DECOMPOSING FUNCTIONS OF BAIRE CLASS 2 ON POLISH SPACES

arXiv:1805.11796v3 [math.LO] 27 Aug 2018

POLISH SPACES

LONGYUN DING, TAKAYUKI KIHARA, BRIAN SEMMES, AND JIAFEI ZHAO

Abstract. We prove the Decomposability Conjecture for functions of Baire class 2 on a Polish space to a separable metrizable space. This partially answers an important open problem in descriptive set theory.

1. Introduction

In descriptive set theory, the study of decomposability of Borel functions originated by a famous question asked by Luzin around a century ago: Is ev- ery Borel function decomposable into countably many continuous functions?

This question was answered negatively. Many counterexamples appeared in the literature (cf. [8, 10]) show that, even a function of Baire class 1 is not necessarily decomposable. Among these counterexamples, the Pawlikowski functionP : (ω+ 1)^ω→ω^ω stands in an important position. Indeed, Solecki [15] proved that:

Let X, Y be separable metrizable spaces with X analytic, and let f : X → Y be of Baire class 1. Then f is not decomposable into countably many continuous functions iff P ⊑f, i.e., there exists embeddings φ: (ω+ 1)^ω → X and ψ:ω^ω →Y such that ψ◦P =f◦φ.

Later, Pawlikowski and Sabok [13] generalized this theorem onto all Borel functions from an analytic space to a separable metrizable space. Motto Ros [11, Lemma 5.6] also gave an elegant proof for all functions of Baire classn withn < ω.

A natural generalization of Luzin’s question is to replace continuous func- tions with Σ⁰_γ-measurable functions. We writef ∈ dec(Σ⁰_γ) if there exists a partition (X_k) of X with each f ↾ X_k is Σ⁰_γ-measurable; and also write f ∈ dec(Σ⁰_γ,∆⁰_δ) if such a partition can be a sequence of ∆⁰_δ subsets of

2010Mathematics Subject Classification. Primary 03E15, 54H05, 26A21.

The Research of the first author was partially supported by the National Natural Science Foundation of China (Grant No. 11725103). The second author was partially supported by JSPS KAKENHI Grant 17H06738 and 15H03634, and also thanks JSPS Core-to-Core Program (A. Advanced Research Networks) for supporting the research.

1

X. It is trivial to see that, for δ ≥ γ, f ∈ dec(Σ⁰_γ,∆⁰_δ) implies the Σ⁰_δ- measurability of f. It is also well known that, for anyΣ⁰_δ-measurable func- tion f withδ > γ, we have f ∈dec(Σ⁰_γ) ⇐⇒ f ∈dec(Σ⁰_γ,∆⁰_δ+1) (cf. [11, Proposition 4.5]).

A slightly more finer notion of Baire hierarchy was essentially introduced by Jayne [3] for studying the Banach space of functions of Baire classα. A functionf :X →Y is called aΣα,β function (or more precisely denoted by f⁻¹Σ⁰_β ⊆Σ⁰_α) if the preimagef⁻¹(A) is Σ⁰_α inX for everyΣ⁰_β subsetA of Y. The following theorem discovers a deep connection between this notion and decomposability:

Theorem 1.1 (Jayne-Rogers [4]). Let X, Y be separable metrizable spaces with X analytic, and let f :X →Y. Then

f⁻¹Σ⁰₂⊆Σ⁰₂ ⇐⇒ f ∈dec(Σ⁰₁,∆⁰₂).

This theorem was generalized in [6] to the case that X is an absolute Souslin-F set and Y is an arbitrary regular topological space.

It is conjectured that the Jayne-Rogers Theorem can be extended to all finite Borel ranks as follows:

The Decomposability Conjecture(cf. [1, 11, 13]). LetX, Y be separable metrizable spaces with X analytic, and let f :X →Y. Then for n≥2 we have

f⁻¹Σ⁰_n⊆Σ⁰_n ⇐⇒ f ∈dec(Σ⁰₁,∆⁰_n).

Furthermore, for 2≤m≤nwe have

f⁻¹Σ⁰_m⊆Σ⁰_n ⇐⇒ f ∈dec(Σ⁰_n−m+1,∆⁰_n).

This conjecture was further generalized to The Full Decomposability Con- jecture (see [2, Section 4]) which covers all infinite Borel ranks. Motto Ros presented an equivalent condition of the decomposability conjecture (see [11, Conjecture 6.1]). Another interesting equivalent condition with some extra restrictions on spaces and on relation betweenm, n, concerning computabil- ity on Borel codes from A to f⁻¹(A), was given by Kihara in [9]. Most recently, Gregoriades-Kihara-Ng [2] proved

f⁻¹Σ⁰_m ⊆Σ⁰_n =⇒ f ∈dec(Σ⁰_n−m+1,∆⁰_n+1).

It is clear that the case m= n= 2 in the decomposability conjecture is just the Jayne-Rogers Theorem. Remarkable progress is due to Semmes, the third author of this article. In his Ph.D. thesis [14], Semmes proved the case m ≤n = 3 for functions f :ω^ω → ω^ω. In his proof, many kinds of games for characterizing Borel functions were widely applied. From the viewpoint of Jayne’s work [3] in functional analysis, the zero-dimensionality constraint on Semmes’ theorem was strongly desired to be removed. In this article, we generalize Semmes’ theorem to arbitrary Polish spaces:

Theorem 1.2. The decomposability conjecture is true for the case that X is Polish space and m≤n= 3.

It is worth noting that in our proof, no game for Borel functions are involved. This is the key point that this proof can be extended to all Ploish spaces. This theorem consists of two cases: (a) m = 2, n = 3, and (b) m=n= 3. We will prove them in sections 3 and 4 respectively.

Following the outline of Semmes’ proof, the proof appearing in this arti- cle was developed by the first and the forth authors. Almost at the same time, the second author independently gave a detailed exposition of Semmes’

strategy. He also asserted that the use of games for Borel functions is mis- leading, and emphasized the use of finite injury priority argument instead.

Soon after reading it, Motto Ros pointed out that the same argument in the second author’s proof also works well, with some minor modifications, for arbitrary Polish spaces.

The authors would like to thank Rapha¨el Carroy and Luca Motto Ros for helpful suggestions. The first author is grateful to Slawomir Solecki for his suggestions and encouragement. The second author also would like to thank Vassilios Gregoriades, Tadatoshi Miyamoto, and Yasuo Yoshinobu for valuable discussions.

2. Preliminaries

All topological spaces considered in this article are separable metrizable.

For any subset Aof a topological spaceX, we denote by Athe closure of A inX and denote A^c =X\Afor brevity.

We recall some basic notations. A topological space is called a Polish space if it is separable and completely metrizable, and is called ananalytic space if it is homeomorphic to an analytic subset of a Polish space. Given a separable metrizable space X, Borel sets ofX can be analyzed into Borel hierarchy, consisting ofΣ⁰_ξ,Π⁰_ξ subsets for 1≤ξ < ω1. As usual, we denote

∆⁰_ξ=Σ⁰_ξ∩Π⁰_ξ.

Let X, Y be two separable metrizable spaces, and f : X → Y. We say f is Σ⁰_α-measurable if f⁻¹(U) ∈ Σ⁰_α for every open set U ⊆ Y. For the definition of the Baire classes of functions, one can see [7, (24.1)]. It is well known that a function is of Baire class ξ iff it is Σ⁰_ξ+1-measurable (cf. [7, (24.3)]).

In the section of introduction, we already presented notion of Σ_α,β func- tions, f⁻¹Σ⁰_β ⊆Σ⁰_α and dec(Σ⁰_γ,∆⁰_δ). The following proposition give some well known properties which will be used again and again in the rest of this article.

Proposition 2.1 (folklore). Let X, Y be two separable metrizable spaces, and let f :X →Y. Then the following are equivalent:

(i) f ∈dec(Σ⁰_γ,∆⁰_δ).

(ii) There exists a sequence (An) of Σ⁰_δ subsets with X = S

nAn such that every f ↾An isΣ⁰_γ-measurable.

(iii) There exists a sequence (An) of Σ⁰_δ subsets with X = S

nAn such that every f ↾An∈dec(Σ⁰_γ,∆⁰_δ).

Proof. (i)⇒(ii) and (ii)⇒(iii) are trivial. We only prove (iii)⇒(i).

For every n < ω, since An ∈Σ⁰_δ, we can choose a sequence (B_n^m)m<ω of

∆⁰_δ sets such that An=S

mB_n^m. Moreover, since f ↾An ∈dec(Σ⁰_γ,∆⁰_δ),

there exist two sequences (C_n^k)k<ω,(D^k_n)k<ω of Σ⁰_δ sets with A_n⊆[

k

C_n^k, A_n∩C_n^k=A_n\D_n^k

such that each f ↾ (C_n^k ∩A_n) is Σ⁰_γ-measurable. Note that B_n^m ∩C_n^k = B^m_n \D_n^k∈∆⁰_δ, and f ↾(B_n^m∩C_n^k) is Σ⁰_γ-measurable for all n, k, m < ω.

Let (Kl)l<ω be an enumeration of allB_n^m∩C_n^k, n, k, m < ω. Then [

l

K_l= [

n,k,m

(B_n^m∩C_n^k) =X.

For each l < ω, putK_l^′ =K_l\(S

i<lK_i). Then the sequence (K_l^′)_l<ω of∆⁰_δ

subsets witnesses thatf ∈dec(Σ⁰_γ,∆⁰_δ).

3. The decomposability conjecture for m= 2, n= 3

We prove Theorem 1.2 form= 2, n= 3 in this section, and form=n= 3 in the next section. The following lemma is the key tool for proving the main theorem of this section, just like the role of Lemma 4.3.3 in [14].

Lemma 3.1. Let X, P be two separable metrizable spaces, and let D⊆X, h :D → P a function of Baire class 2. Let BP be a countable topological basis of P, and for each V ∈ BP, letGV be a countable class of subsets ofD such that

h⁻¹(V) =[ GV.

If h /∈ dec(Σ⁰₂,∆⁰₃), then there exist V ∈ BP, G ∈ GV, and a closed set F ⊆Dsatisfying:

(a) For any open set U with F∩U 6=∅,

h↾(h⁻¹(V^c)∩F ∩U)∈/ dec(Σ⁰₂,∆⁰₃);

(b) G∩F is dense in F; (c) F ∩D6=∅.

Proof. Let{Uk :k < ω} be a topological basis of X. For any V ∈ BP and any closed subsetF ⊆X, we denote

ΓV(F) ={k < ω :h↾(h⁻¹(V^c)∩F ∩Uk)∈dec(Σ⁰₂,∆⁰₃)}, ΘV(F) ={x∈F :∀k < ω(x∈Uk ⇒k /∈ΓV(F))}.

It is trivial to see that ΘV(F)⊆Dis closed.

For anyG∈ GV, we define closed setF_V,G^α forα < ω₁ as follows:

F_V,G⁰ =X, F_V,G^α+1=G∩Θ_V(F_V,G^α ), F_V,G^λ = \

α<λ

F_V,G^α , for limit ordinal λ.

Since X is second countable, there exists a ξ < ω₁ such that F_V,G^α =F_V,G^ξ for each V, Gand α≥ξ.

If there exist V ∈ BP, G ∈ GV such that F_V,G^ξ 6= ∅, then V, G, and F =F_V,G^ξ fulfil clauses (a) and (b). SetU =X in (a), we can see (c) is also fulfilled.

Assume for contradiction that, for any V ∈ BP, G∈ GV, we have F_V,G^ξ =

∅. Forα < ξ and k∈ΓV(F_V,G^α ), put

H_V,G,k^α =F_V,G^α ∩Uk. Note that

h↾(h⁻¹(V^c)∩H_V,G,k^α )∈dec(Σ⁰₂,∆⁰₃).

Now define a subset H of all x satisfying that, there exist V1, V2 ∈ BP

with V₁ ∩ V₂ = ∅, and for i = 1,2, there exist Gi ∈ GVi, αi < ξ, and k_i∈Γ_Vi(F_V^αi

i,Gi) such thatx∈H_V^α1

1,G1,k1 ∩H_V^α2

2,G2,k2. Since h↾(h⁻¹(Vi

c)∩H_V^α₁¹_,G1,k1∩H_V^α₂²_,G2,k2)∈dec(Σ⁰₂,∆⁰₃) (i= 1,2), and h⁻¹(Vi

c) is Σ⁰₃ inDfori= 1,2, by Proposition 2.1, we have h↾(D∩H_V^α₁¹_,G1,k1∩H_V^α₂²_,G2,k2)∈dec(Σ⁰₂,∆⁰₃).

Therefore, h↾(D∩H)∈dec(Σ⁰₂,∆⁰₃).

For anyx∈D,V ∈ BP, andG∈ GV withx∈G, we claim that there exist α < ξ and k∈ΓV(F_V,G^α ) such that x∈H_V,G,k^α . There is uniqueα < ξ such thatx∈(F_V,G^α \F_V,G^α+1). Note thatx /∈(F\G∩F) for any closed setF ⊆X, sox /∈(ΘV(F_V,G^α )\F_V,G^α+1). From the definition of ΘV(F_V,G^α ), we can find a k∈ΓV(F_V,G^α ) such that x∈Uk. Then we havex∈F_V,G^α ∩Uk=H_V,G,k^α .

In the end, we consider h ↾ (D\H). First, for any x ∈ (D\ H), if x ∈ H_V,G,k^α for some V ∈ BP, G ∈ GV, α < ξ, and k ∈ ΓV(F_V,G^α ), we claim that h(x) ∈ V. If not, we can find a V^′ ∈ BP such that h(x) ∈ V^′ and V ∩V^′ = ∅. Since h⁻¹(V^′) = S

GV^′, we can find an G^′ ∈ GV^′ such that x ∈ G^′. Hence x ∈ H_V^α′′,G^′,k^′ for some α^′ < ξ and k^′ ∈ ΓV^′(F_V^α′′,G^′), contradicting x /∈H. Secondly, letdbe a compatible metric on P. For any n < ω, let

(V_mⁿ, Gⁿ_m, kⁿ_m, αⁿ_m)m<ω

be an enumeration of all (V, G, k, α) with diam(V) ≤1/n, G∈ GV,α < ξ, and k∈ΓV(F_V,G^α ). Denote

H_mⁿ =H_V^αnmn m,Gⁿ_m,kⁿ_m.

For any x ∈ D, we can find a V ∈ BP with diam(V) ≤ 1/n such that h(x)∈V and a G∈ GV with x∈G. Hence x∈H_V,G,k^α for someα < ξ and k∈Γ_V(F_V,G^α ). It follows that D⊆S

mH_mⁿ. Put K_mⁿ =H_mⁿ \S

k<mH_kⁿ for each m. Then (K_mⁿ)m<ω is a sequence of pairwise disjoint ∆⁰₂ sets. Fix a y_mⁿ ∈V_mⁿ for eachm. Definegn(x) =yⁿ_m for allx∈K_mⁿ. Thengn is of Baire class 1. Furthermore, we have d(gn(x), h(x)) ≤ 1/n for all x ∈ (D\H).

So (gn ↾ (D\H))n<ω uniformly converges to h ↾ (D\H). It follows that h↾(D\H) is of Baire class 1 also (see [7, (24.4) i)]).

Note that H is an Fσ set from its definition. So D\H isGδ in D, and

henceh∈dec(Σ⁰₂,∆⁰₃). A contradiction!

In the rest of this section, we fix X be a Polish space, Y a separable metrizable space, andf :X →Y a Σ⁰₃-measurable function.

Definition 3.2. LetF =hF0,· · · , Fki be a finite sequence of closed sets of X withF₀ ⊇ · · · ⊇F_k,U an open subset ofX, and let P ⊆Y.

(i) Ifk= 0, i.e.,F =hF0i, then we sayF isP-sharpinU ifU∩F₀ 6=∅, and for any open set U^′ ⊆U withU^′∩F₀ 6=∅, we have

f ↾(f⁻¹(P)∩F₀∩U^′)∈/ dec(Σ⁰₂,∆⁰₃).

We also say F₀ itself isP-sharp inU for brevity.

(ii) If k >0, then we say F is P-sharp inU if F_k is P-sharp in U, and for any open set U^′⊆U withU^′∩Fk6=∅,F ↾k isP-sharp in some open set U^′′⊆U^′.

A similar notion named δ-σ-good was presented in [14]. The following propositions are trivial, we omit the proofs.

Proposition 3.3. Suppose F = hF0,· · ·, Fki is P-sharp in U. Then for any open set U^′ ⊆U with U^′∩Fk 6=∅, we have F is P-sharp in U^′.

Proposition 3.4. Suppose F is P-sharp in U. Then for any m <lh(F), F ↾m isP-sharp in some open set U^′ ⊆U.

The following lemma is modified from [14, Lemma 4.3.6].

Lemma 3.5. Suppose F =hF0,· · · , F_ki is P-sharp in U. Let (C_l)_l<m be a sequence of pairwise disjoint closed subsets of P. Then there exist at most k+ 1many l such that F is notP \Cl-sharp in any open set U^′⊆U. Proof. We begin withk= 0. Without loss of generality, suppose there exists an l < m, say, l = 0, such that F₀ is not P \C₀-sharp in U. Then there exists an open set U0 ⊆U with U0∩F06=∅such that

f ↾(f⁻¹(P\C0)∩F0∩U0)∈dec(Σ⁰₂,∆⁰₃).

Assume for contradiction that there existsl6= 0 such that F₀ is notP \Cl- sharp in U₀, then there is an open set U_l ⊆ U₀ with U_l ∩F₀ 6= ∅ such that

f ↾(f⁻¹(P \Cl)∩F₀∩Ul)∈dec(Σ⁰₂,∆⁰₃).

Since C₀ and Cl are disjoint closed subsets ofP, Proposition 2.1 gives f ↾(f⁻¹(P)∩F₀∩Ul)∈dec(Σ⁰₂,∆⁰₃),

contradicting that F₀ isP-sharp in U.

Fork >0, assume that we have proved for allk^′ < k. SinceF_k isP-sharp inU, from the arguments for k= 0 above, we may assume that there is an open set U₀ ⊆ U with U₀∩F_k 6= ∅ such that F_k is P \C_l-sharp in U₀ for any l 6= 0. Assume for contradiction that there are k+ 1 many l 6= 0, say, l= 1,· · · , k, k+ 1, such thatF is notP\C_l-sharp in any open setU^′ ⊆U₀. Particularly,Fis notP\C1-sharp inU0, so there exists an open setU1⊆U0

withU₁∩Fk6=∅such thatF ↾kis notP\C₁-sharp in any open setU^′ ⊆U₁. Similarly, we can find a sequence of open sets U_k+1 ⊆U_k ⊆ · · · ⊆U₁ ⊆U₀ such that Ul∩Fk 6= ∅ and F ↾ k is not P \Cl-sharp in any U^′ ⊆ Ul for 0< l≤k+ 1. By Definition ofP-sharp, there is a open setU^∗⊆U_k+1 such that F↾kis P-sharp in U^∗, contradicting the induction hypothesis.

Letp·,·qbe the bijection: ω×ω→ω as following:

p0,0q= 0, p0, j+ 1q=pj,0q+ 1, pi+ 1, j−1q=pi, jq+ 1.

Denote

Ω ={z∈2^ω:∃i∃^∞j(z(pi, jq) = 1)}.

It is well known that Ω isΣ⁰₃-complete subset of 2^ω. For anyz∈2^ω and l < ω, we call sequence

z↾(p0, lq+ 1), z↾(p1, l−1q+ 1), · · ·, z ↾(pl,0q+ 1)

thel-thdiagonalof z, and callz↾(pl,0q+ 1) the end ofl-th diagonal. For s∈2^<ω, we denote lh(s) =ithe length ofs. Ifs⊆zand lh(s) =pi, jq+ 1, then s is in (i+j)-th diagonal. Moreover, the l-th diagonal of z is also named the l-th diagonal of swhenpl,0q<lh(s).

For s 6= ∅, let lh(s) = pi, jq+ 1. We denote row(s) = i,col(s) = j. If i+j >0, we calls↾(pi+j−1,0q+ 1)the end of the last diagonal of s, denoted by u(s). If j >0, we call s↾(pi, j−1q+ 1) the left neighbor of s, denoted by v(s).

For proving the following theorem, we need an order on 2^<ω define by ts ⇐⇒ lh(t)<lh(s) or (lh(t) = lh(s), t≤_lexs),

where ≤lex is the usual lexicographical order. We also denote t ≺ s when tsbut nott=s. Denote

sM = max{t:t≺s^a0}.

Theorem 3.6. Let X be a Polish space, Y a separable metrizable space, and let f :X →Y. Then

f⁻¹Σ⁰₂⊆Σ⁰₃ ⇐⇒ f ∈dec(Σ⁰₂,∆⁰₃).

Proof. The “⇐” part is trivial, we only prove the “⇒” part.

Assume for contradiction thatf /∈dec(Σ⁰₂,∆⁰₃). We will define a contin- uous embeddingψ: 2^ω →X and an open set O⊆Y such that

ψ⁻¹(f⁻¹(O)) = Ω.

Thusf⁻¹(Y \O) isΠ⁰₃-complete subset ofX, contradicting f⁻¹Σ⁰₂ ⊆Σ⁰₃. For any open setV ⊆Y, sincef⁻¹(V) is Σ⁰₃, we can fix a system of open set W_n^m(V)⊆X(m, n < ω) with

f⁻¹(V) =[

m

\

n

W_n^m(V).

Denote G^m(V) = T

nW_n^m(V). For G =G^m(V), we also denote Wn(G) = W_n^m(V).

For s∈2^<ω with s6=∅, let lh(s) =pi, jq+ 1. We says is an inheritor if j > 0 and s(pk, i+j−kq) = 0 for any k < i, otherwise we say s is an innovator. Note that sis always an inheritor ifi= 0, j >0, and is always an innovator ifj = 0.

Fix a compatible metricdonXwithd≤1. We will inductively construct, for each s ∈ 2^<ω, an open set Vs of Y, a G_δ set Gs of X, a closed set Fs

of X, an open set Us of X, and a sequence of open sets (U_s^w)_sw≺sa0 of X satisfying the following:

(0) diam(U_s)≤2^−lh(s),U_sa0∩U_sa1 =∅,U_sa0 ⊆U_s^w,U_sa1 ⊆U_s^w; (1) V_sa0=V_sa1,G_sa0 =G_sa1,F_sa0 =F_sa1;

(2) fors, t∈2^<ω, we have Vs=Vt or Vs∩Vt=∅;

(3) there exist m < ω such that Gs=G^m(Vs);

(4) if col(s)>0, thenF_sa0 =F_sa1 ⊆Fs; (5) Fs∩U_s^w6=∅for each w;

(6) Us=U_s^s, and U_s^w1 ⊇U_s^w2 forw1w2; (7) Gs∩Fs is dense inFs;

(8) if sis an inheritor, then we have

Vs =V_v(s), Gs=G_v(s), Fs =F_v(s);

furthermore, (a) if s^a0 is also an inheritor, then U_s^w ∩F_u(s) 6= ∅ for each w; (b) if s^a0 is an innovator, then Us ⊆ Wn(Gs) for all n <lh(s);

(9) ifsis an innovator, thenU_s∩F_s∩G^m(V_t) =∅for allm <lh(s) and all lh(t)<lh(s);

(10) by lettingPs =Y \S

tsVt,

Fs=hFs↾(lh(s)−row(s)),· · · , Fs↾(lh(s)−1), F_si,

forts≺t^a0, we have

(a) if t^a0 is an innovator, thenFt isPs-sharp in U_t^s; (b) if t^a0 is an inheritor, thenF_u(ta0) isPs-sharp inU_t^s.

When we complete the construction, for anyz∈2^ω, we setψ(z) to be the unique element of T

kUz↾k. From (0) and (6), we see thatψ is a continuous embedding from 2^ω to X. Put

O= [

t∈2^<ω

Vt.

If z ∈Ω, let i₀ be the least i such that there are infinitely manyj with z(pi, jq) = 1. Then there isJ₀ < ω such thatz(pi, jq) = 0 for alli < i₀ and all j > J0. Hence for any j > J0, z↾(pi0, jq+ 1) is an inheritor. Denote

V =V_z↾(pi0,J0q+1), G=G_z↾(pi0,J0q+1).

By (8), we have V = V_z↾(pi0,jq+1) and G = G_z↾(pi0,jq+1) for all j ≥ J₀. If j > J₀ with z(pi₀, jq) = 1, by (8)(b), we have ψ(z) ∈ Wn(G) for all n ≤ pi₀, jq. Since there is infinitely many suchj, we have ψ(z) ∈G⊆f⁻¹(V).

Therefore, f(ψ(z))∈V ⊆O.

If z /∈Ω, we show that f(ψ(z)) ∈/ O. If not, there exits t₁ and m₁ such that ψ(z) ∈G^m1(V_t1). Fix an i₁ >max{m1,lh(t₁)}. Since z /∈Ω, for large enough j, we have z(pi, jq) = 0 for all i < i₁, so z ↾ (pi₁, jq+ 1) is an inheritor. Let J₁ be the largest j such thatz↾(pi₁, jq+ 1) is an innovator.

Denote

F =F_z↾(pi1,J1q+1).

By (8), we have F = F_z↾(pi1,jq+1) for all j ≥ J1. From (5) and (6), we see that F ∩U_z↾(pi1,jq+1)) 6= ∅ for any j > J1, and hence ψ(z) ∈ F. It follows from (9) that F ∩U_z↾(pi1,J1q+1)∩G^m1(Vt1) =∅. Thusψ(z) ∈/ G^m1(Vt1). A contradiction!

Now we turn to the construction.

First, set D, P, h,BP,GV as follows:

(i) P =Y,D=X, and h=f; (ii) BP is a countable basis of Y;

(iii) for eachV ∈ BP, let GV ={G^m(V) :m < ω}.

Applying Lemma 3.1 with theseD, P, h,BP,GV, we get an open setV ⊆Y, a Gδ setG∈ GV, and a non-empty closed set F ⊆X. Then put

V∅=V, G∅=G, F∅=F, U∅ =X.

Secondly, assume that we have constructed Vt, Gt, Ft, Ut, and U_t^w for t, w ≺ s^a0. We will define for s^a0 and s^a1. We consider the following two cases:

Case 1. Assume s^a0 is an inheritor. Let v = v(s^ai), u = u(s^ai). For i= 0,1, put

V_sai =Vv, G_sai =Gv, F_sai =Fv.

Note that s=u or s is also an inheritor with u(s) =u. By (5) and (8)(a), U_s^sM ∩Fu 6=∅.

Subcase 1.1. If col(s^a0^a0)>0, then s^a0^a0 is still an inheritor. We can find aU_sa0 such that

U_sa0⊆U_s^sM, U_sa0∩Fu 6=∅, diam(U_sa0)≤2^−(lh(s)+1).

By (10)(b), we see Fu is Ps_M-sharp in U_s^sM. By Proposition 3.4, Fv is PsM-sharp in some open set U ⊆ U_s^sM, and hence U ∩Fv 6= ∅. Denote W =T

n≤lh(s)Wn(Gv). SinceW ⊇Gv, by (7) we haveW∩Fv is open dense inFv, and henceW ∩U ∩Fv 6=∅. We can find U_sa1 such that

U_sa1 ⊆U∩W, U_sa1∩F_v 6=∅, diam(U_sa1)≤2^−(lh(s)+1).

By shrinking we may assume that U_sa0 ∩U_sa1 = ∅. For i = 0,1, we set U_s^s_a^a₀ⁱ =U_sa0,U_s^s_a^a₁¹ =U_sa1, and for other t, setU_t^sai=U_t^sM.

Subcase 1.2. If col(s^a0^a0) = 0, then s^a0^a0 is an innovator. We define U_sai fori= 0,1 similar to U_sa1 in Subcase 1.1 with one more requirement U_sa0∩U_sa1=∅.

It is trivial to check clauses (0)–(9). Note that P_sa0 =P_sa1 =P_sM and F_sa0 =F_sa1=Fv. Since (10) holds for s_M, it also holds fors^a0 and s^a1.

Case 2. Assumes^a0 is an innovator. We inductively defineV^l, G^l, F^l and U^l for each l < ω as the following:

Since ssM ≺s^a0, by (10)(a), we haveFs is Ps_M-sharp inU_s^sM. So f ↾(f⁻¹(P_sM)∩F_s∩U_s^sM)∈/dec(Σ⁰₂,∆⁰₃).

Denote F⁻¹ = Fs, U⁻¹ = U_s^sM. Assume that we have defined V^k, G^k, F^k and p^k fork < l. Set D, P, h,BP,GV as follows:

(i) P =Ps_M \S

k<lV^k, D=f⁻¹(P)∩F^l−1∩U^l−1, and h=f ↾D;

(ii) BP is a countable basis of P such thatV ⊆P for each V ∈ BP; (iii) for eachV ∈ BP, let GV ={D∩G^m(V) :m < ω}.

Applying Lemma 3.1 with theseD, P, h,BP,GV, we get an open setV ⊆Y, aG_δ set G=G^m(V) for somem < ω, and a closed setF ⊆D⊆F^l−1 ⊆Fs

with F ∩U^l−1 ⊇F ∩D6= ∅. Denote V^l = V, G^l = G, F^l =F. IfFs^aF^l is Ps_M \V^l-sharp inU^l−1, setU^l=U^l−1. Otherwise, since Lemma 3.1 implies that F^l isP_sM \V^l-sharp inU^l−1, we can find an open set U^l⊆U^l−1 with U^l∩F^l 6= ∅ such that Fs is not P_sM \V^l-sharp in any open set U ⊆ U^l. This complete the induction.

For s≺ts_M, if t^a0 is an innovator, it follows from (10)(a) that Ft is PsM-sharp in U_t^sM. By Lemma 3.5, we can find a natural number Lt such that, for any l ≥Lt, Ft is Ps_M \V^l-sharp in some open set U_t^l ⊆U_t^sM. If t^a0 is an inheritor, from (10)(b) and Lemma 3.5, we can also find a natural number Lt such that, for any l ≥ Lt, F_u(ta0) is Ps_M \V^l-sharp in some open set U_t^l ⊆ U_t^sM. Moreover, assume for contradiction that there exist l0 <· · ·< lm withm = lh(Fs) such that Fs is not PsM \V^lj-sharp in any

open set U ⊆U^lj for j ≤m. This contradicts Lemma 3.5, because U^lm ⊆ U_s^sM and U^lm∩F^lm 6= ∅ implies that Fs is Ps_M-sharp in U^lm. Therefore, comparing with the definition ofU^l, we can find an natural numberL^′ such that, for any l≥L^′,Fs^aF^l isP_sM \V^l-sharp inU^l. Then we set

L= max{L^′, Lt:s≺tsM}

and U_t^sa0 =U_t^sa1 =U_t^L fort≺s^a0≺t^a0, i.e., fors≺tsM. In the end, denote

A= [

lh(t)≤lh(s)

[

m≤lh(s)

G^m(Vt).

Then A is Gδ set. By (3) and V^L ⊆PsM, we have G^L∩A =∅. It follows from Lemma 3.1 that G^L∩F^L is dense in F^L, soA∩F^L is nowhere dense inF^L. We can find two open setsU_sa0 andU_sa1 such thatU_sa0∩U_sa1 =∅, and for i= 0,1, we have

U_sai⊆U^L, U_sai∩F^L6=∅, U_sai∩F^L∩A=∅, diam(U_sai)≤2^−(lh(s)+1). Now put

V_sai =V^L, G_sai=G^L, F_sai=F^L,

and U_s^s_a^a₀ⁱ =U_sa0, U_s^s_a^a₁¹ =U_sa1. Corollary 3.7. Let X be a Polish space, Y a separable metrizable space, and let f : X → Y. If f /∈ dec(Σ⁰₂,∆⁰₃), then there exists a Cantor set C⊆X such that f ↾C /∈dec(Σ⁰₂,∆⁰₃).

Proof. Let ψ be the continuous embedding defined in Theorem 3.6. Put

C=ψ(2^ω).

4. The decomposability conjecture for m=n= 3

Before proving Theorem 1.2 form=n= 3, we prove a known result first:

for functions of Baire class 1,

f⁻¹Σ⁰₃ ⊆Σ⁰₃⇒f ∈dec(Σ⁰₁,∆⁰₃).

This is an easy corollary of Solecki’s theorem (see [15, Theorem 4.1]), since f ∈dec(Σ⁰₁,∆⁰₃) ⇐⇒ f ∈dec(Σ⁰₁) and P⁻¹Σ⁰₃ 6⊆Σ⁰₃. Furthermore, this result is also a special case of [13, Corollary 1.2], [11, Corollary 5.11], or [2, Theorem 1.1]. In order to show a completely different method of proof, we present a direct proof which follows the same idea as in the previous section.

The readers can skip directly to Theorem 4.7.

Lemma 4.1. Let X, P be two separable metrizable spaces, and let D⊆X, andh:D→P a function of Baire class1. LetBP be a countable topological basis of P. If h /∈dec(Σ⁰₁,∆⁰₃), then there exist a V ∈ BP and two closed sets E ⊆F ⊆D satisfying:

(a) for any open set U withE∩U 6=∅, we have h↾(h⁻¹(V)∩U ∩E)∈/ dec(Σ⁰₁,∆⁰₃);

(b) for any open set U withF ∩U 6=∅, we have h↾(h⁻¹(V^c)∩F ∩U)∈/ dec(Σ⁰₁,∆⁰₃);

(c) E∩D6=∅.

Proof. Let {Uk :k < ω} be a topological basis of X. For any B ∈ BP, we denote

F^B ={x∈X :∀k(x∈U_k⇒h↾(h⁻¹(B^c)∩U_k)∈/ dec(Σ⁰₁,∆⁰₃))}.

It is trivial to see that (i) F^B is closed,

(ii) h↾(h⁻¹(B^c)\F^B)∈dec(Σ⁰₁,∆⁰₃), and (iii) for any open set U withF^B∩U 6=∅,

h↾(h⁻¹(B^c)∩F^B∩U)∈/ dec(Σ⁰₁,∆⁰₃).

Assume for contradiction that, h↾(h⁻¹(B)∩F^B)∈dec(Σ⁰₁,∆⁰₃) for any B ∈ BP. We denote

H₁= [

B∈B_P

(h⁻¹(B)∩F^B),

H₂= [

B∈B_P

(h⁻¹(B^c)\F^B).

It is straightforward to check that, h↾H_i∈dec(Σ⁰₁,∆⁰₃) for i= 1,2.

Denote H₃=D\(H₁∪H₂). For anyx∈H₃ and any B ∈ BP, we have h(x)∈B ⇒x∈h⁻¹(B)⇒x /∈F^B,

h(x) ∈/ B ⇒x∈h⁻¹(B^c)⇒x∈F^B. So h↾H₃ is continuous.

Let ˜Y ⊇Y be a Polish space. By Kuratowski’s theorem (cf. [7, (3.8)]), there is a G_δ set G⊇ H₃ and a continuous function g :G → Y˜ such that g ↾H₃ =h ↾H₃. PutH ={x ∈D∩G:h(x) = g(x)}. Since h is of Baire class 1, we seeH isG_δ subset ofD andH₁∪H₂∪H=D. Note thatH₁ is F_σ subset of Dand H₂ is Σ⁰₃ subset of D. It follows that h∈dec(Σ⁰₁,∆⁰₃).

A contradiction!

Therefore, there exists aB ∈ BP such that

h↾(h⁻¹(B)∩F^B)∈/dec(Σ⁰₁,∆⁰₃).

Since B = S

{V ∈ BP : V ⊆ B}, we can find a V ∈ BP with V ⊆B such that

h↾(h⁻¹(V)∩F^B)∈/dec(Σ⁰₁,∆⁰₃).

In the end, define

E ={x∈F^B :∀k(x∈U_k⇒h↾(h⁻¹(V)∩U_k)∈/ dec(Σ⁰₁,∆⁰₃))}.

Then we haveE∩D6=∅, and

h↾(h⁻¹(V)∩U ∩E)∈/ dec(Σ⁰₁,∆⁰₃)

for any open set with U ∩E 6= ∅. Note that V^c ⊇ B^c. So V, E and F^B

satisfy clauses (a)–(c) as desired.

In the rest of this section, we fix X be a Polish space, Y a separable metrizable space, andf :X→Y a Σ⁰₃-measurable function.

Definition 4.2. LetF =hF0,· · · , F_ki be a finite sequence of closed sets of X withF₀⊇ · · · ⊇Fk,U an open subset ofX, and letP =hP₀,· · · , Pki be a sequence of pairwise disjoint subsets ofY.

(i) Ifk= 0, i.e., F=hF0i,P =hP0i, then we sayF isP-sharp inU if U ∩F₀6=∅, and for any open setU^′ ⊆U withU^′∩F₀6=∅, we have

f ↾(f⁻¹(P0)∩F0∩U^′)∈/ dec(Σ⁰₁,∆⁰₃).

We also say F₀ itself isP₀-sharp inU for brevity.

(ii) Ifk >0, then we say F isP-sharpinU if F_k isP_k-sharp in U, and for any open set U^′ ⊆U withU^′∩Fk, F ↾k isP ↾k-sharp in some open set U^′′⊆U^′.

Proposition 4.3. Suppose F = hF0,· · ·, F_ki is P-sharp in U. Then for any U^′ ⊆U with U^′∩Fk6=∅, we have F is P-sharp in U^′.

Proposition 4.4. Suppose F is P-sharp in U. Then for any m <lh(F), F ↾m isP ↾m-sharp in some open set U^′ ⊆U.

LetP =hP0,· · · , Pki, 0≤j ≤l, and letC ⊆Pj. We denote P \C=hP0,· · ·, Pj\C,· · · , P_ki.

Lemma 4.5. Let F = hF0,· · ·, Fki,P = hP0,· · · , Pki. Suppose F is P- sharp in U. Let 0 ≤ j ≤ k and (C_l)_l<m be a sequence of pairwise disjoint closed subsets of Pj. Then there exist at most one l such that F is not P \C_l-sharp in any open setU^′ ⊆U.

Proof. We begin with k=j= 0. Without loss of generality, suppose there exists an l < m, say, l = 0, such that F0 is not P0 \C0-sharp in U. Then there exists an open set U₀ ⊆U with U₀∩F₀6=∅such that

f ↾(f⁻¹(P₀\C₀)∩F₀∩U₀)∈dec(Σ⁰₁,∆⁰₃).

Assume for contradiction that there existsl6= 0 such thatF₀ is notP₀\Cl- sharp in U₀, then there is an open set U_l ⊆ U₀ with U_l ∩F₀ 6= ∅ such that

f ↾(f⁻¹(P₀\C_l)∩F₀∩U_l)∈dec(Σ⁰₁,∆⁰₃).

Since C₀ and C_l are disjoint closed subsets ofP₀, Proposition 2.1 gives f ↾(f⁻¹(P₀)∩F₀∩Ul)∈dec(Σ⁰₁,∆⁰₃),

contradicting that F₀ isP₀-sharp inU.

Fork >0, assume that we have proved for all k^′ < k.

Case 1. Ifj=k, sinceF_k isP_k-sharp inU, from the arguments fork= 0 above, we may assume that there is an open set U0 ⊆U withU0∩Fk 6=∅

such that Fk is Pk \Cl-sharp in U₀ for any l 6= 0. It follows that F is P \C_l-sharpU₀ for anyl6= 0.

Case 2. Ifj < k, assume for contradiction that there are more than one l, say, l = 0,1, such that F is not P \Cl-sharp in any open set U^′ ⊆ U. Particularly, F is notP \C₀-sharp in U. Note that F_k is P_k\C_l-sharp in U for any l < m, so there exists an U₀ ⊆ U with U₀∩Fk 6= ∅ such that F ↾ k is not (P ↾ k)\C₀-sharp in any open set U^′ ⊆ U₀. Similarly, we can find an open set U₁ ⊆ U₀ with U₁∩Fk 6= ∅ such that F ↾ k is not (P ↾ k)\C₁-sharp in any U^′ ⊆ U₁. By Propositions 4.3 and 4.4, there is an open set U^∗ ⊆ U1 such that F ↾ k is P-sharp in U^∗, contradicting the

induction hypothesis.

Theorem 4.6. LetX be a Polish space,Y a separable metrizable space, and let f :X →Y be of Baire class 1. If f⁻¹Σ⁰₃ ⊆Σ⁰₃, then f ∈dec(Σ⁰₁,∆⁰₃).

Proof. Assume for contradiction that f /∈ dec(Σ⁰₁,∆⁰₃). We will define a continuous embedding ψ : 2^ω → X and an Gδ set G ⊆ Y such that ψ⁻¹(f⁻¹(Y \G)) = Ω. Thus f⁻¹(G) is Π⁰₃-complete subset of X, con- tradicting f⁻¹Σ⁰₃ ⊆Σ⁰₃.

It it well known that Y is homeomorphic to a subspace ofR^ω. Without loss of generality, we may assumeY =R^ω. Granting this assumption, we can fix a sequence of continuous functions fn :X → Y pointwisely converging to f. Fix a compatible metricdon X withd≤1.

Fors6=∅, let lh(s) =pi, jq+ 1. Now weredefineinheritors and innova- tors. We saysis aninheritorifj >0 ands(pk, i+j−kq) = 0 for anyk≤i (note: it was for any k < i in the definition of inheritor in Theorem 3.6), otherwise we say s is an innovator. Note that s is always an innovator if j= 0 or s(pi, jq) = 1.

We will inductively construct for each s∈2^<ω an open set Vs of Y, two closed sets E_s, F_s of X, an open set U_s of X, and a sequence of open sets (U_s^w)_sw≺sa0 of X satisfying the following:

(0) diam(Us)≤2^−lh(s),U_sa0∩U_sa1 =∅,U_sa0, U_sa1 ⊆U_s^w; (1) F_sa1⊆F_sa0;

(2) Vs⊆V∅ and Fs ⊆E∅ for anys6=∅;

(3) for anys, t6=∅with row(s) = row(t), we haveVs=VtorVs∩Vt=∅;

(4) if col(s)>0, thenV_sa0, V_sa1 ⊆Vs and F_sa0 ⊆Fs; (5) E_s ⊆F_s;

(6) Es∩U_s^w 6=∅ for each w;

(7) U_s=U_s^s, and U_s^w1 ⊇U_s^w2 forw₁w₂; (8) if sis an inheritor, then we have

Vs=V_v(s), Fs=F_v(s), Es=E_u(s);

(9) ifsis an innovator, thenVs∩Vt=∅for anytwitht≺sand row(t) = row(s); furthermore, there existsn≥lh(s) such thatfn(Us)⊆Vs;

(10) if s6=∅, by letting V_s⁻=

Vs↾(lh(s)−1), row(s)>0, V_∅, row(s) = 0, P_s^r=V_s⁻\[

{Vt:tr, row(t) = row(s)}, P_s^r =hPs↾(lh(s)−row(s))^r ,· · ·, P_s^r, Vsi, Fs=hFs↾(lh(s)−row(s)),· · · , Fs, Esi, then for any ts≺t^a0, we have

(a) if t^a0 is an innovator, thenFt isP_t^s-sharp inU_t^s;

(b) if t^a0 is an inheritor, thenF_u(ta0) isP_u(t^s _a0)-sharp inU_t^s. When we complete the construction, for anyz∈2^ω, we setψ(z) to be the unique element of T

kUz↾k. From (0) and (7), ψ is continuous embedding from 2^ω to X. Put

G_m= [

row(t)=m

V_t, G= \

m<ω

G_m.

Ifz∈Ω, there existi₀< ω and a strictly increasing sequence jk >0 with z(pi₀, j_kq) = 1 for any k < ω. Sincez↾(pi₀, j_kq+ 1) is an innovator, by (9), there isnk>pi₀, jkqsuch thatfn_k(ψ(z))∈V_z↾(pi0,j_kq+1). It follows from (9) that f(ψ(z))∈/Vt whenever row(t) =i₀. Thus

f(ψ(z))∈/ Gi0 ⊇G.

Ifz /∈Ω, we show that f(ψ(z))∈G. For any m < ω, there existsJm < ω such that z(pi, jq) = 0 for any i≤m and anyj > Jm. So z ↾(pm, jq+ 1) is an inheritor for anyj > Jm. Denote

Vm=V_{z↾(pm,Jmq+1)}, u^m_j =u(z↾(pm, jq+ 1)).

By (8), we haveVm=Vz↾(pm,jq+1)for allj > Jm. Since allu^m_j are innovators, by (9) we can find an n_j ≥lh(u^m_j ) such thatf_nj(ψ(z))∈V_u^m

j . By (4) and (8) we have f_nj(ψ(z)) ∈ V_m for all j > J_m. So f(ψ(z)) ∈ V_m for each m.

Again by (4) we have V_m+1 ⊆V_m for any m < ω. So f(ψ(z))∈V_m+1 ⊆Vm⊆Gm

for all m < ω. It follows thatf(ψ(z))∈G.

Now we turn to the construction.

First, set D, P, h,BP as follows:

(i) P =Y, D=X, h=f;

(ii) BP is a countable basis of Y.

Applying Lemma 4.1 with theseD, P, h,BP, we get an open setV ofY and two closed setsE ⊆F of X. Then put

V_∅ =V, F_∅ =F, E_∅ =E, U_∅ =X.

Secondly, assume that we have constructed V_t, E_t, F_t, U_t, and U_t^w for t, w ≺ s^a0. We will define for s^a0 and s^a1. We consider the following two cases: