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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