\mapsto xL) by
[
\omega:G\to G/L.
]
Every element (s\in G) defines a mapping (s:G/L\to G/L) by the formula
[
s(\omega(x))=\omega(sx)
]
(sometimes we shall omit the parentheses and write (s\omega(x))). It can be proved that there exists a metric (\rho) on (G/L) with the following properties: 1) the mapping (\omega:G\to (G/L,\rho)) is continuous; 2) for (s\in G) and (z_1,z_2\in G/L),
[
\rho(sz_1,sz_2)=\rho(z_1,z_2);
]
3) ({\omega(U_n)}_{n=1}^{\infty}) is a base of neighborhoods of the point (\omega(e)) in ((G/L,\rho)).

Lemma 2. (\omega(H)) is a complete subspace of the metric space ((G/L,\rho)).

Proof of Lemma 2. Let (\xi={A}) be any Cauchy filter in (\omega(H)), and let (\eta) be an ultrafilter in (H) whose image (\omega(\eta)) majorizes (\xi). We shall prove that (\eta) is a Cauchy filter. Let (V) be a neighborhood of the identity in (H); there exists (U_{i_0}\in{U_n}{n=1}^{\infty}) such that (U)) is a neighborhood of (\omega(e)) in ((G/L,\rho)), for some (h_0\in H) the set}\cap H\subset KV). Since (\omega(U_{i_0+1
[
h_0\omega(U_{i_0+1})=\omega(h_0U_{i_0+1})
]
contains an element of the filter (\xi). Consequently,
[
\omega^{-1}\omega(h_0U_{i_0+1})\cap H=h_0U_{i_0+1}L\cap H\in\eta.
]
However
[
h_0U_{i_0+1}L\cap H\subset h_0U_{i_0+1}\cap H\subset h_0(U_{i_0}\cap H)\subset h_0KV,
]
and from the bicompactness of (h_0K) it follows that (h_0K\subset
]

* (\operatorname{int} O_\alpha) is the interior of the set (O_\alpha).

** This theorem adjoins a result of B. A. Pasynkov (((^2),) Theorem 2).

[
\subset \bigcup_{i=1}^{p} h_0 k_i V,\quad \text{where } k_i \in K \ (i=1,2,\ldots,p).
]
Hence,
[
h_0 K V \subset \left(\bigcup_{i=1}^{p} h_0 k_i V\right) \times V \subset \bigcup_{i=1}^{p} h_0 k_i V^2 \in \eta .
]
The ultrafilter (\eta) must contain at least one term of the union
[
\bigcup_{i=1}^{p} h_0 k_i V^2,
]
i.e. (\eta) is a Cauchy filter in (H). Therefore it converges to some (h \in H); (\omega(h)) will be a point of contact of the Cauchy filter (\xi), and hence (\xi) converges to (\omega(h)). The lemma is proved. Since (\omega(xH)=x\omega(H)), the sets (\omega(xH)) are also complete in ((G/L,\rho)).

Let us note that
[
G/L=\bigcup_{x\in G}\omega(xH)=\bigcup_{x\in G}x\omega(H).
]
Moreover, for (x_1,x_2\in G), either (x_1\omega(H)\cap x_2\omega(H)=\varnothing), or (x_1\omega(H)\equiv x_2\omega(H)). Indeed, if (x_1\omega(H)\cap x_2\omega(H)\ne\varnothing), then (x_1HL\cap x_2HL\ne\varnothing). Since (HL=LH) is a subgroup of the group (G) (this follows from the normality of the group (H)), it follows that (x_1HL\equiv x_2HL), i.e. (\omega(x_1H)=\omega(x_2H)).

Denote by (X) the quotient set corresponding to the partition
[
\bigcup_{x\in G}\omega(xH),
]
and the quotient mapping (\omega(x)\mapsto \omega(xH)) by
[
\lambda:\quad G/L\to X.
]
For brevity put (\tilde{x}=\lambda\omega(x)), and define on (X) the metric
[
d(\tilde{x}_1,\tilde{x}_2)=\inf{\rho(h_1\omega(x_1),h_2\omega(x_2)):h_i\in H}
=\inf{\rho(\omega(x_1),h_1^{-1}h_2\omega(x_2)):h_i\in H}
=\inf{\rho(\omega(x_1),h\omega(x_2)):h\in H}.
]

Lemma 3. The mapping (\lambda:(G/L,\rho)\to (X,d)) is open.

Proof of the lemma. Let (O) be an open set in ((G/L,\rho)), and let (\lambda^{-1}\lambda(O)=O). For (\omega(x)\in O) the number
[
\varepsilon=\inf{\rho(\omega(x),\omega(\bar{x})):\omega(\bar{x})\in G/L\setminus O}>0.
]
If (d(\tilde{x},\tilde{x}')<\varepsilon), then there exists (h'\in H) such that
[
\rho(\omega(x),h'\omega(x'))<\varepsilon.
]
Hence (h'\omega(x')\in O), i.e. (\omega(x')\in O). In other words, (\tilde{x}'\in\lambda(O)). The openness of the mapping
[
\lambda:(G/L,\rho)\to (X,d)
]
now follows from the fact that the set
[
\lambda^{-1}\lambda(W)=HW=\bigcup_{h\in H}hW
]
is open whenever (W) is open. The lemma is proved.

As Michael proved (({}^{4}),) Theorem 1), there exists a set (A'\subset G/L) such that (\lambda(A')=X) and the mapping (\lambda|A':A'\to X) is perfect. In what follows we shall see that the set (A=\omega^{-1}(A')) satisfies all the conditions of the theorem.

Consider the group (G/H). The subgroup (\pi(L)) decomposes (G/H) into left cosets. Assigning to each element (y\in G/H) the coset (y\pi(L)), we obtain the quotient mapping
[
\mu:\quad G/H\to Y,
]
where (Y) is the set of left cosets endowed with the quotient topology. Consider the diagram

[
\begin{array}{ccc}
G & \xrightarrow{\ \omega\ } & G/L & \to & X\
\downarrow{\pi} & & \lambda & \uparrow v \
G/H & \xrightarrow{\ \mu\ } & Y
\end{array}
]

Since the mapping (\mu\pi) is quotient, there exists a continuous mapping
[
v:\quad Y\to X
]
such that
[
\lambda\omega=v\mu\pi.
]

Lemma 4. If an ultrafilter (\xi) in (G) is such that (\pi(\xi)) converges to (\pi(x_1)) and (\omega(\xi)) converges to (\omega(x_2)), then (\xi) converges to some (x_0\in G), with
[
\pi(x_0)=\pi(x_1)\quad \text{and}\quad \omega(x_0)=\omega(x_2).
]

Proof of the lemma.
[
\lambda\omega(x_2)=\lambda\lim \omega(\xi)=\lim \lambda\omega(\xi)
=\lim v\mu\pi(\xi)=v\mu\lim \pi(\xi)=\lambda\omega(x_1),
]
i.e. (x_1\in x_2HL). Hence there exist (l\in L) and (h\in H) such that
[
x_1h=x_2l.
]
Put (x=x_1h=x_2l). Then
[
x_1H\cap x_2L=xH\cap xL=x(H\cap L)=xK.
]
We shall prove that, for every neighborhood (U) of the identity of (G), the set (xKU\in\xi). To this end take a neighborhood (V\ni e) such that
[
V^2\subset U.
]
There exists a set (U_i\in {U_n}{n=1}^{\infty}) for which
[
U_i\cap H\subset KV.
]
Put
[
O=U\cap V.
]
The filter (\omega(\xi)) converges to (\omega(x_2)=)

(= \omega(x)), and therefore there is a set (B_2 \in \xi) whose image (\omega(B_2) \subset \omega(xU_{i+2})). Similarly, there exists (B_1 \in \xi) such that (\pi(B_1) \subset \pi(xO)). Then (B_1 \cap B_2 \in \xi), and, moreover,

[
\begin{gathered}
B_1 \cap B_2 \subset xOH \cap xU_{i+2}L = x(OH \cap U_{i+2}L) \subset \
\subset x(OH \cap U_{i+2}^{2}) \subset x(HO \cap U_{i+1}) \subset \
\subset x{[(KVU(H \setminus KV))O]\cap U_{i+1}} \subset \
\subset x[(KVOU(H \setminus KV)O)\cap U_{i+1}] \subset \
\subset x{KV^{2}U[(H \setminus KV)U_{i+1}\cap U_{i+1}]}\subset xKV^{2}\subset xKU\in \xi .
\end{gathered}
]

Since (\xi) is an ultrafilter, it has at least one point of adherence (x_0\in xK). Then (\xi) converges to (x_0). The lemma is proved.

Let us return to the set (A=\omega^{-1}(A')). If for (x\in G) we have (xH\cap A=xH\cap \omega^{-1}(A')=\varnothing), then (\omega(xH)\cap A'=\varnothing), which is impossible, since (\lambda(A')=X). This proves that (\pi(A)=G/H). The perfectness of the mapping (\pi/A: A\to G/H) will follow from the following criterion ([3], p. 156, Theorem 1): a continuous mapping (\varphi: P\to Q) of a separable topological space (P) onto a separable topological space (Q) is perfect if and only if every ultrafilter (\xi) in (P) whose image (\varphi(\xi)) converges in (Q) converges in (P). Let (\xi) be an ultrafilter in (A) for which (\lim \pi(\xi)=\pi(x)). Then (\omega(\xi)) is an ultrafilter in (A'), and moreover

[
\lim \lambda\omega(\xi)=\lim \nu\mu\pi(\xi)=\nu\mu\lim \pi(\xi)=\nu\mu\pi(x).
]

Since the mapping (\lambda/A': A'\to X) is perfect, (\omega(\xi)) converges in (A'). Let (\lim \omega(\xi)=\omega(\bar{x})\in A'), and let (\hat{\xi}) be some ultrafilter in (G) containing the system of sets (\xi). The filter (\pi(\hat{\xi})=\pi(\xi)) converges in (G/H) to (\pi(x)), while (\omega(\hat{\xi})) converges in (G/L) to (\omega(\bar{x})). From Lemma 4 it follows that the filter (\hat{\xi}) converges in (G) to some (x_0\in G), with (\pi(x_0)=\pi(x)) and (\omega(x_0)=\omega(\bar{x})), i.e. (x_0\in \omega^{-1}\omega(x_0)\subset A). Then the filter (\xi) converges in (A) to (x_0). The theorem is proved.

Department of Mechanics and Mathematics
M. V. Lomonosov Moscow State University

Received
27 II 1970

REFERENCES

1. B. A. Pasynkov, DAN, 161, No. 2, 281 (1965).
2. B. A. Pasynkov, DAN, 188, No. 2, 286 (1969).
3. N. Bourbaki, General Topology (Basic Structures), “Nauka,” 1969.
4. E. Michael, Duke Math. J., 26, No. 4, 647 (1959).