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# Exercise 56 - Chapter 1

edited June 2018
• Choose two sets X and Y with at least three elements each.
• Choose a surjective, non-identity function $$f : X \rightarrow Y$$ between them.
• Write down two different partitions $$P$$ and $$Q$$ of $$Y$$, and then find $$f^*(P)$$ and $$f^*(Q)$$.

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1.

Comment Source: ![Figure](https://docs.google.com/drawings/d/e/2PACX-1vTxdNlvHJFkGTfWXBdx20Qtf6jNx1noxIv61gGqqmPkEob9NGNDAD1zChgw4H1iLaR8Ym0oqkb_C4lH/pub?w=470&h=696)
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2.
edited April 2018

Nice, Frederick! It's great that you're posting exercises!

I'm gonna change your formatting of the pullback $$f^\ast Q$$ in your original problem statement: you want

\$$f^\* Q\$$

not

\$$f \* Q\$$

By the way, mathematicians read $$f^\ast$$ as "pullback along $$f$$", and $$f^\ast Q$$ as "$$Q$$ pulled back along $$f$$".

Comment Source:Nice, Frederick! It's great that you're posting exercises! I'm gonna change your formatting of the pullback \$$f^\ast Q\$$ in your original problem statement: you want \$$f^\* Q\$$ not \$$f \* Q\$$ By the way, mathematicians read \$$f^\ast\$$ as "pullback along \$$f\$$", and \$$f^\ast Q\$$ as "\$$Q\$$ pulled back along \$$f\$$".
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3.
edited April 2018

$$X = { a b c }$$

$$Y = { ● ▲ ☐ }$$

$$f(a) = ▲$$

$$f(b) = ▲ \text{or} ☐$$

$$f(c) = ●$$

then

$$P := { { ▲ ☐ } { ● } }$$

$$f*({▲ ◻︎}) = { a b }$$

$$f*({ ● }) = { c }$$

and

$$Q = { { ▲ } { ☐ } { ● } }$$

$$f*({ ▲ }) = { a }$$

$$f*({ ☐ }) = { b }$$

$$f*({ ● }) = { c }$$

Beyond the construction of the examples above, I'm tempted to think of f* as a function, but that feels incorrect and I can't quite say why...

Comment Source:I'm uncertain about the first answer I came up with: \$$X = { a b c } \$$ \$$Y = { ● ▲ ☐ } \$$ \$$f(a) = ▲ \$$ \$$f(b) = ▲ \text{or} ☐ \$$ \$$f(c) = ● \$$ then \$$P := { { ▲ ☐ } { ● } } \$$ \$$f*({▲ ◻︎}) = { a b } \$$ \$$f*({ ● }) = { c } \$$ and \$$Q = { { ▲ } { ☐ } { ● } } \$$ \$$f*({ ▲ }) = { a } \$$ \$$f*({ ☐ }) = { b } \$$ \$$f*({ ● }) = { c } \$$ Beyond the construction of the examples above, I'm tempted to think of f* as a function, but that feels incorrect and I can't quite say why...
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4.
edited April 2018

Jared: $$f^*$$ is indeed a function. Here are some facts about it:

Given any set $$X$$, there is a set $$\mathcal{E}(X)$$ of all partitions of $$X$$. This is a poset - see Example 1.36 of Seven Sketches.

Given any function between sets, say $$f : X \to Y$$, there is a function $$f^* : \mathcal{E}(Y) \to \mathcal{E}(X)$$ sending partitions of $$Y$$ to partitions of $$X$$. See Example 1.49. Here Fong and Spivak point out that this function $$f^*$$ is a monotone function between posets.

Comment Source:Jared: \$$f^*\$$ is indeed a function. Here are some facts about it: Given any set \$$X\$$, there is a set \$$\mathcal{E}(X)\$$ of all partitions of \$$X\$$. This is a poset - see Example 1.36 of _Seven Sketches_. Given any function between sets, say \$$f : X \to Y\$$, there is a function \$$f^* : \mathcal{E}(Y) \to \mathcal{E}(X)\$$ sending partitions of \$$Y\$$ to partitions of \$$X\$$. See Example 1.49. Here Fong and Spivak point out that this function \$$f^*\$$ is a monotone function between posets.
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5.
edited April 2018

Ah, I misunderstood the terminology. It's even more obvious: "The most important sort of relationship between preorders is called a monotone map. These are functions that preserve preorder relations[...]"

says right on the page :)

Comment Source:Ah, I misunderstood the terminology. It's even more obvious: "The most important sort of relationship between preorders is called a _monotone map_. These are functions that preserve preorder relations[...]" says right on the page :)