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* Re: Non-cartesian closedness of Met
@ 2022-12-18 13:04 ptj
  0 siblings, 0 replies; 4+ messages in thread
From: ptj @ 2022-12-18 13:04 UTC (permalink / raw)
  To: Jirí Adámek; +Cc: Categories mailing list

Dear Jirka,

That's an interesting idea, but I'm having difficulty making it work
in practice. It seems to me that the most you can get from considering
morphisms from 2 is that the metric on [X,Y] must satisfy
d(f,g) \geq sup_x d(fx,gx) -- to get the reverse inequality you would 
need to impose the L_1 metric on the product 2 x X. And that
inequality is the wrong way round for showing that the transpose
of addition on R is nonexpansive.

Best regards,
Peter

On Dec 16 2022, Jirí Adámek wrote:

>Hi Peter,
>
>If Met is a CCC, then [X,Y] has as elements all morphisms from X to Y
>(use the adjoint transposes of morphisms from 1 to [X,Y]). And the 
>distance of morphisms f,g is sup_x d(fx,gx) (use the adjoint transposes
>of morphisms from two-element spaces to [X,Y]).
>
>However, addition of real numbers is not nonexpansive from R x R to R, 
>although its curred form from R to [R,R] is. This is a contradiciotn.
>
>Best regards,
>Jiri
>
>On Fri, 16 Dec 2022, ptj@maths.cam.ac.uk wrote:
>
>> Let Met denote the category of metric spaces and nonexpansive maps.
>> It's well known that if we equip the product of two metric spaces
>> with the L_{\infty} metric (the max of the distances in the two
>> coordinates), we get categorical products in Met; alternatively,
>> if we impose the L_1 metric on the product (the sum of the two
>> coordinate distances), we get a monoidal closed structure, at least
>> if we weaken the usual definition of a metric by allowing metrics to
>> take the value \infty.
>>
>> It's intuitively obvious that the cartesian monoidal structure on Met
>> can't be closed. But I've never (until I wrote one down today!) seen
>> a formal proof of this; does anyone know if it exists anywhere in the
>> literature? My proof is not particularly elegant: it amounts to showing
>> that a particular coequalizer in Met is not preserved by a functor of
>> the form (-) x Y.
>>
>>


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^ permalink raw reply	[flat|nested] 4+ messages in thread

* Re: Non-cartesian closedness of Met
  2022-12-16 16:41 ptj
@ 2022-12-19  8:50 ` Jirí Adámek
  0 siblings, 0 replies; 4+ messages in thread
From: Jirí Adámek @ 2022-12-19  8:50 UTC (permalink / raw)
  To: ptj; +Cc: Categories mailing list

Dear Peter,

Thanks for pointing out that my argument has been incomplete. Here
is a proof that Met is not a ccc:

We use that limits in Met are the limits in Set with the supremum
metric (coordinate-wise). Let us denote by Met(X,Y) the hom-sets with the
supremum metric d'(f,g) = sup_x d(fx,gx). The addition of real
numbers from RxR to R is not non-expanding, but its curried from
R to Met(R,R) is; thus all we need to do is to show that if Met were a
ccc, one could take [X,Y]=Met(X,Y).

The underlying set of [X,Y] can be taken to be the hom-set, using
adjoint transposes of morphisms from 1 to [X,Y]. And the universal
morphism eval:[X,Y]xX -> Y can be taken to be the evaluation map
(precompose it with morphisms fxX for f:1->[X,Y]). The metric d of
[X,Y] satisfies d \geq d', using adjoint transposes of morphisms
from 2-element spaces to [X,Y]. To prove d \leq d', we first consider
a finite space X. Let id: n->X be the identity map from the discrete
space on the underlying set of X, a coproduct of n copies of 1. Then
[id,Y]: [X,Y]->[n,Y]= Y^n demonstrates that d= d'. For X arbitrary,
express it as a directed colimit X=colim X_i of all finite subspaces X_i.
Then [X,Y] = lim [X_i,Y] carries the supremum metric.

Best regards,
Jiri


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^ permalink raw reply	[flat|nested] 4+ messages in thread

* Re: Non-cartesian closedness of Met
@ 2022-12-17  9:20 Dirk Hofmann
  0 siblings, 0 replies; 4+ messages in thread
From: Dirk Hofmann @ 2022-12-17  9:20 UTC (permalink / raw)
  To: ptj; +Cc: Maria Manuel Clementino, Categories mailing list

Dear Peter,

we don't have such an example in the paper. Our argument used the 
cartesian closed category of "metric spaces without triangular 
inequality", and showed that, for a metric space X, all 
exponential Y^X (Y metric) taken in that category satisfy the 
triangular inequality if and only if X satisfies that condition.

Best regards
Dirk




On 16 December 2022 at 22:05 GMT+0000, ptj@maths.cam.ac.uk wrote 
...
> Dear Dirk, dear Maria Manuel,
>
> That's very interesting, and I should have remembered it. But 
> did
> your argument come up with an explicit example of a colimit in 
> Met
> not preserved by a functor of the form (-) x Y ?
>
> I suppose I'll have to go and read your paper!
>
> Best regards,
> Peter
>
> On Dec 16 2022, Dirk Hofmann wrote:
>
>>
>>Dear Peter,
>>
>>in our paper
>>
>> - Clementino, M. M., & Hofmann, D. (2006). Exponentiation in
>> $V$-categories. Topology and its Applications, 153(16), 
>> 3113–3128.
>>
>> we give a characterisation of exponentiable metric spaces. The 
>> result
>> essentially states that a metric space (in the sense of 
>> Lawvere) is
>> exponentiable if and only if "there is always a point in the 
>> middle",
>> that is, whenever d(x,z)=u+v, then there is a point y with 
>> d(x,y)≤u+ε
>> and d(y,z)≤v+ε. A finite metric space with a non-trivial 
>> distance
>> cannot be exponentiable.
>>
>>Best regards
>>Dirk
>>
>> On 16 December 2022 at 16:41 GMT+0000, <ptj@maths.cam.ac.uk> 
>> wrote ...
>>> Let Met denote the category of metric spaces and nonexpansive 
>>> maps.
>>> It's well known that if we equip the product of two metric 
>>> spaces
>>> with the L_{\infty} metric (the max of the distances in the 
>>> two
>>> coordinates), we get categorical products in Met; 
>>> alternatively,
>>> if we impose the L_1 metric on the product (the sum of the two
>>> coordinate distances), we get a monoidal closed structure, at 
>>> least
>>> if we weaken the usual definition of a metric by allowing 
>>> metrics to
>>> take the value \infty.
>>>
>>> It's intuitively obvious that the cartesian monoidal structure 
>>> on Met
>>> can't be closed. But I've never (until I wrote one down 
>>> today!) seen
>>> a formal proof of this; does anyone know if it exists anywhere 
>>> in the
>>> literature? My proof is not particularly elegant: it amounts 
>>> to
>>> showing
>>> that a particular coequalizer in Met is not preserved by a 
>>> functor of
>>> the form (-) x Y.
>>>
>>>


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^ permalink raw reply	[flat|nested] 4+ messages in thread

* Non-cartesian closedness of Met
@ 2022-12-16 16:41 ptj
  2022-12-19  8:50 ` Jirí Adámek
  0 siblings, 1 reply; 4+ messages in thread
From: ptj @ 2022-12-16 16:41 UTC (permalink / raw)
  To: Categories mailing list

Let Met denote the category of metric spaces and nonexpansive maps.
It's well known that if we equip the product of two metric spaces
with the L_{\infty} metric (the max of the distances in the two
coordinates), we get categorical products in Met; alternatively,
if we impose the L_1 metric on the product (the sum of the two
coordinate distances), we get a monoidal closed structure, at least
if we weaken the usual definition of a metric by allowing metrics to
take the value \infty.

It's intuitively obvious that the cartesian monoidal structure on Met
can't be closed. But I've never (until I wrote one down today!) seen
a formal proof of this; does anyone know if it exists anywhere in the
literature? My proof is not particularly elegant: it amounts to showing
that a particular coequalizer in Met is not preserved by a functor of
the form (-) x Y.



[For admin and other information see: http://www.mta.ca/~cat-dist/ ]


^ permalink raw reply	[flat|nested] 4+ messages in thread

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