Linear and uniformly continuous surjections between C_p-spaces over metrizable spaces

Definition 1.1

A map T : D(X) → D(Y) is called inversely bounded if for every norm bounded sequence {g_n} ⊂ C^∗(Y) there is a norm bounded sequence {f_n} ⊂ C^∗(X) with T(f_n) = g_n for each n ∈ ℕ.

Evidently, every linear continuous map between D_p(X) and D_p(Y) is uniformly continuous and every c-good map is inversely bounded. Also, every linear continuous surjection T:Cp∗(X)→Cp∗(Y), where X and Y are arbitrary Tychonoff spaces, is inversely bounded, see [11: Proposition 3.3].

Recall that a normal topological space X is countable-dimensional (strongly countable-dimensional) if X can be represented as a countable union of normal finite-dimensional subspaces (resp., closed finite-dimensional subspaces). Note that there are countable-dimensional compact metrizable spaces which are not strongly countable-dimensional, see [10].

Marciszewski [23: Corollary 2.7] observed that, by modifying the Gul’ko’s arguments [14] (cf. [22, 24]), one can show the following theorem:

Theorem 1.2

Let 𝒫 be the property of metrizable spaces such that:

1. If X ∈ 𝒫 and Y is a subset of X, then Y ∈ 𝒫,

2. If X is a metrizable space which is a countable union of closed subsets X_n ∈ 𝒫, then X ∈ 𝒫.

Then, for metrizable spaces X and Y such that C_p(X) and C_p(Y) are uniformly homeomorphic, X ∈ 𝒫 if and only if Y ∈ 𝒫.

It is known that the covering dimension dimX ≤ n satisfies the above conditions (i) and (ii). Much less is known about the case when T : C_p(X) → C_p(Y) is supposed to be only uniformly continuous and surjective. The following open problem has been posed in [13: Question 4.1].

Problem 1.3

Let X be a compact metrizable strongly countable-dimensional [zero-dimensional] space. Suppose that there exists a uniformly continuous surjection T:C_p(X) → C_p(Y). Is Y necessarily strongly countable-dimensional [zero-dimensional]?

The authors of [13] established that the answer to Problem 1.3 is affirmative provided that the uniformly continuous surjection T is c-good for some c > 0. Later, in [11] the same was proved for σ-compact metrizable spaces. In our paper we strengthen this result by proving this remains true for all metrizable spaces and all inversely bounded uniformly continuous surjections. In fact we develop a general scheme for the proof as follows.

We consider the properties 𝒫 of normal spaces such that:

1. if X ∈ 𝒫 and F ⊂ X is closed, then F ∈ 𝒫;

2. 𝒫 is closed under finite products;

3. if X is a countable union of closed subsets each having the property 𝒫, then X ∈ 𝒫;

4. if f : X → Y is a closed map with finite fibers, where Y is a metrizable space with Y ∈ 𝒫, then X ∈ 𝒫.

From the classical results of dimension theory (see [10]) it follows that zero-dimensionality, countable-dimensionality and strongly countable-dimensionality satisfy conditions (a)–(d) above.

Now we formulate one of the main results of our paper.

Theorem 1.4

Let X be a metrizable space and Y be perfectly normal. Suppose that T : D_p(X) → D_p(Y) is a uniformly continuous inversely bounded surjection. For any topological property 𝒫 satisfying conditions (a)–(d) above, if X ∈ 𝒫 then Y ∈ 𝒫.

Corollary 1.5

Let X, Y and T : D_p(X) → D_p(Y) satisfy the hypotheses of Theorem 1.4.

1. If X is either countable-dimensional or strongly countable-dimensional, then so is Y.

2. If X is zero-dimensional, then so is Y.

Note that item (ii) was established in [11: Theorem 1.1] for arbitrary Tychonoff spaces X, Y and c-good surjections T. However, we do not know whether there exists a uniformly continuous inversely bounded map which is not c-good for some c > 0.

A linear continuous version of Theorem 1.4 is also true (for metrizable compact spaces it was implicitly established in [19]).

Theorem 1.6

Let X be a metrizable space and Y be a perfectly normal space. Suppose that T : D_p(X) → D_p(Y) is a linear continuous surjection. For any topological property 𝒫 satisfying conditions (a)–(d) above, if X ∈ 𝒫 then Y ∈ 𝒫.

In the last part of our paper we show that for any two metrizable spaces X and Y such that there is a linear continuous surjection T:Cp(X)→Cp(Y)orT:Cp∗(X)→Cp∗(Y), and X is scattered, then so is Y (Theorem 4.5). Also, we apply Theorem 1.4 to the property 𝒫 of being a strongly σ-scattered space and to the property 𝒫 of being a Δ₁-space. All necessary definitions are given in Section 4.

Our results strengthen and extend several results from [6, 13] and [16].

Claim 1

If M ⊂ M(p, q) is closed in Y for some p, q, then the map ϕ_{pq M} : M → [X]^q is closed and each fiber of ϕ_{pq M} is countably compact.

Because [X]^q is a metrizable space, it suffices to show that if {y_n} is a sequence in M such that ϕ_pq(y_n) converges to some K ∈ [X]^q, then {y_n} has an accumulation point in M. That statement was established in the proof of condition (9) from [24] in case both X and Y are metrizable, but the same proof works when X is metrizable and Y is normal.

Since each Y (p, q) is a closed subset of Y, it follows from (4) that each M(p, q) is a countable union of closed subsets {F_n(p, q) : n ∈ ℕ} of Y. So, by (5), Y = ∪{F_n(p, q) : n, p, q ∈ ℕ}. According to Claim 1, all maps φpqn=φpq↾Fn(p,q):Fn(p,q)→[X]q are closed and have countably compact fibers.

Claim 2

The fibers of φpqn:Fn(p,q)→[X]q are finite.

We follow the arguments from the proof of [13: Theorem 4.2]. Fix z ∈ F_n(p, q) for some n, p, q ∈ ℕ and let A(z) = {y ∈ F_n(p, q) : K_p(y) = K_p(z)}. Suppose that A(z) is infinite, so it contains a sequence S = {y_m} of distinct points. Because, by Claim 1, A(z) is countably compact, there are two possibilities: either {y_m} is closed and discrete or it contains an accumulation point in A(z). Therefore, passing to a subsequence, we may assume that for every y_m there exist a neighborhood U_m in Y and a function g_m: Y → [0, 2p] such that: U_m ∩ S = {y_m}, g_m(y_m) = 2p and g_m(y) = 0 for all y ε̷ U_m. Since T is inversely bounded, there is a norm bounded sequence {f_m} ∈ C^∗(X) with T(f_m) = g_m. Let r > 0 be such that ||f_m|| ≤ r, m ∈ ℕ. So, the sequence {f_m} is contained in the compact set [−r, r]_X. Hence, {f_m} has an accumulation point in [−r, r]_X. This implies the existence of i ≠ j such that |f_i(x) − f_j(x)| < 1 for all x ∈ K_p(z). Consequently, since Kp(yj)=Kp(z),|αyj(fj)−αyj(fi)|≤p. On the other hand, αyj(fj)=T(fj)(yj)=gj(yj)=2p and αyj(fi)=T(fi)(yj)=gi(yj)=0, so |αyj(fj)−αyj(fi)|=2p, a contradiction.

Now we can complete the proof of Theorem 1.4. Suppose that X has a property 𝒫 satisfying conditions (a)–(d). Then so does X^q for each q. The space [X]^q is homeomorphic to the set W_q = {(x₁, x₂,..., x_q) ∈ X^q : x_i ≠ x_j for i ≠ j} which is open in X^q. So, [X]^q ∈ 𝒫 as a countable union of closed subsets of X^q. According to Claim 1, φpqn(Fn(p,q)) is closed in [X]^q. Hence, φpqn(Fn(p,q)) has the property 𝒫. Finally, since the map φpqn:Fn(p,q)→φpqn(Fn(p,q)) is perfect and has finite fibers, we obtain F_n(p, q) ∈ 𝒫. Therefore, by condition (c), Y = ∪{F_n(p, q) : n, p, q ∈ ℕ} also has the property 𝒫.

Claim 3

Let F ⊂ M_k be closed in Y for some k. Then the map S_{k F} : F → [X]^k is closed.

Since X is a metrizable space, it suffices to show that if {y_n} is a sequence in F and S_k(y_n) converges to some K ∈ [X]^k, then {y_n} has an accumulation point in F. Striving for a contradiction, suppose that there is a sequence {y_n} in F such that the set Z = {y_n : n ∈ ℕ} is closed and discrete in Y. Let K = {x₁, x₂,..., x_k} and S_k(y_n) = {x₁(y_n), x₂(y_n),..., x_k(y_n)} for all n. Since S_k(y_n) converges to K in [X]^k, each of the sequences {x_i(y_n)}_n∈N, i = 1, 2,..., k, converges in X to x_i. Therefore, A=⋃i=1k{xi}∪{xi(yn)}n∈N is a compact subset of X. Because X is a metrizable space, according to Dugundji Extension Theorem [9], (see also [26]), there is a continuous linear map Θ:Cp(A)→Cp∗(X) such that Θ(g) _A= g for all g ∈ C(A). Thus, the linear map ϕ: C_p(A) → D_p(Z), ϕ(g) = T(Θ(g)) _Z, is continuous, where D(Z) = C(Z) if D(Y) = C(Y) and D(Z) = C^∗(Z) if D(Y) = C^∗(Y). Moreover, ϕ is surjective. Indeed, take h ∈ D(Z) and its continuous extension h ∈ D(Y). Then T(f) = h for some f ∈ D(X) and the functions f and g = Θ(f _A) have the same restrictions on A. Hence, by (P1), l_y(f) = l_y(g) for all y ∈ Z. Thus, ϕ(f _A) = h. If D(Z) = C(Z), then by (P4), the set Z is bounded. If D(Z) = C^∗(Z), according to (P5), Z is also bounded. Therefore, in both possible cases we have a contradiction.

Since each Y_k is closed in Y and Y is perfectly normal, M_k is the union of a countably many closed subsets M_kn of Y. Let S_kn = S_{k Mkn}. According to Claim 3, each S_kn is a closed map.

Claim 4

The fibers of each map S_kn : M_kn → [X]^k are finite.

Indeed, let z ∈ M_kn, S_kn(z) = {x₁, x₂,..., x_k} and A(z) = {y ∈ M_kn : S_kn(y) = S_kn(z)}. Since supp(l_y) = S_kn(z) for all y ∈ A(z), as in the proof of Claim 3, there is a continuous linear surjection φ: C_p(S_kn(z)) → D_p(A(z)). Because C_p(S_kn(z)) is finite-dimensional, by linearity, so is D_p(A(z)). Thus, A(z) is finite.

Finally, as in the last paragraph from the proof of Theorem 1.4, we can show that Y has the property 𝒫.

Theorem 4.1

([4])|| Let X and Y both be first countable paracompact spaces. Suppose that T : C_p(X) → C_p(Y) is a linear homeomorphism. Then X is scattered if and only if Y is scattered.

Theorem 4.2

([5, 6])|| Let X and Y be metrizable spaces. Suppose that T:Cp∗(X)→Cp∗(Y) is a linear homeomorphism. Then X is scattered if and only if Y is scattered.

It is an open problem whether Theorem 4.2 remains true if both X and Y are assumed to be first countable paracompact spaces ([6: Question 4.8]), despite of the following structural result which apparently is due to Telgársky [31: Theorem 8]: every scattered first countable paracompact space is metrizable and strongly σ-discrete.

Below we strengthen both Theorems 4.1 and 4.2 in case that X and Y are metrizable spaces, assuming only that T is a linear continuous surjection.

We will use several well-known facts about the linear topological spaces, which are dual to D_p(X). In essence, we have already described some properties of the dual to D_p(X) in the proof of Theorem 1.6. We repeat that for any Tychonoff space X the dual space of D_p(X) algebraically can be identified with a linear space of formal linear combinations L(X), where X is a Hamel basis in L(X). For each natural n ∈ ℕ denote by M_n(X) the subspace of L(X) formed by all linear combinations of the reduced length precisely n. Let τ_p be the topology on L(X) when L(X) is considered as a weak topological dual to C_p(X), and let τ_b be the topology on L(X) when L(X) is considered as a weak topological dual to Cp∗(X), respectively. In general, τ_b does not coincide with τ_p on the whole linear space L(X) [8]. However, the analysis of the proof of [2: Proposition 0.5.17] easily shows that the topologies τ_b and τ_p restricted to subsets M_n(X) do coincide. (One need to do some cosmetic changes which are based on the following trivial remark: for any point x ∈ X and open U ⊂ X containing x there is a bounded continuous function f on X such that f(x) = 1, f _XU= 0). Hence, from [2: Proposition 0.5.17] (see also [17: Proposition 2.1]) we can deduce the following result.

Proposition 4.3

(Mn(X),τp)=(Mn(X),τb) is homeomorphic to a subspace of the Tychonoff product (ℝ^∗)ⁿ × Xⁿ, where ℝ^∗ = ℝ \{0}.

Now we have a result for all Tychonoff spaces (everywhere below, except for Theorem 4.5, we suppose that T : D_p(X) → D_p(Y) is a surjection such that all possible four cases are considered).

Proposition 4.4

.Let X and Y be Tychonoff spaces. Suppose that T : D_p(X) → D_p(Y) is a linear continuous surjection. If X is σ-scattered (σ-discrete), then Y also is σ-scattered (σ-discrete, respectively).

Claim 5

Every projection p_i : Y_i → Xⁿ is a finite-to-one mapping.

Evidently, Xⁿ is σ-scattered (resp., σ-discrete) provided so is X. Since p_i is continuous and its fibers are finite, for every Z ⊂ X ⁿ and every isolated point z in Z the fiber pi−1(z) consists of isolated points in pi−1(Z). We conclude, if X is σ-scattered (σ-discrete), then each Y_i, i ∈ ℕ, is σ-scattered (σ-discrete, respectively), and then Y is σ-scattered (σ-discrete, respectively). □

Theorem 4.5

Let X and Y be metrizable spaces. Suppose that T : D_p(X) → D_p(Y) is a linear continuous surjection such that either D(X) = C(X) and D(Y) = C(Y) or D(X) = C^∗(X) and D(Y) = C^∗(Y). If X is scattered, then so is Y.

Problem 4.6

(See [23: 2.18. Problem]). Let X and Y be (separable) metrizable spaces and let T : D_p(X) → D_p(Y) be a a uniformly continuous surjection (uniform homeomorphism). Let X be completely metrizable. Is Y also completely metrizable?

Moreover, the next problem is also open.

Problem 4.7

Let X and Y be (separable) metrizable spaces and let T : D_p(X) → D_p(Y) be an inversely bounded uniformly continuous surjection. Let X be completely metrizable. Is Y also completely metrizable?

We obtain a σ-scattered analogue of Theorem 4.5 for inversely bounded uniformly continuous surjections.

Theorem 4.8

Let X and Y be metrizable spaces. Suppose that T : D_p(X) → D_p(Y) is an inversely bounded uniformly continuous surjection. If X is strongly σ-scattered, then Y also is strongly σ-scattered.

Theorem 4.9

Let X and Y be metrizable spaces. Suppose that T : D_p(X) → D_p(Y) is an inversely bounded uniformly continuous surjection. If X is a Δ₁-space then Y also is a Δ₁-space.