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\\)".
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...
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.
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 :)
Comments
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 :)