Admissible and proper topologies on $C(X,Y)$

Question

Given non-empty sets $A, B, C$, set $B^A$ to be the set of all functions $f:A\to B$ there is a natural bijection $\Lambda: C^{A\times B} \to (C^A)^B$ defined in the following way: for $f:A\times B \to C$ let $\Lambda(f):B\to C^A$ be defined by $[\Lambda(f)(b)](a) = f(a,b) \in C$.

Let $X, Y$ be topological spaces; we denote by $C(X,Y)$ the collection of continuous maps $f: X\to Y$.

A topology on $C(X,Y)$ is said to be admissible if for any space $Z$ and any continuous map $g: Z\to C(X,Y)$ the map $\Lambda^{-1}(g): Z\times X \to Y$ is continuous (where $Z\times X$ carries the product topology.

Dually, a topology on $C(X,Y)$ is said to be proper if for any space $Z$ and any continuous map $g\in C(Z\times X, Y)$ the map $\Lambda(g): Z\to C(X,Y)$ is continuous.

It turns out that a topology on $C(X,Y)$ is admissible if and only if the evaluation map $\text{eval}: C(X,Y)\times X \to Y$ defined by $(f,x)\mapsto f(x)$ is continuous. (This is not difficult to prove.)

Is there a similar criterion for proper topologies on $C(X,Y)$? (By "similar" I mean a criterion not involving $Z$, and possibly the evaluation map.)

Answer 1

First, another name for a proper topology on a function space is splitting topology, and other names for an admissible topology are approximating topology and conjoining topology. This may be of aid in further Google searches.

Here are some basic facts:

• Any topology that is coarser (smaller) than a proper topology is also proper. Any topology that is finer (larger) than an admissible topology is also admissible. (The proofs are easy.)

• Any proper topology on $C(X, Y)$ is coarser than any admissible topology on $C(X, Y)$. See Lemma 2.2 here. It follows that any proper topology is contained in the intersection of all admissible topologies.

• The intersection of all admissible topologies is a proper topology, called the natural topology on $C(X, Y)$. (This is subsumed under Proposition 5.13 that is cited below, but see also here.)

It follows that the natural topology is the unique maximal topology among all proper topologies: a topology is proper iff it is coarser than the natural topology. The natural topology is in general not admissible; it is admissible iff $C(X, Y)$ is an exponential object (i.e., represents the functor $\hom_{\text{Top}}(- \times X, Y): \text{Top} \to \text{Set}$ in the usual sense of category theory: there is an isomorphism $\hom_{\text{Top}}(-, C(X, Y)) \cong \hom_{\text{Top}}(- \times X, Y)$, induced in Yoneda-wise fashion from a map $C(X, Y) \times X \to Y$ which is the evaluation map).

So a reasonable answer to the question is given by understanding a little more concretely the natural topology on $C(X, Y)$. It may be described in terms of filter convergence as follows: let us say that a filter $\Phi$ on $C(X, Y)$ converges continuously to $f \in C(X, Y)$ if, whenever a filter $\mathcal{G}$ on $X$ contains the filter of neighborhoods of $b \in X$, then the filter on $Y$ generated by the filter base $\{F(G): F \in \Phi, G \in \mathcal{G}\}$ contains the filter of neighborhoods of $f(b)$, where we put $F(G) := \bigcup \{f(G): f \in F\}$. As usual when we have a filter convergence relation, we go on to define $U \in C(X, Y)$ to be open with respect to continuous convergence if $U$ belongs to any filter $\Phi$ that converges to a point in $U$.

Theorem (see Proposition 5.13 here): For any spaces $X, Y$, the topology of continuous convergence on $C(X, Y)$ coincides with the natural topology on $C(X, Y)$.

In summary: a topology $T$ on a function space $C(X, Y)$ is proper iff every $T$-open is also an open of every topology for which the evaluation map is continuous; equivalently, is an open with respect to continuous convergence.

Answer 2

This question should also be seen in the light of the notion of a category which is adequate and convenient for the purposes of topology, see the discussion on convenient categories of topological spaces on the ncatlab.