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Let $X,Y,Z$ be three metric spaces. Let $f : X \times Y \to Z$ be a map which is non-expansive in each argument: $$d(f(x,y),f(x',y)) \leq d(x,x')$$ $$d(f(x,y),f(x,y')) \leq d(y,y')$$ Does it follow that $f$ is non-expansive (where we use the sup-metric on the product)? That is, do we have the following? $$d(f(x,y),f(x',y')) \leq \sup(d(x,x'),d(y,y'))$$

I assume the answer is yes, due to a very abstract category-theoretic argument (!), but for some reason I cannot prove it directly.

Martin Brandenburg
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No. For instance, take $X=Y=\{a,b\}$ with $d(a,b)=1$ and let $Z=X\times Y$ with the metric that differs from the sup metric in that $d((a,b),(b,a))=2$. Then the identity function $f$ is non-expansive in each argument (since the metric of $Z$ agrees with the sup metric whenever the two points have a coordinate with a common value) but is not non-expansive.

(The metric you would want on the product to make this true is the $\ell^1$ metric, not the sup metric.)

Eric Wofsey
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  • Thanks! Yes, I confused the tensor product of enriched categories with the product. Here, in the case of the monoidal category $([0,\infty],\geq,+)$ which gives generalized metric spaces. – Martin Brandenburg Dec 18 '22 at 16:05