Multiplicative bijections of semigroups of interval-valued continuous functions

We characterize all compact and Hausdorff spaces $X$ which satisfy that for every multiplicative bijection $\phi$ on $C(X, I)$, there exist a homeomorphism $\mu : X \to X$ and a continuous map $p: X \to (0, +\infty)$ such that $$\phi (f) (x) = f(\mu (x))^{p(x)}$$ for every $f \in C(X,I)$ and $x \in X$. This allows us to disprove a conjecture of Marovt (Proc. Amer. Math. Soc. {\bf 134} (2006), 1065-1075). Some related results on other semigroups of functions are also given.


Introduction
For a compact and Hausdorff space X, let C(X, I) denote the semigroup of all continuous functions on X taking values in the closed unit interval I = [0, 1]. Motivated by a result of L. Molnár (see [6]), J. Marovt studied the form of multiplicative bijections on C(X, I) in a special case, namely, when X satisfies the first axiom of countability (see [4]). He proved that if ϕ is such a map, then there exist a homeomorphism µ : X −→ X and a continuous map p : X −→ (0, +∞) such that ϕ(f )(x) = f (µ(x)) p(x) for every f ∈ C(X, I) and x ∈ X. We call maps of this form standard. Marovt conjectured that every multiplicative bijection on C(X, I) was necessarily standard, and that the assumption on X of being first countable could be dropped.
In this paper we prove that Marovt's conjecture does not hold. In fact we give a characterization of spaces X for which there exists a multiplicative bijection of C(X, I) that is not standard. Roughly speaking, we prove that this is the case if and only if X is the Stone-Čech compactification of a proper subset (see Theorem 2.2).
We study a more general case, as is that of multiplicative bijections between semigroups of functions. Of course, a multiplicative bijection between arbitrary semigroups of C(X, I) need not be of the above form, even if they separate points. A simple example is the following: take X having only one point, so each semigroup of C(X, I) can be identified with a semigroup of I. Consider for instance s = 1/2, t = 1/3, and A := {s n t m : n, m ∈ N}. It is now clear that the map sending each s n t m into s m t n is multiplicative and bijective, but cannot be described as above.
Nevertheless, our technique can be used for semigroups other than C(X, I). In fact, it works if we just require these semigroups A ⊂ C(X, I) to satisfy the following three properties (where, as usual, if f ∈ C(X, I), coz f denotes the set {x ∈ X : f (x) = 0}, and supp f denotes its closure).
Property 1: Given any x ∈ X and any neighborhood U of x, there exists f ∈ A such that f (x) ≡ 1 on a neighborhood of x and supp f ⊂ U .
for every x ∈ X, and supp f ⊂ coz g, then there exists k ∈ A such that f = gk.
It is easy to see that the image by a standard multiplicative bijection of a semigroup satisfying Properties 1, 2, and 3 is a semigroup that also satisfies them. One of such semigroups is, for instance, that of all continuously differentiable functions on I (also taking values in I). The general form of multiplicative bijections between this kind of semigroups will be given in Theorem 2.1. Finally we also give an example where Marovt's result does not hold for multiplicative bijections defined between general semigroups of this type, even when the ground spaces are first countable (see Example 3.5).
Suppose that ϕ : A −→ B is multiplicative and bijective, where A ⊂ C(X, I) and B ⊂ C(Y, I) are semigroups satisfying Properties 1, 2, and 3. We say that y ∈ Y is a standard point for ϕ if there exist a number p ∈ (0, +∞) and a point x ∈ X such that ϕ(f )(y) = f (x) p for every f ∈ A. We denote by R(ϕ) the set of standard points for ϕ.
Throughout we assume that X and Y are compact and Hausdorff spaces. Given a (completely regular) space Z, we denote by βZ its Stone-Čech compactification.
Remark finally that our results are essentially different from those given in [5] for multiplicative bijections between spaces C(X, R) of real-valued continuous functions on X. Even so, there is a similarity in that the multiplicative structure of the spaces of functions determines the space X up to homeomorphism. More recent results in this line, concerning multiplicative maps on C(X, R), are given for instance in [2] and [3].

Main results
We first give a theorem that provides a description of multiplicative bijections between semigroups satisfying the above properties. Recall that a subset Z of Y is a cozero-set if there exists a real-valued continuous function f on Y such that Z = {y ∈ Y : f (y) = 0}.
, is a multiplicative bijection. Nevertheless, R(ϕ) = X, so each point is standard for ϕ, but we cannot find a continuous map p as given in Theorem 2.1.
We next state the theorem characterizing all spaces on which can be defined a multiplicative bijection that is not standard. Recall that a topological space Z is said to be pseudocompact if every real-valued continuous map on Z is bounded. Obviously Theorem 2.2 gives an answer in the negative to Marovt's conjecture. It is enough now to take any completely regular space Z (thus ensuring that its Stone-Čech compactification exists) that is not pseudocompact (as for instance N, R, or any unbounded subset of a normed space), and we have that there are always multiplicative bijections on C(βZ, I) that are not standard.

Some other results and proofs
The following is a key lemma to prove Theorem 2.1.
Proof. Suppose on the contrary that there exist α, β ∈ A 1 such that u(α) = α p and u(β) = β q , where 0 < p < q. By taking integer powers of α if necessary, we may assume without loss of generality that α < β. Define a := − log α and b := − log β. Obviously 0 < b < a, and we can find natural numbers n, m such that Now it is clear that mb < na and npa < mqb. These inequalities lead easily to α n < β m and α np > β mq , that is, u(α n ) > u(β m ), against the fact that u is order preserving.
satisfies the finite intersection property, and we conclude that C x is nonempty.
On the other hand, ϕ −1 is also multiplicative, so given any y ∈ C x , we have that the set C y (defined in a similar way as C x ) is nonempty. Let us see that C y = {x}. Suppose that there exists z ∈ C y , z = x. Then we can find f z ∈ U x such that z / ∈ supp f z . Since z ∈ C y and ϕ(f z )(y) = 1, then we can find k ∈ B with supp k ⊂ coz ϕ(f z ) and ϕ −1 (k)(z) = 0. Clearly, if we now take g ∈ U z with f z g = 0, then gϕ −1 (k) = 0, but ϕ(g)k = 0, which is impossible.
The above process lets us define a map µ : Y −→ X, which turns out to be bijective, such that µ(y) is the only point in C y , and y is the only point in C µ(y) , for each y ∈ Y .
We prove next that µ is continuous at every point of Y (and is consequently a homeomorphism). Take any y ∈ Y , and let U be an open neighborhood of µ(y). We will see that, if f ∈ U µ(y) and supp f ⊂ U , then µ(coz ϕ(f )) is contained in U . Otherwise there exists z ∈ Y such that ϕ(f )(z) = 0 and µ(z) / ∈ supp f , so we can take k ∈ U µ(z) such that kf = 0. Obviously ϕ(k)(z)ϕ(f )(z) = 0, which is absurd.
Finally, we have that, by definition, if y ∈ R(ϕ), then there exist p(y) ∈ (0, +∞) and x ∈ X such that ϕ(f )(y) = f (x) p(y) for every f ∈ A. It is easy to check that x = µ(y). As for the map p : R(ϕ) −→ (0, +∞), we have that for each y ∈ R(ϕ), for every f ∈ A with f (µ (y)) = 0, 1. This implies that p is continuous at y, and consequently on R(ϕ).
We finally see that R(ϕ) is a cozero-set. Suppose that y ∈ Y \R(ϕ). Take any net (z α ) α∈Λ in R(ϕ) converging to y. Clearly, since y / ∈ R 1 (ϕ), then there exists f ∈ A with f (µ(y)) = 0, 1 and either ϕ(f )(y) = 0 or 1. Also for every α ∈ Λ. The conclusion that lim α p(z α ) ∈ {0, +∞} follows easily. Finally, it is clear that R(ϕ) coincides with the cozero-set of the continuous function Y −→ R given by if y ∈ R(ϕ), and y → 0 otherwise. In what follows, when we want to specify the homeomorphism µ and the map p corresponding to a multiplicative and bijective map ϕ, as given in Theorem 2.1, we will write ϕ[µ, p]. It is obvious that if ϕ = ϕ[µ, p], then Proof. It is obvious that if ϕ = ϕ[µ, p] and ϕ −1 = ϕ −1 [µ −1 , q], then both p and q are bounded, and consequently neither +∞ nor 0 are limit points of p(R(ϕ)). The conclusion follows from Theorem 2.1. Proof. We follow the notation given after Remark 3.1. Suppose that ϕ = ϕ[µ, p]. It is clear that if 1 is the map constantly equal to 1 on X, and we consider the multiplicative bijection ψ = ψ[µ −1 , 1] : C(Y, I) −→ C(X, I), then the composition ϕ • ψ : (see Proof of Theorem 2.1). Consequently we can assume without loss of generality that X = Y and µ = id Y .
Taking into account that R(ϕ) = R ϕ −1 , we assume without loss of generality that there exists a continuous function g 0 : Y \p −1 ({0}) −→ I that cannot be continuously extended to Y . We consider the map h ∈ C(Y, I) whose image by ϕ is the constant function 1/e. It is easily seen that, since p(y) log h(y) = log ϕ(h)(y) for every y ∈ R(ϕ), then Since p −1 ({0}) is a zero-set, then there exists a sequence (K n ) of compact subsets of Y with K n+1 contained in the interior of K n for each n ∈ N, and such that p −1 ({0}) = ∞ n=1 K n . For each n ∈ N, we take g n ∈ C(Y, I) satisfying g n ≡ g 0 on Y \ K n . We denote by f n its counterimage by ϕ.
Notice that if we define f 0 (z) = f n (z) whenever z / ∈ K n , then f 0 : Y \ p −1 ({0}) −→ I is continuous. It is also obvious that f 0 h (defined as 0 on p −1 ({0})) belongs to C(Y, I). On the other hand, we have that f 0 h ≡ f n h outside K n for each n ∈ N, and consequently Remark 3.2. We deduce in particular that if the set Y \ R(ϕ) is nonempty, then its cardinality is at least 2 c . Also, if it is endowed with the restricted topology, then no point in Y \ R(ϕ) is a G δ (see [1, Chapter 9]). Remark 3.3. By Remark 3.2, we have that each point of Y having a countable base of neighborhoods belongs to R(ϕ) for every ϕ. In particular, if Y is first countable, then R(ϕ) = Y , which is essentially Marovt's result. But there can be spaces which are not first countable, and for which every point is standard. As an easy example, consider a bijective and multiplicative map ϕ : C([0, ω 1 ], I) −→ C([0, ω 1 ], I), where ω 1 denotes the first noncountable ordinal. Since each point of [0, ω 1 ) has a countable base of neighborhoods, we deduce that [0, ω 1 ] \ R(ϕ) ⊂ {ω 1 }, and again by Remark 3.2, we cannot have [0, ω 1 ] \ R(ϕ) = {ω 1 }. We conclude that R(ϕ) = [0, ω 1 ].
Remark 3.4. The conclusion given in Remark 3.3 is not true for more general semigroups, that is, not every point having a countable base of neighborhoods necessarily belongs to R(ϕ) (see Example 3.5).
Using Theorem 2.1, it is easy to see that every isolated point is standard for every multiplicative bijection. More generally, the next result allows us to identify some points that belong to R(ϕ) for every ϕ. Proof. Suppose that y / ∈ R(ϕ). Thus, taking into account Proposition 3.3 and the fact that R(ϕ) ⊂ Y \ {y}, we deduce that β (Y \ {y}) = Y , and we are finished.
Proof of Theorem 2.2. Suppose that there exists a multiplicative bijection ϕ : C(X, I) −→ C(Y, I) that is not standard. First, by Theorem 2.1, X and Y must be homeomorphic. Also, calling Z := R(ϕ), we have by Corollary 3.2 that Z is not pseudocompact, and by Proposition 3.3 that Y = βZ.
Conversely, suppose that X and Y are homeomorphic, and that there exists a proper subset Z of Y which is not pseudocompact, and such that Y = βZ. It is clear that without loss of generality we may assume that X = Y . Since Z is not pseudocompact, then there exists an unbounded continuous function u : Z −→ [1, +∞). We define a map ϕ : C(Z, I) −→ C(Z, I) as ϕ(g)(z) := g(z) u(z) for each g ∈ C(Z, I) and z ∈ Z.
It is easy to check that ϕ is multiplicative and injective. Let us see that ϕ is surjective. To this end, we take any k ∈ C(Z, I), and consider the map L k : Z −→ [−∞, 0] defined as L k (z) := (log k(z))/u(z) if k(z) = 0, and L k (z) := −∞ if k(z) = 0. It is easy to check that L k is continuous and that ϕ(exp L k ) = k (assuming exp(−∞) = 0).
Obviously ϕ can be seen as a multiplicative bijection on C(βZ, I). On the other hand, if for some homeomorphism µ : βZ −→ βZ and a continuous map v : βZ −→ (0, +∞), ϕ(f )(x) = f (µ(x)) v(x) for every f , then µ must be the identity on βZ, and v(z) = u(z) for every z ∈ Z. This obviously implies that v must attain the value +∞ on some points of βZ \ Z, which is absurd.

Acknowledgements
The author wishes to thank the referee for his/her valuable remarks and, also, Ana M. Ródenas for drawing his attention to this subject.