Re: Non-cartesian closedness of Met

[ptj@maths.cam.ac.uk]

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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