# Spaces $Y$ such that $C(-, Y)$ is always acceptable

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.

A topology on $C(X,Y)$ is acceptable if it is both proper and admissible. (It is a good exercise to show that on $C(X,Y)$ there can be at most one acceptable topology.

We call a space $(X,\tau)$ interesting if it is neither discrete nor indiscrete.

Questions:

1) What is an example of an interesting space $Y$ such that for all spaces $Z$ the set $C(Z,Y)$ allows for an acceptable topology?

2) What is an example of an interesting space $X$ such that for all spaces $Z$ the set $C(X,Z)$ allows for an acceptable topology?

• For $2)$ every locally compact space qualifies (with the compact-open topology on C(X,Z) ). If one replaces spaces by Locales then they are the only one (where local compactness means every open $U$ is the union of the opens way below it, so for non hausdorff spacess it is not exactly the usual notion of local quasi-compactness) I don't know if there is a similar result for the category of topological spaces, but one shouldn't expect there is many examples beside this local compactness assumptions. – Simon Henry Jun 20 '16 at 9:30
• Thanks - I didn't think of this point (obviously) and you can well post it as a partial answer.. – Dominic van der Zypen Jun 20 '16 at 9:49
• My knowledge of general topology is basically restricted to locales theory. So I was waiting to see if someone can point out the precise converse statement for topological spaces. – Simon Henry Jun 20 '16 at 10:18

Following up on Simon's comments on 2): in the literature, one of the usual terms for this is that $X$ is exponentiable. There is in fact quite a lot of literature on this. Categorically, one is asking that there be a right adjoint $C(X, -)$ to $X \times -: \text{Top} \to \text{Top}$. As it turns out, this is equivalent to the seemingly weaker condition that for the Sierpinski space $Z = \mathbf{2}$ in particular (where the underlying set of $C(X, \mathbf{2})$ is in natural bijection with the topology of $X$), there is a topology so that $C(X, \mathbf{2})$ represents the functor $\hom_{\text{Top}}(X \times -, \mathbf{2})$: there is a natural isomorphism

$$\hom_{\text{Top}}(-, C(X, \mathbf{2})) \cong \hom_{\text{Top}}(X \times -, \mathbf{2}).$$

(Of course the topology on $C(X, \mathbf{2}) \cong \text{Open}(X)$ is uniquely determined by this universal property.)

Topologically, such $X$ are exactly core-compact spaces. To define this notion, write $V\ll U$ to mean that any open cover of $U$ admits a finite subcover of $V$; this is read as "$V$ is relatively compact under $U$" or "$V$ is way below $U$". We say that $X$ is core-compact if for every open neighborhood $U$ of a point $x$, there exists an open neighborhood $V$ of $x$ with $V\ll U$. In other words, $X$ is core-compact iff for all open subsets $V$, we have $V = \bigcup \{ U | U\ll V \}$.

This description is synonymous with saying: the topology $\text{Open}(X)$ is a continuous lattice, for which there is also a lot of literature, especially in so-called domain theory (after Dana Scott).

Core-compact spaces are a very mild generalization of locally compact spaces (defined here to mean spaces where the neighborhood filter at any point is generated by compact neighborhoods). All locally compact spaces are core-compact. If $X$ is Hausdorff or even merely sober, then $X$ is core-compact iff it is locally compact (along the lines Simon was suggesting, if we consider spatial locales), according to Theorem 8.3.10 in Non-Hausdorff Topology and Domain Theory by Goubault-Larrecq (partially viewable in Google Books).

It seems to be surprisingly hard to cook up (or even find in the literature!) a core-compact space that is not locally compact, but you can find such an example in this paper by Hofmann and Lawson (section 7).

Another fine compendium of results is Topologies on Spaces of Continuous Functions by Escardó and Heckmann (I believe it addresses aspects of your other recent question on admissible topologies on function spaces, but I've not had time to look into this carefully.)

As for 1), I'm currently somewhat skeptical that a nice answer can be given, but here again I've not had time to investigate this properly.

• For $1)$ my guess was that no such space exists (except the point and more generally indiscrete space of course), which would qualifies as a rather nice answer to the question. Do you have any reasons to think that there exists more examples ? – Simon Henry Jun 21 '16 at 9:49
• @SimonHenry No, that was my guess too; I just didn't express it as I should have. – Todd Trimble Jun 21 '16 at 10:50