UDC 513.831

MATHEMATICS

A. V. ARKHANGELSKII

ON MAPPINGS CONNECTED WITH TOPOLOGICAL GROUPS

(Presented by Academician P. S. Aleksandrov, 25 XII 1967)

The author regards this work as a continuation of the investigations of A. A. Markov \((^1)\) and M. I. Graev \((^2)\) on topological groups. The results are stated; proofs are not given.

I. We shall call a space \(X\): 1) amoebic, if there exists a disjoint covering of this space by open sets, each of which contains no more than one non-isolated point of the space \(X\); 2) \(D\)-bicompact, if it is a free union of bicompacts; 3) standard, if \(X\) is amoebic, \(D\)-bicompact, and metrizable simultaneously.

It is easy to see that condition 3) is equivalent to the following:

\(3')\) \(X\) is representable as a free union of spaces homeomorphic to a convergent sequence completed by its limit.

Theorem 1. Every metrizable topological\(^*\) group is representable as a quotient group of the free topological group of some standard space.

Theorem 2. If the space of a topological group is a \(k\)-space, then this group is isomorphic to a quotient group of the free topological group of some \(D\)-bicompact space.

Theorem 3. Every topological group is isomorphic to a quotient group of the free topological group of some amoebic space.

The basis of Theorems 1, 2, and 3 is the following

Lemma 1. Let \(f : X \to Y\) be a quotient mapping onto, where \(X\) is a completely regular space and \(Y\) is the space of a topological group. Suppose further that \(F(X)\) is the free topological group of the space \(X\), and \(\tilde f : F(X) \to Y\) is an extension of the mapping \(f\) to a continuous homomorphism of the group \(F(X)\) onto the group \(Y\). Then the mapping \(\tilde f\) is open.

Proof of the lemma. Denote by \(\mathfrak T\) the topology given on \(Y\), and by \(\tilde{\mathfrak T}\) the quotient topology induced on \(Y\) by the topology of the group \(F(X)\) via the mapping \(\tilde f\). Then \(\tilde{\mathfrak T} \supset \mathfrak T\). But the mapping \(\tilde f\) of the space \(F(X)\) onto \(Y\), endowed with the topology \(\tilde{\mathfrak T}\), is continuous; hence the mapping \(f\) of the space \(X\) onto \(Y\), endowed with the topology \(\tilde{\mathfrak T}\), is also continuous. Since, with respect to the topology \(\mathfrak T\), the mapping \(f : X \to Y\) is a quotient mapping, \(\mathfrak T\) contains every topology on \(Y\) with respect to which the mapping \(f\) is continuous. Hence, \(\mathfrak T \supset \tilde{\mathfrak T}\) and \(\tilde{\mathfrak T} = \mathfrak T\). The lemma is proved.

Three questions. Let \(X\) be a zero-dimensional \(D\)-bicompact space, or an amoebic space, or a standard space. Is it true that the free topological group of the space \(X\) is zero-dimensional in the sense of ind?

\(^*\) We consider here only Hausdorff topological groups. By isomorphisms and quotient groups are meant topological isomorphisms and topological quotient groups.

Connected with them is the question:

Is every topological group whose space is a \(k\)-space representable as a quotient group of a zero-dimensional topological group?

The author does not know the answer to these questions, but he has proved

Theorem 4. Every topological group with a countable base is a quotient group of some zero-dimensional topological group with a countable base.

A more general assertion:

Theorem \(4'\). If the space of a group is finally compact and paracompact, then this group is a quotient group of some zero-dimensional finally compact group.

Theorem 5. Every bicompact group is a quotient group of some zero-dimensional finally compact topological group (of the same weight).

Corollary. Every totally bounded group is a quotient group of some zero-dimensional (in the sense of \(\operatorname{ind}\)) group.

Probably these results should not be regarded as intuitively the most natural: A. Weil \((^3)\) stipulates that, on passing to a quotient group, the dimension cannot increase\({}^{**}\), and C. Kaplan then constructs a counterexample (see \((^3)\), p. 170); we now see that this example is a manifestation of a general law.

The proofs of Theorems 4, \(4'\), and 5 are based, in addition to Lemma 1 and the like, on the following assertions.

Proposition 1. The free topological group of a zero-dimensional bicompactum is zero-dimensional.

Proposition 2. The free topological group of a zero-dimensional space with a countable base is zero-dimensional.

The proof of the cited assertions is based on

General assertion 1. Let \(X\) be a topological space, \(F(X)\) the free topological group of this space, \(F_n(X)\), where \(n=1,2,\ldots,\infty\), the set of all those elements of \(F(X)\), i.e. those irreducible words, whose length does not exceed \(n\). Put \(A_n(X)=F_n(X)\setminus F_{n-1}(X)\), \(n=1,2,\ldots,\infty\)*. Then all \(F_n(X)\) are closed in \(F(X)\), and all natural mappings \(A_n(X)\to (X\cup X^{-1})^n\) (where \((X\cup X^{-1})^n\) is the \(n\)-th Tikhonov power of the free union of two copies of the space \(X\)—for the notation see \((^2)\)) are homeomorphisms into.

The author does not know whether every metrizable group is a quotient group of a zero-dimensional metrizable group.

II. Here we touch upon questions of embedding topological spaces into spaces of groups. The well-known result of A. A. Markov is supplemented by

Theorem 6. Every completely regular space of weight \(\tau\)** is homeomorphic to a closed subspace of a topological group whose weight does not exceed \(\tau\).

Corollary. Every regular space with a countable base is homeomorphic to a closed subspace of a topological group with a countable base.

Our main goal in this section is to consider the question of the possibility of representing a bicompactum in the form of a (closed) subset of type \(G_\delta\) of some topological group. We shall call such representations \(\aleph_0\)-embeddings of this bicompactum.

The author’s basic hypothesis: every bicompactum admitting an \(\aleph_0\)-embed-

* Zero-dimensionality may be understood both in the sense of \(\operatorname{ind}\) and in the sense of \(\dim\).

** This is indeed so when the group is locally bicompact (see \((^4)\)).

*** \(F_0(X)\) is the identity element of the group \(F(X)\).

**** Here \(\tau\) is any infinite cardinal number.

...embedding*, is a dyadic bicompactum, remains unproved. But the following holds.

Theorem 7. If \(\mathcal P\) is a property of a topological space that is hereditary with respect to passage to a closed subspace, and every dyadic bicompactum possessing the property \(\mathcal P\) is metrizable, then every \(\aleph_0\)-bicompactum with the property \(\mathcal P\) is also metrizable.

The question of metrizability of dyadic bicompacta has been well studied (see \(({}^{6,7})\)); for example, we obtain the following corollaries:

Corollary 1. If an \(\aleph_0\)-bicompactum satisfies the first axiom of countability or, more generally, is a sequential space—the latter being equivalent to the assertion that this \(\aleph_0\)-bicompactum is a quotient space of a metric space—then it is metrizable.

Corollary 2. If every closed subspace of some \(\aleph_0\)-bicompactum is an \(\aleph_0\)-bicompactum, then this \(\aleph_0\)-bicompactum is metrizable.

Let us note that each of the conditions formulated so far that are sufficient for metrizability of an \(\aleph_0\)-bicompactum is a necessary condition for metrizability of any bicompactum. The following is of a different nature.

Theorem 8. The set of extremally disconnected points of an \(\aleph_0\)-bicompactum is finite (or empty).

Theorems 7 and 8 are proved on the basis of the following considerations. Let \(\varphi: G \to H\) be a continuous homomorphism from one topological group onto another, let \(Y \subset H\) be some subspace, and let \(X=\varphi^{-1}Y\) be a subspace of the group \(G\). By \(f\) we denote the continuous mapping \(X \to Y\) induced by \(\varphi\).

Any mapping of one space \(X\) onto another topological space which can be represented, in the manner indicated above, by means of some open homomorphism \(\varphi\), we shall call a skew mapping.

Theorem 9. Every \(\aleph_0\)-bicompactum can be skew mapped onto a compactum.

Let us note that a skew mapping is always open, and that the inverse images of points under it are homeomorphic to one another and to some group.

Proposition 3. If the inverse images of points under a skew mapping of a bicompactum \(X\) onto a compactum \(Y\) are compacta, then \(X\) is metrizable.

For the proof of the last assertion we need the following

Lemma. If the space of a group has a countable pseudocharacter, then every bicompactum lying in it is metrizable.

The general features of the structure of nonmetrizable \(\aleph_0\)-bicompacta are revealed by

Theorem 10. Every nonmetrizable \(\aleph_0\)-bicompactum is homogeneous with respect to character and weight, and the weight of the whole bicompactum is equal to its character at an arbitrary point.

Theorem 11. The Suslin number of an arbitrary \(\aleph_0\)-bicompactum does not exceed \(\aleph_0\).

Theorem 12. A bicompactum is \(\aleph_0\)-embedded in its free topological group if and only if it is metrizable.

Hence (or from Theorem 6) it follows that

Corollary. Every compactum is an \(\aleph_0\)-bicompactum.

On the other hand, it follows from Theorem 10 that not every dyadic bicompactum is an \(\aleph_0\)-bicompactum; consequently, the property of being an \(\aleph_0\)-bicompactum is not invariant under arbitrary continuous mappings (invariance under open mappings is possible).

Theorem 13. If the cardinality of an \(\aleph_0\)-bicompactum is less than \(2^{\aleph_1}\), then it is metrizable.

* In what follows we call bicompacta admitting \(\aleph_0\)-embeddings \(\aleph_0\)-bicompacta.

Assuming the continuum hypothesis, we conclude that every $\aleph_0$-bicompactum of cardinality continuum is metrizable.

Theorem 14. If a bicompactum is $\aleph_0$-embedded in the space of a completely bounded topological group, then it is dyadic.

In connection with Theorems 9, 12, and 14, and the hypothesis on the dyadicity of an arbitrary $\aleph_0$-bicompactum, the following question is of interest:

Question. Is the preimage of a compactum under a bicompact homomorphism necessarily a dyadic bicompactum?

III. Questions concerning the location of bicompacta lead to the consideration of bicompactly generated groups.*

Theorem 15. If a bicompactly generated group is extremally disconnected, then it is countable and finitely generated.

Theorem 15 is based on

Theorem 16. Every bicompact subset of an extremally disconnected topological group is finite (or empty).

It remains unclear whether there exists a nondiscrete finitely generated extremally disconnected topological group. The space of the latter, by Theorem 15, cannot be a $k$-space.

Theorem 17. If a bicompact space with the first axiom of countability can be topologically embedded in some bicompact group as a set of algebraic generators of the latter, then both this space and the group are metrizable.

Corollary. Not every bicompactum can be represented as an algebraically generating subspace of a bicompact group.

As is known, every bicompactum can be represented as the set of algebraic generators of some completely bounded group and can be embedded in a bicompact group. This means that the last assertions are of a definitive character.

Lemma 2. If a topological group $X$ is algebraically generated both by a subspace $Y$ and by a bicompact subspace $\Phi$, then $\operatorname{weight}\Phi \leq \operatorname{weight}Y$*.

Theorem 18. If a topological group $X$ is algebraically generated by its bicompact subspaces $\Phi_1$ and $\Phi_2$, then the weight of $\Phi_1$ is equal to the weight of $\Phi_2$*.

REFERENCES CITED

1. A. A. Markov, Izv. Akad. Nauk SSSR, Ser. Mat., 9, No. 1, 3 (1945).
2. M. I. Graev, Uspekhi Mat. Nauk, 5, issue 2 (36), 3 (1950).
3. A. Weil, Integration in Topological Groups and Its Applications, IL, 1950.
4. T. Yamanoshita, J. Math. Soc. Japan, 6, 151 (1951).
5. B. A. Efimov, Tr. Mosk. Mat. Obshch., 14, 211 (1965).
6. A. V. Arhangel’skii, Dokl. Akad. Nauk SSSR, 175, No. 4 (1967).
7. A. V. Arhangel’skii, Dokl. Akad. Nauk SSSR, 126, No. 2, 239 (1959).
8. B. A. Pasynkov, Dokl. Akad. Nauk SSSR, 161, No. 2 (1965).

* Groups are meant which are algebraically generated by their bicompact subspace.

** A more general assertion: if, in the space of a topological group, the closures of any two disjoint open subsets of type $F_\sigma$ do not intersect, then every bicompactum lying in it is finite. I am grateful to W. W. Comfort, who drew my attention to the named class of spaces.

*** If these weights are infinite.