UDAYAN B. DARJI AND ´ETIENNE MATHERON

Abstract. We prove some “universality” results for topological dynamical
systems. In particular, we show that for any continuous self-mapT of a perfect
Polish space, one can find a dense,T-invariant set homeomorphic to the Baire
spaceN^{N}; that there exists a bounded linear operatorU :`1→`1 such that
any linear operatorT from a separable Banach space into itself withkTk ≤1
is a linear factor ofU; and that given anyσ-compact familyF of continuous
self-maps of a compact metric space, there is a continuous self-mapUF ofN^{N}
such that eachT∈ F is a factor ofUF.

1. Introduction

It is well known that the Cantor space 2^{N}, the Baire spaceN^{N} and the Banach
space `1 = `1(N) have some interesting universality properties: every compact
metric space is a continuous image of 2^{N}, every Polish space is a continuous image
ofN^{N}, and every separable Banach space is a quotient of`1.

In this note, we will mainly be concerned with the following kind of generalization of these classical facts: instead of just lifting the points of a given spaceX into one of the above spaces, one would like to lift continuous self-maps ofX.

Throughout the paper, by amapwe always mean a continuous mapping between topological spaces. For us, adynamical system will be a pair (X, T), whereX is a topological space andT is a self-map ofX.

A dynamical system is (X, T) is a factor of a dynamical system (E, S), and (E, S) is an extension of (X, T), if there is a map π from E onto X such that T π=πS,i.e.the following diagram commutes:

E ^{S} //

π

E

π

X ^{T} //X

WhenT and S are linear operators acting on Banach spaces X and E, we say that T is a linear factor of S if the above diagram can be realized with a linear factoring mapπ.

Among other things, we intend to prove the following two “common extension”

results.

2000Mathematics Subject Classification. Primary: 37B99,54HB20. Secondary: 54C20,47A99.

Key words and phrases. universal, factor,`1, Cantor space, Baire space.

The first author would like to acknowledge the hospitality and financial support of Universit´e d’Artois.

1

• There exists a bounded linear operator U : `1 →`1 such that every linear operator T from a separable Banach space into itself with kTk ≤ 1 is a linear factor ofU. (This is proved in Section 3.)

• Given anyσ-compact familyFof self-maps of a compact metric space, there
is a map U_{F} :N^{N}→N^{N} such that each T ∈ F is a factor of U_{F}. (This is
proved in Section 4.2.)

Regarding the Baire spaceN^{N}, we will also show that any Polish dynamical the
dynamics of any continuous self-mapT of a perfect Polish spaceX is in some sense

“captured” by a self-map ofN^{N}; namely, there is a dense,T-invariant set insideX
which is homeomorphic toN^{N}. This result is proved in Section 2.

The reader may wonder why, in the second result quoted above, the domain of
the mapU_{F}isN^{N}and not the Cantor space 2^{N}. The reason is that, even for simple
compact families F, one cannot obtain a common extension defined on a compact
metrizable space; see Remark 4.6. We have not been able to characterize those
σ-compact familiesF admitting a common extension defined on 2^{N}. However, we
give a simple sufficient condition which implies in particular that this holds true for
the family of all contractive self-maps of a compact metric space. This is proved
in Section 4.3. We also show that for an arbitraryσ-compact family F, a rather
natural relaxation in the definition of an extension allows this; see Section 5.

The kind of question we are considering here may be formulated in a very general
way. One has at hand a certain categoryCand, given a collection of objectsF ⊆C,
one wants to know if there exists an objectU_{F} ∈Cwhich is projectively universal
forF, in the following sense: for anyT∈ F, there is an epimorphismπ:U_{F} →T
from U_{F} to T. Less ambitiously, one may just want to find a “small” family of
objectsU which is projectively universal forF, i.e.for anyT∈ F, there exists an
epimorphsim from someU∈ U toT.

In this paper, we have been concerned with categoriesCwhose objects are topo-
logical dynamical systems and whose morphisms are factoring maps. For example,
if C is the category of all compact metrizable dynamical systems then, as men-
tioned above, there is no projectively universal object for C, but there is one for
the family of all contractive dynamical systems defined on a given compact metric
space. Moreover, it is well known that every self-map of a compact metrizable
space is a factor of some self-map of 2^{N}(see Section 4.1); in other words, the fam-
ily of all dynamical systems defined on 2^{N} is projectively universal for the whole
category. Likewise, if C is the category all linear dynamical systems on separable
Banach spaces then (as annouced above) there is an objectUliving on`1 which is
projectively universal for the family all (X, T)∈CwithkTk ≤1.

Many other categories may be worth studying from this point of view. We point out three of them.

One is the category of Polish dynamical systems, for which the family of all
dynamical systems defined onN^{N}happens to be projectively universal (see Section
4.1).

The second category we have in mind is that of dynamical systems defined on arc-like continua. An exotic space called thepseudo-arc plays a central role in the theory of arc-like continua. As it turns out, the family of all dynamical systems defined on the pseudo-arc is projectively universal for this category (this is proved

in [10]). The pseudo-arc has a rich and long history; seee.g.[9] for a detailed survey and [5] for an interesting model theoretic construction.

A third interesting example is the category of allminimal G-flows, for a given topological group G. Recall that a G-flow (X, G) is a compact topological space X endowed with a continuous action ofG, and that aG-flow (X, G) is said to be minimal if everyG-orbit{g·x; g∈G} is dense inX. It is well known that there exists a projectively universal object (M(G), G) for this category, which is unique up to isomorphism. There is a vast literature on universal minimal flows (see [14]

and the references therein, for example [3], [4], [7], [12], [13], [15]). In particular, the problem of determining exactly whenM(G) ismetrizablehas been well investigated;

see [12]. Very much in the spirit of the present paper, it seems quite natural to ask for which families F of minimalG-flows one can find a projectively universal G-flow defined on a metrizable compact space.

To conclude this introduction, we would like to thank the anonymous referee who kindly pointed out to us that the paper [11] by M. Lupini contains several very interesting examples of “universal” maps, all of them put inside a quite general and far-reaching model-theoretic framework.

2. Dense subsystems homeomorphic toN^{N}
In this section, our aim is to prove the following result.

Theorem 2.1. Let X be a perfect Polish space. For any self-map T : X → X,
there is a dense Gδ setY ⊆X homeomorphic to N^{N}such that T(Y)⊆Y.

This will follow from Lemmas 2.6 and 2.7 below. Also, we will make use of the
well-known topological characterization ofN^{N}due to Alexandrov and Urysohn:

Theorem 2.2. A (nonempty) Polish space spaceY is homeomorphic toN^{N}if and
only if it is0-dimensional and nowhere compact.

Here and afterwards, a space E is said to be 0-dimensional if it has a basis consisting of clopen sets, and nowhere compact if all compact subsets of E have empty interior inE. For a proof of Theorem 2.2, see e. g. [6, Theorem 7.7].

We start with a lemma where no assumption is made on the Polish spaceX. Lemma 2.3. Let X be a Polish space, T : X →X be continuous and Z ⊆X be 0-dimensional with T(Z)⊆Z. Then, there is Y such that

(1) Z⊆Y ⊆X andT(Y)⊆Y;

(2) Y is0-dimensional and Polish (i.e. aG_{δ} subset ofX).

Proof. We need the following simple and well known fact, which easily follows from basic definitions.

Fact 2.4. Let E be Polish and letF ⊆E be 0-dimensional. Then, for anyε >0,
there is a sequence of pairwise disjoint open sets{Un}inEsuch that diam(U_{n})< ε
andF ⊆S∞

n=1U_{n}.

By Fact 2.4 we may choose a sequence {U_{n}^{1}} of pairwise disjoint open subsets
of X such that the diameter of each U_{n}^{1} is less than 1 and Z ⊆ S∞

n=1U_{n}^{1}. Now
letV_{n}^{1} :=T^{−1}(U_{n}^{1}). Then the {V_{n}^{1}} are pairwise disjoint open sets. Moreover, as
T(Z) ⊆ Z, we have that Z ⊆ S∞

n=1V_{n}^{1}. Now we apply Fact 2.4 with E := V_{m}^{1}
andF :=V_{m}^{1} ∩Z, for eachm∈N. This produces a collection of pairwise disjoint

open sets{U_{n}^{2}}inX such that, for eachn, the diameter ofU_{n}^{2}is less than 1/2 and
U_{n}^{2}⊆V_{m}^{1} for somem. Hence the collection{U_{n}^{2}} satisfies the following properties:

• Z⊆S∞
n=1U_{n}^{2},

• diam(U_{n}^{2})<1/2 for eachn,

• T(U_{n}^{2})⊆U_{m}^{1} for some m.

Continuing in this fashion, we get for each j ∈N a collection of pairwise disjoint
open sets{U_{n}^{j}; n∈N}such that the following properties hold:

• Z⊆S∞
n=1U_{n}^{j},

• diam(U_{n}^{j})<1/2^{j} for eachn,

• T(U_{n}^{j})⊆U_{m}^{j−1} for somem.

Now we simply letY :=T∞ j=1

S∞

n=1U_{n}^{j}. ThisY has the required properties.

Remark 2.5. In the above lemma, one cannot makeY dense inX. For example,
takeX to be the disjoint union ofRand a countable discrete set{an; n∈N}. The
topology is the usual one. Let the mapT onX defined as follows: it is the identity
onRand it mapsa_{n} to then^{th}rational inR(under some fixed enumeration). Now
takeZ to be the set of irrationals on the line. There is no way to extend this Z to
a dense,T-invariant and still 0-dimensional setY.

Lemma 2.6. LetX be a perfect Polish space andT :X→X be continuous. Then,
there is a PolishX0⊆X dense inX and nowhere compact such thatT(X0)⊆X0.
Proof. We will find a meager,F_{σ}set Z⊆X dense inX such thatT^{−1}(Z)⊆Z. It
will then suffice to let X_{0} :=X\Z. AsX is Polish andZ anF_{σ} meager set, we
have thatX_{0} is Polish and dense inX. Moreover,X_{0}is nowhere compact. Indeed,
assume that for some nonempty open set U in X, the set X_{0}∩U has compact
closure inX0. Then, (X0∩U)∩X0is compact and hence is closed inX, and it has
empty interior in X sinceX\X0 is dense inX. In particular,X0∩U is nowhere
dense inX, a contradiction becauseX0 is dense inX.

To complete the proof, let us construct the desired Z. Let B1, B2, . . . be an enumeration of some countable basis of open sets forX. Let

S={(n, m)∈N×N : T^{m}(Bn) is meager}.

Let U =S

(n,m)∈SB_{n}. Let A =S{T^{m}(B_{n}); (n, m)∈S}. Then, A is meager in
X. LetD be a countable dense subset ofX\A. Clearly, each of T^{−m}(D),m≥0
isFσ. We claim, moreover, thatT^{−m}(D) is meager. This is clear form= 0 asD
is countable. To obtain a contradiction, assume that for some m≥1, T^{−m}(D) is
not meager. As T^{−m}(D) is Fσ, it must contain a nonempty open set. Letn∈N
be such that Bn ⊆T^{−m}(D). Then, we have that T^{m}(Bn) is meager since it is a
subset of the countable setD. Hence,T^{m}(B_{n})⊆A, contradicting thatD∩A=∅.

Let Z = S

m≥0T^{−m}(D). Then, Z is an F_{σ}, dense, and meager subset of X.
Moreover,T^{−1}(Z)⊆Z, concluding the proof of the lemma.

Lemma 2.7. Let X be a nowhere compact Polish space and let T : X → X be
continuous. Then there is a dense Gδ set Y ⊆X such that T(Y) ⊆Y and Y is
homeomorphic to N^{N}. Moreover, one may require that Y contains any prescribed
countable setC⊆X.

Proof. LetCbe any countable subset ofX. AsX is separable, we may expandCto be a countable set which is dense inX. LetZ =S∞

n=0T^{n}(C). Clearly,T(Z)⊆Z.

Moreover, Z is 0-dimensional as it is countable. Let Y be given by Lemma 2.3.

Only thatY is homeomorphic toN^{N}needs an argument. We already know thatY
is 0-dimensional and Polish. Finally,Y is nowhere compact asX is andY is dense

inX. Hence,Y is homeomorphic toN^{N}.

Remark 2.8. It follows in particular that if T : G→ G is a self-map of a non-
locally compact Polish group Gthen, for any given countable set C⊆G, there is
Y ⊆Ghomeomorphic toN^{N} such thatC⊆Y andT(Y)⊆Y.

Proof of Theorem 2.1. This is now immediate by applying Lemma 2.6 and then

Lemma 2.7.

3. A universal linear operator

In this section, we prove the following result, which says essentially that any bounded linear operator on a separable Banach space is a linear factor of a single

“universal” operator acting on`1.

Theorem 3.1. There exists a bounded linear operator U :`1 →`1 with kUk = 1 such that any bounded linear operatorT :X →X on a separable Banach spaceX is a linear factor ofρ·U for everyρ≥ kTk.

Proof. We need the following fact. This is probably well known but we include an outline of the proof for completeness.

Fact 3.2. There exists a 1-1 map µ : N → N such that, for any other 1-1 map
σ : N → N, one can find a set A ⊆ N and a bijection πA : N A such that
µ(A) =Aandσ=π^{−1}_{A} µπ_{A}.

Proof of Fact 3.2. For any 1-1 mapσ:N→N, letGσbe the digraph onNinduced byσ; that is : −→

ij is a directed edge ofG_{σ} iffσ(i) =j. Then, each componentC of
G_{σ} has one of the following forms:

• C is a cycle of lengthnfor somen∈N,

• there isi0∈Csuch that C={i0, i1, i2, . . .}where allik’s are distinct and the only directed edges are of the form−−−−→

i_{k}i_{k+1}, or

• C = {. . . , i_{−1}, i_{0}, i_{1}, . . .} where all i_{k}’s are distinct and the only directed
edges are of the form−−−−→

ikik+1.

Now, consider any µ whose digraph contains infinitely components of each type

described above.

To prove the theorem, we first observe that it suffices to construct U so that every bounded linear operator with norm at most 1 (on a separable Banach space) is a linear factor ofU.

Let µ be as in Fact 3.2. Let {en} be the standard basis of `1, i.e., en is 0 in
every coordinate except then^{th} coordinate where it takes the value 1. We define
U :`1→`1simply byU(en) =e_{µ(n)}. It is clear thatU is a bounded linear operator
of norm 1. Let T : X → X be a bounded linear operator acting on a separable
Banach space X, with kTk ≤ 1. Since X is separable and kTk ≤ 1, one can find
a countable dense subsetD of BX, the unit ball in X, so that T(D)⊆D. This
may be done by starting with any countable set dense in the unit ball and taking
its union with its forward orbit. We let{zn} be a enumeration ofD so that each
element ofD appears infinitely often in{zn}. Now we defineσ:N→N. For any

i∈N, we haveT(zi) =zj for somej. Moreover, since every element of D appears infinitely often in the sequence{zn}, we may choose a 1-1 mapσ:N→Nsuch that T(zi) =zσ(i) for alli∈N. Now, letA⊆NandπAbe as in Fact 3.2. Then there is a uniquely defined bounded linear operatorπ:`1→X (withkπk ≤1) such that

π(ei) =
z_{π}−1

A (i) ifi∈A, 0 ifi /∈A.

The operatorπisonto becauseπ(B`_{1}) contains all ofD, which is dense inBX;
see e.g. [2, Lemma 2.24].

Finally, let us show show thatT is a factor ofU with witness π,i.e.. πU =T π.

It is enough to check thatπU andT π are equal on{ei} as all maps are linear and continuous and {ei} spans`1. Let i /∈A. Then, by Fact 3.2 µ(i)∈/ A. Hence, we have that neitherinorµ(i) is inA, so that

π(U(ei)) =π(eµ(i)) = 0 and T(π(ei)) =T(0) = 0.

Now supposei∈A. Then,
T(π(e_{i})) =T(z_{π}−1

A (i)) =z_{σ(π}−1

A (i)) =z_{π}−1

A (µ(i)) =π(e_{µ(i)}) =π(U(e_{i})).

Remark 3.3. One really has to consider all multiples of U: it is not possible
to find a single bounded linear operatorU so that every bounded linear operator
T :X →X is a linear factor of it. Indeed, if an operatorT is a linear factor of some
operatorU, then there is a constantC such thatkT^{n}k ≤CkU^{n}k for alln∈N; in
particular,λIdcannot be a factor ofU if|λ|>kUk. To see this, letπwitness the
fact that T :X →X is a factor ofU :Y →Y. By the Open Mapping Theorem,
there a constant k such that BX ⊆ kπ(BY). Let A ≥1 be such that kπk ≤ A.

Then,C=kAhas the property thatkT^{n}k ≤CkU^{n}kfor alln∈N.

In spite of this last remark, one can find a single Fr´echet space operator which is universal for separable Banach space operators. (Interestingly, it is not known if there exist quotient universal Fr´echet spaces.)

Corollary 3.4. Let E := (`1)^{N}, the product of countably many copies of `1, en-
dowed with the product topology. There exists a continuous linear operator UE :
E→E such that any bounded operatorT acting on a separable Banach space is a
linear factor ofUE.

Proof. LetU :`1 → `1 be the operator given by Theorem 3.1. Then defineUE : E→E by

UE(x1, x2, x3, . . .) := (U(x1),2·U(x2),3·U(x3), . . .).

This is indeed a continuous linear operator, and it is readily checked that UE has
the required property. Indeed, if T : X →X is a bounded linear operator (X a
separable Banach space) then one can find n∈Nsuch that T is a linear factor of
n·U, with factoring map π :`_{1} →X. Then the operator eπ: E →X defined by
eπ(x_{1}, x_{2}, . . .) :=π(x_{n}) shows thatT is a linear factor ofU_{E}.

4. Common extensions of compact dynamical systems

In this section, we prove some “common extension” results for self-maps ofcom- pact metric spaces. In what follows, we denote byDthe collection of all dynamical systems (X, T) where X is a compact metric space; or, equivalently, the collection of all self-maps of compact metric spaces.

4.1. The extension property. A mapπ:E→X between topological spaces has theextension propertyif, for every mapφ:E→X, there is a mapS =Sφ:E→E such thatφ=πS. Note that such a mapπ is necessarily onto (consider constant mapsφ). The following theorem can be found in [8, p. 110], where it is attributed to Weihrauch [16].

Theorem 4.1. For any compact metric space X, there exists a map π: 2^{N} →X
with the extension property.

This is a strengthening of the very well known result saying that every compact
metric space is a continuous image of 2^{N}. In fact, from this theorem one gets
immediately the following result from [1]. (It seems that the two results were
proved independently at about the same time.)

Corollary 4.2. Every (X, T) ∈ D is a factor of some dynamical system of the
form(2^{N}, S).

Proof. Given π : 2^{N} → X with the extension property, just consider the map

φ:=T π: 2^{N}→X.

At this point, we digress a little bit by showing that a result of the same kind
holds true for Polish dynamical systems, the “universal” space being (as should
be expected) the Baire spaceN^{N}. This may be quite well known, but we couldn’t
locate a reference.

Proposition 4.3. For any Polish space X, there exists a map π: N^{N} →X with
the extension property. Consequently, any self-map T : X → X is a factor of a
self-map ofN^{N}.

Proof. We show that the “usual” construction of a continuous surjection fromN^{N}
ontoX produces a map with the extension property.

Denoting by N^{<}^{N} the set of all finite sequences of integers, let (Vt)_{t∈N}<N be a
family of non-empty open subsets ofX satisfying the following properties:

• V_{∅}=X and diam(V_{t})<2^{−|s|}for allt6=∅;

• Vti⊆Vs for eachtand alli∈N;

• S

i∈NVti=Vt for allt.

Let π: N^{N} → X be the map defined by {π(α)} :=T

k≥1V_{α|}_{k}. We show that
this mapπhas the extension property.

Let us fix a mapφ:N^{N}→X. Sinceφis continuous, one can find, for eachk∈N,
a family Sk ⊆N^{<}^{N} such thatS

{[s]; s∈ Sk} =N^{N} and each setφ([s]), s∈ Sk is
contained in some open setV_{t(s)}with|t(s)|=k. Moreover, we may do this in such
a way that the following properties hold true:

• ifk≥2, everys∈ Sk is an extension of somees∈ S_{k−1};

• ifs∈ Sk is an extension ofes∈ S

ek, (with ek≤k), thent(s) is an extension oft(es);

• Sk is an antichain,i.e.nos∈ Sk is a strict extension of ans^{0}∈ Sk.
(To get the third property, just remove from Sk the unnecessary s, i.e. those
extending somes^{0}∈ Sk withs^{0}6=s.)

Note that for each fixed k∈N, the family{[s]; s∈ Sk} is actually a partition
ofN^{N}because Sk is an antichain; hence, for anyα∈N^{N}, we may denote bysk(α)
the uniques∈ Sk such thats⊆α. Now, defineS :N^{N}→N^{N}as follows: ifα∈N^{N}
then S(α)|k :=t(s_{k}(α)) for all k ∈ N. It is readily checked that S is continuous

andφ=πS.

Let us now come back to compact metrizable dynamical systems. From Corollary 4.2, one can easily prove that there is a “universal extension” for all dynamical systems (X, T)∈D. However, this map is defined on a huge compactnonmetrizable space. In what follows, for any compact metric spacesX andY, we endowC(X, Y) with the topology of uniform convergence, which turns it into a Polish space.

Proposition 4.4. There exists a compact 0-dimensional space E and a map U : E→Esuch that every(X, T)∈D is a factor of(E, U).

Proof. By Corollary 4.2, it is enough to show that there exists a 0-dimensional
compact dynamical system (E, U) which is an extension of everyCantordynamical
system (2^{N}, S).

Consider the spaceE:= (2^{N})^{C(2}^{N}^{,2}^{N}^{)}endowed with the product topology. This is
a 0-dimensional compact space. For anyz= (zS)_{S∈C(2}N,2^{N})∈E, we defineU(z)∈E
as follows:

U(z)S =S(zS) for every S∈ C(2^{N},2^{N}).

Then, U is continuous because eachz 7→ U(z)S is continuous. Moreover, each
S∈ C(2^{N},2^{N}) is indeed a factor ofU by the very definition ofEandU: just consider
the projection mapπS :E→2^{N}fromEonto the S-coordinate.

Remark 4.5. If one just wants to lift those dynamical systems (X, T)∈Dwhich
aretransitive,i.e.have dense orbits, then one can take (E, U) := (βZ+, σ), where
βZ+ is the Stone- ˇCech compactification of Z+ and σ:βZ+ →βZ+ is the unique
continuous extension of the mapn7→1+n. Indeed, given a pointx0∈X with dense
T-orbit, the mapn7→T^{n}(x0) can be extended to a continuous mapπT :βZ+→X
showing that (X, T) is a factor of (βZ+, σ).

Remark 4.6. If (E, U) is a compact dynamical system which is universal forDin the sense of Proposition 4.4, then the spaceEcannot be metrizable.

Proof. For anyg ∈T, let us denote by Rg : T→ T the associated rotation ofT. We show that if (E, U) is a compact dynamical system that is “only” a common extension of all rotationsRg, then Ealready cannot be metrizable.

Let us fix a point e ∈ E. Since E is assumed to be compact and metrizable,
one can find an increasing sequence of integers (i_{k}) such that the sequenceU^{i}^{k}(e)
is convergent. Now, given g ∈ T, let π_{g} : E → T be a map witnessing that U
is an extension of R_{g}. Setting ξ_{g} := π_{g}(e), we then have g^{i}^{k}ξ_{g} = π_{g}U^{i}^{k}(e) for
all k; and since π_{g} is continuous, it follows that for every g ∈ T, the sequence
g^{i}^{k} is convergent. But it is well known that this is not possible. (One can prove
this as follows: if g^{i}^{k} → φ(g) pointwise, then R

Tφ(g)g^{−i}^{k}dg → 1 by dominated
convergence, contradicting the Riemann-Lebesgue lemma.)

4.2. Lifting to the Baire space. As observed above, if one wants to lift too many maps at the same time then one cannot do it with a compact metrizable space. On the other hand, our next result says that for “small families” of maps, one can find a common extension defined on aPolish space.

In what follows, for any compact metric spaces X and Y, we endow C(X, Y) with the topology of uniform convergence, which turns it into a Polish space.

Theorem 4.7. Let X be a compact metric space. Given any σ-compact family
F ⊆ C(X, X), one can find a mapU_{F} :N^{N}→N^{N}such that everyT ∈ F is a factor
of U_{F}.

This will follow from the next three lemmas, the first of which is a refinement of Theorem 4.1.

Let us say that a mapπ:E→X between compact metric spaces has thestrong
extension property if for every given map Φ : 2^{N} → C(E, X), there exists a map
F : 2^{N}→ C(X, X) such that for allp∈2^{N} we have that Φ(p) =πF(p).

Lemma 4.8. Let X be a compact metric space. Then, there is a mapπ: 2^{N}→X
which has the strong extension property.

Proof. We follow the proof of [8, Theorem 3.8]. The construction of the map πis the same as in that proof. We only need to verify thatπhas the additional stronger property.

We recall the construction ofπ. As in [8], one can easily construct inductively a sequence{Ji}of finite sets of cardinality at least 2 and collections{Vs}and {Ws} of nonempty open subsets ofXsuch that the following conditions hold for allk≥1 and for alls∈Qk

i=1Ji:

• the diameters ofV_{s}andW_{s} are less than 2^{−k};

• n

Vt; t∈Qk i=1Ji

o

is a cover ofX;

• {W_{sj}; j∈J_{k+1}} is a cover ofV_{s};

• V_{sn}=V_{s}∩W_{sn}.
We let Λ :=Q∞

i=1Ji, which is homeomorphic to 2^{N}. Forα∈Λ, we define π(α)
in the obvious way: {π(α)} = ∩^{∞}_{k=1}V_{α|}_{k}. Then, π is a well-defined map from Λ
ontoX.

Now let Φ : 2^{N} → C(Λ, X) be a map. With each p ∈ 2^{N} and Φ(p) we will
associate a mapF(p) : Λ→Λ as in [8]. However, we must do this in a continuous
fashion.

First, for eachk ≥1, we let 2^{−k} > εk >0 be sufficiently small so that εk is a
Lebesgue number for the covering{Wsj; j∈Jk+1}ofVsfor alls∈Qk

i=1Ji.
For any fixed p ∈ 2^{N}, the uniform continuity of Φ(p) allows us to choose an
increasing sequence of positive integers {m_{k}} such that for each k ≥ 1 and s ∈
Qmk

i=1J_{i}, the diameter of the set Φ(p)([s]) is less thanε_{k}/4. (Here [s] denotes those
elements of Λ which begin with s.) Moreover, as the map Φ is continuous, Φ(2^{N})
is a uniformly equicontinuous subset of C(Λ, X); so we may in fact choose {mk}
independent ofp. Hence, we have an increasing sequence of positive integers{mk}
such that for all p∈ 2^{N} and all k ≥ 1 and s ∈ Qm_{k}

i=1Ji, the diameter of the set Φ(p)([s]) is less thanεk/4.

By the uniform continuity of Φ we can choose an increasing sequence of integers
{lk} such that if p, p^{0} ∈2^{N} satisfyp|lk =p^{0}|lk, thend(Φ(p),Φ(p^{0}))< ε_{k}/4. (Here,

dis the “sup” metric onC(Λ, X).) Note that this implies in particular that for all
q∈ {0,1}^{l}^{k} and s∈Qm_{k}

i=1Ji, the diameter of the setEs,q:=

Φ(p)([s]); p|l_{k} =q
is less thanεk. This, in turn, implies (by the choice ofεk) thatEs,q is contained
in an open set Vt for somet ∈ Qk

i=1Ji; and, moreover, that if Es,q was already known to be contained inV

etfor someet∈Qk−1

i=1 Ji, then tmay be chosen to be an extension ofet. Let us summarize what we have:

(i) For all k≥1, for all q∈ {0,1}^{l}^{k} and for alls∈Qmk

i=1J_{i}, there
existst=t(q, s)∈Qk

i=1Jisuch that for allp∈2^{N} extendingq, we
have that Φ(p)([s])⊂Vt.

(ii) Everything can be done recursively in such a way that ifk^{0}≥k,
q^{0} ∈ {0,1}^{l}^{k}^{0} is an extension ofq ands^{0} ∈Qm_{k}0

i=1Ji is an extension
ofs, thent^{0}=t^{0}(q^{0}, s^{0}) is an extension oft.

Let us now define the desiredF : 2^{N}→ C(Λ,Λ). Fixp∈2^{N}. Then,F(p) : Λ→Λ
is defined as follows. For each α ∈ Λ, we define F(p)(α) so that for all k ≥ 1,
F(p)(α)|_{k} =t(p|_{l}_{k}, α|_{m}_{k}). By the above, F(p) is a well-defined function, and it is
continuous. Moreover, F itself is continuous: if p, p^{0} ∈2^{N} with p|_{l}_{k} = p^{0}|_{l}_{k}, then
d(F(p), F(p^{0}))< ε_{k} <2^{−k}. That Φ(p) =πF(p) for all p∈2^{N} follows in the same

manner as in [8].

Corollary 4.9. Let X be a compact metric space and let M be a compact subset
of C(X, X). Then, there is a compact setN ⊆ C(2^{N},2^{N})such that everyT ∈ Mis
a factor of someS∈ N.

Proof. SinceMis compact and metrizable, there is a mapG: 2^{N}→ C(X, X) such
that M =G(2^{N}). Now, choose a map π : 2^{N} → X having the strong extension
property, and consider the map Φ : 2^{N} → C(2^{N}, X) defined by Φ(p) = G(p)π.

Let F : 2^{N} → C(2^{N},2^{N}) be a map such that Φ(p) = πF(p) for all p∈ 2^{N}. Then
G(p)π= Φ(p) =πF(p), so thatG(p) is a factor ofF(p) for everyp∈2^{N} (because

πis onto). Hence, we may takeN :=F(2^{N}).

Lemma 4.10. Let N be a compact subset of C(2^{N},2^{N}). Then, there exists map
U_{N} :N^{N}→N^{N} such that everyS∈ N is factor ofU_{N}.

Proof. Recall first that the Polish spaceC(2^{N},2^{N}) is homeomorphic toN^{N}. Indeed,
it is easy to check thatC(2^{N},2^{N}) is 0-dimensional (because 2^{N}is), and it is also easy
to convince oneself that every equicontinuous family of self-maps of 2^{N}has empty
interior inC(2^{N},2^{N}), so thatC(2^{N},2^{N}) is nowhere compact.

AsN is a compact subset ofC(2^{N},2^{N})'N^{N}, we may enlarge it if necessary and
assume thatN is homeomorphic to 2^{N}. ThenZ:=C(N,2^{N}) is homeomorphic to
N^{N}. Now we define U_{N} : Z→ Zas in the proof of Proposition 4.4 (changing the
notation slightly): if z ∈ Z then U_{N}(z)(S) := S(z(S)) for every S ∈ N. Then
U_{N} is easily seen to be continuous (recall that Z=C(N,2^{N}) is endowed with the
topology of uniform convergence), and the evaluation mapπS atSshows that each
S∈ N is a factor ofUN. (To see thatπS :Z→2^{N}is onto, consider constant maps

z∈Z.)

Lemma 4.11. Let Z:= 2^{N} orN^{N}, and letU_{1}, U_{2}, . . .be maps fromZtoZ. Then,
there is a mapU :Z→Zsuch that eachU_{n} is a factor ofU.

Proof. The only property needed on the space Z is that Z is homeomorphic to
Z^{N}. Define U : Z^{N} → Z^{N} by U(z1, z2, . . .) := (U1(z1), U2(z2), . . .). Then, T is
continuous. Moreover, if πn : Z^{N} → Z is the projection of Z^{N} onto the n-th
coordinate, we have that π_{n}U =U_{n}π_{n} for all n≥1. Identifying Z^{N} withZ, this

shows thatU has the required property.

Proof of Theorem 4.7. LetFbe aσ-compact ofC(X, X), and writeF=S∞ n=1Mn

whereMn are compact sets. By Corollary 4.9, we may choose compact setsNn⊆
C(2^{N},2^{N}) such that everyT ∈ Mn is a factor of someS ∈ Nn. By Lemma 4.10,
we may choose for each n a map U_{n} : N^{N} → N^{N} such that every S ∈ Nn is a
factor ofU_{n}. Now, by Lemma 4.11, there is a map U :N^{N} →N^{N} such that each
mapU_{n} is factor ofU. Since factors of factors are again factors, we may thus take

U_{F} :=U.

4.3. Lifting to the Cantor space. The appearance of the Baire space N^{N} in
Theorem 4.7 may look rather surprising since we are dealing with self-maps of
compact metric spaces. This suggests the following question.

Question 4.12. LetX be a compact metric space. For which familiesF ⊆ C(X, X)
is it possible to find a common extensionU defined on the Cantor space 2^{N}?

Let us denote by I the family of all subsets F of C(X, X) having the above property. It follows from Lemma 4.11 thatIis closed under countable unions; and clearly, Iis hereditary for inclusion. Hence, I is aσ-ideal of subsets of C(X, X).

Sinceσ-ideals of sets are quite classical objects of study, it might be interesting to investigate the structural properties of this particular example.

Having said that, we are quite far from being able to answer Question 4.12 in a satisfactory way. However, in this section we give a rather special sufficient condition for belonging toI.

In what follows, we fix a compact metric space (X, d), and we also denote byd the associated “sup” metric onC(X, X).

For eachi∈Z+, let us denote byei:C(X, X)→ C(X, X) the map defined by
e_{i}(S) :=S^{i} =S◦ · · · ◦S

| {z }

i times

.

We shall say that a family N ⊆ C(X, X) has uniformly controlled powers if the
sequence {ei|_{N}; i ≥ 0} ⊆ C(N,C(X, X)) is pointwise equicontinuous; in other
words, if whenever a sequence (Sk)⊆ N converges to someS∈ N, it holds that
(1) S_{k}^{i}(x)−^{k→∞}−−−→S^{i}(x) uniformly with respect toxandi.

Example 4.13. Obviously, every finite family N ⊆ C(X, X) has uniformly con- trolled powers.

Example 4.14. Let N be a compact subset of C(X, X), and assume the each
map S ∈ N has the following properties: S is 1-Lipschitz with a unique fixed
point α(S) in X, and S^{i}(x)→α(S) for every x∈X. (This holds for example if
d(S(x), S(y))< d(x, y)wheneverx6=y.) ThenN has uniformly controlled powers.

Proof. Observe first that the mapS7→α(S) is continuous onN (because it has a
closed graph and acts between compact spaces). Moreover,S^{i}(x)→α(S)uniformly
with respect to S and xas i → ∞. (This follows from Dini’s theorem applied to

the continuous functions Φi : N ×X →R defined by Φi(S, x) :=d(S^{i}(x), α(S)).

The sequence (Φi) is nonincreasing because eachS∈ N is 1-Lipschitz.) So there is a sequence (εi) tending to 0 such that

d(S^{i}(x), α(S))≤ε_{i} for allx,iandS∈ N.

Now, let (Sk) be a sequence inN converging (uniformly) to some mapS. Set αk:=α(Sk) andα:=α(S). For everyxandi, we have on the one hand

d(S_{k}^{i}(x), S^{i}(x))≤d(S_{k}^{i}, S^{i}),
and on the other hand

d(S_{k}^{i}(x), S^{i}(x)) ≤ d(S_{k}^{i}(x), α_{k}) +d(α_{k}, α) +d(α, S^{i}(x))

≤ 2εi+d(αk, α).

Sinceαk→αandS_{k}^{i} →S^{i} for each fixedi∈Z^{+}, this gives (1).

Proposition 4.15. Let F ⊆ C(X, X). Assume that F can be covered by count-
ably many compact sets N having uniformly controlled powers. ThenF admits a
common extension U_{F} defined on2^{N}.

Proof. By Lemma 4.11, it is enough to prove the result for each one of the compact
familiesN. Moreover, since every self-map of a compact metric space is a factor of
some self-map of 2^{N}, it is enough to show that N admits a common extensionU
defined on some compact metric spaceE.

For any i ∈ Z+ and a ∈ X, denote by ei,a : N → X the map defined by
ei,a(S) :=S^{i}(a). Since N has uniformly controlled powers, the family of all such
mapsei,a is pointwise equicontinuous onN; so its uniform closureE is a compact
subset ofZ=C(N, X), by Ascoli’s theorem. Note also thatEcontains all constant
maps because{ei,a}already does (takei= 0 and letavary).

Now, consider the mapUN :Z→Zdefined in the proof of Lemma 4.10,
U_{N}(z)(S) =S(z(S)).

The family{ei,a} is invariant under UN, becauseUN(ei,a) =ei+1,a. Hence, E
is also invariant under U_{N}. Moreover, since E contains all constant maps, each
evaluation map π_{S} : E → X is onto. So the map U := U_{N}|E has the required

property.

Corollary 4.16. The familyF of all d-contractive self-maps of X has a common
extension defined on2^{N}.

Proof. For eachc∈(0,1), let us denote byNcthe family of allc-Lipschitz self-maps of X. Then Nc is compact by Ascoli’s theorem, and it has uniformly controlled powers by Example 4.14. SinceF=S

n∈NN_{1−}1

n, the resut follows.

Remark 4.17. In general, the family of all 1-Lipschitz self-maps ofX, or even the
family of all isometries ofX, does not have a common extension to 2^{N}. Consider
for example the rotations ofX :=T(see the proof of Remark 4.6).

We conclude this section with a kind of abstract nonsense characterizing the belonging to a subclass of theσ-idealI.

Fact 4.18. Let N be a compact subset of C(X, X), and let UN : C(N, X) →
C(N, X) be the (usual) map defined by U_{N}(z)(S) = S(z(S)). The following are
equivalent.

(i) There is a map U: 2^{N}→ 2^{N} and a continuous map S 7→qS from N into
C(2^{N}, X) such that U is an extension of each S ∈ N, with factoring map
qS.

(ii) There is a compact set E ⊆ C(N, X) invariant under U_{N} such that, for
each S∈ N, the evalution map z7→z(S) mapsEontoX.

Proof. Since every compact metrizable dynamical system has an extension to 2^{N},
the implication (ii) =⇒(i) follows easily from the proof of Proposition 4.15.

Conversely, assume that (i) holds true, and denote by q : N → C(2^{N}, X) the
map S 7→ qS. For any p∈2^{N}, denote byδp : C(2^{N}, X)→X the evaluation map
at p. Then, the map p 7→ δ_{p}◦ q is continuous from 2^{N} into C(N, X). So the
set E := {δp◦q; p ∈ 2^{N}} is a compact subset of C(N, X). Moreover, for any
z=δp◦q∈Eand allS∈ N we have

U_{N}(z) (S) =S(z(S)) =S(q_{S}(p)) =q_{S}(U(p)) =δ_{U(p)}◦q(S);

and this show thatE is invariant under the mapU_{N}. Finally, the evaluation map

at anyS∈ N mapsE ontoX becauseq_{S} is onto.

5. Common generalized extensions

In this section, we show that something similar to Theorem 4.7 does hold true
with 2^{N}in place ofN^{N}, but with a weaker notion of “factor”.

For any compact metric spaceX, letK(X) be the space of all nonempty compact subsets of X endowed with the Haudorff metric. If T : X → X is a map, we denote again by T : K(X) → K(X) the map induced by T on K(X); that is, T(M) ={T(x); x∈M}for everyM ∈ K(X). Obviously, (X, T) is equivalent to a subsystem of (K(X), T) thanks to the embeddingX3x7→ {x} ∈ K(X). However, in general (X, T) is not afactor of (K(X), T). For example, this cannot be the case ifT is onto (so thatX is a fixed point ofT in K(X)) andT has no fixed point in X.

We shall say that (X, T) is a ageneralized factor of a dynamical system (E, S) if there exists an upper semi-continuous mapping π : E → K(X) such that the following conditions hold:

• {x} ∈π(E) for every x∈X;

• π(S(u)) =T(π(u)) for allu∈E.

(Recall thatπ: E → K(X) is upper semicontinuous if the relation x∈π(u) is closed inE×X; equivalently, if for any open setO⊆X, the set{u∈E : π(u)⊆ O}is open inE.)

It is obvious from the definition that “true” factors are generalized factors. Also, we have the expected transitivity property:

Remark 5.1. If (X, T) is a generalized factor of (Y, S) and (Y, S) is a generalized factor of (Z, R), then (X, T) is a generalized factor of (Z, R).

Proof. Letπ:Y → K(X) witness thatT is a generalized factor ofSand letπ^{0}:Z→
K(Y) witness the fact thatSis a generalized factor ofR. Defineπ^{∗}:K(Y)→ K(X)
byπ^{∗}(M) :=S

{π(y) :y ∈ M}. Then, π^{∗} is a well-defined upper semicontinuous
mapping. Now, the mapping γ : Z → K(X) defined by γ := π^{∗}◦π^{0} is is easily
checked to be again upper semicontinuous, and shows thatT is a generalized factor

ofR.

Theorem 5.2. Let X be a compact metric space. Given any σ-compact family
F ⊆ C(X, X), one can find a map UF : 2^{N} → 2^{N} such that every T ∈ F is a
generalized factor of UF.

The proof will use Corollary 4.9, Lemma 4.11 and the following variant of Lemma 4.10.

Lemma 5.3. Let N be a compact subset of C(2^{N},2^{N}). Then, there exists map
U_{N} : 2^{N}→2^{N} such that everyS∈ N is generalized factor of U_{N}.

Proof. We assume of course that N 6= ∅. Let us denote by Z be the set of all
compact subsets of N ×2^{N} whose projection on the first coordinate is all of N.
This is a closed subset ofK(N ×2^{N}). Moreover,Zis 0-dimensional becauseN ×2^{N}
is, and it is easily seen to have no isolated points. Hence, Z is homeomorphic to
2^{N}.

We defineU_{N} :Z→Zin the following manner. LetA∈Zand for eachS∈ N,
letA_{S} be the vertical cross section ofAatS, i.e.,A_{S} ={x∈2^{N} : (S, x)∈A}. We
define the set U_{N}(A)⊆ N ×2^{N} by its vertical cross sections: U_{N}(A)_{S} :=S(A_{S})
for everyS ∈ N. In other words,

U_{N}(A) = [

S∈N

{S} ×S(A_{S}) ={(S, S(x)); (S, x)∈A}.

Let us check thatU_{N}(A)∈Z. That the projection ofUN(A) on the first coor-
dinate is all ofN is obvious becauseA_{S}6=∅for every S∈ N. Let

(S_{n}, u_{n}) be a
sequence inUN(A) converging to some point (S, u)∈ N ×2^{N}. Thenun=Sn(xn)
for some xn ∈ 2^{N} such that (Sn, xn)∈ A. By compactness, we may assume that
xn converges to some x ∈ 2^{N}. Then (S, x) ∈ A because A is closed in N ×2^{N};
and u = S(x) because u_{n} → u and S_{n} → S uniformly. So u ∈ S(A_{S}), i.e.

(S, u)∈U_{N}(A). Thus, we see thatU_{N}(A) is a closed and hence compact subset of
N ×2^{N}.

The same kind of reasoning shows thatU_{N} :Z→Zis continuous. One has to
check that if A_{n} →A in K(N ×2^{N}) then: (i) for every (S, u)∈ U_{N}(A), one can
find a sequence

(S_{n}, u_{n}) with (S_{n}, u_{n}) ∈ U_{N}(A_{n}) such that (S_{n}, u_{n}) → (S, u);

and (ii) for every sequence

(Sn_{k}, un_{k}) with (Sn_{k}, un_{k})∈UN(An_{k}) converging to
some (S, u)∈ N ×2^{N}, we have (S, u)∈U_{N}(A). Both points follow easily from the
definition ofU_{N} and the fact thatAn→A.

Now, fixS ∈ N and considerπS :Z→ K(2^{N}) defined byπS(A) :=AS. This is
an upper semicontinuous mapping; andπSU_{N} =SπS by the very definition ofU_{N}.
Moreover,{x}=πS(N × {x}) for everyx∈2^{N}. This shows thatS is a generalized

factor ofUN.

Proof of Theorem 5.2. One can now proceed exactly as in the proof of Theorem 4.7 (using Lemma 5.3 instead of Lemma 4.10), keeping in mind that factors are generalized factors and that generalized factors of generalized factors are again

generalized factors.

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E-mail address, Darji:ubdarj01@louisville.edu

E-mail address, Matheron: etienne.matheron@univ-artois.fr

(Darji)Department of Mathematics, University of Louisville, Louisville, KY 40292, USA.

(Matheron) Laboratoire de Math´ematiques de Lens, Universit´e d’Artois, Rue Jean Souvraz S. P. 18, 62307 Lens (France).