# 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. 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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$\copyright \ 2014 \text{ by Dan Ma}$