John Baez wrote:

>Keith: _don't forget the fiendish reversal of inequality in the definition of \\(\mathbf{Cost}\\)._

Luckily, such a fix is easy.

>**Definition.** Let \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) be \\(\mathbf{Cost}\\)-categories. A \\(\mathbf{Cost}\\)-functor from \\(\mathcal{X}\\) to \\(\mathcal{Y}\\), denoted \\(F\colon\mathcal{X}\to\mathcal{Y}\\), is a function

>\[ F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y}) \]

>such that

>\[ \mathcal{d}(x,x') \geq \mathcal{d'}(F(x),F(x')) \]

>for all \\(x,x' \in\mathrm{Ob}(\mathcal{X})\\).

>I would call this is a ***non-expanding*** map, since the generalized distance or cost metric may only stay the same or ***contract***.

>Keith: _don't forget the fiendish reversal of inequality in the definition of \\(\mathbf{Cost}\\)._

Luckily, such a fix is easy.

>**Definition.** Let \\(\mathcal{X}\\) and \\(\mathcal{Y}\\) be \\(\mathbf{Cost}\\)-categories. A \\(\mathbf{Cost}\\)-functor from \\(\mathcal{X}\\) to \\(\mathcal{Y}\\), denoted \\(F\colon\mathcal{X}\to\mathcal{Y}\\), is a function

>\[ F\colon\mathrm{Ob}(\mathcal{X})\to \mathrm{Ob}(\mathcal{Y}) \]

>such that

>\[ \mathcal{d}(x,x') \geq \mathcal{d'}(F(x),F(x')) \]

>for all \\(x,x' \in\mathrm{Ob}(\mathcal{X})\\).

>I would call this is a ***non-expanding*** map, since the generalized distance or cost metric may only stay the same or ***contract***.