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Options

- 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 \( f^*(Q) \).

## Comments

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

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 \(f\)".

`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\\)".`

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...`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...`

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.

`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.`

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 :)

`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 :)`