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Is there a natural reason for defining the compact-open topology on the set of continuous functions between two locally compact spaces. For example "to make ... functions continuous". Or in another way of asking this, is there an adjoint functor of the functor, say F, which assigns the topological space $F(X,Y):=Hom_C(X,Y)$ (with the compact-open topology on it) to the couple X,Y.

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On natural reason is that with that topology things that we want to converge converge, and things that we want to not converge do not converge. It is a very natural abstraction of convergence notions lik e "uniform convergence on compact sets" and others which show up all over the place in nature. One can probably wrap the topology in the elegant clothes of a functorial explanation (this is done in Munkres, iirc) but the real reason the topology is useful is in my first sentence. –  Mariano Suárez-Alvarez Oct 31 '10 at 17:56
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One: this makes composition a continuous map. Two (this is sort of the same as one): this functor is right adjoint to the product in the compactly-generated category, i.e. $\operatorname{Hom}(X\times Y, Z)\simeq \operatorname{Hom}(X, \operatorname{Hom}(Y, Z))$. –  Daniel Litt Oct 31 '10 at 18:25

3 Answers 3

up vote 6 down vote accepted

In regard to your question I recommend Topologies on spaces of continuous functions, Topology Proceedings, volume 26, number 2, pp. 545-564, 2001-2002 by Martin Escardo and Reinhold Heckmann.

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I always liked the following reason:

Let's call a topology on a space "admissible" if the evaluation function $e: Hom(X,Y) \times X \rightarrow Y$ is continuous. Then the compact-open topology is coarser than any other admissible topology. In particular, in any case where the compact-open topology is admissible, it is the smallest possible topology that does this.

EDIT: See comments for some references. I don't claim any originality here :)

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This reason which you like, including the word "admissible", appears in one of the first papers on the compact-open topology. See R. Arens, A Topology for Spaces of Transformations, Annals of Math. 47 (1946), 480--495. –  KConrad Oct 31 '10 at 20:51
    
Yes- sorry I forgot to put the reference. I got it from the book "Algebraic topology from a homotopical viewpoint" –  Dylan Wilson Nov 1 '10 at 1:56
    
Is there a connection to the notion of an admissible representations? –  plusepsilon.de Nov 19 '10 at 17:50
    
I have absolutely no idea- but it seems to me like there isn't. –  Dylan Wilson Nov 19 '10 at 20:33

Exactly as you say, adjoint functors are the answer! (Or at least, they're one possible answer.) In particular, for reasonable spaces $X,Y,Z$, there is a natural isomorphism

$\mathrm{Hom} (X \times Y, Z) \cong \mathrm{Hom} (X, [Y,Z])$

where $[Y,Z]$ denotes $\mathrm{Hom}(Y,Z)$ with the compact-open topology. This is exactly the categorical characterisation of an exponential object.

This certainly holds when $X,Y,Z$ are compactly-generated Hausdorff spaces, so the category of such spaces is Cartesian closed. It actually also holds under rather weaker conditions on $X,Y$ and $Z$ individually — I can't recall them off the top of my head though and am in a bit of a rush, so am wiki'ing this answer in hopes someone can fill them in. Now, this also holds if we take weaker conditions on $X$ and $Y$ (just Hausdorffness), and a slightly stronger condition on $Y$ (local compactness). $Z$ may be arbitrary.

Re Mariano's comment: yes, in some sense this is just fancy language for “things we want to converge, converge; things we don't, don't”. But I think this helps explain why we want the things we want to converge, to converge. ☺

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OR: this explains why we turn to the compactly generated $T_2$ spaces: they are nice enough to behave when we use the topology we want to use :) –  Mariano Suárez-Alvarez Oct 31 '10 at 18:38
    
Well, they're nice enough to behave when we use some topology — we can appreciate that even if we didn't decide on this one in advance! I don't know of any other topology on the function space which works well with a such a wide class of spaces? But IANAToplogist, so there may be other options I'm ignorant of… –  Peter LeFanu Lumsdaine Oct 31 '10 at 18:51
    
You just need compactly generated, no separation axioms- but the proof of this is more difficult. –  David Carchedi Oct 31 '10 at 19:12
    
Well, from what I've seen if you give $Map(X,Y)$ the limit topology of $Map(K,Y)$- each with its compact-open topology, ranging over all maps $K \to X$ with compact Hausdorff domain. Whether or not this agrees with the compact open on $Map(X,Y)$, I'd have to check. –  David Carchedi Oct 31 '10 at 19:14
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Also, I'm not sure, but I think you may have to take the k-ification of the function spaces (and of the product, certainly) if you want to stay in the category of compactly generated Hausdorff spaces. –  Dylan Wilson Oct 31 '10 at 19:32

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