The compact-open topology

A.E. Brouwer

2002-02-28

Abstract

Elementary properties of the compact-open topology of function spaces.

1 Introduction

For sets X and Y, the set Y^{X} is the set of all functions from X into Y. If Y is a topological space, then Y^{X} can be given the product topology, where subbasic open sets are {f ∈ Y^{X}|f(a) ∈ U}, for given a ∈ X and open U ⊆ Y.

This is the smallest topology that will make the projections continuous. (The projection on the a-coordinate is evaluation at a, that is, is the map f ↦ f(a).) This topology does not depend on a topology on X.

Now, let X and Y be topological spaces, and let Y^{X} (or C(X, Y)) stand for the set of all maps (continuous functions) from X into Y.

One would like to have a continuous evaluation map Y^{X} × X → Y defined by (f, x) ↦ f(x): when f and x change a little then f(x) changes a little. That is, we want to restrict f not at finitely many points, but at a neighbourhood of a given point.

This is achieved by taking subbasic open sets {f ∈ Y^{X}|f(V) ⊆ U}, with U open in Y and V open in X, but that would be far too strong a topology. For example, it would be nice if f_{a} : R → R defined by f_{a}(x) = ax depended continuously on a, but for this topology f_{0} would be isolated in the collection {f_{a}|a ∈ R} (take V = R and U = (−1, 1)). This shows that one wants to restrict f only at ”small” neighbourhoods of a given point.

It turns out that a theory with nice functorial properties is obtained by letting ”small” mean ”compact” (or, more precisely, ”with compact closure”).

Of course such a definition works especially well in case X is locally compact.

2 The compact-open topology

Let X and Y be topological spaces, and let C(X, Y) = Y^{X} be the space of continuous maps from X into Y, provided with the compact-open topology, where subbasic open sets are W_{K,U} := {f ∈ Y^{X}|f(K) ⊆ U} for compact sets K ⊆ X and open U ⊆ Y.

When X is empty, Y^{X} consists of a single point only. Below we shall assume that X is nonempty.

2.1 Y is embedded in Y^{X}

For each y ∈ Y, let c_{y} : X → Y be the constant map defined by c_{y}(x) = y for all x ∈ X. Now the map y ↦ c_{y} is a homeomorphic embedding of Y in Y^{X}.

(Indeed, y ∈ U if and only if c_{y} ∈ W_{K,U}, for any compact K.)

2.2 Separation

The space Y^{X} is Hausdorff if and only if Y is Hausdorff.

(Indeed, if Y^{X} is Hausdorff then Y, being homeomorphic to a subspace, also is Hausdorff. Conversely, given f, g ∈ Y^{X}, let f(x) ≠ g(x), and let U_{1}, U_{2} be disjoint neighbourhoods of f(x), g(x) in Y. Then W_{{x},U_{1}} and W_{{x},U_{2}} are disjoint neighbourhoods of f and g.)

2.3 Adjunction

Theorem 2.1 Let X, Y and Z be topological spaces. If X and Y are Hausdorff spaces, and Y is locally compact, then Z^{X×Y} and (Z^{Y})^{X} are canonically homeomorphic.

Proof: (We use ”function” for set-theoretic function, and ”map” for continuous function.)

Given a function f : X × Y → Z, let f_{x} : Y → Z be the function defined by f_{x}(y) = f(x, y), and let f̄ be the function from X into the set of maps from Y to Z defined by f̄(x) = f_{x}.

We show that the map f ↦ f̄ is a homeomorphism from Z^{X×Y} onto (Z^{Y})^{X}.

First, we have to check that this map is well-defined. We have to check that if f is continuous then (i) each f_{x} is continuous, and (ii) f̄ is continuous. Part (i) is clear by the definition of product topology. For part (ii): if f̄(x) ∈ W_{K,U} then {x} × K ⊆ f^{−1}(U), and hence, since K is compact and f^{−1}(U) open, there is a neighbourhood V of x such that V × K ⊆ f^{−1}(U), that is, x ∈ V ⊆ f̄^{−1}(W_{K,U}). This shows that f̄ is continuous, so that f ↦ f̄ indeed maps Z^{X×Y} into (Z^{Y})^{X}.

Next, we have to check that this map is a bijection, that is, that continuity of f̄ implies continuity of f, defined by f(x, y) = f̄(x)(y). For fixed x, y, let U be a neighbourhood of f(x, y) in Z. Since f̄(x) is continuous, and Y is locally compact, there is a compact neighbourhood K of y in Y such that f({x} × K) ⊆ U, that is, f̄(x) ∈ W_{K,U}. Since f̄ is continuous there is a neighbourhood V of x in X with x ∈ V ⊆ f̄^{−1}(W_{K,U}), that is, with (x, y) ∈ V × K ⊆ f^{−1}(U). This shows that f is continuous, and hence that f ↦ f̄ is a bijection.

Next, we check that f ↦ f̄ is continuous. We must describe the topology of (Z^{Y})^{X}. To this end first an auxiliary result.

Let S be a topological space and X a Hausdorff space. A subbasis for the topology of S^{X} is given by the sets W_{K,U} where K runs through the compact sets in X and U through a subbasis S of open sets of S.

(Indeed, we have to show that if V is open in S, and f ∈ W_{K,V} then there are finitely many compact K_{i} and U_{i} ∈ S such that f ∈ ⋂ W_{K_{i},U_{i}} ⊆ W_{K,V}. Now since S is a subbasis for the topology of S, we find a basis B by taking finite intersections of elements of S. An arbitrary open set, like V, is a union of elements of B, say V = ⋃_{i∈I} B_{i}. Now shrink the open cover {f^{−1}(B_{i})|i ∈ I} of K, that is, for each k ∈ K choose a neighbourhood (in K) of which the closure K_{k} is entirely contained in some f^{−1}(B_{i}). (That is possible, since a compact Hausdorff space is locally compact.) Finitely many of the K_{k} cover K, so after relabeling we find a finite cover {K_{j}|1 ≤ j ≤ n} of K such that f(K_{j}) ⊆ B_{j} (1 ≤ j ≤ n). Now f ∈ ⋂ W_{K_{j},B_{j}} ⊆ W_{K,V}, and each W_{K,B} is an intersection of sets W_{K,U_{i}} with U_{i} ∈ S.)

OK, so a subbasis for the topology of (Z^{Y})^{X} is given by the sets W_{K,W_{L,V}}, where K is compact in X, L is compact in Y, and V is open in Z. The inverse image of W_{K,W_{L,V}} for the map f ↦ f̄ is the open set W_{K×L,V} in Z^{X×Y}, so our map is indeed continuous (and deserves to be called ”map”).

Finally we have to check that the inverse of our map is continuous. Now a basis for the topology of Z^{X×Y} is formed by the sets W_{K×L,V}, and W_{K×L,V} is mapped to W_{K,W_{L,V}}. This proves everything.

2.4 Composition

Let X, Y and Z be Hausdorff spaces. If Y is locally compact, then the composition map Z^{Y} ×Y^{X} → Z^{X} defined by (f, g) ↦ h with h(x) = f(g(x)), is continuous.

2.5 Evaluation

Let Y and Z be Hausdorff spaces. If Y is locally compact, then the evaluation map Z^{Y} × Y → Z defined by (f, y) ↦ f(y), is continuous.