1 Introduction
All topological spaces under consideration are assumed to be Tychonoff and infinite. In 1945, Markov posed the following problem (see [Reference Markov20]). Let
$X$
and
$Y$
be topological spaces for which the free (Markov) topological groups
$F(X)$
and
$F(Y)$
are topologically isomorphic. Are then
$X$
and
$Y$
homeomorphic?
Spaces with topologically isomorphic free topological groups are said to be
$M$
-equivalent. Denote by
$I$
the closed unit interval
$[0,1]$
;
$\mathbb{R}$
stands for the real line. Answering Markov’s question, Graev [Reference Graev9] showed that the segment
$I$
and the letter
$T$
considered as a subspace of the plane
$\mathbb{R}^{2}$
are
$M$
-equivalent but not homeomorphic. Note also that the segment
$I$
and the unit circle are not
$M$
-equivalent (see [Reference Arhangel’skii and Tkachenko3, Examples 7.10.5 and 7.10.8]).
Topological spaces
$X$
and
$Y$
are called
$A$
-equivalent (
$B$
-equivalent) if the free abelian (boolean) topological groups
$A(X)$
and
$A(Y)$
(
$B(X)$
and
$B(Y)$
, respectively) are topologically isomorphic. Similarly, topological spaces
$X$
and
$Y$
are called
$L$
-equivalent if the free locally convex spaces
$L(X)$
and
$L(Y)$
are linearly isomorphic, and, finally,
$X$
and
$Y$
are called
$l$
-equivalent if the spaces of continuous functions equipped with the topology of pointwise convergence
$C_{p}(X)$
and
$C_{p}(Y)$
are linearly homeomorphic. It is known that for all Tychonoff topological spaces
$X$
and
$Y$
, the following implications hold:
$X$
is
$M$
-equivalent to
$Y$
$\Longrightarrow$
$X$
is
$A$
-equivalent to
$Y$
$\Longrightarrow$
$X$
is
$B$
-equivalent to
$Y$
;
$X$
is
$A$
-equivalent to
$Y$
$\Longrightarrow$
$X$
is
$L$
-equivalent to
$Y$
$\Longrightarrow$
$X$
is
$l$
-equivalent to
$Y$
. In fact, for
$k$
-spaces (in particular, for all compact spaces)
$L$
- and
$l$
-equivalences coincide. Nevertheless, all other implications cannot be reversed even for compact metrizable spaces. For instance,
$I$
and the free union
$I\oplus I$
are
$l$
-equivalent but not
$M$
-equivalent, and
$B$
-equivalence does not imply
$l$
-equivalence as well (see [Reference Sipacheva30]).
Graev [Reference Graev9] proved that if
$X$
and
$Y$
are
$M$
-equivalent metrizable compact spaces, then their dimensions are equal:
$\dim X=\dim Y$
. This theorem has been a subject of many subsequent generalizations. Firstly, the same statement was proved to be true for
$l$
-equivalent complete separable metrizable spaces [Reference Pavlovskiǐ26] and later Pestov [Reference Pestov27] showed that
$\dim X=\dim Y$
for all
$l$
-equivalent Tychonoff spaces
$X$
and
$Y$
.
One may ask in which cases the converse is true:
$\dim X=\dim Y=n$
for metrizable compact spaces
$X$
and
$Y$
implies that
$X$
and
$Y$
are
$M$
-equivalent. For
$n=0$
, the complete topological classification of all
$M$
-equivalent metrizable locally compact spaces was obtained by Baars [Reference Baars4]. In particular, zero-dimensional metrizable compact space
$X$
and the Cantor discontinuum
$C$
are
$M$
-equivalent if only if
$X$
is uncountable. If
$X$
is an
$n$
-dimensional compact polyhedron, where finite
$n\geqslant 1$
, then
$X$
is
$l$
-equivalent to the cube
$I^{n}$
[Reference Pavlovskiǐ26]. However, the last result is not true for more general metrizable
$n$
-dimensional compact spaces; the complete characterization of those
$X$
which are
$l$
-equivalent to the segment
$I$
was obtained by Matrai [Reference Matrai21].
A natural question arises whether the inequality
$\dim Y\leqslant \dim X$
holds for compact spaces
$X$
and
$Y$
whenever there exists a linear continuous (and open) surjection
$C_{p}(X)\rightarrow C_{p}(Y)$
[Reference Arkhangel’skii, van Mill and Reed2, Problems 1046 and 1047]. These problems were answered negatively in [Reference Leiderman, Levin and Pestov16, Reference Leiderman, Morris and Pestov17].
First, we note the duality between linear surjections of
$C_{p}(X)$
-spaces and isomorphic embeddings of
$L(X)$
.
Proposition 1.1 [Reference Leiderman, Morris and Pestov17].
For every Tychonoff space
$Y$
and metrizable compact space
$X$
, there exists a linear continuous surjection
$C_{p}(X)\rightarrow C_{p}(Y)$
if and only if the free locally convex space
$L(Y)$
embeds into
$L(X)$
as a linear subspace.
A topological space
$X$
is called a
$k_{\unicode[STIX]{x1D714}}$
-space if
$X$
is the union of an increasing sequence
$\{X_{n}\}$
of compact subspaces, with the property that a subset
$A\subset X$
is closed if and only if intersections
$A\cap X_{n}$
are closed for each
$n\in \mathbb{N}$
, where
$\mathbb{N}$
denotes the set of natural numbers. The following surprising results obtained in [Reference Leiderman, Morris and Pestov17] are relevant for our paper.
Theorem 1.2 [Reference Leiderman, Morris and Pestov17].
For a Tychonoff space
$X$
, the following are equivalent:
(i)
$A(X)$
isomorphically embeds into
$A(I)$
;(ii)
$F(X)$
isomorphically embeds into
$F(I)$
;(iii)
$X$
is a
$k_{\unicode[STIX]{x1D714}}$
-space such that every compact subspace of
$X$
is finite-dimensional and metrizable.
Theorem 1.3 [Reference Leiderman, Morris and Pestov17].
(i) Let
$X$
be a Tychonoff space. If
$L(X)$
isomorphically embeds into
$L(I)$
, then
$X$
is a metrizable compact space which is a countable union of finite-dimensional closed subspaces.(ii) Let
$X$
be a finite-dimensional metrizable compact space. Then
$L(X)$
isomorphically embeds into
$L(I)$
.
So, as an easy corollary, we have the following claim: for every finite-dimensional metrizable compact space
$X$
containing the segment
$I$
,
$A(X\times X)$
is topologically isomorphic to a subgroup of
$A(X)$
; and, similarly,
$L(X\times X)$
is topologically isomorphic to a linear subspace of
$L(X)$
.
Being motivated by these results, now we are ready to formulate the main question investigated in this paper.
Question 1.4. Let
$O$
be any of the free topological functors
$F,\,A,\,B,\,L$
.
(1) Is it true that for every metrizable compact (metrizable separable) space
$X$
there exists an isomorphic embedding of
$O(X\times X)$
into
$O(X)$
?(2) Is it true that for every metrizable compact (metrizable separable) space
$X$
there exists an isomorphic embedding of
$O(X\oplus X)$
into
$O(X)$
?
Question 1.4 for particular
$O$
has been asked many times in several informal seminars, though its general formulation did not appear in print. Observe that since the free topological sum
$X\oplus X$
is homeomorphic to a closed subspace of the square
$X\times X$
, then
$O(X\oplus X)$
isomorphically embeds into
$O(X\times X)$
for all aforementioned functors
$O$
and therefore the positive answer to the first item would imply the positive answer to the second item.
In fact, for
$O=F$
the positive answer to Question 1.4 has been known for many years.
Theorem 1.5 [Reference Nickolas24].
Let
$X$
be a
$k_{\unicode[STIX]{x1D714}}$
-space. Then
$F(X^{n})$
isomorphically embeds into
$F(X)$
for each
$n\in \mathbb{N}$
.
We will show here that the non-abelian case is an exception. For
$O=L$
, the negative answer to Question 1.4 easily follows from the recent result obtained in [Reference Krupski and Marciszewski14].
Theorem 1.6 [Reference Krupski and Marciszewski14].
Let
$X$
be a Cook continuum
$M$
or
$X$
be a rigid Bernstein set
$B$
. Then there is no linear continuous surjection from
$C_{p}(X)$
onto
$C_{p}(X\oplus X)$
.
Note that around 20 years earlier Pol [Reference Pol28] showed that both for
$X=M$
and for
$X=B$
, the space
$X$
is not
$l$
-equivalent to
$X\oplus X$
. So, Theorem 1.6 in view of Proposition 1.1 immediately implies that both for a Cook continuum
$X=M$
and for a rigid Bernstein set
$X=B$
the free locally convex space
$L(X\oplus X)$
does not embed into
$L(X)$
as a linear subspace.
The proof of Theorem 1.6 presented in [Reference Krupski and Marciszewski14] does not work automatically for embedding into the free abelian topological group
$A(X)$
or the free boolean topological group
$B(X)$
. Moreover, it is unknown whether for all metrizable compact spaces
$X$
,
$Y$
, “
$A(Y)$
isomorphically embeds into
$A(X)$
” implies that “
$L(Y)$
isomorphically embeds into
$L(X)$
”. In the case that
$Y$
is a non-compact space, such an implication does not hold.
Our Theorem 3.1, which is the main result of this paper, settles Question 1.4 for
$O=A$
and
$O=B$
. We prove that if
$X$
is a Cook continuum or
$X$
is a rigid Bernstein set, then the free abelian topological group
$A(X\oplus X)$
does not embed into
$A(X)$
as a topological subgroup. Analogously, the free boolean topological group
$B(X\oplus X)$
for such
$X$
does not embed into
$B(X)$
as a topological subgroup (Theorem 3.4).
The proofs are given in §3. We conclude the paper with some additional comments and open questions.
2 Preliminaries
Definition 2.1 [Reference Graev9, Reference Markov20].
Let
$X$
be a Tychonoff space. A topological group
$F(X)$
(
$A(X)$
, respectively,
$B(X)$
) is called the (Markov) free (abelian, respectively, boolean) topological group over
$X$
if
$F(X)$
(
$A(X)$
, respectively,
$B(X)$
) satisfies the following conditions:
(i) there is a continuous embedding
$\unicode[STIX]{x1D6FE}:X\rightarrow F(X)$
(
$\unicode[STIX]{x1D6FE}:X\rightarrow A(X)$
, respectively,
$\unicode[STIX]{x1D6FE}:X\rightarrow B(X)$
) such that
$\unicode[STIX]{x1D6FE}(X)$
algebraically generates
$F(X)$
(
$A(X)$
, respectively,
$B(X)$
);(ii) if
$f:X\rightarrow G$
is a continuous mapping to an (abelian, respectively, boolean) topological group
$G$
, then there exists a continuous homomorphism
$\bar{f}:F(X)\rightarrow G$
(
$\bar{f}:A(X)\rightarrow G$
, respectively,
$\bar{f}:B(X)\rightarrow G$
) such that
$f=\bar{f}\circ \unicode[STIX]{x1D6FE}$
.
Analogously one can define the free locally convex space.
Definition 2.2 [Reference Markov20, Reference Raikov29].
Let
$X$
be a Tychonoff space. The free locally convex space
$L(X)$
on
$X$
is a pair consisting of a locally convex space
$L(X)$
and a continuous mapping
$\unicode[STIX]{x1D6FE}:X\rightarrow L(X)$
such that
$\unicode[STIX]{x1D6FE}(X)$
forms a Hamel basis for
$L(X)$
, and every continuous mapping
$f$
from
$X$
to a locally convex space
$E$
gives rise to a unique continuous linear operator
$\bar{f}:L(X)\rightarrow E$
with
$f=\bar{f}\circ \unicode[STIX]{x1D6FE}$
.
The free topological groups
$F(X)$
,
$A(X)$
,
$B(X)$
, as well as the free locally convex space
$L(X)$
, always exist by the adjoint functor theorem [Reference Morris22] and are unique up to isomorphism. In what follows we will identify
$X$
with its homeomorphic copy
$\unicode[STIX]{x1D6FE}(X)$
.
For any space
$X$
, the group
$A(X)$
is the quotient topological group of
$F(X)$
by the commutator subgroup; and the group
$B(X)$
is the quotient topological group of
$A(X)$
by the subgroup of squares
$A(2X)$
. Thus,
$B(X)$
is the image of
$A(X)$
(and of
$F(X)$
) under a continuous open homomorphism.
Below we collect several known results about the structure of the free abelian group
$A(X)$
(see [Reference Arkhangel’skii1, Reference Hardy, Morris and Thompson10, Reference Joiner12, Reference Mack, Morris and Ordman19, Reference Morris23, Reference Thomas31]) which will be used in the sequel. Despite the fact that
$A(X)$
is a commutative group, we prefer to use the multiplicative notation. The neutral element of
$A(X)$
is denoted by
$e$
.
Theorem 2.3. Let
$X$
be any Tychonoff space,
$A(X)$
the free abelian topological group on
$X$
, and
$A_{n}(X)$
the subset of
$A(X)$
consisting of all words of length
${\leqslant}n$
with respect to
$X$
. Then there is a base
${\mathcal{B}}$
of open neighborhoods in the subspace
$A_{n}(X)$
of the reduced words
$x_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots \,x_{n}^{\unicode[STIX]{x1D716}_{n}}$
, for
$\unicode[STIX]{x1D716}_{i}=+1$
or
$-1$
, such that
${\mathcal{B}}$
consists of all subsets in the form
$U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots \,U_{n}^{\unicode[STIX]{x1D716}_{n}}$
, where
$U_{i}$
are open neighborhoods of
$x_{i}$
in
$X$
, and
$U_{i}$
,
$U_{j}$
are disjoint whenever
$x_{i}\neq x_{j}$
. Also, each
$A_{n}(X)$
is closed in
$A(X)$
and
$A_{n}(X)\setminus A_{n-1}(X)$
is an open subset of
$A_{n}(X)$
. Further, if
$X$
is a
$k_{\unicode[STIX]{x1D714}}$
-space with
$k_{\unicode[STIX]{x1D714}}$
-decomposition
$X=\bigcup _{n=1}^{\infty }X_{n}$
, then
$A(X)$
is a
$k_{\unicode[STIX]{x1D714}}$
-space with
$k_{\unicode[STIX]{x1D714}}$
-decomposition
$A(X)=\bigcup _{n=1}^{\infty }A_{n}(X_{n})$
and every compact subspace
$K$
of
$A(X)$
satisfies
$K\subseteq A_{n}(X_{n})$
for some natural number
$n$
.
Corollary 2.4. In the notation of Theorem 2.3, the projection maps
$\unicode[STIX]{x1D719}_{i}:U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}\cap (A_{n}(X)\setminus A_{n-1}(X))\rightarrow U_{i}\subseteq X$
given by
$\unicode[STIX]{x1D719}_{i}(z_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots z_{n}^{\unicode[STIX]{x1D716}_{n}})=z_{i}$
, where each
$z_{i}\in U_{i}$
, are continuous.
Sipacheva [Reference Sipacheva30] observed that the statements which are completely analogous to Theorem 2.3 and Corollary 2.4 are valid also for the free boolean group. One should everywhere replace
$A_{n}(X)$
by
$B_{n}(X)$
and consider
$\unicode[STIX]{x1D716}_{i}=+1$
only.
Remark 2.5. It is not true however that all known properties of
$A(X)$
are transferred automatically to
$B(X)$
. Namely, the finite power
$X^{n}$
is homeomorphic to a closed subset of
$A(X)$
for each natural number
$n$
, but there exists (under the Continuum Hypothesis) a countable space
$X$
such that
$X^{2}$
is not contained in
$B(X)$
as a subspace [Reference Sipacheva30].
Definition 2.6. A topological space is said to be a continuum if it is a compact metrizable connected space. All continua in this paper are considered to be non-degenerate, meaning that they contain at least two points.
Definition 2.7 [Reference Cook5, Reference Hart, Nagata and Vaughan11].
A continuum
$X$
is said to be a Cook continuum if for every subcontinuum
$K\subseteq X$
, the only continuous maps of
$K$
into
$X$
are the identity map and the constant maps.
Proposition 2.8 [Reference Kuratowski15, §47, III.1].
Every non-empty open subset of a continuum
$X$
contains a non-degenerate subcontinuum.
Corollary 2.9. Every non-empty open subset of a Cook continuum
$X$
contains a non-degenerate subset
$M$
such that
$M$
is a Cook continuum.
Clearly, a non-degenerate continuum is not zero-dimensional. It is easy to deduce from the main result of [Reference Baars4] that
$F(X)$
and
$F(X\oplus X)$
(and therefore also
$A(X)$
and
$A(X\oplus X)$
) are isomorphic for every infinite zero-dimensional metrizable locally compact space
$X$
. For this reason, we are forced to look at more complicated zero-dimensional separable metrizable spaces.
Similarly to many others published in the past works, for our purposes we will consider a rigid Bernstein set
$B$
in the real line
$\mathbb{R}$
. A detailed description of the properties of
$B$
and its construction can be found in [Reference van Mill32, Example 6.13.1] (see also [Reference Krupski and Marciszewski14, Reference Pol28]). According to Pol, the standard method of construction of
$B$
was originated by Kuratowski.
In our paper we make extensive use of the following strong rigidity property of
$B$
, the proof of which in fact is a straightforward consequence of [Reference Krupski and Marciszewski14, Corollary 3.3].
Proposition 2.10. If
$G$
is an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset of the rigid Bernstein set
$B$
, then for each finite collection of continuous functions
$f_{j}:G\rightarrow B,j\in \{1,2,\ldots ,n\}$
there exists an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset
$G^{\prime }$
of
$G$
such that the restriction
$f_{j}\restriction G^{\prime }$
is either the identity or constant for every
$j\in \{1,2,\ldots ,n\}$
.
3 Main results
Theorem 3.1. Let:
(a)
$X$
be a Cook continuum; or(b)
$X$
be a rigid Bernstein set.
Then the free abelian topological group
$A(X\oplus X)$
does not embed into
$A(X)$
as a topological subgroup.
Proof. We start with the following crucial lemma.
Lemma 3.2. Let
$\unicode[STIX]{x1D6FC}:M\rightarrow A(X)$
be a homeomorphic embedding, where:
(a)
$M\subseteq X$
and both
$X$
and
$M$
are Cook continua; or(b)
$X$
is a rigid Bernstein set and
$M$
is an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset of
$X$
.
Then in the case (a) there exists a Cook continuum
$M_{1}\subseteq M$
; in the case (b) there exists an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset
$M_{1}$
of
$M$
such that:
(i) for some natural number
$n$
,
$\unicode[STIX]{x1D6FC}(M_{1})\subseteq A_{n}(X)\setminus A_{n-1}(X)$
;(ii)
$\unicode[STIX]{x1D6FC}(M_{1})\subseteq U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}$
, where the latter set
$U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}$
is some fixed member of the base
${\mathcal{B}}$
, as in the notation of Theorem 2.3;(iii) for at least one
$i\in \{1,2,\ldots ,n\}$
and all
$x\in M_{1}$
, we have
$x_{i}=x$
, where
$\unicode[STIX]{x1D6FC}(x)=x_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots x_{n}^{\unicode[STIX]{x1D716}_{n}}\in U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}$
. Without loss of generality, this is the case for
$i=1,\ldots ,r$
for some natural number
$1\leqslant r\leqslant n$
;(iv) for each
$i\in \{r+1,\ldots ,n\}$
, there exists an element
$z_{i}\in X$
such that for all
$x\in M_{1}$
, we have
$x_{i}=z_{i}$
;(v) there is a fixed integer
$-n\leqslant s\leqslant n$
such that
$\unicode[STIX]{x1D6FC}(x)=x^{s}u$
for all
$x\in M_{1}$
; and
$u=e$
if
$r=n$
, otherwise
$u=z_{r+1}^{\unicode[STIX]{x1D716}_{r+1}}\ldots z_{n}^{\unicode[STIX]{x1D716}_{n}}\in A_{n}(X)$
, where all
$z_{i}$
are fixed by the previous item (iv).
Proof of Lemma 3.2.
Case (a). By Theorem 2.3, as
$\unicode[STIX]{x1D6FC}(M)$
is compact, we can choose the smallest natural number
$n$
such that
$\unicode[STIX]{x1D6FC}(M)\subseteq A_{n}(X)$
. Put
$V_{n}=\unicode[STIX]{x1D6FC}^{-1}(\unicode[STIX]{x1D6FC}(M)\cap (A_{n}(X)\setminus A_{n-1}(X)))$
. Since
$\unicode[STIX]{x1D6FC}(M)\cap (A_{n}(X)\setminus A_{n-1}(X))$
is an open subspace of the connected compact space
$\unicode[STIX]{x1D6FC}(M)$
, the set
$V_{n}$
is uncountable and open in
$M$
. Further, consider the open cover of
$V_{n}$
formed by the sets
$V_{n}\cap \unicode[STIX]{x1D6FC}^{-1}(\unicode[STIX]{x1D6FC}(M)\cap U)$
, where open neighborhoods
$U$
run over the base
${\mathcal{B}}$
, as in the notation of Theorem 2.3.
$V_{n}$
is a separable metrizable space; therefore, this cover has a countable subcover. Since
$V_{n}$
is uncountable, at least one member
$G$
of the subcover must also be uncountable. So,
$G$
is a non-trivial open subset of the Cook continuum
$M$
.
By Corollary 2.9, there exists a subspace
$M_{1}$
of
$G\subseteq M$
such that
$M_{1}$
is a Cook continuum. Clearly, (i) and (ii) are true, namely
$\unicode[STIX]{x1D6FC}(M_{1})\subseteq U\cap (A_{n}(X)\setminus A_{n-1}(X))$
, and
$U=U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}\in {\mathcal{B}}$
.
Let
$\unicode[STIX]{x1D719}_{j}:U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}\rightarrow X$
, for each
$j\in \{1,2,\ldots ,n\}$
, be defined by
$\unicode[STIX]{x1D719}_{j}(x_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots x_{n}^{\unicode[STIX]{x1D716}_{n}})=x_{j}$
. So, every composition map
$\unicode[STIX]{x1D719}_{j}\circ \unicode[STIX]{x1D6FC}\restriction _{M_{1}}:M_{1}\rightarrow X$
is a continuous map of the Cook continuum
$M_{1}\subseteq X$
into
$X$
, where
$\unicode[STIX]{x1D6FC}\restriction _{M_{1}}$
is the restriction of
$\unicode[STIX]{x1D6FC}$
to
$M_{1}$
.
Since
$M_{1}$
is a Cook continuum, each
$\unicode[STIX]{x1D719}_{j}\circ \unicode[STIX]{x1D6FC}\restriction _{M_{1}}$
is either the identity map or a constant map. Further, as
$\unicode[STIX]{x1D6FC}$
is a homeomorphism, so too
$\unicode[STIX]{x1D6FC}\restriction _{M_{1}}$
is a homeomorphism of
$M_{1}$
onto its image. Thus, not all
$\unicode[STIX]{x1D719}_{j}\circ \unicode[STIX]{x1D6FC}\restriction _{M_{1}}$
can be constant maps. We conclude that for at least one
$i\in \{1,2,\ldots ,n\}$
,
$\unicode[STIX]{x1D719}_{i}\,\circ \,\unicode[STIX]{x1D6FC}\restriction _{M_{1}}$
is the identity map; that is,
$x_{i}=x$
for all
$x\in M_{1}$
. This may happen for up to
$n$
values of
$i$
. Without loss of generality, since
$A(X)$
is an abelian group, we can assume that
$x_{i}=x$
for all
$x\in M_{1}$
, where
$i=1,2,\ldots ,r$
. This proves (iii).
Now, for each
$j\in \{r+1,\ldots ,n\}$
,
$\unicode[STIX]{x1D719}_{j}\circ \unicode[STIX]{x1D6FC}\restriction _{M_{1}}(M_{1})=\{z_{j}\}$
for some fixed
$z_{j}\in X$
. This proves (iv).
We see that for all
$x\in M_{1}$
,
$\unicode[STIX]{x1D6FC}(x)=x^{\unicode[STIX]{x1D716}_{1}}x^{\unicode[STIX]{x1D716}_{2}}\ldots x^{\unicode[STIX]{x1D716}_{r}}z_{r+1}^{\unicode[STIX]{x1D716}_{r+1}}z_{r+2}^{\unicode[STIX]{x1D716}_{r+2}}\ldots z_{n}^{\unicode[STIX]{x1D716}_{n}}$
. Put
$u=e$
if
$r=n$
, otherwise
$u=z_{r+1}^{\unicode[STIX]{x1D716}_{r+1}}z_{r+2}^{\unicode[STIX]{x1D716}_{r+2}}\ldots z_{n}^{\unicode[STIX]{x1D716}_{n}}$
, and define
$s=\unicode[STIX]{x1D716}_{1}+\unicode[STIX]{x1D716}_{2}+\cdots +\unicode[STIX]{x1D716}_{r}$
. Thus, we obtain (v), which finishes the proof of Lemma 3.2 in the case (a).
Case (b). For every natural number
$n$
, consider
$V_{n}=\unicode[STIX]{x1D6FC}^{-1}(\unicode[STIX]{x1D6FC}(M)\cap (A_{n}(X)\setminus A_{n-1}(X)))$
. The set
$V_{n}$
can be expressed as the difference of two closed subsets in a separable metrizable set; therefore, every
$V_{n}$
is a
$G_{\unicode[STIX]{x1D6FF}}$
-subset of
$M$
. Since
$M$
is uncountable, we can choose
$n$
such that
$V_{n}$
is uncountable. Exactly as in case (a), we find an open uncountable
$G\subseteq V_{n}$
such that
$\unicode[STIX]{x1D6FC}(G)$
is contained in some
$U\in {\mathcal{B}}$
, that is,
$U=U_{1}^{\unicode[STIX]{x1D716}_{1}}\ldots U_{n}^{\unicode[STIX]{x1D716}_{n}}\in {\mathcal{B}}$
, as in the notation of Theorem 2.3. On this stage the finite collection of continuous functions
$\unicode[STIX]{x1D719}_{j}\circ \unicode[STIX]{x1D6FC}\restriction _{G}:G\rightarrow X,j\in \{1,2,\ldots ,n\}$
is correctly defined; therefore, by Proposition 2.10, there exists an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset
$M_{1}$
of
$G\subseteq M$
such that every
$\unicode[STIX]{x1D719}_{j}\circ \unicode[STIX]{x1D6FC}\restriction _{M_{1}}$
is either the identity map or a constant map. It is easy to see that (i) and (ii) are true, and we complete the proof of (iii) exactly as in the case (a). Items (iv) and (v) also can be proved without any change in the arguments, which finishes the proof of Lemma 3.2 in the case (b).◻
Now we complete the proof of Theorem 3.1. (a)
$X$
is a Cook continuum or (b)
$X$
is a rigid Bernstein set. In both cases let
$Y^{\prime }$
and
$Y^{\prime \prime }$
be two copies of the space
$X$
; denote by
$h^{\prime }:X\rightarrow Y^{\prime }$
and
$h^{\prime \prime }:X\rightarrow Y^{\prime \prime }$
the witnessing homeomorphisms. Assume that
$Z=Y^{\prime }\cup Y^{\prime \prime }$
is a subspace of
$A(X)$
with
$Y^{\prime }\cap Y^{\prime \prime }=\emptyset$
; that is,
$Z$
is homeomorphic to the free sum
$X\oplus X$
. We shall prove that
$Z$
is not a free algebraic basis for the subgroup of
$A(X)$
that it generates.
Firstly, we apply Lemma 3.2 with
$M=X$
. So, there are a natural number
$n$
and a Cook continuum
$M_{1}\subseteq M$
(in case (a)); an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset
$M_{1}$
of
$M$
(in case (b)) such that for all
$x\in M_{1}$
,
$h^{\prime }(x)=x^{s}u$
, where
$-n\leqslant s\leqslant n$
is a fixed integer and
$u$
is a fixed element of
$A_{n}(X)$
.
Next we apply Lemma 3.2 with
$M=M_{1}$
. So, there are a natural number
$k$
and a Cook continuum
$M_{2}\subseteq M_{1}$
(case (a)); an uncountable
$G_{\unicode[STIX]{x1D6FF}}$
-subset
$M_{2}$
of
$M_{1}$
(case (b)) such that for all
$x\in M_{2}$
,
$h^{\prime \prime }(x)=x^{t}v$
, where
$-k\leqslant t\leqslant k$
is a fixed integer and
$v$
is a fixed element of
$A_{k}(X)$
.
Let
$m_{1}$
and
$m_{2}$
be two distinct points in
$M_{2}$
and put
$w_{1}=h^{\prime }(m_{1})\in gp\{Y^{\prime }\}$
,
$w_{2}=h^{\prime }(m_{2})\in gp\{Y^{\prime }\}$
,
$w_{3}=h^{\prime \prime }(m_{1})\in gp\{Y^{\prime \prime }\}$
, and
$w_{4}=h^{\prime \prime }(m_{2})\in gp\{Y^{\prime \prime }\}$
, where
$gp\{S\}$
denotes the subgroup of
$A(X)$
generated algebraically by a subset
$S$
of
$A(X)$
.
Then
$w_{1}w_{2}^{-1}=m_{1}^{s}um_{2}^{-s}u^{-1}=m_{1}^{s}m_{2}^{-s}$
and
$w_{3}w_{4}^{-1}=m_{1}^{t}vm_{2}^{-t}v^{-1}=m_{1}^{t}m_{2}^{-t}$
.
Thus,
$(w_{1}w_{2}^{-1})^{t}=m_{1}^{st}m_{2}^{-st}$
and
$(w_{3}w_{4}^{-1})^{s}=m_{1}^{st}m_{2}^{-st}$
; that is,
$w_{1}^{t}w_{2}^{-t}w_{3}^{-s}w_{4}^{s}=e$
. But
$w_{1},w_{2},w_{3}$
, and
$w_{4}$
are distinct points in
$Z=Y^{\prime }\cup Y^{\prime \prime }$
. Hence,
$Z$
is not a free algebraic basis for the subgroup of
$A(X)$
that it generates. So,
$gp\{Y^{\prime }\cup Y^{\prime \prime }\}$
is not topologically isomorphic to
$A(Y^{\prime }\oplus Y^{\prime \prime })$
.◻
Since
$X\oplus X$
is a closed subset of the square
$X\times X$
, we obtain the following result.
Corollary 3.3. Let
$X$
be a Cook continuum or a rigid Bernstein set. Then the free abelian topological group
$A(X\times X)$
does not embed into
$A(X)$
as a topological subgroup.
Theorem 3.4. Let
$X$
be a Cook continuum or a rigid Bernstein set. Then the free boolean topological group
$B(X\oplus X)$
(and therefore also
$B(X\times X)$
) does not embed into
$B(X)$
as a topological subgroup.
Proof. We repeat word by word the proof of Theorem 3.1. Everywhere one should assume that each
$\unicode[STIX]{x1D716}_{i}=+1$
only, and all other integer-valued coefficients are taken modulo dividing by
$2$
.◻
4 Remarks and problems
What is the relationship between existence of embeddings of
$O(X)$
for different free topological functors
$O$
?
Let
$X$
and
$Y$
be any metrizable compact spaces. For brevity let us enumerate the following conditions:
(1) “
$A(Y)$
isomorphically embeds into
$A(X)$
”;(2) “
$B(Y)$
isomorphically embeds into
$B(X)$
”;(3) “
$L(Y)$
isomorphically embeds into
$L(X)$
”.
We do not know the answer to the following questions.
Problem 1. Does (1) imply (2)? Does (2) imply (1)? Does (1) imply (3)?
Remark 4.1. We note that
$A(\mathbb{R})$
isomorphically embeds into
$A(I)$
, but
$L(\mathbb{R})$
does not isomorphically embed into
$L(I)$
by Theorems 1.2 and 1.3, so condition (1) does not imply condition (3) if
$X$
is non-compact. On the other hand, let
$Y$
be the one-point compactification of the disjoint union of the Euclidean cubes
$I^{n}$
for
$n\in \mathbb{N}$
. Then
$L(Y)$
isomorphically embeds into
$L(I)$
(see [Reference Leiderman, Morris and Pestov17, Remark 4.6], cf. [Reference Gartside and Feng7, Reference Górak, Krupski and Marciszewski8]), but
$A(Y)$
does not isomorphically embed into
$A(I)$
by Theorem 1.2. So, condition (3) does not imply condition (1).
Our next open problem is motivated by [Reference Krupski and Marciszewski14, Question 5.8].
Problem 2. Suppose that
$X$
is a Polish zero-dimensional
$\unicode[STIX]{x1D70E}$
-compact space. Does
$A(X\oplus X)$
embed into
$A(X)$
as a topological subgroup?
Remark 4.2. Now we consider another remarkable metrizable continuum—the pseudo-arc
$P$
. The pseudo-arc
$P$
is a non-generate hereditarily indecomposable chainable continuum (see the survey article [Reference Lewis18]). Like the arc
$[0,1]$
, the pseudo-arc is homeomorphic to each of its non-generate subcontinua. Clearly,
$P$
does not contain any copy of
$[0,1]$
. A much stronger statement has been proved in [Reference Kawamura and Leiderman13]: there exists no linear continuous surjection from
$C_{p}(P)$
onto
$C_{p}[0,1]$
.
Therefore, the free locally convex space
$L[0,1]$
is not embeddable into
$L(P)$
as a linear subspace. However,
$L(P)$
and
$L(P\oplus P)$
are shown to be isomorphic [Reference Kawamura and Leiderman13]. Since
$P$
is connected, and the relation of
$A$
-equivalence preserves connectedness [Reference Graev9] (see [Reference Arhangel’skii and Tkachenko3, 7.10.3]), the free abelian topological groups
$A(P)$
and
$A(P\oplus P)$
are not isomorphic. This makes the next proposition interesting.
Proposition 4.3. Let
$P$
be the pseudo-arc and
$\{\ast \}$
be any abstract point not in
$P$
. Then
$A(P\oplus \{\ast \})$
and
$A(P\oplus P)$
are isomorphic.
Proof. Take any proper non-generate subcontinuum
$Q$
of
$P$
and consider the quotient space
$P/Q$
obtained from
$P$
by shrinking
$Q$
into a point. Then
$Q$
is homeomorphic to
$P$
and the quotient space
$P/Q$
is also homeomorphic to
$P$
(for the details, see [Reference Kawamura and Leiderman13]) and, moreover,
$Q$
is a retract of
$P$
[Reference Cornette6].
By [Reference Okunev25], the free abelian topological groups
$A(P\oplus \{\ast \})$
and
$A(P/Q\oplus Q)$
are isomorphic, but
$P/Q\oplus Q$
is homeomorphic to
$P\oplus P$
, and the claim follows.◻
Since
$P\oplus \{\ast \}$
is homeomorphic to a closed subspace of
$P$
, we immediately deduce that
$A(P\oplus P)$
isomorphically embeds into
$A(P)$
as a topological subgroup.
Note that this statement can be proved in an even shorter way: it is enough to take two disjoint non-generate subcontinua
$Q_{1}$
and
$Q_{2}$
in
$P$
.
However, the following question remains open.
Problem 3. Let
$X$
be the pseudo-arc
$P$
.
Does
$A(X\times X)$
embed into
$A(X)$
as a topological subgroup?
Does
$L(X\times X)$
embed into
$L(X)$
as a linear space?
A negative answer to Problem 3 would show that the items (1) and (2) in Question 1.4 are not equivalent.
Remark 4.4. As has been noticed by many authors, there is no principal difference between free (abelian) topological objects in the sense of Markov or Graev (see for details [Reference Morris22, Reference Sipacheva30]). For instance, denote the free abelian topological group in the sense of Graev by
$AG(X)$
. It is known that for every
$X$
, the free abelian topological group in the sense of Markov
$A(X)$
is isomorphic to the product of
$AG(X)$
and the group of integers
$\mathbb{Z}$
. If
$AG(Y)$
embeds isomorphically into
$AG(X)$
, then evidently
$AG(Y)\times \mathbb{Z}\cong A(Y)$
embeds isomorphically into
$AG(X)\times \mathbb{Z}\cong A(X)$
. So, if
$A(Y)$
does not embed isomorphically into
$A(X)$
, then also
$AG(Y)$
does not embed isomorphically into
$AG(X)$
and all the main results of the article are valid also for embeddings of the free abelian topological groups in the sense of Graev.
Acknowledgement
The third author thanks Ben-Gurion University of the Negev for hospitality during which much of the research for this paper was completed.
