Could you have invented the compact–open topology on function spaces?

Question

Let $X$ and $Y$ be topological spaces and write $\textrm{Hom} (X, Y)$ for the set of continuous maps $X \to Y$. There are two canonical ways of topologising $\textrm{Hom} (X, Y)$:

• the coarsest topology such that the evaluation map $\epsilon : \textrm{Hom} (X, Y) \times X \to Y$ is continuous, or

• the finest topology such that, for every topological space $T$ and every continuous map $f : T \times X \to Y$, the transposed map $\tilde{f} : T \to \textrm{Hom} (X, Y)$ defined by $\tilde{f} (t) (x) = f (t, x)$ is continuous.

Notice that the first topology is finer than the second topology: writing $\textrm{Hom}' (X, Y)$ and $\textrm{Hom}'' (X, Y)$ for the respective topological spaces, continuity of $\epsilon : \textrm{Hom}' (X, Y) \times X \to Y$ implies the transposed map $\textrm{Hom}' (X, Y) \to \textrm{Hom}'' (X, Y)$ is continuous; but the transposed map is simply the identity map on points, so every open subspace of $\textrm{Hom}'' (X, Y)$ is open in $\textrm{Hom}' (X, Y)$ too.

The exponential topology on $\textrm{Hom} (X, Y)$ is the unique topology with both properties, if such a topology exists. To avoid confusion, write $\textrm{Map} (X, Y)$ for $\textrm{Hom} (X, Y)$ with the exponential topology. The name is justified by the natural bijection so induced: $$\textrm{Hom} (T, \textrm{Map} (X, Y)) \cong \textrm{Hom} (T \times X, Y)$$

We have the following:

Theorem. If $X$ is locally compact (i.e. for every open $U \subseteq X$ and every $x \in U$, there exist a compact $K \subseteq U$ and open $U' \subseteq U$ with $x \in U'$ and $U' \subseteq K$) then the compact–open topology on $\textrm{Hom} (X, Y)$ is exponential.

There is a sharper result, but it relies on a less familar concept. For open subspaces $U'$ and $U$ of a topological space $X$, say $U'$ is way below $U$ and write $U' \ll U$ just if every open cover of $U$ has a finite subcover of $U'$. (It follows that $U' \subseteq U$, of course.) For example, if there is a compact $K \subseteq U$ such that $U' \subseteq K$, then $U' \ll U$.

Theorem. The following are equivalent for a topological space $X$:

• $X$ is core-compact, i.e. for every open $U \subseteq X$ and every $x \in U$, there exists an open $U'$ such that $x \in U'$ and $U' \ll U$.

• For every topological space $Y$, the exponential topology on $\textrm{Hom} (X, Y)$ exists and is the topology generated by the subsets $$\{ f \in \textrm{Hom} (X, Y) : U \ll f^{-1} V \}$$ for all open $U \subseteq X$ and open $V \subseteq Y$.

Question. Is there any reason to expect that local compactness (or core-compactness) plays any role in exponentiability, or having guessed so, that the compact–open topology is the exponential topology?

Taking a few steps back, consider the case where $Y$ is a metric space and we have a sequence of continuous functions $f_0, f_1, f_2, \ldots : X \to Y$. Already in classical analysis there are numerous modes of convergence, e.g.:

• Uniform convergence: $$\forall (\epsilon > 0) \exists (T \ge 0) \forall (t > T) \forall (x \in X) (d (f_t (x), f (x)) < \epsilon)$$

• Local uniform convergence: $$\forall (x \in X) \exists (U \in \textrm{Nbd} (x)) \forall (\epsilon > 0) \exists (T \ge 0) \forall (t > T) \forall (u \in U) (d (f_t (u), f (u)) < \epsilon)$$

• Continuous convergence: $$\forall (\epsilon > 0) \forall (x \in X) \exists (U \in \textrm{Nbd} (x)) \exists (T \ge 0) \forall (t > T) \forall (u \in U) (d (f_t (u), f (u)) < \epsilon)$$

• Compact convergence: $$\forall (K \subseteq X, K \text{ is compact}) ( f_t \to f \text{ uniformly on } K )$$

• Pointwise convergence: $$\forall (\epsilon > 0) \forall (x \in X) \exists (T \ge 0) \forall (t > T) (d (f_t (x), f (x)) < \epsilon)$$

The compact–open topology corresponds to compact convergence, whereas topologising $\textrm{Hom} (X, Y)$ as a subspace of the product $\prod_{x \in X} Y$ corresponds to pointwise convergence. But why is compact convergence the right mode of convergence if we want the exponential law?

---

To give an idea of what I am looking for, here is a story that makes the compact–open topology look plausible and natural.

Suppose we have topologised $\textrm{Hom} (X, Y)$ so that the evaluation map $\epsilon : \textrm{Hom} (X, Y) \times X \to Y$ is continuous. Then:

• For every open $V \subseteq Y$, $\{ (f, x) \in \textrm{Hom} (X, Y) \times X : f (x) \in V \}$ is open.
• Hence for every $x \in X$, $\{ f \in \textrm{Hom} (X, Y) : f (x) \in V \}$ is open.
• Open sets are closed under finitary intersection, so for every finite $K \subseteq X$, $\{ f \in \textrm{Hom} (X, Y) : K \subseteq f^{-1} V \}$ is also open.
• Generalising the above significantly, for every compact $K \subseteq X$, the projection $\textrm{Hom} (X, Y) \times K \to \textrm{Hom} (X, Y)$ is a closed map, so $\{ f \in \textrm{Hom} (X, Y) : K \subseteq f^{-1} V \}$ is open.

Therefore the compact–open topology is coarser than any topology making the evaluation map continuous. This story also shows how it is a generalisation of the product topology. On the other hand, there is no hint here as to why this topology should be compatible with the transposition operation, and to go from here to guessing that $X$ should be locally compact seems tenuous...

Answer 1

Definitely not a complete answer, I haven't really thought about this before so apologies if I say anything trivial or wrong.

We can reason intuitively as follows: say $f_j : X \to Y$ is a convergent sequence of functions, so a convergent sequence of points in $\text{Hom}(X, Y)$ wrt the exponential topology; if you like this means we take $T = \mathbb{N} \cup \{ \infty \}$ in the universal property. What should this mean?

At the very least it should mean that each $f_j(x)$ converges, so each evaluation map $e_x : \text{Hom}(X, Y) \to Y$ should be continuous. So we want something at least as strong as pointwise convergence. If we think of these sequences as each converging at some rate, then intuitively the exponential topology should involve the "rate" of convergence of $f_j(x)$ itself "changing continuously in $x$."

So, what would it mean for the rate of convergence of $f_j(x)$ itself to change continuously in $x$? This means it only changes a little if $x$ changes a little. So this naturally draws our attention to the idea of a small change in $x$, which in the metric space case looks like a small ball, and in general might look like a small compact (i.e. a compact contained in some arbitrarily small open). So metrically we are drawn to locally uniform or compact convergence. Pointwise convergence is insensitive to the topology of $X$, and uniform convergence is "too global"; the rate of convergence should in general be allowed to get bigger if we make bigger changes in $x$, as already happens in the very simple case of a convergent sequence of linear maps $f_j : \mathbb{R}^n \to \mathbb{R}^m$, where we want to recover the Euclidean topology / the topology induced by any norm, including operator norms. Actually I think the operator norm is a good intuitive special case to keep in mind here.

To add some substance to this argument, if $X \cong \text{colim}_i K_i$ is a colimit of compact spaces (is this what compactly generated means?) then the exponential topology must satisfy

$$\text{Hom}(X, Y) \cong \lim_i \text{Hom}(K_i, Y)$$

which says in particular that a sequence $f_j : X \to Y$ should converge iff its restriction to $K_i$ converges for each $i$. And if everything in sight is a metric space, we know that continuous functions $K_i \to Y$ are uniformly continuous and that $\mathbb{N} \cup \{ \infty \}$ is compact, so

$$(\mathbb{N} \cup \{ \infty \}) \times K_i$$

is also compact, hence continuous functions from it to $Y$ are also uniformly continuous. It's been a long time since I've had to think about uniform continuity but I think this at least strongly suggests that the exponential topology on $\text{Hom}(K_i, Y)$ should be uniform convergence, which suggests that on $\text{Hom}(X, Y)$ it should be locally uniform convergence.