Re: Canonical quotients

Re: Canonical quotients

[Thorsten Palm]

Michael Barr hat am 13.09.10 geschrieben:

> But that's only an equivalent category; everyone knows that can be done.

Yes, as far as that category of partitions is concerned. But note my
last sentence:

> On Sun, 12 Sep 2010, Thorsten Palm wrote:
>
>> For the remaining sets as sources, additionally choose the
>> identity in case of the trivial quotient, the canonical map otherwise.

Perhaps I should have put the emphasis differently: for a proper
quotient of a non-partition, choose the canonical map (its range is a
partition); for a quotient of a partition, use the union construction.
Of course the case distinction is somewhat unsatisfactory, and so I
was trying to hint that there would be a nicer solution if all sets
were partitions in the first place.

It might be worth mentioning that a similar idea serves to equip the
very category of sets with a (binary-)product functor to make it
strictly monoidal.

Thorsten








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Re: Canonical quotients

[Michael Barr]

But that's only an equivalent category; everyone knows that can be done.
Even easier is the dual category of complete atomic boolean algebras,
which has canonical subobjects.

On Sun, 12 Sep 2010, Thorsten Palm wrote:

>
> Robert Pare hat am 12.09.10 geschrieben:
>
>>
>> Peter Freyd's and John Kennison's examples definitively settled
>> Mike Barr's question about canonical subobjects that compose. But
>> I had started thinking about it and had what I thought would be a
>> nice example. The category of sets has canonical quotients (equivalence
>> classes) but they don't compose. I think there is no choice that do,
>> but so far I haven't been able to prove or disprove this. Anybody?
>
> There is. First consider the full subcategory of partitions; that is,
> sets whose elements happen to be non-empty, pairwise disjoint sets. It
> has an obvious choice of quotient maps that does the trick, namely
> those maps for which each element of the target is the union of its
> fibre. For the remaining sets as sources, additionally choose the
> identity in case of the trivial quotient, the canonical map otherwise.
>
> Thorsten
>


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Re: Canonical quotients

[Peter Selinger]

[note from moderator: post delayed from Monday by a transmission error]


Hi Bob,

how about this: using a sufficiently strong version of the axiom of
choice, assume a well-ordering on the class of all sets (the "global
well-ordering"). (Someone will be able to tell me whether this is
strictly stronger than the ordinary axiom of choice, or equivalent to
it. In any case, assuming set theory is consistent in the first place,
and sufficiently large cardinals exist, there certainly exist models
of set theory with such a property).

Given any equivalence relation ~ on a set X, say that an element x of
X is a "canonical representative" of ~ if x is the (unique) least
element in its equivalence class under the global well-ordering. Let
X/~ be the set of canonical representatives, and define the map X ->
X/~ to pick out the canonical representative of each class.  I think
this gives canonical quotients on (this version of) Set.

-- Peter

Robert Pare wrote:
>
>
> Peter Freyd's and John Kennison's examples definitively settled
> Mike Barr's question about canonical subobjects that compose. But
> I had started thinking about it and had what I thought would be a
> nice example. The category of sets has canonical quotients (equivalence
> classes) but they don't compose. I think there is no choice that do,
> but so far I haven't been able to prove or disprove this. Anybody?
>
> Bob




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Re: Canonical quotients

[Thorsten Palm]

Robert Pare hat am 12.09.10 geschrieben:

>
> Peter Freyd's and John Kennison's examples definitively settled
> Mike Barr's question about canonical subobjects that compose. But
> I had started thinking about it and had what I thought would be a
> nice example. The category of sets has canonical quotients (equivalence
> classes) but they don't compose. I think there is no choice that do,
> but so far I haven't been able to prove or disprove this. Anybody?

There is. First consider the full subcategory of partitions; that is,
sets whose elements happen to be non-empty, pairwise disjoint sets. It
has an obvious choice of quotient maps that does the trick, namely
those maps for which each element of the target is the union of its
fibre. For the remaining sets as sources, additionally choose the
identity in case of the trivial quotient, the canonical map otherwise.

Thorsten



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Canonical quotients

[Robert Pare]

Peter Freyd's and John Kennison's examples definitively settled
Mike Barr's question about canonical subobjects that compose. But
I had started thinking about it and had what I thought would be a
nice example. The category of sets has canonical quotients (equivalence
classes) but they don't compose. I think there is no choice that do,
but so far I haven't been able to prove or disprove this. Anybody?

Bob

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