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Exercise 63 - Chapter 1

edited June 2018 in Exercises

Show that the map \(\Phi\) from the Introduction, which was roughly given by ‘Is • connected to ∗?’ is a monotone map from the preorder shown in Eq. (1.2) to \( \mathbb{B}.\)

Equation 1.2 B

Comments

  • 1.

    If we apply the map "is \(\bullet\) connected to \(\star\)" to the diagram on the left we obtain the following diagram:

    Phi applied to partition poset

    All the paths above are valid since they exist in the Boolean poset (as indicated by the colors in the diagram below); hence the map \(\phi\) is a monotone map.

    The Boolean poset

    Comment Source:If we apply the map "is \\(\bullet\\) connected to \\(\star\\)" to the diagram on the left we obtain the following diagram: ![Phi applied to partition poset](https://doneata.bitbucket.io/applied-category-theory/ch1-ex54-a.png) All the paths above are valid since they exist in the Boolean poset (as indicated by the colors in the diagram below); hence the map \\(\phi\\) is a monotone map. ![The Boolean poset](https://doneata.bitbucket.io/applied-category-theory/ch1-ex54-b.png)
  • 2.
    edited April 2018

    Since \(\mathrm{false} \le a\) for all \(a \in \mathbb{B}\) and \(\mathrm{true} \le \mathrm{true}\), it is sufficient to check that if \(a \le a'\) and \(f(a) = \mathrm{true}\), then \(f(a') = \mathrm{true}\). Because \(a\) is finer than \(a'\), every part in \(a\) is a subset of a part in \(b\). Therefore, since \(\bullet\) and \(\ast\) are in the same part in \(a\), they will also be in the same part in \(a'\). Hence, \(f(a') = \mathrm{true}\).

    Comment Source:Since \\(\mathrm{false} \le a\\) for all \\(a \in \mathbb{B}\\) and \\(\mathrm{true} \le \mathrm{true}\\), it is sufficient to check that if \\(a \le a'\\) and \\(f(a) = \mathrm{true}\\), then \\(f(a') = \mathrm{true}\\). Because \\(a\\) is finer than \\(a'\\), every part in \\(a\\) is a subset of a part in \\(b\\). Therefore, since \\(\bullet\\) and \\(\ast\\) are in the same part in \\(a\\), they will also be in the same part in \\(a'\\). Hence, \\(f(a') = \mathrm{true}\\).
  • 3.

    We want to show that x≤y implies Φ(x)≤ Φ(y). If x≤y, that means that if • and ∗ are connected in x then they are connected in y. So • and ∗ are either connected in both x and y, disconnected in x and y, or disconnected in x and connected in y. Then (Φ(x), Φ(y)) is either (true, true), (false, false), or (false, true). In all cases Φ(x)≤Φ(y).

    Comment Source:We want to show that x≤y implies Φ(x)≤ Φ(y). If x≤y, that means that if • and ∗ are connected in x then they are connected in y. So • and ∗ are either connected in both x and y, disconnected in x and y, or disconnected in x and connected in y. Then (Φ(x), Φ(y)) is either (true, true), (false, false), or (false, true). In all cases Φ(x)≤Φ(y).
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