Re: Compact open topology



sanchopancho80@xxxxxx writes:

Yes, that's what I wanted to say: The TOP map hom(X x Y, Z) --->
hom(Y, hom(X,Z)) does not EXIST in general in TOP, because it is not
continuous, right?

For a category C, I denote by C(X,Y) the set of morphisms from X to
Y. So in Top, the set of continuous maps from X to Y is denoted by
Top(X,Y). Of course, Top(X,Y) is included in Set(X,Y). And from now
on, hom(X,Y) is the set Top(X,Y) equipped with the compact-open
topology.

The only set map which always exists is Top(XxY,Z)-->Set(Y,Set(X,Z)).
If X is locally compact, the image of the set map
Top(XxY,Z)-->Set(Y,Set(X,Z)) is included in Set(Y,Top(X,Z)), and even
in Top(Y,Hom(X,Z)). In categorical language, Xx-:Top-->Top admits
Hom(X,-):Top-->Top as right adjoint. So if X is locally compact, there
is a natural set bijection Top(XxY,Z)=Top(Y,Hom(X,Z)).

The next step is to prove that this set bijection induces a
homeomorphism Hom(XxY,Z)=Hom(Y,Hom(X,Z)). I don't think that it is
true in the category of general topological spaces (in fact I don't
know whether it is true). If both X and Y are locally compact, then
XxY is locally compact. And then, here is a proof: pick a topological
space T. Then Top(T,Hom(XxY,Z)) = Top(TxXxY,Z) = Top(TxX,Hom(Y,Z)) =
Top(T,Hom(X,Hom(Y,Z))). Apply the Yoneda Lemma: one obtains the
homeomorphism Hom(XxY,Z) = Hom(X,Hom(Y,Z)).

pg.
.



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