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# Exercise 63 - Chapter 1

edited June 2018

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 $$\mathbb{B}.$$

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1.

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

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.

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)
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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}\$$.
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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).