# Contents

## Introduction

Recall that a space $X$ is locally compact if each point has a neighborhood basis consisting of compact neighborhoods.

Somewhat less known but slightly more general is the class of spaces called “core-compact spaces”:

###### Definition

A topological space is core-compact if, for every open neighborhood $V$ of a point $x$, there is a smaller open neighborhood $U$ of $x$, such that every open covering of $V$ admits a finite subcover of $U$. We write $U \ll V$ (pronounced “$U$ is well below $V$”) to indicate that this covering condition holds.

Core-compactness is equivalent to the statement that every open $V$ is the union of open sets $U$ such that $U \ll V$.

The goal of this article is to give a succinct proof that core-compact spaces are exponentiable in $Top$, so that in particular locally compact spaces (with no further regularity conditions) are exponentiable in $Top$. This proof is extracted from a nice article by Escardó and Heckmann, who show moreover that exponentiable spaces in $Top$ are precisely the core-compact spaces (which is a matter we won’t go into here).

## Direct description of the exponential topology

For given topological spaces $X$ and $Y$, we let $Top(X, Y)$ be the set of continuous functions $f: X \to Y$. The theory developed by Escardó and Heckmann shows that there is a “best” candidate for an exponential topology on $Top(X, Y)$, which they call the Isbell topology but which is also known as the “natural topology” or the “topology of continuous convergence”. Whatever it is called, it is uniquely characterized by the following property:

• The natural topology is maximal among all topologies $\tau$ on $Top(X, Y)$ with the property that whenever $g: A \times X \to Y$ is continuous, the transform $g^\wedge: A \to Top(X, Y)$ defined by $g^\wedge(a)(x) = g(a, x)$ is continuous with respect to $\tau$.

In the case where $Y$ is the Sierpinski space $\mathbf{2} = \{0, 1\}$ (where $1$ is open but $0$ is not), $Top(X, \mathbf{2})$ is in natural bijection with the set $\mathcal{O}(X)$ of open sets of $X$, i.e., the topology of $X$. Under this isomorphism the natural topology on $Top(X, \mathbf{2})$ corresponds to a topology also known as the Scott topology on $\mathcal{O}(X)$:

###### Definition

A subset $\mathcal{U} \subseteq \mathcal{O}(X)$ is called Scott-open if

• it is upward-closed: if $U \subseteq V$ are open sets of $X$ and $U \in \mathcal{U}$, then $V \in \mathcal{U}$;

• for every $V \in \mathcal{U}$ there exists $U \in \mathcal{U}$ such that $U \ll V$.

Then for arbitrary spaces $Y$, the natural topology on $Top(X, Y)$ may be explicitly described as the topology generated by sets of the form

$F(\mathcal{U}, V) \coloneqq \{X \stackrel{f}{\to} Y: f^{-1}(V) \in \mathcal{U}\}$

where $V$ ranges over open sets of $Y$ and $\mathcal{U}$ over Scott-open subsets $\mathcal{U}$ of $\mathcal{O}(X)$.

We let $Y^X$ denote the space whose underlying set is $Top(X, Y)$ and whose topology is the natural topology thus described. The theorem we prove may be stated as follows:

###### Theorem

If $X$ is core-compact and $Y$ is any space, then the evaluation map $Y^X \times X \to Y$ is continuous, and realizes (via the Yoneda lemma) a natural isomorphism $Top(-, Y^X) \cong Top(- \times X, Y): Top^{op} \to Set$. Thus, $Y^X$ satisfies the universal property of an exponential in $Top$.

As we said, the proof we give is a very whittled-down rendition extracted from the much more thorough account given by Escardó and Heckmann. It is based on two lemmas:

###### Lemma

For any spaces $A, Y$, and $X$ (without assuming $X$ is locally compact), if $g: A \times X \to Y$ is continuous, then its transpose $g^\wedge: A \to Y^X$ is continuous.

###### Lemma

For any space $Y$, the evaluation map $ev: Y^X \times X \to Y$ is continuous if $X$ is core-compact.

It follows that if $h: A \to Y^X$ is continuous, then so is the inverse transpose

$h^\vee = \left(A \times X \stackrel{h \times 1}{\to} Y^X \times X \stackrel{ev}{\to} Y\right).$

As the operations $g \mapsto g^\wedge$ and $h \mapsto h^\vee$ are mutually inverse, this establishes the natural isomorphism $Top(A, Y^X) \cong Top(A \times X, Y)$ of the theorem.

## Proofs of the lemmas

First we establish Lemma 1.

###### Proof

Suppose $g: A \times X \to Y$ is continuous, and let $F(\mathcal{U}, V)$ be a typical generating element of the topology of $Y^X$. We are to show that

$\array{ (g^\wedge)^{-1}(F(\mathcal{U}, V)) & = & \{a: g^\wedge(a) \in F(\mathcal{U}, V)\} \\ & = & \{a: g^\wedge(a)^{-1}(V) \in \mathcal{U}\} }$

is open in $A$. Fix an element $a \in (g^\wedge)^{-1}(F(\mathcal{U}, V))$; then

$g^\wedge(a)^{-1}(V) = \{x: (a, x) \in g^{-1}(V)\}.$

Consider the collection of all pairs $(W', U')$ where $U'$ is open in $X$ and $W'$ is an open neighborhood of $a$ such that $W' \times U' \subseteq g^{-1}(V)$. The collection of such $U'$ covers $g^\wedge(a)^{-1}(V)$. Since $g^\wedge(a)^{-1}(V) \in \mathcal{U}$ and $\mathcal{U}$ is Scott-open, there are finitely many such $U'$, say $U_1, \ldots, U_n$, that cover some $U \in \mathcal{U}$, and we have corresponding open neighborhoods $W_1, \ldots, W_n$ of $a$ such that $W_i \times U_i \subseteq g^{-1}(V)$ for $i = 1, \ldots, n$.

We claim the neighborhood $W = W_1 \cap \ldots \cap W_n$ of the point $a$ is included in $(g^\wedge)^{-1}(F(\mathcal{U}, V))$. Indeed, suppose $b \in W$; we must show $g^\wedge(b)^{-1}(V) \in \mathcal{U}$. But since $\mathcal{U}$ is Scott-open, it is upward closed, and so it suffices to verify that

$U \subseteq g^\wedge(b)^{-1}(V) = \{x: (b, x) \in g^{-1}(V)\}.$

But if $x \in U$, then $x \in U_i$ for some $i$ and of course $b \in W_i$, so $(b, x) \in W_i \times U_i \subseteq g^{-1}(V)$ and the claim is proven.

Now we prove Lemma 2.

###### Proof

Let $f \in Y^X, x \in X$; let $V$ be an open neighborhood of $f(x)$. Then $f^{-1}(V)$ is an open neighborhood of $x$, and since $X$ is core-compact, there is an open neighborhood $U$ of $x$ such that $U \ll f^{-1}(V)$. Put

$\mathcal{U} = \{W\; open\; in \; X: U \ll W\}$

Then it is enough to show

1. $\mathcal{U}$ is Scott-open (and hence $F(\mathcal{U}, V)$ is open in $Y^X$);

2. $(f, x) \in F(\mathcal{U}, V) \times U$;

3. $F(\mathcal{U}, V) \times U \subseteq ev^{-1}(V)$.

The only tricky part is 1. For 2. it is obvious that $f \in F(\mathcal{U}, V)$ since $f^{-1}(V) \in \mathcal{U}$, i.e., $U \ll f^{-1}(V)$. For 3., if $g \in F(\mathcal{U}, V)$ and $x \in U$, then $U \ll g^{-1}(V)$, whence $x \in U \subseteq g^{-1}(V)$ and then $ev(g, x) = g(x) \in V$, as required.

For 1., it is obvious that if $U \ll W$ and $W \subseteq W'$, then $U \ll W'$, so $\mathcal{U}$ is upward-closed. Now we need to check that for all $W'$ in $\mathcal{U}$ there exists $W$ in $\mathcal{U}$ such that $W \ll W'$. By core-compactness,

$W' = \bigcup_{W \ll W'} W = \bigcup_{W \ll W'} \bigcup_{V \ll W} V = \bigcup_{\exists W: V \ll W \ll W'} V$

The covering of $W'$ by open sets $V$ such that $\exists W: V \ll W \ll W'$ is a collection that is closed under finite unions, and so since $U \ll W'$, we have that $U \subseteq V$ for some such $V$. If $U \subseteq V \ll W$, then $U \ll W$, so indeed there is $W \in \mathcal{U}$ such that $W \ll W'$. This completes the proof that $\mathcal{U}$ is Scott-open and the proof of the lemma.

###### Remark

It is not hard to deduce that if $X$ is core-compact, then the collections of the form

$\mathcal{U}_U = \{W: U \ll W\}$

form a base of the Scott topology on $\mathcal{O}(X)$. In that case, the exponential = natural topology on $Y^X$ may be described as generated by sets of the form

$F_{U, V} \coloneqq \{X \stackrel{f}{\to} Y: U \ll f^{-1}(V)\}$

where $U, V$ range over open sets of $X, Y$, respectively. This bears a closer resemblance to the classical compact-open topology than our first abstract description of the natural topology on $Top(X, Y)$ in terms of the Scott topology on $\mathcal{O}(X)$.

## Reference

• Martín Escardó and Reinhold Heckmann, On topologies for general function spaces, 1999. (pdf file)
Created on July 31, 2017 at 11:57:28 by Todd Trimble