PRESERVATION OF THE BOREL CLASS UNDER COUNTABLE-COMPACT-COVERING MAPPINGS

JEAN SAINT RAYMOND

Abstract. The aim of this note is to prove the following result : “Assume that X is a metric Borel space of class ξ, that f : X → Y is continuous, that every fiber f^{−1}(y) is complete and that every countable compact subset of Y is the image by f of some compact subset of X. Then Y is Borel and moreover of class ξ”. We give also an extension to the case where the fibers are only assumed to be Polish.

All spaces under consideration are metrizable and separable. Recall that the additive Borel classes Σ_{ξ}^{0} and multiplicative Borel classes Π_{ξ}^{0} in a space X are inductively defined by “ Σ_{1}^{0} is the set of open subsets of X ”, “ Π_{1}^{0} is the set of closed subsets of X ”, “ Σ_{ξ}^{0} is the set of countable unions of Borel subsets of class < ξ ”, “ Π_{ξ}^{0} is the set of countable intersections of Borel subsets of class < ξ ”. We denote also by D(Σ_{ξ}^{0}) the class of those sets which are the difference of two Σ_{ξ}^{0} sets.

A space X is said to be Π_{ξ}^{0} (resp. Σ_{ξ}^{0}, resp. D(Σ_{ξ}^{0})) if it is so in some (hence in every) metrizable compactification X̂. Indeed a space X is Π_{1}^{0} iff it is compact, Π_{2}^{0} iff it is Polish, Σ_{2}^{0} iff it is σ-compact.

A continuous mapping f : X → Y is said to be compact-covering if for every compact subset L of Y there is a compact subset K of X such that f(K) = L. Of course a compact-covering mapping is onto. It is known for a long time that if X is a Polish space and f : X → Y is compact-covering, then Y is Polish too (see [1] or [11]). This can be viewed as a result of preservation of the Borel class: “ The compact-covering image of a Π_{2}^{0} space is Π_{2}^{0} ”. And the following question was raised by A.V. Ostrovsky : “ Is the compact-covering image of a Π_{ξ}^{0} space also Π_{ξ}^{0} ? ” This question was recently solved in the positive way by G. Debs and the author in [5].

In fact the proof for Polish spaces in [11] uses only a weaker property of the mapping f : it is enough that countable compact subsets of Y can be lifted into (non-necessarily countable) compact subsets of X so that the argument works. And we shall say that a continuous mapping f : X → Y is countable-compact-covering if every countable compact subset of Y is the image under f of some compact subset of X. So the image of a Polish space under a countable-compact-covering mapping is Polish. Unfortunately this cannot be extended for non-Polish spaces. It is easily shown that continuous open onto mappings are countable-compact-covering and that every analytic set Y is the image of a simple Borel space X (namely a D(Σ_{2}^{0}) space) under a continuous open mapping. Even every Borel space Y is the image of such a D(Σ_{2}^{0}) space X under a continuous open countable-to-one mapping.

Date: January 12, 2007.

2000 Mathematics Subject Classification. Primary 03E15; Secondary 03E45, 54H05.

Key words and phrases. Borel classes, compact covering, countable compact covering.

Nevertheless one can hope that if we assume some more regularity on the fibers of f (the sets f^{−1}(y) for y in Y) such a pathology could not arise. It was conjectured by E. Michael that countable-compact-covering mappings with compact fibers were inductively perfect, hence compact-covering. In fact this conjecture happened to be false, as shown by G. Debs and the author, who constructed in [2] a counterexample with compact fibers having at most one cluster point, then in [4] another counterexample with finite fibers. In particular the notion of countablecompact-covering with complete fibers is strictly weaker than compact-covering with complete fibers. And the question of preservation of the Borel class under countable-compact-covering mappings with complete fibers still makes sense.

The first result in this direction was given in [9] by A.V. Ostrovsky who proved that the image of a Π_{ξ}^{0} space X under a countable-compact-covering mapping with complete fibers is Borel and even Π_{3+ξ}^{0}. In fact, using his main argument, it is not very hard to prove the existence of a Baire-one section of f (it is a mapping g : Y → X such that f ∘g is the identity map of Y), and to deduce that Y is in fact Π_{1+ξ}^{0}. Very recently the same author ([10]) gave a proof that the image of a Π_{ξ}^{0} space X under a countable-compact-covering with finite fibers is Π_{ξ}^{0}.

In this note we prove that for ξ ≥ 2 the image of a Π_{ξ}^{0} (resp. Σ_{ξ}^{0}) space X under a countable-compact-covering mapping with complete fibers is Π_{ξ}^{0} (resp. Σ_{ξ}^{0}). Indeed we may assume ξ ≥ 3 since for the class Π_{2}^{0} the result is already known, and the continuous image of a σ-compact space is itself σ-compact.

There is a stronger property than being compact-covering for which it is known for long that the Borel class is preserved: it was shown by the author in [12] that the image of a Π_{ξ}^{0} space X under a perfect mapping f is Π_{ξ}^{0} too. The main idea of this paper, which was used again for the proof of Theorem 8.6 in [2], was to construct special Baire-one sections of f, namely for every Baire-one function h on X to construct a Baire-one section g of f such that h∘g is Baire-one (and not only Baire-two as the composition of two Baire-one mappings). And again we will apply here this idea. A variation of the notion of perfect mapping is this of inductively perfect mapping : a mapping f : X → Y is said to be inductively perfect if there is a (necessarily closed) subset X′ of X such that the restriction f_{|X′} of f to X′ is perfect and onto. It is quite clear that inductively perfect mappings are compactcovering, and a fortiori countable-compact-covering, and it follows readily from what precedes that the inductively perfect image of a Π_{ξ}^{0} space is itself Π_{ξ}^{0}.

It was noticed above that such a result of class preservation cannot be expected for countable-compact-covering mappings with σ-compact (or even countable) fibers. Nevertheless it is quite natural to ask whether the result on countablecompact-covering mappings with complete fibers can be extended to the case of Polish fibers. And we will show that this is the case. Of course this last statement is stronger than the former. But we also will show that the result about Polish fibers cannot be obtained by methods of Baire-one sections. And since the existence of special Baire-one sections could be of some interest by iteself, we will give both proofs.

Recall that for every space X the hyperspace 𝒦 (X) is the set of non-empty compact subsets of X equipped with the Vietoris topology (induced by the Hausdorff metric). We consider also the subspace 𝒦_{ω}(X) of 𝒦 (X) constituted of countable compact subsets of X.

In [4] G. Debs and the author introduced the notion of quasi-upper semi-cont-inuous mappings : if X and Y are spaces, we call a mapping Φ from Y to the set 𝒫(𝒦 (X)) of non-empty subsets of 𝒦 (X) quasi-upper semi-continuous (abbreviated in quasi-u.s.c.) if for every open subset U of X the set

{y : ∃K ∈ Φ(y) K ⊂ U}

is open in Y. This notion appears implicitly in [6], and the following proposition is easily deduced from Theorem 3.1 of this last paper.

Proposition 1. Let f : X → Y be a countable-compact-covering mapping. Then there exists a quasi-u.s.c. mapping Φ : Y → 𝒫(𝒦 (X)) such that

(i) for all y ∈ Y and all K ∈ Φ(y), K ⊂ f^{−1}(y).

(ii) for all y ∈ Y, all K ∈ Φ(y) and all countable compact subset S of Y containing y, there exists a (non-necessarily countable) compact subset S′ of X such that f(S′) = S and S′ ∩ f^{−1}(y) = K.

It follows easily from this Proposition that there is a maximal such quasi-u.s.c. mapping Φ_{m} defined by

“Φ_{m}(y) is the set of all compact subsets of f^{−1}(y) such that

∀L ∈ 𝒦_{ω}(Y) ∃K′ ∈ 𝒦 (X) : f(K′) = L and K′ ∩ f^{−1}(y) ⊂ K ”

For any open set U of X we denote by W (U) and H(U) the sets

W (U) := {K ∈ 𝒦 (X) : K ⊂ U}
H(U) := {K ∈ 𝒦 (X) : K ∩ U ≠ ∅}

which are both open in 𝒦 (X). It is clear that Φ_{m}(y) is closed in 𝒦 (X) and monotone ( K ∈ Φ_{m}(y) and K ⊂ K′ ∈ 𝒦 (f^{−1}(y)) imply K′ ∈ Φ_{m}(y)). And replacing if necessary Φ by Φ_{m} we will assume from now on that Φ(y) is monotone and closed. Clearly every element of Φ(y) contains a minimal one and we denote by Φ_{μ}(y) the set of minimal elements of Φ(y) that is

Φ_{μ}(y) = {K ∈ Φ(y) : ∀L ∈ Φ(y) L ⊂ K ⟹ L = K}

Lemma 2. If K ∈ Φ_{μ}(y) every neighborhood 𝒱 of K in Φ(y) contains the set W (V^{∗}) ∩ Φ(y) for some open set V^{∗} of X containing K.

Proof. By definition of the Vietoris topology, there are open sets V, V_{1}, V_{2},…, V_{m} such that

K ∈ W (V) ∩ ⋂_{j=1}^{m} H(V_{j}) ⊂ 𝒱

Since K is minimal and meets V_{j}, K \ V_{j} ∉ Φ(y), and since Φ(y) is closed there is an open set V′_{j} containing K \ V_{j} and open sets U′_{1}, U′_{2}, …, U′_{k} meeting K \ V_{j} such that

W (V′_{j}) ∩ ⋂_{ℓ≤k} H(U′_{ℓ}) ∩ Φ(y) = ∅

If there were some L ∈ Φ(y) ∩ W (V′_{j}) then (K \ V_{j}) ∪ L would belong to Φ(y) hence

to (W (V′_{j}) ∩ ⋂_{ℓ≤k} H(U′_{ℓ})) ∩ Φ(y), and this is impossible. Thus W (V′_{j}) ∩ Φ(y) = ∅, and we conclude that every L ∈ Φ(y) such that L ⊂ V_{j} ∪ V′_{j} belongs to H(V_{j}).

Thus the open subset V^{∗} = V ∩ ⋂_{j=1}^{k}(V_{ℓ} ∪ V′_{ℓ}) of X satisfies K ⊂ V^{∗} and W (V^{∗}) ∩ Φ(y) ⊂ 𝒱 . □

Lemma 3. The set Φ_{μ}(y) is a Π_{2}^{0} subset of 𝒦 (f^{−1}(y)).

Proof. Since Φ(y) is closed in 𝒦 (X) hence in 𝒦 (f^{−1}(y)), we have only to prove that Φ_{μ}(y) is Π_{2}^{0} in Φ(y). Let (U_{n})_{n∈ω} be a basis of the topology of X. Then for K ∈ Φ(y) we have K ∈ Φ_{μ}(y) if and only if

∀n ∈ ω K ∩ U_{n} = ∅ or ∃V open : K ⊂ U_{n} ∪ V and W (V) ∩ Φ(y) = ∅.

Indeed if K is not minimal, there is a L ∈ Φ_{μ}(y) such that L ⫋ K. And since L ≠ K there is a U_{n} disjoint from L such that U_{n} ∩ K ≠ ∅. Then any open set V containing K \ U_{n} contains L, and this implies that W (V) meets Φ(y).

Conversely, if K is minimal and U_{n} ∩ K ≠ ∅, L := K \ U_{n} does not belong to Φ(y), hence possesses by the proof of Lemma 2 a neighborhood V such that W (V) is disjoint from Φ(y).

The set {K : K ∩ U_{n} = ∅} is the complement of H(U_{n}) hence closed, and the union of the open sets W (U_{n} ∪ V) for V such that W (V) is disjoint from Φ(y) is open. It follows that

Φ_{μ}(y) = ⋂_{n∈ω}(H(U_{n})^{c} ∪ ⋃_{W (V)∩Φ(y)=∅} W (U_{n} ∪ V))

that is the countable intersection of sets each of which is the union of a closed and an open subset of Φ(y), hence a Π_{2}^{0} subset of Φ(y). □

In all the sequel f will denote a countable-compact-covering mapping from the Borel space X to the space Y and Φ will be a quasi-u.s.c. mapping from Y to 𝒫(𝒦 (X)) as above.

Since Φ is quasi-u.s.c. the set

τ(U) := {y ∈ Y : ∃K ∈ Φ(y) K ⊂ U}

is open in Y for any open subset U of X. Moreover the function τ is clearly nondecreasing : indeed, if U ⊂ U′ and K ∈ Φ(y) is contained in U then K ⊂ U′, hence y ∈ τ(U′).

Definition 4. For any closed subset F of X we shall call the essential image of F under f and denote by f^{∗}(F) the set

f^{∗}(F) := {y ∈ Y : ∃U open in X such that y ∈ τ(U) and y ∉ τ(U \ F)}

The image under f of a closed subset F of X is analytic and in general non-Borel. So the essential image of F is a good substitute for f(F), as stated in the following lemma.

Lemma 5. For all closed subset F of X its essential image f^{∗}(F) is a Σ_{2}^{0} subset of Y. Moreover f^{∗}(X) = Y, f^{∗}(F) ⊂ f(F), and f^{∗}(F) ⊂ f^{∗}(F′) whenever F ⊂ F′.

Proof. Let (U_{n})_{n∈ω} be a basis of the topology of X. If y ∈ f^{∗}(F) there is an open set U of X such that y ∈ τ(U) \ τ(U \ F), hence a compact set K ∈ Φ(y) such that K ⊂ U. Then there is a finite cover (U_{n})_{n∈J} of K by basic open sets contained in U, and we get K ⊂ U^{∗} := ⋃_{n∈J} U_{n} ⊂ U, hence y ∈ τ(U^{∗}) but y ∉ τ(U^{∗} \ F) since τ(U^{∗} \ F) ⊂ τ(U \ F).

It follows that

f^{∗}(F) = ⋃{τ(U^{∗}) \ τ(U^{∗} \ F) : U^{∗} finite union of basic open sets}

which is a countable union of differences of open sets, hence a Σ_{2}^{0} set. Moreover if y ∉ f(F) we have y ∈ τ(U \ F) for every open set U such that y ∈ τ(U), hence y ∉ f^{∗}(F).

For every y ∈ Y we have τ(X) = Y and τ(X \ X) = τ(∅) = ∅, hence y ∈ f^{∗}(X). Finally, if F ⊂ F′ and y ∈ f^{∗}(F) there is an open set U such that y ∈ τ(U) and y ∉ τ(U \ F), hence y ∉ τ(U \ F′) ; and it follows that y ∈ f^{∗}(F′). □

Theorem 6. Let F be a closed subset of X and y belong to Y. Then y ∉ f^{∗}(F) if and only if W (F^{c}) ∩ Φ_{μ}(y) is dense in Φ_{μ}(y).

Proof. Assume first that y ∉ f^{∗}(F) and let K ∈ Φ_{μ}(y). If 𝒱 is any neighborhhod of K in Φ_{μ}(y), it follows from Lemma 2 that there is some open U such that K ∈ W (U) ⊂ 𝒱 . Then since K ⊂ U we have y ∈ τ(U) and since y ∉ f^{∗}(F) we have y ∈ τ(U \ F). Thus there is some K′ ∈ Φ_{μ}(y) such that K′ ⊂ U \ F. So K′ ∈ W (U) ∩ W (F^{c}) ∩ Φ_{μ}(y) ⊂ W (F^{c}) ∩ 𝒱 ∩ Φ_{μ}(y). And this shows that W (F^{c}) meets all nonempty open subsets of Φ_{μ}(y).

Conversely, if W (F^{c}) ∩ Φ_{μ}(y) is dense and y ∈ τ(U) for some open set U then W (U) ∩ Φ_{μ}(y) is a non-empty open subset of Φ_{μ}(y). Thus W (F^{c}) ∩ W (U) ∩ Φ_{μ}(y) is non-empty, and this shows that y ∈ τ(U \ F) whenever y ∈ τ(U), hence that y ∉ f^{∗}(F). □

Lemma 7. Assume the fibers of f are all Polish. Let F be a closed subset of X and (F_{k})_{k∈ω} a covering of F by closed subsets of F. Then

f^{∗}(F) = ⋃_{k∈ω} f^{∗}(F_{k})

Proof. We have F_{k} ⊂ F , hence f^{∗}(F_{k}) ⊂ f^{∗}(F) for all k. Thus

⋃_{k∈ω} f^{∗}(F_{k}) ⊂ f^{∗}(F)

Conversely, if y ∉ ⋃_{k} f^{∗}(F_{k}) then the set {K ∈ Φ_{μ}(y) : K ∩ F_{k} ≠ ∅} is by Theorem 6 closed and nowhere dense in Φ_{μ}(y) for every k. Moreover since f^{−1}(y) is Polish the space 𝒦 (f^{−1}(y)) is Polish itself and it follows then from Lemma 3 that Φ_{μ}(y) is Polish hence a Baire space. Thus M(y) := ⋃_{k}{K ∈ Φ_{μ}(y) : K ∩ F_{k} ≠ ∅} is meager in Φ_{μ}(y), and W (F^{c}) ∩ Φ_{μ}(y) = Φ_{μ}(y) \ M(y) is dense in Φ_{μ}(y), whence we conclude by Theorem 6 that y ∉ f^{∗}(F). □

As usual we denote by ω^{<ω} the (countable) set of finite sequences of integers. For s = (n_{0}, n_{1}, …, n_{k−1}) we call k its length and write |s| = k. For s = (n_{0}, n_{1}, …, n_{k−1}) and t = (m_{0}, m_{1}, …, m_{p−1}) we write s ≺ t if |s| < |t| and moreover s(j) = t(j) for i < k. Finally for s = (n_{0}, n_{1}, …, n_{k−1}) and n ∈ ω we denote by s^{⌢}n the sequence (n_{0}, n_{1}, …, n_{k−1}, n) whose length is k + 1.

Lemma 8. Assume all fibers of f are complete. Let (H_{j})_{j∈ω} be a sequence of closed subsets of X. Then there exists a Baire-one section g : Y → X of f such that g^{−1}(H_{j}) ∈ Σ_{2}^{0} for every j.

Proof. We have Y = f^{∗}(X). We construct by induction for all s ∈ ω^{<ω} a closed subset A_{s} of X and a Σ_{2}^{0} subset B_{s} of Y in such a way that (i) A_{∅} = X and B_{∅} = Y (ii) (B_{s^{⌢}n})_{n∈ω} is a partition of B_{s} into Σ_{2}^{0} sets. (iii) A_{s^{⌢}n} ⊂ A_{s} for all s ∈ ω^{<ω} and all n ∈ ω. (iv) diam(A_{s}) ≤ 1/ |s|

(v) B_{s} ⊂ f^{∗}(A_{s})

(vi) for all s such that |s| = k ≥ 1, A_{s} ⊂ H_{k−1} or A_{s} ∩ H_{k−1} = ∅.

Assume A_{s} and B_{s} are defined for |s| = k. One can find a countable covering

of A_{s} ∩ H_{k} into closed subsets with diameter less than \frac{1}{k + 1}, and a countable of

A_{s} \H_{k} into closed subsets of A_{s}, disjoint from H_{k} and of diameter less than \frac{1}{k + 1}.

Index then the union of these two coverings by ω and denote it by (A_{s^{⌢}n})_{n∈ω}. It is clear that (A_{s^{⌢}n})_{n∈ω} is a covering of A_{s} into closed sets.

Applying Lemma 7 we get B_{s} ⊂ f^{∗}(A_{s}) = ⋃_{n} f^{∗}(A_{s^{⌢}n}), and by reduction of the family (B_{s} ∩ f^{∗}(A_{s^{⌢}n}))_{n} of Σ_{2}^{0} sets we obtain a partition (B_{s^{⌢}n})_{n} of B_{s} into Σ_{2}^{0} sets satisfying B_{s^{⌢}n} ⊂ f^{∗}(A_{s^{⌢}n}) for all n. And this completes the inductive construction of the sets (A_{s})_{s∈ω^{<ω}} and (B_{s})_{s∈ω^{<ω}}.

For all y ∈ Y and all integer k there exists a unique s(y, k) ∈ ω^{<ω} of length k such that y ∈ B_{s(y,k)}. Clearly we have s(y, k) ≺ s(y, k + 1). Then the set

E_{k}(y) := A_{s(k,y)} ∩ f^{−1}(y) is a closed nonempty set of diameter less than \frac{1}{k}. And

since E_{k+1}(y) ⊂ E_{k}(y) the completeness of the fiber f^{−1}(y) implies that ⋂_{k} E_{k}(y) is a singleton {g(y)} contained in f^{−1}(y). Hence g : Y → X is a section of f. Moreover for every open set W of X we have

g(y) ∈ W ⟺ ∃k ∈ ω E_{k}(y) ⊂ W ⟺ ∃s ∈ ω^{<ω} (A_{s} ⊂ W and y ∈ B_{s})

thus g^{−1}(W) = ⋃{B_{s} : A_{s} ⊂ W } ∈ Σ_{2}^{0}. And this shows that g is of Baire class 1. Similarly we have

g(y) ∈ H_{k} ⟺ E_{k+1}(y) ⊂ H_{k} ⟺ ∃s (|s| = k + 1 and y ∈ B_{s} and A_{s} ⊂ H_{k})

whence we get g^{−1}(H_{k}) = ⋃{B_{s} : |s| = k + 1 and A_{s} ⊂ H_{k}} ∈ Σ_{2}^{0}. □

Corollary 9. Let Ẑ be a compact space and h : X → Ẑ be a Baire-one mapping. Then there exists a Baire-one section g : Y → X of f such that h∘g is Baire-one too.

Proof. Let (W_{j})_{j∈ω} be a basis for the topology of Ẑ. Then there is a countable family (H_{k})_{k∈ω} of closed subsets of X̂ such that each h^{−1}(W_{j}) is the union of some subfamily of (H_{k}). By Lemma 8 there exists a Baire-one section g of f such that g^{−1}(X ∩ H_{k}) is Σ_{2}^{0} for all k, and this implies that (h∘g)^{−1}(W_{j}) is Σ_{2}^{0} in Y for all j, hence that h∘g : Y → Ẑ is Baire-one. □

Theorem 10. Assume that ξ ≥ 3, that X is Π_{ξ}^{0} and that f : X → Y is countablecompact-covering with complete fibers. Then Y is Π_{ξ}^{0}.

Proof. One can find compactifications X̂ of X and Ŷ of Y such that f extends to a continuous mapping f̂ : X̂ → Ŷ. There exists η ≥ 2 such that 1 + η = ξ, a Baire-one mapping h from X̂ to a compact space Ẑ and a Π_{η}^{0} subset Z of Ẑ such that h^{−1}(Z) = X.

By Corollary 9 there is a Baire-one section g : Y → X of f such that h∘g is Baireone. Then there is a Π_{3}^{0} subset Y^{∗} of Ŷ containing Y such that g and h∘g extend into Baire-one mappings g_{1} : Y^{∗} → X̂ and h_{1} : Y^{∗} → Ẑ. Since h∘g_{1} is Baire-two the set Y_{1} := {y^{∗} ∈ Y^{∗} : h_{1}(y^{∗}) = h∘g_{1}(y^{∗})} is a Π_{3}^{0} subset of Y^{∗} containing Y, and replacing if necessary Y^{∗} by Y_{1} we can and do assume that h_{1} and h∘g_{1} agree on Y^{∗}.

Then

Y = {y^{∗} ∈ Y^{∗} : g_{1}(y^{∗}) ∈ X and f̂∘g_{1}(y^{∗}) = y^{∗}}

= {h^{∗} ∈ Y^{∗} : h_{1}(y^{∗}) ∈ Z} ∩ {y^{∗} ∈ Y^{∗} : f̂∘g_{1}(y^{∗}) = y^{∗}}

is the intersection of a Π_{1+η}^{0} subset and a Π_{2}^{0} subset of Y^{∗} since h_{1} and f̂∘g_{1} are Baire-one, hence a Π_{ξ}^{0} subset of Ŷ. □

We now study the case of Σ_{ξ}^{0} spaces.

Theorem 11. Assume that ξ ≥ 4, that X is Σ_{ξ}^{0} and that f : X → Y is countablecompact-covering with complete fibers. Then Y is Σ_{ξ}^{0}.

Proof. The same proof as in Theorem 10 shows that if X is Σ_{ξ}^{0} for ξ ≥ 4 and f countable-compact-covering with complete fibers then Y is Σ_{ξ}^{0} too : replacing the hypothesis “ X is Π_{ξ}^{0} and Z is Π_{η}^{0} ” by “ X is Σ_{ξ}^{0} and Z is Σ_{η}^{0} ”, Y appears as a Σ_{ξ}^{0} subset of the Π_{3}^{0} set Y^{∗}, hence Σ_{ξ}^{0} in Ŷ. □

Nevertheless if X is Σ_{3}^{0} the above proof shows only that Y is Σ_{3}^{0} in Y^{∗}, hence D(Σ_{3}^{0}) in Ŷ. So we need the following theorem.

Theorem 12. Assume X is a metric space and f : X → Y is countable-compact-covering. Then there exist a space X^{∗} containing X, a Polish space Y^{∗} containing Y and a countable-compact-covering mapping f^{∗} : X^{∗} → Y^{∗} extending f such that X = f^{∗}^{−1}(Y) and that the fibers f^{∗}^{−1}(y) are complete for y ∈ Y^{∗} \ Y.

Proof. Replacing if necessary the distance on X by an equivalent distance we can and do assume that diam(X) ≤ 1 and that f is 1-Lipschitz, hence extends to a continuous mapping f̂ from the completion X̂ of X to the completion Ŷ of Y. Since f is countable-compact-covering there exists a quasi-u.s.c. mapping Φ from Y to 𝒫(𝒦 (X)) as in Proposition 1, and we define as above the mapping τ from the topology of X to the topology of Y by

τ(U) = {y ∈ Y : ∃K ∈ Φ(y) K ⊂ U}

Let (U_{n})_{n∈ω} be a basis of the topology of X̂ such that U_{0} = X̂, ℱ be the set of finite subsets of ω and

ℱ_{k} = {J ∈ ℱ : ∀n ∈ J diam(U_{n}) ≤ 2^{−k}}

For every J ∈ ℱ we denote by U(J) the open subset U(J) = ⋃_{n∈J} U_{n} of X̂ and choose an open subset V (J) of Ŷ such that τ(X ∩ U(J)) = V (J) ∩ Y. Since Y is dense in Ŷ, we have V (J) ⊂ \overline{V (J) ∩ Y} = \overline{τ(X ∩ U(J))} ⊂ \overline{f̂(U(J))}. Then let

Y^{∗} := {y ∈ Ŷ : ∀k ∈ ω ∀J ∈ ℱ such that y ∈ V (J)

∃J′ ∈ ℱ_{k} \overline{U(J′)} ⊂ U(J) and y ∈ V (J′) and diam(V (J′)) ≤ 2^{1−k}\} Lemma 13. Y^{∗} is a Π_{2}^{0} subset of Ŷ containing Y.

Proof. Since

Y^{∗} = ⋂_{k∈ω,J∈ℱ} ((Ŷ \ V (J))

∪ ⋃\{V (J′) : J′ ∈ ℱ_{k}, \overline{U(J′)} ⊂ U(J), diam(V (J′)) ≤ 2^{1−k}\})

the right hand side is a countable intersection of sets which are each the union of an open and a closed set in Ŷ. Hence Y^{∗} is Π_{2}^{0}. Moreover, if y ∈ Y ∩ V (J) and k ∈ ω, we have y ∈ τ(X ∩ U(J)), hence there is a compact K ∈ Φ(y) such that K ⊂ U(J). Then the open sets U_{n} such that \overline{U_{n}} ⊂ U(J) and diam(U_{n}) < 2^{−k} cover K. It follows that there is some J′ ∈ ℱ_{k} such that K ⊂ U(J′) ⊂ \overline{U(J′)} ⊂ U(J).

Replacing if necessary J′ by \{n ∈ J′ : U_{n} ∩ K ≠ ∅\} we get for all n ∈ J′ : \overline{f̂(U_{n})} ⊂

B(y, 2^{−k}) hence V (J′) ⊂ \overline{f̂(U(J′))} ⊂ B(y, 2^{−k}) and y ∈ τ(X ∩U(J′)) ⊂ V (J′), and finally y ∈ Y^{∗}. □

Now let

X^{∗} = X ∪ {x ∈ X̂ : f̂(x) ∈ Y^{∗} \ Y }

and f^{∗} be the restriction of f̂ to X^{∗}. Then it is clear that X = f^{∗}^{−1}(Y), that f^{∗}^{−1}(y) is f^{−1}(y) if y ∈ Y, and f̂^{−1}(y) if y ∈ Y^{∗} \ Y, which is closed in the complete space X̂ hence complete.

Lemma 14. The mapping f^{∗} is countable-compact-covering.

Proof. Let L be a countable compact subset of Y^{∗}. We have to produce a compact subset K of X^{∗} such that f^{∗}(K) = L. Let φ be a bijection from L × ω onto ω increasing with respect to the second variable. We define inductively for each y ∈ L ∩ Y a compact set T_{y} ∈ Φ(y) and for each integer m open sets W_{m} in X̂ and V_{m} in Y^{∗} and J_{m} ∈ ℱ satisfying

(i) W_{0} = X̂, V_{0} = Y^{∗} and J_{0} = {0}.

(ii) for m < m′, \overline{V_{m′}} ⊂ V_{m} or V_{m} ∩ V_{m′} = ∅.

(iii) if m = φ(y, ℓ) then y ∈ V_{m}.

(iv) if m = φ(y, ℓ) then J_{m} ∈ ℱ_{ℓ}, W_{m} = U(J_{m}) and V_{m} ⊂ V (J_{m}).

(v) if m < m′ and V_{m′} ⊂ V_{m} then \overline{W_{m′}} ⊂ W_{m}.

(vi) if m = φ(y, ℓ) < m′ = φ(y′, ℓ′) and y ≠ y′ then y ∉ V_{m′}.

(vii) if y ∈ L ∩ Y then T_{y} ⊂ W_{φ(y,ℓ)} ⊂ {x ∈ X̂ : d(x, T_{y}) ≤ 2^{−ℓ}}. Indeed assume W_{m}, V_{m} and J_{m} are defined for m < p = φ(y, ℓ).

There is a largest m < p such that y belongs to V_{m}. Then W_{m} = U(J_{m}) and y ∈ V (J_{m}). If y ∈ L ∩ Y and ℓ = 0 we have y ∈ τ(X ∩ U_{m}). Thus there is some T_{y} ∈ Φ(y) such that T_{y} ⊂ X ∩ U_{m}.

If y ∈ L ∩ Y the open basic sets U_{n} such that \overline{U_{n}} ⊂ W_{m} and diam(U_{n}) < 2^{−ℓ} cover T_{y}. Thus there exists a minimal finite cover (U_{n})_{n∈J_{p}} of T_{y} by these sets, and by minimality each of these sets meets T_{y}. Choose W_{p} = U(J_{p}). So T_{y} ⊂ W_{p} ⊂ \{x ∈ X̂ : d(x, T_{y}) < 2^{−ℓ}\} and V (J_{p}) ⊂ ⋃_{n∈J_{p}} \overline{f̂(U_{n})} ⊂ B(y, r) with r < 2^{−ℓ}, hence diam(V (J_{p})) < 2^{1−ℓ}.

If y ∈ L \ Y then since y ∈ Y^{∗} there is some J′ ∈ ℱ_{ℓ} such that y ∈ V (J′), diam(V (J′)) < 2^{1−ℓ} and \overline{U(J′)} ⊂ W_{m}. We then choose J_{p} = J′ and W_{p} = U(J′).

In either case V_{p} will be a ball B(y, r) whose closure is contained in V_{m} ∩ V (J′) and disjoint from ⋃\{\overline{V_{q}} : q < p and y_{k} ∉ V_{q}\} ∪ \{y′ ≠ y : φ(y′, 0) < p\}, and such that the sphere S(y, r) contains no point of L.

Then for each y ∈ L the space f̂^{−1}(y) is complete and K_{y} := ⋂_{ℓ} \overline{W_{φ(y,ℓ)}} is a nonempty compact subset of the fiber f̂^{−1}(y). Moreover if y ∈ L ∩ Y then K_{y} = T_{y} ⊂ X by condition vii).

Let K := ⋃{K_{y} : y ∈ L}. Then K ⊂ X^{∗} and f^{∗}(K) = L. We claim that K is a compact subset of X^{∗}. Indeed let (x_{j}) be a sequence of points of K. Passing if necessary to a subsequence we can assume that the sequence (f(x_{j})) of points of L converges to some point y of L.

We construct inductively infinite subsets H_{ℓ} of ω such that

(i) for j and j′ in H_{ℓ}, d(x_{j}, x_{j′}) ≤ 2^{−ℓ} (ii) H_{ℓ} ⊂ H_{ℓ−1}

Let H_{0} = ω. If H_{ℓ−1} is defined the set V_{φ(y,ℓ)} is a neighborhood of y and contains all points f(x_{j}) for j ≥ N_{ℓ} and j ∈ H_{ℓ−1}. Fix j ∈ H_{ℓ−1} such that j ≥ N_{ℓ}. and let y′ := f(x_{j}). There is ℓ′ such that m′ := φ(y′, ℓ′) > m := φ(y, ℓ). Then V_{m′} ∩ V_{m} ≠ ∅, hence \overline{V_{m′}} ⊂ V_{m}, and

x_{j} ∈ K_{y′} ⊂ W_{m′} ⊂ \overline{W_{m′}} ⊂ W_{m}

Since W_{m} is a finite union of basic open sets of diameter less than 2^{−ℓ}, there is some n ∈ J_{m} such that H_{ℓ} := {j ∈ H_{ℓ−1} : x_{j} ∈ U_{n}} is infinite. Then for j and j′ in H_{ℓ}, we have d(x_{j}, x_{j′}) ≤ diam(U_{n}) ≤ 2^{−ℓ}.

Then a standard diagonal argument shows that (x_{j}) has a Cauchy subsequence, which converges in X̂ to a point x^{∗}. By continuity of f̂, we have f̂(x^{∗}) = y. And by construction we also have x^{∗} ∈ ⋂_{ℓ} \overline{W_{φ(k,ℓ)}}, hence x^{∗} ∈ K_{y} ⊂ K. This shows that K is a compact subset of X^{∗}. □

This completes the proof of Theorem 12. □

Theorem 15. Assume that ξ ≥ 3, that X is Σ_{ξ}^{0} and that f : X → Y is countablecompact-covering with complete fibers. Then Y is Σ_{ξ}^{0}.

Proof. Because of Theorem 11, it is enough to prove the theorem for ξ = 3. It follows from Theorem 12 that there exist X^{∗} containing X, Y^{∗} Polish containing Y and f^{∗} = X^{∗} → Y^{∗} countable-compact-covering with complete fibers extending f such that f^{∗}^{−1}(Y) = X. Let X̃ be a compactification of X̂. Since X is Σ_{3}^{0} in any compactification of X, it is Σ_{3}^{0} in X̃ hence in X̂ and in X^{∗}. As in the proof of Theorem 10 there is some Baire-one function h defined on X^{∗} with values in some compact Ẑ and a Σ_{2}^{0} subset Z of Ẑ such that X = h^{−1}(Z). We find by Corollary 9 a Baire-one section g : Y^{∗} → X^{∗} of f^{∗} such that h∘g is Baire-one. Moreover we have Y = g^{−1}(X) = (h∘g)^{−1}(Z), and this shows that Y is Σ_{3}^{0}, hence completes the proof of Theorem 15. □

We now extend the previous results to the case where all fibers are Polish. Let us first show by an example that the methods of “Baire-one sections type” do not work anymore in this context.

Theorem 16. There exists a Σ_{3}^{0} space X, a compact space Y and a countablecompact-covering mapping f : X → Y with locally compact fibers such that no Baire-one section of f can exist.

Proof. Let Y = 2^{ω}, X_{0} = 2^{ω} × [0, 1], f_{0} : X_{0} → Y be the first projection : f_{0}(y, z) = y and 𝒞 denote the space of continuous functions from 2^{ω} to the interval [0, 1], equipped with the metric of uniform convergence. Fix a dense sequence (φ_{k}) in 𝒞 and a homeomorphism j from ω^{ω} to some Π_{2}^{0} subset P of Y, e.g. the complement of some countable dense subset.

For any α ∈ ω^{ω} let ψ_{k}(α) = φ_{α(k)}∘j(α). The functions ψ_{k} are clearly continuous from ω^{ω} to [0, 1]. Finally let ψ(α) := lim inf_{k→∞} ψ_{k}(α), denote by G the graph of ψ∘j^{−1} : G := {(y, z) ∈ P × [0, 1] : z = ψ∘j^{−1}(y)}, let X := X_{0} \ G and denote by f the restriction of f_{0} to X.

The function ψ∘j^{−1} is of Baire class 2 on P. So its graph G is Π_{3}^{0} in P × [0, 1], hence in X_{0}, and X is Σ_{3}^{0}. The fiber f^{−1}(y) is homeomorphic to [0, 1], hence compact for y ∉ P, and homeomorphic to the complement of the single point ψ∘j^{−1}(y) in [0, 1], hence locally compact, for y ∈ P.

If L is any countable compact subset of Y, the set ψ∘j^{−1}(L) is countable in [0, 1]. Thus there is some z ∈ [0, 1] \ ψ∘j^{−1}(L) and K := L × {z} is a compact subset of X such that f(K) = L. This shows that f is countable-compact-covering. One could also notice that f is open.

To finish we claim that there is no Baire-one section of f. Assume towards a contradiction that the section s : y ↦ (y, g(y)) is Baire-one. Then there would be a sequence (g_{k}) of continuous functions from Y to [0, 1] such that g(y) = lim_{k→∞} g_{k}(y) for all y ∈ Y. There would then be a sequence α = (n_{k}) of integers such that d(g_{k}, φ_{n_{k}}) < 2^{−k}.

So we would have for y = j(α) : ψ_{k}(α) → g(y) hence ψ∘j^{−1}(y) = g(y) and s(y) ∈ G. Thus s(y) ∉ X, which is a contradiction. □

The main additional argument of the proof in the case of Polish fibers is the following lemma whose proof is inspired by Vaught’s theorem.

Lemma 17. Let φ : X_{0} → Y_{0} be continuous with Polish fibers, Y be a subset of Y_{0}, X = φ^{−1}(Y). Assume that φ(U) is Σ_{2}^{0} in Y_{0} for all open subset U of X_{0}. Then, for all ζ ≥ 1,

(i) if X is Σ_{ζ}^{0} then Y is Σ_{1+ζ}^{0}.

(ii) if X is Π_{ζ}^{0} then Y is Π_{1+ζ}^{0}.

Proof. For all Borel subset Z of X_{0} we define

N_{0}(Z) := {y ∈ Y_{0} : Z ∩ φ^{−1}(y) is non-meager in φ^{−1}(y)}
N_{1}(Z) := {y ∈ Y_{0} : Z ∩ φ^{−1}(y) is comeager in φ^{−1}(y)}

Clearly N_{1}(Z^{c}) is the complement in Y_{0} of N_{0}(Z) for every Borel set Z. Denote by (O_{k}) a countable basis of the topology of X_{0}. We show by induction on ζ ≥ 1 that if Z is Σ_{ζ}^{0} so is N_{0}(Z) and that if Z is Π_{ζ}^{0} then so is N_{1}(Z).

The statements for Σ_{ζ}^{0} and for Π_{ζ}^{0} are clearly equivalent. For ζ = 1 and Z ∈ Σ_{1}^{0}, we know by hypothesis that φ(Z) is Σ_{2}^{0}, and since all fibers are Baire spaces the open subset φ^{−1}(y)∩Z of φ^{−1}(y) is non-meager if and only if it is non-empty. Then N_{0}(Z) = φ(Z) ∈ Σ_{2}^{0}.

Assume now that the statement is true for every ordinal ζ′ < ζ and that Z is Σ_{ζ}^{0}. Then Z is the union of a countable family (Z_{n}) with Z_{n} ∈ Π_{ζ_{n}}^{0} for some ζ_{n} < ζ. For any y ∈ Y_{0} the set Z∩φ^{−1}(y) is non-meager in φ^{−1}(y) if and only if some Z_{n}∩φ^{−1}(y) is non-meager in φ^{−1}(y), hence if and only if some Z_{n} ∩φ^{−1}(y) is comeager in some non-empty open subset O_{k} ∩ φ^{−1}(y) of φ^{−1}(y), it is (Z_{n} ∪ O_{k}^{c}) ∩ φ^{−1}(y) is comeager in φ^{−1}(y) and O_{k} ∩ φ^{−1}(y) ≠ ∅. So

N_{0}(Z) = ⋃_{n,k} N_{1}(Z_{n} ∪ O_{k}^{c}) ∩ φ(O_{k})

Since φ(O_{k}) is Σ_{2}^{0} hence Σ_{1+ζ}^{0} and N_{1}(Z_{n} ∪ O_{k}^{c}) is Π_{1+ζ_{n}}^{0} by the hypothesis of induction, hence Σ_{1+ζ}^{0}, we get N_{0}(Z) ∈ Σ_{1+ζ}^{0}, and this completes the inductive proof.

Since X = φ^{−1}(Y), we have N_{0}(X) = N_{1}(X) = Y, and applying the previous statement to Z = X we get the desired conclusion. □

Theorem 18. Assume that ξ ≥ 3, that X is Π_{ξ}^{0} (resp. Σ_{ξ}^{0}) and that f : X → Y is countable-compact-covering with Polish fibers. Then Y is Π_{ξ}^{0} (resp. Σ_{ξ}^{0}).

Proof. Let η be such that ξ = 1 + η, X be Π_{ξ}^{0} and X̂ be the completion of X. Using Theorem 12 we obtain a Polish space Ỹ containing Y, a space X̃ containing X and contained in X̂, and a countable-compact-covering mapping f̃ extending f such that f̃^{−1}(Y) = X and that the fibers f̃^{−1}(y) are closed in X̂ for y ∈ Ỹ \ Y. Then all fibers of f̃ are Polish. By a classical result of Kuratowski there exists on X̂ a finer Polish topology for which X becomes Π_{η}^{0} and the identity map is Baire-one from the old to the new topology. Denote by X^{+} the space X̂ equipped with this new topology, by j : X^{+} → X̂ the identity map, by X_{1} the subset j^{−1}(X̃) of X^{+}, by φ_{1} : X_{1} → Ỹ the composition f̃∘j_{|X_{1}} and by (O_{n})_{n∈ω} a basis for the topology of X_{1}. Since all fibers f̃^{−1}(y) are Polish, they are Π_{2}^{0} in X̂ hence a fortiori φ_{1}^{−1}(y) is Π_{2}^{0} in X^{+}, hence Polish for all y ∈ Ỹ

Since j^{−1} is Baire-one, every j(O_{n}) is Σ_{2}^{0} in X̃. Thus there exists a countable family (F_{k})_{k∈ω} of closed subsets of X̃ such that F_{0} = X̃ and j(O_{n}) is the union of a subfamily of (F_{k})_{k∈ω} for all n. Define Λ(y) := {ℓ : y ∉ f^{∗}(F_{ℓ})} and denote for every k

G_{k} := {x ∈ F_{k} : f(x) ∉ f^{∗}(F_{k})}

and finally let

X_{0} := X_{1} \ ⋃_{k} j^{−1}(G_{k})

and denote by φ the restriction φ_{1}_{|X_{0}} of φ_{1} to X_{0}.

For every y ∈ Ỹ the fiber of φ at y is

φ^{−1}(y) = φ_{1}^{−1}(y) ∩ X_{0} = φ_{1}^{−1}(y) \ ⋃_{k} j^{−1}(G_{k}) = φ_{1}^{−1}(y) \ ⋃{j^{−1}(F_{k}) : k ∈ Λ(y)}

which is Π_{2}^{0} in the Polish space φ_{1}^{−1}(y) hence itself Polish.

Lemma 19. For all k we have f(F_{k} ∩ j(X_{0})) = f^{∗}(F_{k}).

Proof. Let x be a point of F_{k} such that f(x) ∉ f^{∗}(F_{k}). Then x ∈ G_{k} hence x ∉ j(X_{0}). Thus f(F_{k} ∩ j(X_{0})) ⊂ f^{∗}(F_{k}).

Conversely let y ∈ f^{∗}(F_{k}). Then the set Λ(y) is countable and, by Theorem 6, Φ_{μ}(y) ∩ W (F_{ℓ}^{c}) is a dense open set of the Polish space Φ_{μ}(y) for all ℓ ∈ Λ(y). Moreover, since k ∉ Λ(y), the closed subset Φ_{μ}(y) ∩ {K : F ∩ F_{k} ≠ ∅} is nonmeager in Φ_{μ}(y). Thus there is some K ∈ Φ_{μ}(y) such that K ∩ F_{k} ≠ ∅ and K ∩ F_{ℓ} = ∅ for all ℓ ∈ Λ(y). It follows that K ⊂ j(X_{0}). Then if x is any element of K ∩ F_{k}, we have x ∈ F_{k} ∩ j(X_{0}) and f̃(x) = y. □

Lemma 20. The function φ : X_{0} → Ỹ is onto and φ(U) is Σ_{2}^{0} in Ỹ for any open set U of X_{0}.

Proof. We have X̃ = F_{0}. We also have X_{0} = j^{−1}(F_{0}) ∩ X_{0}, hence using Lemma 19:

φ(X_{0}) = f(F_{0} ∩ j(X_{0})) = f^{∗}(X̃) = Ỹ

Moreover, for every n there exists A_{n} ⊂ ω such that j(O_{n}) = ⋃_{k∈A_{n}} F_{k}. Thus

φ(O_{n}) = ⋃_{k∈A_{n}} f̃(F_{k} ∩ j(X_{0})) = ⋃_{k∈A_{n}} f^{∗}(F_{k}) ∈ Σ_{2}^{0}

And since any open set U of X_{0} is a countable union of basic open sets, the proof is complete. □

If follows from the previous lemmas that φ : X_{0} → Ỹ satifies the assumptions of Lemma 17. Since φ_{1}^{−1}(Y) = j^{−1}(X) is Π_{η}^{0} in X_{1}, hence φ^{−1}(Y) is Π_{η}^{0} in X_{0}, it follows from Lemma 17 that Y is Π_{1+η}^{0} = Π_{ξ}^{0} in Ỹ.

The proof goes in a similar way if X is assumed to be Σ_{ξ}^{0} instead Π_{ξ}^{0}.

□

It would be tempting to use the previous theorem to get an alternative proof of the result of class preservation under compact-covering mappings given in [5] and quoted in the introduction. This could be done if we were able to prove the following statement :

(∗∗) For any Borel space X and any compact-covering mapping f : X → Y there exists some Borel subspace X′ of X of the same class as X such that the restriction of f to X′ is countable-compact-covering with Polish fibers.

Obviously the conclusion holds whenever f is inductively perfect. So if every compact-covering mapping with Borel domain is inductively perfect the statement (∗∗) is true. Unfortunately (∗∗) is not provable in ZF C as shown by the following counterexample, which is essentially the same as the one constructed in [3].

Theorem 21 (¬DOM). There exists a D(Σ_{2}^{0}) space X, a Polish space Y and a compact-covering mapping f : X → Y such that for no Borel subset B of X with f(B) = X the restriction f_{|B} is countable-compact-covering with Polish fibers.

Proof. Denote as usual by ≤^{∗} the preordering on ω^{ω} defined by

α ≤^{∗} β ⟺ ∃n ∈ ω ∀p ≥ n α(p) ≤ β(p)

Let C be the largest Π_{1}^{1} thin subset of ω^{ω} × 2^{ω}. It is a well-known fact that its projection on ω^{ω} is the set ω^{ω} ∩ L of constructible elements of ω^{ω}. And the axiom ¬DOM asserts that this set is not dominated in ω^{ω}, that is

∀α ∈ ω^{ω} ∃β ∈ ω^{ω} ∩ L β ≰^{∗} α

and in particular ω^{ω} ∩ L is uncountable.

Let T be a compact space and 𝒜 be a closed subset of 𝒦 (T). We denote by 𝒜^{∩} the set {K ∈ 𝒦 (T) : ∀A ∈ 𝒜 A ∩ K ≠ ∅}. The following simple lemma is proved in [2] :

Lemma 22. The set 𝒜^{∩} is closed and monotone in 𝒦 (T) and

𝒜^{∩∩} = {B ∈ 𝒦 (T) : ∃A ∈ 𝒜 A ⊂ B}

Let (y_{n})_{n∈ω} be a dense sequence of ω^{ω} and P be a compactification of ω^{ω}. Since (P × 2^{ω}) \ C is analytic, it is the projection of some Π_{2}^{0} subset of (P × 2^{ω}) × 2^{ω} whose complement H is a Σ_{2}^{0} subset of the compact space P × 2^{ω} × 2^{ω} satisfying

(z, t) ∈ C ⟺ ∀α ∈ 2^{ω} (z, t, α) ∈ H

Let 𝒜 be the compact subset of 𝒦 (P × 2^{ω} × 2^{ω}) defined by

𝒜 := {{(z, t)} × 2^{ω} : (z, t) ∈ P × 2^{ω}}

Then ℬ := 𝒜^{∩} is the set of compact subsets of P × 2^{ω} × 2^{ω} having total projection on P × 2^{ω}. And we fix a dense sequence (T_{k}) in ℬ.

The space Y is the union of ω^{ω} and a countable family (y_{n,k,p})_{(n,k,p)∈ω^{3}} of isolated points such that d(y_{n}, y_{n,k,p}) ≤ 2^{−p} for all (n, k, p), hence a Polish space. The space X is the subset of Y × P × 2^{ω} × 2^{ω} defined by

X := {(y, z, t, α) : (y ∈ ω^{ω} and (z, t, α) ∈ H and z ≰^{∗} y)
or (y = y_{n,p,k} and (z, t, α) ∈ T_{k})}

Finally f is the first projection : (y, z, t, α) ↦ y.

It is easily checked that X is D(Σ_{2}^{0}). So the theorem will follow from the two next lemmas.

Lemma 23. The mapping f is compact-covering.

Proof. Let M be a compact subset ot Y. If M ∩ ω^{ω} = ∅ then M is finite and K = f^{−1}(M) is compact with f(K) = M. If M′ := M ∩ ω^{ω} is non-empty, it is compact, hence bounded in ω^{ω}. Thus there is some y^{∗} ∈ ω^{ω} such that y(j) ≤ y^{∗}(j) for all j ∈ ω and all y ∈ M′. By assumption there is some z ∈ ω^{ω} ∩ L such that z ≰^{∗} y^{∗} hence t ∈ 2^{ω} such that (z, t) ∈ C. For every y = y_{n,p,k} ∈ M \ ω^{ω}, the set T_{k} belongs to 𝒜^{∩}, and there exists some α_{y} ∈ 2^{ω} such that (z, t, α_{y}) ∈ T_{k}. Then define

K = (M′ × {(z, t)} × 2^{ω}) ∪ {(y, z, t, α_{y}) : y ∈ M \ ω^{ω}}

Then it is easy to check that K is a compact subset of X and that f(K) = M. □

Lemma 24. There is no Borel set B ⊂ X such that f_{|B} is countable-compact-covering with Polish fibers.

Proof. Assume towards a contradiction that B is a Borel subset of X such that f_{|B} is countable-compact-covering and that B(y) := {(z, t, α) : (y, z, t, α) ∈ B} is Π_{2}^{0} in P × 2^{ω} × 2^{ω} for every y ∈ Y. Then there is a quasi-u.s.c. mapping Φ : Y → 𝒫(𝒦 (X)) as in Proposition 1. For every y ∈ ω^{ω} and every k ∈ ω there exists some sequence (y^{(j)}) of points of Y \ ω^{ω} converging to y and such that f^{−1}(y^{(j)}) = T_{k} for all j. If K is any element of Φ(y) then there exists a compact subset K′ of X such that f(K′) = {y} ∪ {y^{(j)} : j ∈ ω} and that K′ ∩ f^{−1}(y) ⊂ K hence for every j a point (z_{j}, t_{j}, α_{j}) ∈ T_{k} such that (y^{(j)}, z_{j}, t_{j}, α_{j}) ∈ K′. The sequence (z_{j}, t_{j}, α_{j}) has a cluster point (z, t, α) in T_{k} and it follows from what precedes that (y, z, t, α) ∈ K hence that K ∩ ({y} × T_{k}) ≠ ∅. By density of (T_{k}) in ℬ we conclude that K ∈ ℬ^{∩} = 𝒜^{∩∩} and by Lemma 22 that there is some A = {(z_{0}, t_{0}, α) : α ∈ 2^{ω}} in 𝒜 such that K ⊃ A. So

∀α ∈ 2^{ω} (z_{0}, t_{0}, α) ∈ B(y) ⊂ X(y) ⊂ H

and this implies that (z_{0}, t_{0}) ∈ C.

Moreover since B(y) is Π_{2}^{0} in P × 2^{ω} × 2^{ω} for every y ∈ Y, so is for every (y, z, t) the set {α ∈ 2^{ω} : (y, z, t, α) ∈ B}. It follows that the Borel set (Y ×P ×2^{ω} ×2^{ω})\B has σ-compact sections hence by Arsenin’s theorem that its projection on Y ×P ×2^{ω} is Borel. And the complement

B_{0} = {(y, z, t) ∈ Y × P × 2^{ω} : ∀α ∈ 2^{ω} (y, z, t, α) ∈ B}

of this projection is Borel too. It follows from the previous arguments that for every (y, z, t) ∈ B_{0} such that y ∈ ω^{ω}, we have (z, t) ∈ C and that for every y ∈ ω^{ω}

B_{0}(y) := {(z, t) : (y, z, t) ∈ B_{0}} ≠ ∅

Thus the projection E of B_{0} ∩ (ω^{ω} × P × 2^{ω}) on P × 2^{ω} is an analytic subset of the thin set C. We conclude that E is countable hence that D := {z : ∃t (z, t) ∈ E} is countable too. So we can find an y^{∗} ∈ ω^{ω} such that z ≤^{∗} y^{∗} for every z ∈ D. And for this y^{∗} we have B_{0}(y^{∗}) = ∅, a contradiction. □

This completes the proof of Theorem 21. □

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7. A. Louveau and J. Saint Raymond, Borel classes and closed games : Wadge-type and Hurewicz-type results, Trans. Amer. Math. Soc. 304 vol. 2 (1987) 431-467

8. A.V. Ostrovsky, New class of maps connected with compact-covering maps, Vestnik Moskov. Gos. Univ. 1994 n^{o} 4, 24–27

9. A.V. Ostrovsky, s-covering maps with complete fibers, Topology Appl. 102 (2000), 1–11

10. A.V. Ostrovsky, Stability of Borel classes relative to stable maps, preprint

11. J. Saint Raymond, Caractérisation d’espaces polonais, Séminaire d’Initiation à l’Analyse, 11–12èmes années, n^{o} 5 (1971-73)

12. J. Saint Raymond, Fonctions boréliennes sur un quotient, Bull. Sc. Math. 100 (1976), 141-147

Analyse Fonctionnelle, Institut de Mathématique de Jussieu, Boîte 186, 4, place Jussieu, 75252 Paris CEDEX 05

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