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# Exercise 59 - Chapter 2

edited June 2018

Let $$M$$ be a set and let $$\mathcal{M} := ( \mathbb{P}(M), \subseteq, M, \cap )$$ be the monoidal preorder whose elements are subsets of $$M$$.

Someone gives the following interpretation, “for any set $$M$$, imagine it as the set of modes of transportation (e.g. car, boat, foot). Then a category $$\mathcal{X}$$ enriched in $$\mathcal{M}$$ tells you all the modes that will get you from $$a$$ all the way to $$b$$, for any two points $$a, b \in Ob(\mathcal{X})$$.”

1. Draw a graph with four vertices and four or five edges, each labeled with a subset of $$M = \{car, boat, foot\}$$.

2. This corresponds to a $$\mathcal{M}$$-category; call it X. Write out the corresponding four-by-four matrix of hom-objects.

3. Does the person’s interpretation look right, or is it subtly mistaken somehow?

This is confusing because the construction mimics the diagrams in (2.53) but it's not a metric for which the weighted graphs are shorthand. Each edge between each pair of vertices (therefore all 6, and undirected, edges, rather than just 4 or 5, directed or not) must be associated with a value in $$\mathcal{M}$$.
Now, we can rescue this by claiming that missing edges are implicitly labeled by $$\varnothing$$, but in order for the graph to be an $$\mathcal{M}$$-category we need property b of Definition 2.43 to hold, that is, $$\forall x, y, z \in \mathcal{X}, \mathcal{M}(x, y) \cap \mathcal{M}(y, z) \subseteq \mathcal{M}(x, z)$$. So the way we've labeled those edges needs to be consistent with that rule; we can't just arbitrarily label them.
Comment Source:This is confusing because the construction mimics the diagrams in (2.53) but it's not a metric for which the weighted graphs are shorthand. Each edge between each pair of vertices (therefore all 6, and undirected, edges, rather than just 4 or 5, directed or not) must be associated with a value in \$$\mathcal{M} \$$. Now, we can rescue this by claiming that missing edges are implicitly labeled by \$$\varnothing \$$, but in order for the graph to be an \$$\mathcal{M} \$$-category we need property b of Definition 2.43 to hold, that is, \$$\forall x, y, z \in \mathcal{X}, \mathcal{M}(x, y) \cap \mathcal{M}(y, z) \subseteq \mathcal{M}(x, z) \$$. So the way we've labeled those edges needs to be consistent with that rule; we can't just arbitrarily label them. (Using numberings of the book as of the Version 15 June 2018.)