The canonical evaluation map with a function space perspective

The evaluation map is a useful tool for embedding a space into a product space and plays an important role in many theorems and problems in topology. See here for a previous discussion. In this post, we present the evaluation map with the perspective that the map can be used for embedding a space into a function space of continuous functions. This post will be useful background for subsequent posts on C_p(X). In this post, we take a leisurely approach in setting up the scene. Once the map is defined properly, we show what additional conditions will make the evaluation map a homeomorphism. Then a function space perspective is presented as indicated above. After presenting an application, we conclude with some special cases for evaluation maps.

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The general setting

Let X be a set (later we will add a topology). Let \mathcal{F} be a set of real-valued functions defined on X. Another way to view \mathcal{F} is that it is a subspace of the product space \mathbb{R}^X. For each x \in X, consider the map \pi_x: \mathbb{R}^X \longrightarrow \mathbb{R} defined by \pi_x(f)=f(x) for all f \in \mathbb{R}^X. One way to look at \pi_x is that it is the projection map from the product space \mathbb{R}^X to one of the factors. Thus \pi_x is continuous when \mathbb{R}^X has the product topology. In fact, the product topology is the smallest topology that can be defined on \mathbb{R}^X that would make the \pi_x continuous. When we restrict the map \pi_x to the subspace \mathcal{F}, the map \pi_x: \mathcal{F} \longrightarrow \mathbb{R} is still continuous.

One more comment before defining the evaluation map. The set \mathcal{F} of functions is a subspace of the product space \mathbb{R}^X. Therefore the set \mathcal{F} inherits the subspace topology from the product space. It makes sense to consider the function space C_p(\mathcal{F}), the space of all continuous real-valued functions defined on \mathcal{F} endowed with the pointwise convergence topology. Thus we can write \pi_x \in C_p(\mathcal{F}).

We now define the evaluation map. Define the map E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F}) by letting E_\mathcal{F}(x)=\pi_x for each x \in X, or more explicitly, by letting, for each x \in X, E_\mathcal{F}(x) be the map such that E_\mathcal{F}(x)(f)=f(x) for all f \in \mathcal{F}.

The map E_\mathcal{F} is called the evaluation map defined by the family \mathcal{F}. When the set \mathcal{F} is understood, we can omit the subscript and denote the evaluation map by E. We say E_\mathcal{F} is the canonical evaluation map.

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What makes the evaluation map works

The goal of the evaluation map is that it be a homeomorphism. For that to happen, we need to make a few more additional assumptions. In defining the evaluation map above, the functions in the family \mathcal{F} are not required to be continuous. In fact, the set X is just a set in the above section. Now we require that X is a topological space (it must be a completely regular space) and that all functions in \mathcal{F} are continuous. Thus we have \mathcal{F} \subset C_p(X). With this assumption, the evaluation map is then a continuous function. We have the following theorem.

Theorem 1
Let X be a space. Let \mathcal{F} \subset C_p(X). Then the evaluation map E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F}) is always continuous.

Proof of Theorem 1
Let x \in X. Let U be open in C_p(\mathcal{F}) with E_\mathcal{F}(x)=\pi_x \in U such that

    U=\left\{q \in C_p(\mathcal{F}): \forall \ i=1,\cdots,n, \ q(f_i) \in U_i \right\}

where f_1,\cdots,f_n are arbitrary points of \mathcal{F} and each U_i is an open interval of \mathbb{R}. For each i=1,\cdots,n, \pi_x(f_i)=f_i(x) \in U_i. Let V=\bigcap_{i=1}^n f_i^{-1}(U_i), which is open in X since each f_i is a continuous function. We show that E_\mathcal{F}(V) \subset U. For each y \in V and for each i=1,\cdots,n, E_\mathcal{F}(y)(f_i)=\pi_y(f_i)=f_i(y) \in U_i. This means that for each y \in V, E_\mathcal{F}(y)=\pi_y \in U. The continuity of the evaluation map is established. \blacksquare

In order to make the evaluation map a homeomorphism, we consider two more definitions. A family \mathcal{F} \subset \mathbb{R}^X is said to separate points of X if for any x,y \in X with x \ne y, there exists an f \in \mathcal{F} such that f(x) \ne f(y). A family \mathcal{F} \subset \mathbb{R}^X is said to separate points from closed subsets of X if for each x \in X and for each closed subset C of X with x \notin C, there exists an f \in \mathcal{F} such that f(x) \notin \overline{f(C)}. We have the following theorem.

Theorem 2
Let X be a space. Let \mathcal{F} \subset C_p(X). Then the following are true about the evaluation map E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F}).

  1. If \mathcal{F} separates the points of X, then the evaluation map \mathcal{F} is a one-to-one.
  2. If \mathcal{F} separates the points from closed subsets of X, then the evaluation map \mathcal{F} is a homeomorphism.

Proof of Theorem 2
To prove the bullet point 1, suppose that \mathcal{F} separates the points of X. Let x,y \in X with x \ne y. Then there is some f \in \mathcal{F} such that f(x) \ne f(y). It follows that the functions E_\mathcal{F}(x)=\pi_x and E_\mathcal{F}(y)=\pi_y differ at the point f \in \mathcal{F}. This completes the proof for the bullet 1 of Theorem 2.

To prove the bullet point 2, suppose that the family \mathcal{F} separates the points from closed subsets of X. It suffices to show that the evaluation map E_\mathcal{F} is an open map. Let U \subset X be a non-empty open set. We show that E_\mathcal{F}(U) is open in image E_\mathcal{F}(X). Let E_\mathcal{F}(x)=\pi_x \in E_\mathcal{F}(U) where x \in U. Since \mathcal{F} separates the points from closed subsets of X, there exists an f \in \mathcal{F} such that f(x) \notin \overline{f(X \backslash U}). Let V=\mathbb{R}-\overline{f(X \backslash U}). Consider the following open set.

    W=\left\{q \in E_\mathcal{F}(X): q(f) \in V \right\}

Clearly E_\mathcal{F}(x)=\pi_x \in W. We show that W \subset E_\mathcal{F}(U). Choose q \in W. Then q=E_\mathcal{F}(y)=\pi_y for some y \in X. It is also the case that q(f)=\pi_y(f)=f(y) \in V. Thus f(y) \notin \overline{f(X \backslash U}). This means that y \in U and that q=E_\mathcal{F}(y)=\pi_y \in E_\mathcal{F}(U). This completes the proof for the bullet 2 of Theorem 2. \blacksquare

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Embedding every space into a function space

In defining the evaluation map, we start with a space X. Then take a family of continuous maps \mathcal{F} \subset C_p(X). As long as the family of functions \mathcal{F} separates points from closed sets, we know for sure that the evaluation map is a homeomorphism from X into a subspace of C_p(\mathcal{F}). We now look at some choices for \mathcal{F}. One is that \mathcal{F} = C_p(X). Then we have the following corollary.

Corollary 3a
Any space X is homeomorphic to a subspace of the function space C_p(C_p(X)).

Because X is a completely regular space, the family \mathcal{F} = C_p(X) clearly separates points from closed sets. Thus Corollary 3a is valid. In fact, the complete regularity of X only requires that we use \mathcal{F} = C_p(X,I), the set of all continuous functions from X into I where I=[0,1]. We have the following corollary.

Corollary 3b
Any space X is homeomorphic to a subspace of the function space C_p(C_p(X,I)).

We can also let \mathcal{F} = C_p^0(X), the set of all bounded real-valued continuous functions defined on X. It is clear that C_p^0(X) separates points from closed sets. So we also have:

Corollary 3c
Any space X is homeomorphic to a subspace of the function space C_p(C_p^0(X)).

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One application

We demonstrate one application of Corollary 3a. When X is a separable metric space, C_p(X) has a countable network (see this previous post). It is natural to ask whether every space with a countable network can be embedded in a C_p(Y) for some separable metric space Y? The answer is yes. We have the following theorem. One direction of the theorem is Theorem III.1.13 in [1].

Theorem 4
Let X be a space. Then the following conditions are equivalent.

  1. The space X has a countable network.
  2. The space X can be embedded in a C_p(Y) for some separable metric space Y.

Proof of Theorem 4
The direction 2 \longrightarrow 1 is clear. As shown here, C_p(Y) has a countable network whenever Y has a countable base. Having a countable network carries over to subspaces. The direction 1 \longrightarrow 2 is the one that uses evaluation map.

1 \longrightarrow 2
Suppose that \mathcal{M} is a countable network for X. Then C_p(X) has a countable network, e.g., the set of all [M,V] where M \in \mathcal{M}, V is any open interval with rational endpoints and [M,V] is the set of all f \in C_p(X) such that f(M) \subset V.

Any space with a countable network is the continuous image of a separable metric space. Thus there exists a separable metric space Y such that C_p(X) is the continuous image of Y. Let g: Y \longrightarrow C_p(X) be a continuous surjection. Then C_p(C_p(X)) can be embedded into C_p(Y). The embedding \rho:C_p(C_p(X)) \longrightarrow C_p(Y) is defined by \rho(f)=f \circ g.

By Corollary 3a, X is embedded into C_p(C_p(X)). Then X is embedded into C_p(Y). \blacksquare

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More on the evaluation map

In this section, we consider some special cases. As shown in Theorem 2, what makes the evaluation map a one-to-one map is that the family \mathcal{F} \subset C_p(X) separates points of X (for short, the family is point separating). What makes the evaluation map a homeomorphism is that the family \mathcal{F} separates points from closed subsets of X. In this section, we present one property that implies the property of separating points from closed sets. It is clear that if \mathcal{F} is dense in C_p(X), then \mathcal{F} separates points of X. In general, the fact that \mathcal{F} is point separating does not mean it separates point from closed sets. We show that whenever X is compact, the fact that \mathcal{F} is dense in C_p(X) does imply that \mathcal{F} separates points from closed sets.

The family \mathcal{F} \subset C_p(X) is said to be a generating set of functions if it determines the topology of X, i.e., the following set is a base for the topology of X.

    \mathcal{B}_{\mathcal{F}}=\left\{f^{-1}(U) \subset X: f \in \mathcal{F} \text{ and } U \text{ is open in } \mathbb{R} \right\}

Since X is assumed to be a completely regular space, we observe that if \mathcal{B}_{\mathcal{F}} is a base for X, then the family \mathcal{F} separates points from closed subsets of X. The following theorem captures the observations we make.

Theorem 5
Let X be a space. Let \mathcal{F} \subset C_p(X). Then if \mathcal{F} is a generating set of functions, then \mathcal{F} separates points from closed subsets of X, hence the evaluation map E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F}) as defined above is a homeomorphism.

We now show that if X is compact and if \mathcal{F} is dense in C_p(X), then \mathcal{F} separates points from closed subsets of X, making the evaluation map a homeomorphism.

First one definition. Let X be a space. For any finite F=\left\{f_1,\cdots,f_n \right\} consisting of functions in C_p(X), define the maximum of F to be the function f:X \longrightarrow \mathbb{R} such that for each x \in X, f(x) is the maximum of the real values in \left\{f_1(x),\cdots,f_n(x) \right\}. In other words, the maximum of F is the pointwise maximum of the functions in F. It is not too difficult to show that the pointwise maximum of finitely many continuous real-valued functions is also continuous. We have the following lemma and corollary.

Lemma 6
Let the space X be compact. Suppose the family \mathcal{F} is dense in C_p(X) such that the pointwise maximum of any finite set of functions in \mathcal{F} is also in \mathcal{F}. Then \mathcal{F} separates points from closed subsets of X.

Proof of Lemma 6
Let x \in X and let C be a closed subset of X such that x \notin C. For each y \in C, consider the open set:

    U_y=\left\{f \in C_p(X): f(x) \in O_1 \text{ and } f(y) \in O_2 \right\}

where O_1 is the open interval (-0.1,0.1) and O_2 is the open interval (2,\infty). For each y \in C, choose f_y \in \mathcal{F} \cap U_y. The set of all f_y^{-1}(O_2) is an open cover of the compact set C. Choose y_1,y_2,\cdots,y_n \in C such that V_{1},V_{2},\cdots,V_{n} cover C where each V_i=f_{y_i}^{-1}(O_2). Let g:X \longrightarrow \mathbb{R} be the pointwise maximum of \left\{f_{y_1},\cdots, f_{y_n} \right\}. By assumption, g \in \mathcal{F}. It is clear that for all y \in C, 2<g(y). Thus \overline{g(C)} \subset [2,\infty).

Let W=\bigcap_{i=1}^n f_{y_i}^{-1}(O_1). It is also clear that f_{y_i}(x) \in O_1=(-0.1,0.1) for all i, implying g(x) <0.1<1. Thus g(x) \notin \overline{g(C)}. Thus completes the proof that \mathcal{F} separates points from closed subsets of X. \blacksquare

Corollary 7
Let the space X be compact. If \mathcal{F} is a dense subspace of C_p(X), then the evaluation map E_\mathcal{F}: X \longrightarrow C_p(\mathcal{F}) as defined above is a one-to-one map.

In Corollary 7, even if \mathcal{F} is not closed under taking pointwise maximum of finitely many functions, then throw all pointwise maxima of all finite subsets of \mathcal{F} into \mathcal{F} and then apply Lemma 6. Throwing in all pointwise maxima will not increase the cardinality of \mathcal{F}. For example, suppose that X is compact, C_p(X) is separable and \mathcal{F} is a countable dense subspace of C_p(X). Even if \mathcal{F} does not contain all the pointwise maxima of finite subspaces, we can then throw in all pointwise maxima and the subspace \mathcal{F} is still countable. Then the compact space X is homeomorphic to a subspace of C_p(\mathcal{F}). Since C_p(\mathcal{F}) \subset \mathbb{R}^\omega, C_p(\mathcal{F}) is separable and metrizable. Thus the compact space X is separable and metrizable. The following corollary captures this observation.

Corollary 8
If X is a compact space and the function space C_p(X) is separable, then X is metrizable.

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Reference

  1. Arkhangelskii, A. V., Topological Function Spaces, Mathematics and Its Applications Series, Kluwer Academic Publishers, Dordrecht, 1992.

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\copyright \ 2014 \text{ by Dan Ma}

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