There is a close relationship between Frölicher spaces and the subcategory of , the Isbell envelope of , of those objects satisfying Isbell duality.
An object of that satisfies Isbell duality is a Frölicher space.
Let be a generalised -object satisfying Isbell duality. From the page about the Isbell envelope, is concrete. Let denote the set of constant elements of . By concreteness, injects into and injects into . For clarity, we shall distinguish between an element of and its image in writing for the former and for the latter. We shall do similarly for elements of .
Suppose that is such that for all . Then the map is a natural transformation (it obvious commutes with the left -actions). Hence it is an element of . That is, there is an element of such that for all .
Let us compare with . Let us put as the constant function at then for all ,
But if and differ, then they differ at some . Now for , as satisfies Isbell duality, there must be some such that . Hence and so is in the image of in .
For the other part, let be such that for all . Then, exactly as above, we define a natural transformation by . This corresponds to some and it remains to compare with . This is simpler since is an element of and so .
However, not all Frölicher spaces can be obtained in this manner. The simplest example is the following:
where this is taken to mean that all curves are smooth and only the constant functionals are smooth. The problem here is that there are far more curves than the functionals warrant. Put another way, the functionals cannot distinguish between the points of the set.
All Frölicher spaces satisfy half of the requirements for Isbell duality: the functions are always the natural transformations of the curves.
The object of corresponding to a Frölicher space is -saturated.
Let be a Frölicher space. Recall from the page about the Isbell envelope that -saturated means that is precisely the set of -homs .
We know that every element of gives a map which commutes with the right actions. As contains the constant functions at the points of , this assignment is injective.
Let commute with the right actions. Define by where is the constant function at . Then for , . Hence for all and so is in .
Hence is -saturated.
The other half is more complicated. Using concreteness one can see that the essence depends on comparing the set with the set of constant natural transformations . That is, those natural transformations with the property that is a constant function in for all .
A constant natural transformation is a function with the property that .
Let us write for this set.
Now every point of defines a constant natural transformation via evaluation but the map need be neither injective nor surjective. However, we can determine conditions on when it is injective or surjective. Injectivity is related to a fairly simple condition (as indicated by the example earlier).
A Frölicher space is Hausdorff if and only if the map is injective.
The key point here is that an element of is completely determined by its effect on functions in . Thus are such that they define the same element of if and only if for all . This means that the smooth functions do not separate and . Hence is Hausdorff if and only if is injective.
It is simple to construct non-Hausdorff Frölicher spaces. Indeed, the example earlier was one.
Surjectivity is more complicated. As currently stated, not even very simple Frölicher spaces satisfy the surjectivity condition.
The Frölicher space defined by the usual structure on does not satisfy the surjectivity condition.
We need to construct a map such that but which does not correspond to a point in .
A non-zero point defines an element of by composing orthogonal projection with the map defined by . If are not collinear then the only situation in which is if and are constant functions. To see this, let be orthogonal to and orthogonal to be such that the orthogonal projection of to the line spanned by is , and similarly maps to . Then and span and
As this holds for all we see that and are constant functions.
Moreover, the functions are “initial” in in the sense that if for some then is of the form . We thus conclude that in defining a map such that we have free choice on the values . However, the maps which come from evaluation do not have this free choice: their value on is completely determined by the values on and .
However, all is not lost. The set of functions in a Frölicher space has much more structure than simply composition by functions from .
The set of functions in a Frölicher space is a commutative -algebra.
Let be a Frölicher space. Let . Then , , and lie in . Thus as is a ring,
is in . As this holds for all , . It is commutative because is commutative. Finally we note that there is an obvious ring homomorphism sending to the function .
This suggests that we should consider a Frölicher space not as a pair of functors but as a pair of functors and .
There is yet more structure on . Not only can we compose element of with elements of but if is a particular element then we can compose with an element of . This suggests that we ought to enlarge the category so that its objects are the power set .
(An alternative to this extension is to insist that the natural transformations be continuous with respect to a topology on compatible with the compact-open topology on .)
With these two augmentations, a constant natural transformation from , with , defines an algebra homomorphism with the property that for all .
Such a natural transformation, , has the property that for any pair there is a point such that and .
Consider the function
As is an algebra homomorphism, . Since , there is thus some such that . For this we therefore have that and .
This clearly extends to any finite family. Indeed, from the proof of this result we deduce that the family
is directed and thus defines a filter on . If our natural transformation is not represented by a point on then this filter will have empty intersection.
Now consider the subfamily of consisting of those functions with the property that and . It is clear that if we restrict to this family then we still get all of . We can order this family; there are two equivalent (but not isomorphic) orderings:
This family is directed (downwards) since (and ). In any reasonable topology on then this net converges to the zero function: on any compact subset of then we have uniformly. Therefore if we simply add the condition that our natural transformation be continuous (something we might have been ready to do anyway) we see that it must be represented by an element of since otherwise we have .
Thus is the set of continuous algebra homomorphisms and we finally see the relationship between Hausdorff Frölicher spaces as objects in the Isbell envelope of satisfying Isbell duality.