re **Puzzle 185**...

According to the [definition](https://forum.azimuthproject.org/discussion/2169/lecture-32-chapter-2-enriched-functors/p1) a \\(\mathcal{V}\\)-functor \\(F : \mathcal{X}\rightarrow \mathcal{Y}\\) is a function from objects of \\(\mathcal{X}\\) to objects of \\(\mathcal{Y}\\) such that \\(\mathcal{X}(x_0, x_1) \leq \mathcal{Y}(F(x_0), F(x_1))\\) for all \\(\mathcal{X}\\)-objects \\(x_0, x_1\\).

Applying this in the case \\(\mathcal{V} = \textbf{Cost}\\), and bearing in mind the reverse ordering on \\(\textbf{Cost}\\), this says that a \\(\textbf{Cost}\\)-functor is a function between Lawvere metric spaces that "shrinks all journeys", in that the cost of getting from \\(x_0\\) to \\(x_1\\) in \\(\mathcal{X}\\) is always \\(\geq\\) the cost of getting from \\(F(x_0)\\) to \\(F(x_1)\\) in \\(\mathcal{Y}\\).

OK, what does this mean when \\(\mathcal{X} = C^\text{op}\times D\\) and \\(\mathcal{Y} = \textbf{Cost}\\)?

It says the cost of getting from \\((c_0, d_0)\\) to \\((c_1, d_1)\\) is \\(\geq\\) the cost of getting from \\(\Phi(c_0, d_0)\\) to \\(\Phi(c_1, d_1)\\).

ie the cost of getting from \\(c_1\\) to \\(c_0\\) plus the cost of getting from \\(d_0\\) to \\(d_1\\) is \\(\geq\\) the maximum of \\(\Phi(c_1, d_1) - \Phi(c_0, d_0)\\) and zero – thanks @Simon for explaining how \\(\textbf{Cost}\\) is a \\(\textbf{Cost}\\)-enriched category!

ie \\(C(c_1, c_0) + D(d_0, d_1) \geq \text{max}(\Phi(c_1, d_1) - \Phi(c_0, d_0), 0)\\)

ie \\(\Phi(c_1, d_1) \leq C(c_1, c_0) + \Phi(c_0, d_0) + D(d_0, d_1)\\)

ie flying directly from \\(c_1\\) to \\(d_1\\) never costs more than driving from \\(c_1\\) to \\(c_0\\), flying from \\(c_0\\) to \\(d_0\\), then driving from \\(d_0\\) to \\(d_1\\).

According to the [definition](https://forum.azimuthproject.org/discussion/2169/lecture-32-chapter-2-enriched-functors/p1) a \\(\mathcal{V}\\)-functor \\(F : \mathcal{X}\rightarrow \mathcal{Y}\\) is a function from objects of \\(\mathcal{X}\\) to objects of \\(\mathcal{Y}\\) such that \\(\mathcal{X}(x_0, x_1) \leq \mathcal{Y}(F(x_0), F(x_1))\\) for all \\(\mathcal{X}\\)-objects \\(x_0, x_1\\).

Applying this in the case \\(\mathcal{V} = \textbf{Cost}\\), and bearing in mind the reverse ordering on \\(\textbf{Cost}\\), this says that a \\(\textbf{Cost}\\)-functor is a function between Lawvere metric spaces that "shrinks all journeys", in that the cost of getting from \\(x_0\\) to \\(x_1\\) in \\(\mathcal{X}\\) is always \\(\geq\\) the cost of getting from \\(F(x_0)\\) to \\(F(x_1)\\) in \\(\mathcal{Y}\\).

OK, what does this mean when \\(\mathcal{X} = C^\text{op}\times D\\) and \\(\mathcal{Y} = \textbf{Cost}\\)?

It says the cost of getting from \\((c_0, d_0)\\) to \\((c_1, d_1)\\) is \\(\geq\\) the cost of getting from \\(\Phi(c_0, d_0)\\) to \\(\Phi(c_1, d_1)\\).

ie the cost of getting from \\(c_1\\) to \\(c_0\\) plus the cost of getting from \\(d_0\\) to \\(d_1\\) is \\(\geq\\) the maximum of \\(\Phi(c_1, d_1) - \Phi(c_0, d_0)\\) and zero – thanks @Simon for explaining how \\(\textbf{Cost}\\) is a \\(\textbf{Cost}\\)-enriched category!

ie \\(C(c_1, c_0) + D(d_0, d_1) \geq \text{max}(\Phi(c_1, d_1) - \Phi(c_0, d_0), 0)\\)

ie \\(\Phi(c_1, d_1) \leq C(c_1, c_0) + \Phi(c_0, d_0) + D(d_0, d_1)\\)

ie flying directly from \\(c_1\\) to \\(d_1\\) never costs more than driving from \\(c_1\\) to \\(c_0\\), flying from \\(c_0\\) to \\(d_0\\), then driving from \\(d_0\\) to \\(d_1\\).