Re: Cartesian closed without products

Re: Cartesian closed without products

[Toby Bartels]

Richard Garner wrote:

>On reflection, [...]
>EK-closed requires the unit object to be representable as
>well.

This is a reasonable requirement anyway,
since the unit is the 0fold iterated hom,
just as it's the 0fold iterated product.
So if you think <closed under binary products>
should be accompanied by <with a unit object>,
then so should <closed under internal homs>.

OTOH, the version without unit probably has some use,
just as non-monoid semigroups occasionally do.


--Toby

Re: Cartesian closed without products

[Toby Bartels]

Peter wrote:

>Toby wrote:

>>This is a reasonable requirement anyway,
>>since the unit is the 0fold iterated hom,
>>just as it's the 0fold iterated product.

>What is the "0fold iterated" of a non-associative operation?

OK, admittedly it's a bit ambiguous,
and I should not have used the adverb "just".

But given a finite list (X_i : i < n) of n sets (or other objects),
the n-fold iterated hom may be defined by recursion on n as follows:
* The 0fold iterated hom of the empty list is the unit object; and
* the (n + 1)fold iterated hom of the list (X_i : i < n + 1)
  is hom[Y,X_n], where Y is the n-fold iterated hom of (X_i : i < n).
Of course, you might use hom[X_n,Y] instead,
but (even if you change the base case) this will not work.
It would work if you make the 1fold version the base case,
so there are (at least) two notions of iterated hom,
but only one of these has a 0fold case.

If you use exponential notation for the hom,
then the n-fold iterated hom has a well-known notation
$ X_{n-1} ^ {X_{n-2} ^ {X_{n-3} ^ {\rddots ^ {X_2 ^ {X_1 ^ {X_0}}}}}} $
(in TeX, where |rddots| is the reverse of the standard |ddots|).
The version using hom[X_n,Y], although written in a more pleasant order,
required parentheses (or can be rewritten using multiplication).
On the other hand, logicians (using implication for the hom)
require parentheses for my iterated hom.

Note that the 0fold case relies on the property hom[1,X] = X,
so a ~general~ nonassociative operation doesn't have a 0fold iteration.
A nonassociative operation both distinct left and right units
would have two (a priori) equally justified 0fold iterations.


--Toby

Re: Cartesian closed without products

[Peter Selinger]

Toby Bartels wrote:
>
> Richard Garner wrote:
>
> >On reflection, [...]
> >EK-closed requires the unit object to be representable as
> >well.
>
> This is a reasonable requirement anyway,
> since the unit is the 0fold iterated hom,
> just as it's the 0fold iterated product.
> So if you think <closed under binary products>
> should be accompanied by <with a unit object>,
> then so should <closed under internal homs>.
>
> OTOH, the version without unit probably has some use,
> just as non-monoid semigroups occasionally do.

What is the "0fold iterated" of a non-associative operation?

-- Peter

Re: Cartesian closed without products

[Ronnie]

Toby Bartels wrote:

>
> OTOH, the version without unit probably has some use,
> just as non-monoid semigroups occasionally do.

Just a thought that this interest may be more than occasional:
inverse semigroups are closely related to pseudogroups and so are part of
Charles Ehresmann local-to-global view and interests, with a deep relation
to ordered groupoids (see further work by Mark Lawson). So all these partial
units express something on the *local* structure, even if there is a global
unit.  My paper with Aof (based on ideas of Pradines) uses an inverse
semigroup generated by continuous local admissible sections.
(with M. E.-S. A.-F. AOF), ``The holonomy groupoid of a
locally  topological groupoid'', {\em Top. and its  Appl.}, 47 (1992)
97-113.

Is this a rash view that one should look more at inverse categories?

Ronnie Brown

Re: Cartesian closed without products

[Richard Garner]

A minor correction:

> A small category C has finite products just when the
> representables in [C^op, Set] are closed under finite
> products. It is cartesian closed just when the
> representables are closed under finite products and
> internal hom.
>
> It seems natural, therefore, to consider a notion of
> "cartesian closedness without finite products": categories
> in which the representables are closed in [C^op, Set] under
> internal hom but not necessarily under finite products.
> This amounts to giving, for each pair of objects X and Y,
> an object [X, Y] together with a universal natural
> transformation C(-, [X, Y]) x C(-, X) -> C(-, Y). Such
> categories will be closed in the sense of Eilenberg-Kelly
> without necessarily being monoidal:

On reflection, this bit isn't necessarily true, since
EK-closed requires the unit object to be representable as
well.

Richard

Cartesian closed without products

[Richard Garner]

Dear categorists,

A small category C has finite products just when the
representables in [C^op, Set] are closed under finite
products. It is cartesian closed just when the representables
are closed under finite products and internal hom.

It seems natural, therefore, to consider a notion of
"cartesian closedness without finite products": categories in
which the representables are closed in [C^op, Set] under
internal hom but not necessarily under finite products. This
amounts to giving, for each pair of objects X and Y, an
object [X, Y] together with a universal natural
transformation C(-, [X, Y]) x C(-, X) -> C(-, Y). Such
categories will be closed in the sense of Eilenberg-Kelly
without necessarily being monoidal: let us call them
"universally closed" for now.

Obviously, any cartesian closed category is universally
closed; and categorical proof theory gives us a class of
non-degenerate examples built from the syntax of (classical)
sequent calculi with implication but no product.

The question now arises as to whether there are any
non-syntactic examples of universally closed categories which
are not cartesian closed. The most likely place seems to me
to be domain theory, but I have been unable to track anything
down. Does anyone have any pointers?

Thanks,

Richard.