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Exercise 59 - Chapter 2

edited June 2018 in Exercises

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?

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Comments

  • 1.
    edited June 2018

    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.)

    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.)
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