# nLab compact-open topology

Contents

### Context

#### Mapping space

internal hom/mapping space

# Contents

## Idea

A natural topology on mapping spaces of continuous functions, important because of its role in exhibiting locally compact topological spaces to be exponentiable, as demonstrated below, culminating in Corollary .

The compact-open topology on the set of continuous functions $X \to Y$ is generated by the subbasis of subsets $U^K \subset C(X,Y)$ that map a given compact subspace $K \subset X$ to a given open subset $U \subset Y$, whence the name.

When restricting to continuous functions between compactly generated topological spaces one usually modifies this definition to a subbase of open subsets $U^{\phi(K)}$, where now $\phi(K)$ is the image of a compact topological space under any continuous function $\phi \colon K \to X$. This definition gives a cartesian internal hom in the category of compactly generated topological spaces (see also at convenient category of topological spaces).

The two definitions agree when the domain $X$ is a compactly generated Hausdorff space, but not in general. Beware that texts on compactly generated spaces nevertheless commonly say “compact-open topology” for the second definition.

## Definition

###### Notation

Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. We denote by $Y^{X}$ the set of continuous maps from $(X,\mathcal{O}_{X})$ to $(Y, \mathcal{O}_{Y})$.

###### Remark

Other common notations for $Y^{X}$ are $Map(X,Y)$ or $C(X,Y)$.

###### Notation

Let $(X, \mathcal{O}_{X})$ be a topological space. We denote by $\mathcal{O}^{c}_{X}$ the set of subsets of $X$ which are compact with respect to $\mathcal{O}_{X}$.

###### Notation

Let $(X, \mathcal{O}_{X})$ be a topological space. Let $A$ be a subset of $X$. We denote by $\overline{A}$ the topological closure of $A$ in $X$ with respect to $\mathcal{O}_{X}$.

###### Definition

A topological space $(X, \mathcal{O}_{X})$ is locally compact if the set of compact neighbourhoods of points of $X$ form a neighbourhood basis of $X$. That is to say: for every $x \in X$ and every neighbourhood $N$ of $x$, there is a $V \in \mathcal{O}^{c}_{X}$ and a $U \in \mathcal{O}_{X}$ such that $x \in U \subset V \subset N$.

###### Remark

There are many variations on this definition, which can be found at locally compact space. These are all equivalent if $(X, \mathcal{O}_{X})$ is Hausdorff. We do not however make the assumption that $(X, \mathcal{O}_{X})$ is Hausdorff.

###### Notation

Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. Given $A \in \mathcal{O}^{c}_{X}$ and $U \in \mathcal{O}_{Y}$, we denote by $M_{A,U}$ the set of continuous maps $f : X \rightarrow Y$ such that $f(A) \subset U$.

###### Definition

Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. The compact-open topology on $Y^{X}$ is that with sub-basis given by the set of sets $M_{A,U}$ such that $A \in \mathcal{O}^{c}_{X}$ and $U \in \mathcal{O}_{Y}$.

###### Notation

Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. We shall denote the compact-open topology (def. ) on $Y^{X}$ by $\mathcal{O}_{Y^{X}}$.

## Exponentiability

###### Proposition

Let $(X, \mathcal{O}_{X})$ be a locally compact topological space, def. , and let $(Y, \mathcal{O}_{Y})$ be a topological space. The map $ev : X \times Y^{X} \rightarrow Y$ given by $(x,f) \mapsto f(x)$ is continuous, where $X \times Y^{X}$ is equipped with the product topology $\mathcal{O}_{X \times Y^{X}}$ with respect to $\mathcal{O}_{X}$ and $\mathcal{O}_{Y^{X}}$ (notation ).

###### Proof

By an elementary fact concerning continuous maps, it suffices to show that for any $(x,f) \in X \times Y^{X}$, and any $U \in \mathcal{O}_{Y}$ such that $f(x) \in U$, there is a $U' \in \mathcal{O}_{X \times Y^{X}}$ such that $(x,f) \in U'$, and such that $ev(U') \subset U$.

To demonstrate this, we make the following observations.

1) Since $f$ is continuous, we have that $f^{-1}(U) \in \mathcal{O}_{X}$.

2) Since $(X, \mathcal{O}_{X})$ is locally compact, we deduce from 1) that there is a $V \in \mathcal{O}^{c}_{X}$ and a $U'' \in \mathcal{O}_{X}$ such that $x \in U'' \subset V \subset f^{-1}(U)$.

3) By 2) and by definition of $\mathcal{O}_{X \times Y^{X}}$ and $\mathcal{O}_{Y^{X}}$, we have that $U'' \times M_{V, U} \in \mathcal{O}_{X \times Y^{X}}$, and that $(x,f) \in U'' \times M_{V, U}$.

4) Let $(x',f') \in U'' \times M_{V, U}$, Since $x' \in U''$, and since $f'(U'') \subset f'(V) \subset U$, the latter inclusion holding by definition of $M_{V, U}$, we have that $f'(x) \in U$. We deduce that $ev(U'' \times M_{ V, U}) \subset U$.

By 3) and 4), we see that we can take $U'$ as the beginning of the proof to be $U'' \times M_{V , U}$.

###### Proposition

Let $(X, \mathcal{O}_{X})$, $(Y, \mathcal{O}_{Y})$, and $(Z, \mathcal{O}_{Z})$ be topological spaces. Let $f : X \times Y \rightarrow Z$ be a continuous map, where $X \times Y$ is equipped with the product topology $\mathcal{O}_{X \times Y}$ with respect to $\mathcal{O}_{X}$ and $\mathcal{O}_{Y}$. Then the map $\alpha_{f} : X \rightarrow Z^{Y}$ given by $x \mapsto f_{x}$ is continuous, where $f_{x} : Y \rightarrow Z$ is given by $y \mapsto f(x,y)$, and where $Z^{Y}$ is equipped with the compact-open topology $\mathcal{O}_{Z^{Y}}$.

###### Proof

By an elementary fact concerning continuous maps, it suffices to show that for any $x \in X$, and any $M_{A,U} \in \mathcal{O}_{Z^{Y}}$ such that $f_{x} \in M_{A,U}$, there is a $U' \in \mathcal{O}_{X}$ such that $x \in U'$, and such that $\alpha_{f}( U' ) \subset M_{A,U}$.

To demonstrate this, we make the following observations.

1) Since $f_{x} \in M_{A,U}$, we have, by definition of $f_{x}$ and by definition of $M_{A,U}$, that $f(x,a) \in U$ for all $a \in A$.

2) Since $f$ is continuous, we have that $f^{-1}(U) \in \mathcal{O}_{X \times Y}$.

3) By 1), we have that $\{x\} \times A \subset f^{-1}(U)$.

4) Let $\mathcal{O}_{X \times A}$ denote the subspace topology on $X \times A$ with respect to $\mathcal{O}_{X \times Y}$. By 2), we have that $f^{-1}(U) \cap (X \times A) \in \mathcal{O}_{X \times A}$.

5) By 3), we have that $\{x \} \times A \subset f^{-1}(U) \cap (X \times A)$.

6) Since $A \in \mathcal{O}^{c}_{Y}$, it follows from 4), 5), and the tube lemma that there is a $U' \in \mathcal{O}_{X}$ such that $x \in U'$ and such that $U' \times A \subset f^{-1}(U) \cap (X \times A) \subset f^{-1}(U)$.

7) We deduce from 6) that $f(U' \times A) \subset U$. This is the same as to say that $f_{x'}(a) \in U$ for all $x' \in U'$. Thus $f_{x'}$ belongs to $M_{A,U}$ for all $x' \in U'$, which is the same as to say that $\alpha_{f}(U') \subset M_{A,U}$.

We conclude that we can take the required $U'$ of the beginning of the proof to be the $U'$ of 6).

###### Proposition

Let $(X, \mathcal{O}_{X})$, $(Y, \mathcal{O}_{Y})$, and $(Z, \mathcal{O}_{Z})$ be topological spaces. Suppose that $(Y, \mathcal{O}_{Y})$ is locally compact. Let $f : X \rightarrow Z^{Y}$ be a continuous map, where $Z^{Y}$ is equipped with the compact-open topology $\mathcal{O}_{Z^{Y}}$. Then the map $\alpha^{-1}_{f} : X \times Y \rightarrow Z$ given by $(x,y) \mapsto \big( f(x) \big)(y)$ is continuous, where $X \times Y$ is equipped with the product topology $\mathcal{O}_{X \times Y}$.

###### Proof

We make the following observations.

1) We have that $\alpha^{-1}_{f} = ev \circ \tau \circ (f \times id)$, where $\tau : Z^{Y} \times Y \rightarrow Y \times Z^{Y}$ is given by $(f,y) \mapsto (y,f)$.

2) By Proposition , we have that $ev$ is continuous.

3) It is an elementary fact that $\tau$ is continuous.

4) Since $f$ is continuous, it follows by an elementary fact that $f \times id$ is continuous.

5) We deduce from 2) - 4) that $ev \circ \tau \circ (f \times id)$ is continuous. By 1), we conclude that $\alpha^{-1}_{f}$ is continuous.

###### Corollary

Let $(X, \mathcal{O}_{X})$ and $(Y, \mathcal{O}_{Y})$ be topological spaces. Suppose that $(X, \mathcal{O}_{X})$ is locally compact. Then $(Y^{X}, \mathcal{O}_{Y^{X}})$ together with the corresponding map $ev$ defines an exponential object in the category Top of all topological spaces.

###### Proof

Follows immediately from Proposition , Proposition , and the fact that $Y^{X}$ and $ev$ are exhibited by the corresponding maps $\alpha_{-}$ and $\alpha^{-1}_{-}$ of Proposition and Proposition to define an exponential object in the category $\mathsf{Set}$ of sets.

###### Remark

A proof can also be found in Aguilar-Gitler-Prieto 02, prop. 1.3.1, or just about any half-decent textbook on point-set topology!

However, the result is almost universally stated with an assumption that $(X,\mathcal{O}_{X})$ is Hausdorff which, as the proof we have given illustrates, is not needed.

###### Remark

We moreover have a homeomorphism $Map(X,Map(Y,Z))\cong Map(X\times Y,Z)$ if in addition $X$ is Hausdorff. See also convenient category of topological spaces.

## Examples

### Maps out of the point space

###### Example

(mapping space construction out of the point space is the identity)

The point space $\ast$ is clearly a locally compact topological space. Hence for every topological space $(X,\tau)$ the mapping space $Maps(\ast, (X,\tau))$ exists. This is homeomorphic to the space $(x,\tau)$ itself:

$Maps(\ast, (X,\tau)) \simeq (X,\tau) \,.$

### Maps into metric spaces

If $Y$ is a metric space then the compact-open topology on $Map(X,Y)$ is the topology of uniform convergence on compact subsets in the sense that $f_n \to f$ in $Map(X,Y)$ with the compact-open topology iff for every compact subset $K\subset X$, $f_n \to f$ uniformly on $K$. If (in addition) the domain $X$ is compact then this is the topology of uniform convergence.

### Loop spaces and path spaces

###### Example

(loop space and path space)

Let $(X,\tau)$ be any topological space.

1. The standard circle $S^1$ is a compact Hausdorff space hence a locally compact topological space. Accordingly the mapping space

$\mathcal{L} X \coloneqq Maps( S^1, (X,\tau) )$

exists. This is called the free loop space of $(X,\tau)$.

If both $S^1$ and $X$ are equipped with a choice of point (“basepoint”) $s_0 \in S^1$, $x_0 \in X$, then the topological subspace

$\Omega X \subset \mathcal{L}X$

on those functions which take the basepoint of $S^1$ to that of $X$, is called the loop space of $X$, or sometimes based loop space, for emphasis.

2. Similarly the closed interval is a compact Hausdorff space hence a locally compact topological space (def. ). Accordingly the mapping space

$Maps( [0,1], (X,\tau) )$

exists. Again if $X$ is equipped with a choice of basepoint $x_0 \in X$, then the topological subspace of those functions that take $0 \in [0,1]$ to that chosen basepoint is called the path space of $(X\tau)$:

$P X \subset Maps( [0,1], (X,\tau) ) \,.$

Notice that we may encode these subspaces more abstractly in terms of universal properties:

The path space and the loop space are characterized, up to homeomorphisms, as being the limiting cones in the following pullback diagrams of topological spaces:

1. $\array{ \Omega X &\longrightarrow& Maps(S^1, (X,\tau)) \\ \downarrow &(pb)& \downarrow^{\mathrlap{Maps(const_{s_0}, id_{(X,\tau)})}} \\ \ast &\underset{const_{x_0}}{\longrightarrow}& X \simeq Maps(\ast,(X,\tau)) } \,.$
2. $\array{ P X &\longrightarrow& Maps([0,1], (X,\tau)) \\ \downarrow &(pb)& \downarrow^{\mathrlap{Maps(const_x, id_{(X,\tau)})}} \\ \ast &\underset{const_{x_0}}{\longrightarrow}& X \simeq Maps(\ast,(X,\tau)) }$

Here on the right we are using that the mapping space construction is a functor and we are using example in the identification on the bottom right mapping space out of the point space.

Textbook accounts: