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Exercise 56 - Chapter 1

edited June 2018 in Exercises
  • 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

  • 1.

    Figure

    Comment Source: ![Figure](https://docs.google.com/drawings/d/e/2PACX-1vTxdNlvHJFkGTfWXBdx20Qtf6jNx1noxIv61gGqqmPkEob9NGNDAD1zChgw4H1iLaR8Ym0oqkb_C4lH/pub?w=470&h=696)
  • 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\\)".
  • 3.
    edited April 2018

    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...

    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...
  • 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.
  • 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 :)
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