Options

Lecture 4 - Chapter 1: Galois Connections

2»

Comments

  • 51.
    edited April 2018

    Thank you John. WRT puzzle:

    Puzzle. What's the left adjoint \(f_2 \) of this function \(f(x)=10x \)?

    Let me try:

    Working through that, it seems to be, still $$ f_2(x) = \lfloor \frac{x}{10} \rfloor $$ So the right adjoint of f is the same as left adjoint of f. is that right? Seems like that right adjoint and left adjoint are things that just work opposite to the 'original' map f (but then why the distinction in naming ?).

    Here is the thinking process.. given \(f(x)=10x \), we observe that this function increases the value of the input. because our task is to find \(f_2\), the f's left adjoint, we have to find a function that works in opposite direction of f. Why, because the result of \(f_2\) has to be decreased or, made less, than input parameter of f (this follows from \(f_2(a) \le b\) ).

    Say b=50. \(f(b)=10 *50 = 500\)

    Our task is to find \(f_2\), such that \(f_2( 500, or, 501,or,502... or, 509)\) is \(\le\) 50. That is a function that produces something that's les or equal to b.

    A good candidate is, of course for \(f_2\) to be \(\lfloor \frac{x}{10} \rfloor \)

    Comment Source:Thank you John. WRT puzzle: **Puzzle.** What's the left adjoint \\(f_2 \\) of this function \\(f(x)=10x \\)? Let me try: Working through that, it seems to be, still $$ f_2(x) = \lfloor \frac{x}{10} \rfloor $$ So the right adjoint of f is the same as left adjoint of f. is that right? Seems like that right adjoint and left adjoint are things that just work opposite to the 'original' map f (but then why the distinction in naming ?). Here is the thinking process.. given \\(f(x)=10x \\), we observe that this function *increases* the value of the input. because our task is to find \\(f_2\\), the f's left adjoint, we have to find a function that works in opposite direction of f. Why, because the result of \\(f_2\\) has to be *decreased* or, *made less*, than input parameter of f (this follows from \\(f_2(a) \le b\\) ). Say b=50. \\(f(b)=10 *50 = 500\\) Our task is to find \\(f_2\\), such that \\(f_2( 500, or, 501,or,502... or, 509)\\) is \\(\le\\) 50. That is a function that produces something that's les or equal to b. A good candidate is, of course for \\(f_2\\) to be \\(\lfloor \frac{x}{10} \rfloor \\)
  • 52.
    edited April 2018

    There is a one to one correspondence between monads (T, unit, mult) on a skeletal preorder \(P\) and partitions of \(P\) into sets with a maximum.

    \(T\) is just an endofunctor, which on preorders is simply a monotone function, but the natural transformations unit and mult are more mysterious. It turns out that all the data for a natural transformation \(F \to G\) reduces to the brute fact that $$ \forall a \in P, F a \le G a .$$

    Therefor, unit is the proposition that T is upward, that is $$ \forall x, x \le T x ,$$ and mult is $$ \forall x, T (T x) \le T x .$$

    Note that substituting \(T x\) for \(x\) in unit, produces \( T x \le T (T x) \) so T is idempotent.

    By idempotency and unit, \( T a \) is the maximum of the inverse image of \(T a\).

    So T determines a partition into preorders with a maximum. The inverse correspondence is obvious.

    I don't think non-skeletality actually makes things much more complicated. You just have to add (up to isomorphisms on elements) to everything.

    Comment Source:There is a one to one correspondence between monads (T, unit, mult) on a skeletal preorder \\(P\\) and partitions of \\(P\\) into sets with a maximum. \\(T\\) is just an endofunctor, which on preorders is simply a monotone function, but the natural transformations unit and mult are more mysterious. It turns out that all the data for a natural transformation \\(F \to G\\) reduces to the brute fact that $$ \forall a \in P, F a \le G a .$$ Therefor, unit is the proposition that T is upward, that is $$ \forall x, x \le T x ,$$ and mult is $$ \forall x, T (T x) \le T x .$$ Note that substituting \\(T x\\) for \\(x\\) in unit, produces \\( T x \le T (T x) \\) so T is idempotent. By idempotency and unit, \\\( T a \\) is the maximum of the inverse image of \\(T a\\\). So T determines a partition into preorders with a maximum. The inverse correspondence is obvious. I don't think non-skeletality actually makes things much more complicated. You just have to add (up to isomorphisms on elements) to everything.
  • 53.
    edited April 2018

    Vladislav wrote:

    So the right adjoint of \(f\) is the same as left adjoint of \(f\). Is that right?

    No. You didn't correctly use the definition of "left adjoint". Try again: look at the definition, and use it in a systematic way to compute the left adjoint of \(f\).

    Comment Source:[Vladislav wrote](https://forum.azimuthproject.org/discussion/comment/16871/#Comment_16871): > So the right adjoint of \\(f\\) is the same as left adjoint of \\(f\\). Is that right? No. You didn't correctly use the definition of "left adjoint". Try again: look at the definition, and use it in a systematic way to compute the left adjoint of \\(f\\).
  • 54.

    Christopher Upshaw: welcome to the conversation!

    As you note, on posets natural transformations between functors become mere properties (namely, inequalities) rather than extra structure. So, a monad just becomes a monotone map with two extra properties! Such a monotone map is called a closure operator, and plenty of people study them while completely unaware that they're studying monads!

    I fixed up your LaTeX a bit, e.g. using \\( and \\) rather than the non-working \$ to enclose inline equations, and using \forall rather than the non-working \all.

    Comment Source:[Christopher Upshaw](https://forum.azimuthproject.org/discussion/comment/16890/#Comment_16890): welcome to the conversation! As you note, on posets natural transformations between functors become mere _properties_ (namely, inequalities) rather than _extra structure_. So, a monad just becomes a monotone map with two extra properties! Such a monotone map is called a [closure operator](https://en.wikipedia.org/wiki/Closure_operator#Closure_operators_on_partially_ordered_sets), and plenty of people study them while completely unaware that they're studying monads! I fixed up your LaTeX a bit, e.g. using `\\(` and `\\)` rather than the non-working `\$` to enclose inline equations, and using `\forall` rather than the non-working `\all`.
  • 55.

    No. You didn't correctly use the definition of "left adjoint". Try again: look at the definition, and use it in a systematic way to compute the left adjoint of

    John, I seem to be stuck.

    For Puzzle. What's the left adjoint \(g \) of this function \(f(x)=10x \)?

    I am still coming up with $$ g(x) = \lfloor \frac{x}{10} \rfloor $$ I am looking for
    \(g : \mathbb{N} \to \mathbb{N}\) such that

    $$ g(m) \le n \textrm{ if and only if } m \le f(n) $$ for all \(m,n \in \mathbb{N}\).

    Here was my thinking process First, we are given the definition of f, therefore looking at the right side of the Galois connection definition ( \(m \le f(n) \) ), to establish my 'test values' for m and n.

    A) f is monotone increasing function, therefore my m, n should be such that \(\frac{m}{10} \le n \)

    B) next step, I am switching to the left side of the connection definition, and that's where I am doing my tests for g.

    When choosing m,n I am choosing sort of corner cases.

    Specifically, I am looking for minimal m. the least m can be is 1. if m=1, n needs to be at least 1, according to A. With that constraint, I need to find g, such that it takes m=1 and makes 1 out of it (or 0) (but g also needs to be monotone increasing function too).

    \(g(1) \le 1 \)

    \(g(x) = \lfloor \frac{x}{10} \rfloor\) fits the bill, so to speak. Actually.., \(g(x) = \lceil \frac{x}{10} \rceil\) would work too.

    But since these are wrong, I am a bit at a loss , of where I am making a mistake.

    Comment Source:>No. You didn't correctly use the definition of "left adjoint". Try again: look at the definition, and use it in a systematic way to compute the left adjoint of John, I seem to be stuck. For **Puzzle.** What's the left adjoint \\(g \\) of this function \\(f(x)=10x \\)? I am still coming up with $$ g(x) = \lfloor \frac{x}{10} \rfloor $$ I am looking for \\(g : \mathbb{N} \to \mathbb{N}\\) such that $$ g(m) \le n \textrm{ if and only if } m \le f(n) $$ for all \\(m,n \in \mathbb{N}\\). Here was my thinking process First, we are given the definition of f, therefore looking at the right side of the Galois connection definition ( \\(m \le f(n) \\) ), to establish my 'test values' for m and n. A) f is monotone increasing function, therefore my m, n should be such that \\(\frac{m}{10} \le n \\) B) next step, I am switching to the left side of the connection definition, and that's where I am doing my tests for g. When choosing m,n I am choosing sort of corner cases. Specifically, I am looking for minimal m. the least m can be is 1. if m=1, n needs to be at least 1, according to A. With that constraint, I need to find g, such that it takes m=1 and makes 1 out of it (or 0) (but g also needs to be monotone increasing function too). \\(g(1) \le 1 \\) \\(g(x) = \lfloor \frac{x}{10} \rfloor\\) fits the bill, so to speak. Actually.., \\(g(x) = \lceil \frac{x}{10} \rceil\\) would work too. But since these are wrong, I am a bit at a loss , of where I am making a mistake.
  • 56.

    Puzzle 12

    How many right adjoints can you find?

    Isn't there essentially an infinite number of linear functions taken to the roof or floor in this case?

    Comment Source:Puzzle 12 > How many right adjoints can you find? Isn't there essentially an infinite number of linear functions taken to the roof or floor in this case?
  • 57.
    edited April 2018

    Puzzle 12:

    Let's call \(m\) the elements in which each side of $$ 2m \le n \textrm{ and } m \le g(n) $$ is true.

    What happen when one of the sides is not true, well the other neither. Let's call \(m'\) such numbers.

    Thus, for each \(n\) $$ 2 m \le n \textrm{ and } m \le g(n) $$ $$ 2m' > n \textrm{ and } m' > g(n) $$ We can see that for each \(n\) we separate the \(A\) set in two parts, \(m\) and \(m'\).

    We are interested in finding constraints for \(g(n)\). Let's rewrite part of the previous equations in the suggestive form $$ m \le g(n) < m'$$ What values of \(m\) and \(m'\) most constraint the equation?

    For a fix \(n\), we seek the maximum possible \(m\) and the minimum possible \(m'\).

    Since \( m \le n/2\), we also have that \(m \le \lfloor \frac{n}{2} \rfloor\) and its highest value as a lower bound is \(\lfloor \frac{n}{2} \rfloor\).

    Similarly the lowest \(m'\) that \( m' > n/2 \) is given by \( \lfloor \frac{n}{2} \rfloor +1 \),

    Therefore, the most restrictive value of \(m\) and \(m'\) we have that $$ \lfloor \frac{n}{2} \rfloor \le g(n) < \lfloor \frac{n}{2} \rfloor + 1 $$ whose only possible solution is $$ g(n) = \lfloor \frac{n}{2} \rfloor $$ There is only one right adjoint.

    Comment Source:**Puzzle 12:** Let's call \\(m\\) the elements in which each side of $$ 2m \le n \textrm{ and } m \le g(n) $$ is true. What happen when one of the sides is not true, well the other neither. Let's call \\(m'\\) such numbers. Thus, for each \\(n\\) $$ 2 m \le n \textrm{ and } m \le g(n) $$ $$ 2m' > n \textrm{ and } m' > g(n) $$ We can see that for each \\(n\\) we separate the \\(A\\) set in two parts, \\(m\\) and \\(m'\\). We are interested in finding constraints for \\(g(n)\\). Let's rewrite part of the previous equations in the suggestive form $$ m \le g(n) < m'$$ What values of \\(m\\) and \\(m'\\) most constraint the equation? For a fix \\(n\\), we seek the maximum possible \\(m\\) and the minimum possible \\(m'\\). Since \\( m \le n/2\\), we also have that \\(m \le \lfloor \frac{n}{2} \rfloor\\) and its highest value as a lower bound is \\(\lfloor \frac{n}{2} \rfloor\\). Similarly the lowest \\(m'\\) that \\( m' > n/2 \\) is given by \\( \lfloor \frac{n}{2} \rfloor +1 \\), Therefore, the most restrictive value of \\(m\\) and \\(m'\\) we have that $$ \lfloor \frac{n}{2} \rfloor \le g(n) < \lfloor \frac{n}{2} \rfloor + 1 $$ whose only possible solution is $$ g(n) = \lfloor \frac{n}{2} \rfloor $$ There is only one **right adjoint**.
  • 58.
    edited April 2018

    Puzzle 12 @IgnacioPeixoto Thanks for the nice explanation, I want to emphasize the mistake, implicit in your proof, that is so easy to make.

    The definition for adjunction is $$ f(p) \le q \iff q \le g(p) $$ and not simply $$ f(p) \le q \implies q \le g(p) $$ nor $$ q \le g(p) \implies f(p) \le q $$ but $$ [ f(p) \le q \implies q \le g(p) ] \wedge [ q \le g(p) \implies f(p) \le q ] $$.

    To my mind this is easiest to verify by checking the two "round trips" $$ [ f(g(q)) \le q ] \wedge [ p \le g(f(p)) ] $$.

    Comment Source:**Puzzle 12** @IgnacioPeixoto Thanks for the nice explanation, I want to emphasize the mistake, implicit in your proof, that is so easy to make. The definition for adjunction is $$ f(p) \le q \iff q \le g(p) $$ and **not** simply $$ f(p) \le q \implies q \le g(p) $$ **nor** $$ q \le g(p) \implies f(p) \le q $$ but $$ [ f(p) \le q \implies q \le g(p) ] \wedge [ q \le g(p) \implies f(p) \le q ] $$. To my mind this is easiest to verify by checking the two "round trips" $$ [ f(g(q)) \le q ] \wedge [ p \le g(f(p)) ] $$.
  • 59.

    @FrederickEisele, thanks, it was a typo. I'm using the adjuntion definition, and I separate the elements that make both sides true, from the elements that make both sides false. The \(m\) make both sides true, whereas the \(m'\) make both sides false.

    Comment Source:@FrederickEisele, thanks, it was a typo. I'm using the adjuntion definition, and I separate the elements that make both sides true, from the elements that make both sides false. The \\(m\\) make both sides true, whereas the \\(m'\\) make both sides false.
  • 60.
    edited April 2018

    Vladislav Papayan wrote:

    Puzzle. What's the left adjoint \(g \) of this function \(f(x)=10x \)?

    I am still coming up with

    $$ g(x) = \lfloor \frac{x}{10} \rfloor $$ I am looking for \(g : \mathbb{N} \to \mathbb{N}\) such that $$ g(m) \le n \textrm{ if and only if } m \le f(n) $$ for all \(m,n \in \mathbb{N}\).

    Let's take \(m = 11, n = 1\). Then you're claiming that

    $$ \lfloor \frac{11}{10} \rfloor \le 1 \textrm{ if and only if } 11 \le 10 . $$ Is that right?

    Comment Source:[Vladislav Papayan wrote:](https://forum.azimuthproject.org/discussion/comment/16931/#Comment_16931) > **Puzzle.** What's the left adjoint \\(g \\) of this function \\(f(x)=10x \\)? > I am still coming up with >$$ g(x) = \lfloor \frac{x}{10} \rfloor $$ > I am looking for \\(g : \mathbb{N} \to \mathbb{N}\\) such that > $$ g(m) \le n \textrm{ if and only if } m \le f(n) $$ > for all \\(m,n \in \mathbb{N}\\). Let's take \\(m = 11, n = 1\\). Then you're claiming that $$ \lfloor \frac{11}{10} \rfloor \le 1 \textrm{ if and only if } 11 \le 10 . $$ Is that right?
  • 61.
    edited April 2018

    So for the toy example here is some ruby code, which you can run from the terminal after pasting into a file with .rb extension using: ruby right_adjoint.rb


    X = 1..3 # an integer range

    XY = [] # pairs of (x, f(x)); (x,f(x)-1); (x,f(x)+1)

    RA = [] # collection of adjoint pairs

    def f(x) 2*x end

    def g(y) y/2.0.floor end

    X.each { |x|

    XY << [ x, f(x)-1] if (f(x) - 1 > 0)

    XY << [ x, f(x) ]

    XY << [ x, f(x)+1] if (f(x) + 1 < f(X.last)) }

    XY.each { |pair|

    x = pair.first

    y = pair.last

    RA << pair if ( (f(x) <= y) && (x <= g(y)) ) }

    puts "XY pairs: " + XY.to_s

    puts "RA pairs: " + RA.to_s


    If this the correct way to compute right adjoints for this example (which it very well may not be as I'm a bit tired right now), why is it I can vary the 2.0 parameter in g(y), i.e. 2.001, 2.0001 etc and get the same set of adjoint pairings? Why is there only 1 unique right adjoint if this is the case?

    Comment Source:So for the toy example here is some ruby code, which you can run from the terminal after pasting into a file with .rb extension using: **ruby right_adjoint.rb** --- X = 1..3 # an integer range XY = [] # pairs of (x, f(x)); (x,f(x)-1); (x,f(x)+1) RA = [] # collection of adjoint pairs def f(x) 2*x end def g(y) y/2.0.floor end X.each { |x| XY << [ x, f(x)-1] if (f(x) - 1 > 0) XY << [ x, f(x) ] XY << [ x, f(x)+1] if (f(x) + 1 < f(X.last)) } XY.each { |pair| x = pair.first y = pair.last RA << pair if ( (f(x) <= y) && (x <= g(y)) ) } puts "XY pairs: " + XY.to_s puts "RA pairs: " + RA.to_s --- If this the correct way to compute right adjoints for this example (which it very well may not be as I'm a bit tired right now), why is it I can vary the 2.0 parameter in g(y), i.e. 2.001, 2.0001 etc and get the same set of adjoint pairings? Why is there only 1 unique right adjoint if this is the case?
  • 62.

    Thank you John.
    WRT

    Let's take \(m = 11, n = 1\). Then you're claiming that

    $$ \lfloor \frac{11}{10} \rfloor \le 1 \textrm{ if and only if } 11 \le 10 . $$ Is that right?

    With your help, I can see where at least one of my logic mistakes was.

    I did not correctly understand what if-and-only-if meant in $$ g(m) \le n \textrm{ if and only if } m \le f(n) $$ I thought, if right side evaluates to false (eg like in your choice n and m), then the left side -- does not matter. I was mentally writing this:

    If (right-side is false ) then { ignore left side, and go to next choice of m,n pair to test with}

    But what iff means both sides had to be either true or false at the same time. I should have realized that, sorry about this.

    With that, clearly my choice of \(g(x) = \lfloor \frac{x}{10} \rfloor\) was wrong.

    What I should have been searching, was g such that:

    \(g(11) \gt 1 \) (so that the \(g(m) \le n \) would evaluate to false for any number m above 10 (while keeping n=1).

    While \(g(10,9...,0) \le 1 \) (so that the \(g(m) \le n \) would evaluate to true for any number 10 and below (while keeping n=1).

    is the above correct framework of thinking?

    Comment Source:Thank you John. WRT > Let's take \\(m = 11, n = 1\\). Then you're claiming that >$$ \lfloor \frac{11}{10} \rfloor \le 1 \textrm{ if and only if } 11 \le 10 . $$ >Is that right? With your help, I can see where at least one of my logic mistakes was. I did not correctly understand what if-and-only-if meant in $$ g(m) \le n \textrm{ if and only if } m \le f(n) $$ I thought, if right side evaluates to false (eg like in your choice n and m), then the left side -- does not matter. I was mentally writing this: > If (right-side is false ) then { ignore left side, and go to next choice of m,n pair to test with} But what iff means *both* sides had to be either true or false at the same time. I should have realized that, sorry about this. With that, clearly my choice of \\(g(x) = \lfloor \frac{x}{10} \rfloor\\) was wrong. What I should have been searching, was *g* such that: \\(g(11) \gt 1 \\) (so that the \\(g(m) \le n \\) would evaluate to *false* for any number m above 10 (while keeping n=1). While \\(g(10,9...,0) \le 1 \\) (so that the \\(g(m) \le n \\) would evaluate to *true* for any number 10 and below (while keeping n=1). is the above correct framework of thinking?
  • 63.
    edited April 2018

    Vladislav - yes, that's the correct framework. I'm glad you figured out what "if and only if" means. In classical logic, which is what mathematicians normally use, the logical connectives "not", "and", "or", "implies", and "if and only iff" are defined as follows:

    image
    Comment Source:Vladislav - yes, that's the correct framework. I'm glad you figured out what "if and only if" means. In classical logic, which is what mathematicians normally use, the logical connectives "not", "and", "or", "implies", and "if and only iff" are defined as follows: <center><img src = "https://ethicalrealism.files.wordpress.com/2012/10/master-table-small.png"></center>
  • 64.
    edited April 2018

    Thank you John. I have seen 'not', 'and' and 'or' as a built-in operators, in every programming language I used. But I do not think I came across 'implies' and 'if-and-only-iff' as built-in operators. Perhaps others on the forum had different experience. I have seen XOR (which appears to be opposite of if-and-only-iff'.

    But I should have realized sooner, how to read -- 'if-and-only-if', still, this was a bad on my part...

    Going back to your Puzzle, that I still have not solved :-)

    Puzzle. What's the left adjoint \(g \) of this function \(f(x)=10x \)?

    would that be \(g(x) = \lceil \frac{x}{10} \rceil\) ?

    In my checks it satisfies the two constraints:

    C1) \(g(11) \gt 1 \) (so that the \(g(m) \le n \) would evaluate to false for any number m above 10 (while keeping n=1).

    C2) While \(g(10,9...,0) \le 1 \) (so that the \(g(m) \le n \) would evaluate to true for any number 10 and below (while keeping n=1).

    Specifically for m=11, n=1 it would produce 'false' on the left side as well (given that the right side is false too) $$ \lceil \frac{11}{10} \rceil \le 1 \textrm{ if and only if } 11 \le 10 . $$ So it produces 'false' on both sides, which is needed for 'if-and-only-if'.
    While, at the same time, it produces 'true' on the left side, every time m \(\le\) \(f(n=1)=1*10 \)

    Comment Source:Thank you John. I have seen 'not', 'and' and 'or' as a built-in operators, in every programming language I used. But I do not think I came across 'implies' and 'if-and-only-iff' as built-in operators. Perhaps others on the forum had different experience. I have seen XOR (which appears to be opposite of if-and-only-iff'. But I should have realized sooner, how to read -- 'if-and-only-if', still, this was a bad on my part... Going back to your Puzzle, that I still have not solved :-) **Puzzle.** What's the left adjoint \\(g \\) of this function \\(f(x)=10x \\)? would that be \\(g(x) = \lceil \frac{x}{10} \rceil\\) ? In my checks it satisfies the two constraints: C1) \\(g(11) \gt 1 \\) (so that the \\(g(m) \le n \\) would evaluate to *false* for any number m above 10 (while keeping n=1). C2) While \\(g(10,9...,0) \le 1 \\) (so that the \\(g(m) \le n \\) would evaluate to *true* for any number 10 and below (while keeping n=1). Specifically for m=11, n=1 it would produce 'false' on the left side as well (given that the right side is false too) $$ \lceil \frac{11}{10} \rceil \le 1 \textrm{ if and only if } 11 \le 10 . $$ So it produces 'false' on both sides, which is needed for 'if-and-only-if'. While, at the same time, it produces 'true' on the left side, every time *m* \\(\le\\) \\(f(n=1)=1*10 \\)
  • 65.
    edited April 2018

    Vladislav wrote:

    Puzzle. What's the left adjoint \(g \) of this function \(f(x)=10x \)?

    would that be \(g(x) = \lceil \frac{x}{10} \rceil\) ?

    Yes! Excellent! Now I hope you read, or reread, Lecture 6. It should make a lot of sense now that you've fought through these issues yourself. In particular, you should believe this now:

    The left adjoint comes as close as possible to the (perhaps nonexistent) correct answer while making sure to never choose a number that's too small. The right adjoint comes as close as possible while making sure to never choose a number that's too big.

    Comment Source:Vladislav wrote: > **Puzzle.** What's the left adjoint \\(g \\) of this function \\(f(x)=10x \\)? > would that be \\(g(x) = \lceil \frac{x}{10} \rceil\\) ? Yes! Excellent! Now I hope you read, or reread, [Lecture 6](https://forum.azimuthproject.org/discussion/1901/lecture-6-chapter-1-computing-adjoints/p1). It should make a lot of sense now that you've fought through these issues yourself. In particular, you should believe this now: > The left adjoint comes as close as possible to the (perhaps nonexistent) correct answer while making sure to never choose a number that's _too small_. The right adjoint comes as close as possible while making sure to never choose a number that's _too big_.
  • 66.
    edited April 2018

    I guess my question fundamentally is why \( \lfloor \frac{y}{\lim_{1 \to 2} } \rfloor \) doesn't work as a right adjoint for \( f(x)=2x \). Can't you approximate \( \lfloor \frac{y}{2} \rfloor \) arbitrarily precisely in such a way that a floor will always yield the same natural number? I know you want to pick the smallest value possible, does the function definition itself need to be restricted to \( \mathbb{N} \)? I apologize if this is a silly question, I guess I'm missing something obvious.

    Comment Source:I guess my question fundamentally is why \\( \lfloor \frac{y}{\lim_{1 \to 2} } \rfloor \\) doesn't work as a right adjoint for \\( f(x)=2x \\). Can't you approximate \\( \lfloor \frac{y}{2} \rfloor \\) arbitrarily precisely in such a way that a floor will always yield the same natural number? I know you want to pick the smallest value possible, does the function definition itself need to be restricted to \\( \mathbb{N} \\)? I apologize if this is a silly question, I guess I'm missing something obvious.
  • 67.
    edited April 2018

    @JohnBaez #65. Oh, finally! John thx for your guidance and patience. I will be moving to the next lectures now. I felt like I needed to pass that puzzle, to move on past Lecture 4.

    Comment Source:@JohnBaez #65. Oh, finally! John thx for your guidance and patience. I will be moving to the next lectures now. I felt like I needed to pass that puzzle, to move on past Lecture 4.
  • 68.
    edited April 2018

    @GrantRoy #66. Grant, I am not sure if I will be answering your question, right.. But I think the definition of a function actually does not matter, while it is numeric (or object) result is what matters. So \( \lfloor \frac{y}{2} \rfloor \) numerically (that is, it output) is the same as \( \lfloor \frac{y}{2 * 1.1} \rfloor \) . Therefore, these functions are the same, not different (while their implementation can be different, of course). Same thing as, say, x*2 can be computed by x+x, or by representing x in a binary form and shifting left by one position.

    Comment Source:@GrantRoy #66. Grant, I am not sure if I will be answering your question, right.. But I think the definition of a function actually does not matter, while it is numeric (or object) result is what matters. So \\( \lfloor \frac{y}{2} \rfloor \\) numerically (that is, it output) is the same as \\( \lfloor \frac{y}{2 * 1.1} \rfloor \\) . Therefore, these functions are the same, not different (while their implementation can be different, of course). Same thing as, say, x*2 can be computed by x+x, or by representing x in a binary form and shifting left by one position.
  • 69.
    edited April 2018

    Grant Roy wrote:

    I guess my question fundamentally is why \( \lfloor \frac{y}{\lim_{1 \to 2} } \rfloor \) doesn't work as a right adjoint for \( f(x)=2x \).

    What does \(\lim_{1 \to 2} \) mean, and how can you divide a number by this thing? Limits usually show up in expressions like this:

    $$ \lim_{x \to 2} \frac{x-2}{x^2 - 4} $$ We call this the limit of \( \frac{x-2}{x^2 - 4}\) as \(x\) approaches \(2\). The idea is that you compute \( \frac{x-2}{x^2 - 4}\) and see what, if anything, it gets closer to as \(x\) gets closer and closer to \(2\).

    But \(\lim_{1 \to 2} \) is a complete mystery to me, since you can't let \(1\) get closer and closer to \(2\) - or at least I can't. Furthermore, in the expression \(\lim_{1 \to 2} \) you're also not taking the limit of anything!

    Comment Source:Grant Roy wrote: > I guess my question fundamentally is why \\( \lfloor \frac{y}{\lim_{1 \to 2} } \rfloor \\) doesn't work as a right adjoint for \\( f(x)=2x \\). What does \\(\lim_{1 \to 2} \\) mean, and how can you divide a number by this thing? Limits usually show up in expressions like this: $$ \lim_{x \to 2} \frac{x-2}{x^2 - 4} $$ We call this the limit _of_ \\( \frac{x-2}{x^2 - 4}\\) _as_ \\(x\\) approaches \\(2\\). The idea is that you compute \\( \frac{x-2}{x^2 - 4}\\) and see what, if anything, it gets closer to as \\(x\\) gets closer and closer to \\(2\\). But \\(\lim_{1 \to 2} \\) is a complete mystery to me, since you can't let \\(1\\) get closer and closer to \\(2\\) - or at least I can't. Furthermore, in the expression \\(\lim_{1 \to 2} \\) you're also not taking the limit _of_ anything!
  • 70.
    edited April 2018

    So if I did \( \lim_{x \to 2^{+}} \frac{y}{4-x} \), I'm guessing maybe that won't work? I guess what I'm asking is, if you give me a natural number of some size, I say I can match it's halfway approximation to a natural number in the same ratio (using the floor) by adding another 9 to 1.999..... etc, for as many natural numbers as you want to give me, you give me a natural number n+1, and I add another 9 to the end of my sequence 1.999...etc. Can I do this forever? In other words, can I keep pace with you so that we always end up with the same ratio?

    Comment Source:So if I did \\( \lim_{x \to 2^{+}} \frac{y}{4-x} \\), I'm guessing maybe that won't work? I guess what I'm asking is, if you give me a natural number of some size, I say I can match it's halfway approximation to a natural number in the same ratio (using the floor) by adding another 9 to 1.999..... etc, for as many natural numbers as you want to give me, you give me a natural number n+1, and I add another 9 to the end of my sequence 1.999...etc. Can I do this forever? In other words, can I keep pace with you so that we always end up with the same ratio?
  • 71.
    edited April 2018

    I have tried to illustrate Patrick O'Neill's "no wires crossed" condition with \( f(x)=2x \) (in goldenrod) and a left adjoint \(g = \lceil\frac{n}{2}\rceil\) (in cyan).

    adjoint_wires

    Comment Source:I have tried to illustrate Patrick O'Neill's "no wires crossed" condition with \\( f(x)=2x \\) (in goldenrod) and a left adjoint \\(g = \lceil\frac{n}{2}\rceil\\) (in cyan). ![adjoint_wires](https://i.imgur.com/P20c4UB.png)
  • 72.

    Nice, Matthew!

    Comment Source:Nice, Matthew!
  • 73.
    edited April 2018

    I think I realize what I was struggling with in #70

    There is a distinction between functional equivalence and formulaic equivalence. In the mathematical sense, if two functions produce the same output, they are considered to be identical. However, in algorithmic information theory and computational complexity theory, this is not the case at all, and I think this is what caused me such grief. Two functions can have the same inputs and outputs, yet as a result of their formulaic differences, vary wildly in terms of their time, space, and information requirements.

    It is interesting to me that in this case, constructing a formula like \(\lfloor \frac{y}{1.99999} \rfloor \) might yield the same result as \( \lfloor \frac{y}{2} \rfloor \), however, the Kolmogorov complexity is much higher for the former as compared to the latter. If we were to view these two functions as black boxes, their time and space complexity asymptotically would be quite different also, with the latter function preferable in all cases.

    It seems to me that we are throwing this information away when we consider only functional equivalence and not formulaic equivalence. I don't know if there are situations where it might be interesting to be more restrictive on adjunctions (maybe only with respect to computation), where you consider formulaic equivalence in $$ f(p) \le q \iff q \le g(p) $$ choosing \(f\) and \(g \) such that they also minimize some information theoretic or complexity lower bound.

    Whether or not the above is interesting or not I don't know, but I at least had fun thinking about it, and maybe someone else will! Thanks, Vladislav, Ignacio, and John for your answers.

    Comment Source:I think I realize what I was struggling with in #70 There is a distinction between _functional equivalence_ and _formulaic equivalence_. In the mathematical sense, if two functions produce the same output, they are considered to be identical. However, in algorithmic information theory and computational complexity theory, this is not the case at all, and I think this is what caused me such grief. Two functions can have the same inputs and outputs, yet as a result of their formulaic differences, vary wildly in terms of their time, space, and information requirements. It is interesting to me that in this case, constructing a formula like \\(\lfloor \frac{y}{1.99999} \rfloor \\) might yield the same result as \\( \lfloor \frac{y}{2} \rfloor \\), however, the Kolmogorov complexity is much higher for the former as compared to the latter. If we were to view these two functions as black boxes, their time and space complexity asymptotically would be quite different also, with the latter function preferable in all cases. It seems to me that we are throwing this information away when we consider only functional equivalence and not formulaic equivalence. I don't know if there are situations where it might be interesting to be more restrictive on adjunctions (maybe only with respect to computation), where you consider formulaic equivalence in $$ f(p) \le q \iff q \le g(p) $$ choosing \\(f\\) and \\(g \\) such that they also minimize some information theoretic or complexity lower bound. Whether or not the above is interesting or not I don't know, but I at least had fun thinking about it, and maybe someone else will! Thanks, Vladislav, Ignacio, and John for your answers.
  • 74.
    edited April 2018

    @Matthew Piziak # 71

    Matthew, I think what you have diagrammed is the right adjoint, \( \lfloor \frac{n}{2} \rfloor \). N prime is n in the formula? I'm not sure to be honest, still wrapping my head around these adjoints.

    Comment Source:@Matthew Piziak # 71 Matthew, I think what you have diagrammed is the right adjoint, \\( \lfloor \frac{n}{2} \rfloor \\). N prime is n in the formula? I'm not sure to be honest, still wrapping my head around these adjoints.
  • 75.
    edited April 2018

    40, 41, 73 Is this the difference between intensional and extensional equality?

    \(\lceil \frac{n}{2} \rceil\) is extensionally equal to \( \lfloor \frac{n+1}{2} \rfloor \) but not intensionally equal.

    Comment Source:40, 41, 73 Is this the difference between intensional and extensional equality? \\(\lceil \frac{n}{2} \rceil\\) is extensionally equal to \\( \lfloor \frac{n+1}{2} \\rfloor \\) but not intensionally equal.
  • 76.
    edited April 2018

    JohnBeattie - intuitively that's right. To formalize it one needs to choose a framework to do so. In the context of functions between sets there is no concept of intensional equality, just extensional, but we can create various frameworks to formalize methods of computing functions, and then we can talk about two methods that compute the same function in different ways, which would be extensionally equal but not intensionally equal. One nice framework for this is the lambda-calculus, but there are many others.

    Comment Source:JohnBeattie - intuitively that's right. To formalize it one needs to choose a framework to do so. In the context of _functions between sets_ there is no concept of intensional equality, just extensional, but we can create various frameworks to formalize _methods of computing functions_, and then we can talk about two methods that compute the same function in different ways, which would be extensionally equal but not intensionally equal. One nice framework for this is the lambda-calculus, but there are many others.
  • 77.
    edited April 2018

    Does the following construction make any sense? Is it useful for anything?

    Given preorders \(P, Q\), consider the set \(F\) of all monotone maps \(P \to Q\) and \(Q \to P\). Then for \(f, g \in F\), we say \(f \le_F g\) iff \(f\) is a left adjoint for \(g\) (symmetrically, \(g\) is a right adjoint for \(f\)). The relation \(\le_F\) is neither transitive nor reflexive -- it's just a directed graph -- but we can make it a preorder by taking the transitive reflexive closure of the relation. Monotone maps related by this preorder have a (possibly empty) path of adjoints connecting them.

    If \(f\) and \(g\) are inverses (and \(f \neq g\)), then they are both left and right adjoints of each other, giving a cycle; so if we have invertible maps, this preorder cannot be a poset.

    A graph-theoretic observation: if \(P \neq Q\) (and before we take the reflexive transitive closure), \((F, \le_F)\) is a bipartite directed graph.

    Is there always at most one simple path between monotone maps in this graph? That would seem to give \(F\) an interesting structure -- somewhat reminiscent of a pseudoforest, but perhaps not quite the same.

    Comment Source:Does the following construction make any sense? Is it useful for anything? Given preorders \\(P, Q\\), consider the set \\(F\\) of all monotone maps \\(P \to Q\\) and \\(Q \to P\\). Then for \\(f, g \in F\\), we say \\(f \le_F g\\) iff \\(f\\) is a left adjoint for \\(g\\) (symmetrically, \\(g\\) is a right adjoint for \\(f\\)). The relation \\(\le_F\\) is neither transitive nor reflexive -- it's just a directed graph -- but we can make it a preorder by taking the transitive reflexive closure of the relation. Monotone maps related by this preorder have a (possibly empty) path of adjoints connecting them. If \\(f\\) and \\(g\\) are inverses (and \\(f \neq g\\)), then they are both left and right adjoints of each other, giving a cycle; so if we have invertible maps, this preorder cannot be a poset. A graph-theoretic observation: if \\(P \neq Q\\) (and before we take the reflexive transitive closure), \\((F, \le_F)\\) is a bipartite directed graph. Is there always at most one simple path between monotone maps in this graph? That would seem to give \\(F\\) an interesting structure -- somewhat reminiscent of a [pseudoforest](https://en.wikipedia.org/wiki/Pseudoforest), but perhaps not quite the same.
  • 78.
    edited April 2018

    Jonathan - to fit your thoughts into the usual structures we category theorists like, I'd do this:

    Consider not just two posets but any collection of them you like - even all of them. Create a category where a morphism from the poset \(P\) to the poset \(Q\) is a left adjoint monotone function \(f : P \to Q\). By saying this is a category, I'm efficiently saying a number of things, but the most interesting is this: if\(f : P \to Q \) and \(g : Q \to R\) are left adjoints, so is their composite \(gf: P \to R\).

    Much of what one wants to know about left adjoints is contained in this category.

    Comment Source:Jonathan - to fit your thoughts into the usual structures we category theorists like, I'd do this: Consider not just two posets but any collection of them you like - even all of them. Create a category where a morphism from the poset \\(P\\) to the poset \\(Q\\) is a left adjoint monotone function \\(f : P \to Q\\). By saying this is a category, I'm efficiently saying a number of things, but the most interesting is this: if\\(f : P \to Q \\) and \\(g : Q \to R\\) are left adjoints, so is their composite \\(gf: P \to R\\). Much of what one wants to know about left adjoints is contained in this category.
  • 79.
    edited April 2018

    That led to an enjoyable little puzzle!

    JMC3: Show that the composition of two left adjoints is a left adjoint.

    Comment Source:That led to an enjoyable little puzzle! **JMC3:** Show that the composition of two left adjoints is a left adjoint.
  • 80.

    Yes, that's incredibly important and I don't know why I haven't mentioned it!

    Comment Source:Yes, that's incredibly important and I don't know why I haven't mentioned it!
  • 81.

    John, re intensional, thank you. So a function between sets is, formally, a subset of the the cross product of domain and codomain, with at least one obvious condition which must be satisfied. Yet functions are very often defined by a formula and the formula tends to take precedence, at least in my intuition. Very interesting.

    Comment Source:John, re intensional, thank you. So a function between sets is, formally, a subset of the the cross product of domain and codomain, with at least one obvious condition which must be satisfied. Yet functions are very often defined by a formula and the formula tends to take precedence, at least in my intuition. Very interesting.
  • 82.

    Here are my answers to puzzle's 12 and 13. I include some tedious thinking that got me going, and then it gets clearer.

    Puzzle 12: What does it mean to find a right adjoint? It means finding a \(g\) such that \(f(n) \leq m \implies n \leq g(m)\) and \(n \leq g(m) \implies f(n) \leq m\). Let's try and see what this means for \(m = 0\). Then \(f(n) \leq 0\) holds only if \(f(n) = 0\). Thus, this only holds when \(n = 0\). This implies that \(n = 0 \leq g(0)\). This doesn't help at all. The second part requires that if \(n \leq g(0)\), then \(f(n) \leq 0\). This means that \(n\) can at most be \(0\), since \(f(n) \leq 0 \implies 2n \leq 0 \implies n = 0\). Thus we cannot have \(1 \leq g(0)\). So \(g(0) = 0\).

    That was tedious and long. So I tried to figure a faster pattern on the whiteboard.

    $$ m=0: f(n) \leq 0 \iff n \leq g(0) $$ Since the left hand side, \(f(n) \leq 0\), only holds (and does hold) when \(n \leq 0\), the right hand side can only hold (and must hold) when \(n \leq 0\). This means that we cannot have \(1 \leq g(0)\) -- thus we have \(g(0) < 1\) (where I use the fact that all numbers can be compared in \(\mathbb{N}\)).

    More generally though, \(f(n) \leq m\) holds exactly for those \(n\) (and only those) where \(n \leq \lfloor m/2 \rfloor\) after some thinking. Clearly, if \(n\) is strictly larger than this, then the double of it will be strictly larger than \(m\). And if it is smaller than this, then the double will be smaller than \(m\). Therefore, the condition \(n \leq g(m)\) must hold exactly for all \(n \leq \lfloor m/2 \rfloor\), but not for any other \(n\). Thus \(g(m)\) must be equal to the largest of those which is \(\lfloor m/2 \rfloor\).

    So \(g(m) = \lfloor m/2 \rfloor\) and there is only one solution.

    Puzzle 13 (Azimuth): Same as Puzzle 12, but find the left adjoints. How many are there?

    Answer: Given that I solved puzzle 12, I want to figure out for which \(n\) the condition holds about which we know a lot (fixing \(m\)). Once I know the set of \(n\) for which \(m \leq f(n)\) holds, I know this is exactly the set for which \(g(m) \leq n\) holds, which gives us the value of \(g(m)\) as being the smallest of all those \(n\).

    \(m \leq f(n)\) holds for all those \(n\) that are at least as large as \(m/2\). Since they must be integers, they must be larger than or equal to \(\lceil m/2 \rceil\). Thus \(g(m) \leq n\) holds exactly (and only) when \(n \leq \lceil m/2 \rceil\). So \(g(m) = \lceil m/2 \rceil\).

    Comment Source:Here are my answers to puzzle's 12 and 13. I include some tedious thinking that got me going, and then it gets clearer. **Puzzle 12**: What does it mean to find a right adjoint? It means finding a \\(g\\) such that \\(f(n) \leq m \implies n \leq g(m)\\) *and* \\(n \leq g(m) \implies f(n) \leq m\\). Let's try and see what this means for \\(m = 0\\). Then \\(f(n) \leq 0\\) holds only if \\(f(n) = 0\\). Thus, this only holds when \\(n = 0\\). This implies that \\(n = 0 \leq g(0)\\). This doesn't help at all. The second part requires that if \\(n \leq g(0)\\), then \\(f(n) \leq 0\\). This means that \\(n\\) can at most be \\(0\\), since \\(f(n) \leq 0 \implies 2n \leq 0 \implies n = 0\\). Thus we cannot have \\(1 \leq g(0)\\). So \\(g(0) = 0\\). That was tedious and long. So I tried to figure a faster pattern on the whiteboard. \[ m=0: f(n) \leq 0 \iff n \leq g(0) \] Since the left hand side, \\(f(n) \leq 0\\), only holds (and does hold) when \\(n \leq 0\\), the right hand side can only hold (and must hold) when \\(n \leq 0\\). This means that we cannot have \\(1 \leq g(0)\\) -- thus we have \\(g(0) < 1\\) (where I use the fact that all numbers can be compared in \\(\mathbb{N}\\)). More generally though, \\(f(n) \leq m\\) holds exactly for those \\(n\\) (and only those) where \\(n \leq \lfloor m/2 \rfloor\\) after some thinking. Clearly, if \\(n\\) is strictly larger than this, then the double of it will be strictly larger than \\(m\\). And if it is smaller than this, then the double will be smaller than \\(m\\). Therefore, the condition \\(n \leq g(m)\\) must hold exactly for all \\(n \leq \lfloor m/2 \rfloor\\), but not for any other \\(n\\). Thus \\(g(m)\\) must be equal to the largest of those which is \\(\lfloor m/2 \rfloor\\). So \\(g(m) = \lfloor m/2 \rfloor\\) and there is only one solution. **Puzzle 13 (Azimuth):** Same as Puzzle 12, but find the **left** adjoints. How many are there? Answer: Given that I solved puzzle 12, I want to figure out for which \\(n\\) the condition holds about which we know a lot (fixing \\(m\\)). Once I know the set of \\(n\\) for which \\(m \leq f(n)\\) holds, I know this is exactly the set for which \\(g(m) \leq n\\) holds, which gives us the value of \\(g(m)\\) as being the smallest of all those \\(n\\). \\(m \leq f(n)\\) holds for all those \\(n\\) that are at least as large as \\(m/2\\). Since they must be integers, they must be larger than or equal to \\(\lceil m/2 \rceil\\). Thus \\(g(m) \leq n\\) holds exactly (and only) when \\(n \leq \lceil m/2 \rceil\\). So \\(g(m) = \lceil m/2 \rceil\\).
  • 83.
    edited April 2018

    A question regarding the initially wrongly stated puzzle 11, is it the case that if f is monotone and has an inverse g, and if either (i) f is also the inverse of g or (ii) f is surjective, then it is the case that g must be monotone? Hence, it is both the right and left adjoint (and so is f for g)? The problem is thus with injections or rather not mapping to all values in the image space (which I guess is called codomain or something).

    Comment Source:A question regarding the initially wrongly stated puzzle 11, is it the case that if f is monotone and has an inverse g, and if either (i) f is also the inverse of g or (ii) f is surjective, then it is the case that g must be monotone? Hence, it is both the right and left adjoint (and so is f for g)? The problem is thus with injections or rather not mapping to all values in the image space (which I guess is called codomain or something).
  • 84.

    @Marc The answer to your first question is no.

    Let \(X\) to be a poset with the discrete ordering (ie \(x\leq y\implies x = y\)).

    Let \(Y\) be the same set but with the codiscrete ordering (ie \(x\leq y\) for all \(x\), \(y\)).

    Then the identity map \(X\rightarrow Y\) is monotone, but its inverse \(Y\rightarrow X\) is not.

    Comment Source:@Marc The answer to your first question is no. Let \\(X\\) to be a poset with the discrete ordering (ie \\(x\leq y\implies x = y\\)). Let \\(Y\\) be the same set but with the codiscrete ordering (ie \\(x\leq y\\) for all \\(x\\), \\(y\\)). Then the identity map \\(X\rightarrow Y\\) is monotone, but its inverse \\(Y\rightarrow X\\) is not.
  • 85.

    @Anindya Ha, that's great, thanks for the answer. I have to figure out why I thought that injections matter - I think it is because I thought "If g(f(x)) = x, then clearly if g(f(.)) is monotone (since it is the identify function), and f is, then g must be too". Some part of my model of monotone functions is off. Thanks for the clarifying answer.

    Comment Source:[@Anindya](https://forum.azimuthproject.org/profile/1950/Anindya%20Bhattacharyya) Ha, that's great, thanks for the answer. I have to figure out why I thought that injections matter - I think it is because I thought "If g(f(x)) = x, then clearly if g(f(.)) is monotone (since it is the identify function), and f is, then g must be too". Some part of my model of monotone functions is off. Thanks for the clarifying answer.
  • 86.
    edited April 2018

    @MarcKaufman WRT

    Puzzle 12: What does it mean to find a right adjoint? It means finding a \(g\) such that \(f(n) \leq m \implies n \leq g(m)\) and \(n \leq g(m) \implies f(n) \leq m\).

    Marc, I think there might be a small problem with your re-wording of the Puzzle 12. You use and in place of if-and-only-if. Those are not equivalent.

    For example, when using and -- if left side is evaluates to false, then it does not matter what the right side is, as false and < something > is going to be false.

    But that's not the case for if-and-only-if (or iff). With iff, if one side evaluates to 'false', another side 'has to' evaluate to false too.

    This was at least for, me as I war re-reading your answers.

    Another small thing, these examples operate on a number plane, which is a totally ordered set (therefore, they are also posets) - so in those scenarios we get one left and one right adjoint, if they exist (John talks about this in following lectures). However, generically, if we would be working off a pre-order, this might not be these case. In pre order, there could be more than one of those.

    Comment Source:@MarcKaufman WRT >**Puzzle 12**: What does it mean to find a right adjoint? It means finding a \\(g\\) such that \\(f(n) \leq m \implies n \leq g(m)\\) *and* \\(n \leq g(m) \implies f(n) \leq m\\). Marc, I think there might be a small problem with your re-wording of the Puzzle 12. You use **and** in place of **if-and-only-if**. Those are not equivalent. For example, when using *and* -- if left side is evaluates to false, then it does not matter what the right side is, as *false* **and** < something > is going to be *false*. But that's not the case for **if-and-only-if** (or iff). With iff, if one side evaluates to 'false', another side 'has to' evaluate to false too. This was at least for, me as I war re-reading your answers. Another small thing, these examples operate on a number plane, which is a totally ordered set (therefore, they are also posets) - so in those scenarios we get one left and one right adjoint, if they exist (John talks about this in following lectures). However, generically, if we would be working off a pre-order, this might not be these case. In pre order, there could be more than one of those.
  • 87.

    Vladislav Papayan #86, I see what you are saying, but I'm pretty sure what I wrote is true. The right adjoint is an iff statement of the form \(A \iff B\). I am saying that this is the same as proving that \(A \implies B\) and \(B \implies A\). So I use the and merely to break the iff into two implies statements.

    Regarding the second part on uniqueness: I am not sure though whether I was saying that the left or right adjoint is unique in general - only that it is in this case (which follows from features of the numbers).

    Having said that, I do think that I made that mistake when I was asking

    A question regarding the initially wrongly stated puzzle 11, is it the case that if f is monotone and has an inverse g, and if either (i) f is also the inverse of g or (ii) f is surjective, then it is the case that g must be monotone? Hence, it is both the right and left adjoint (and so is f for g)? The problem is thus with injections or rather not mapping to all values in the image space (which I guess is called codomain or something).

    I think my logic there was something like this: "If \(x \neq y\), and \(x \leq y\), then \(f(x) \leq f(y)\). Then, since \(f(x) \neq f(y)\) (true, otherwise how are you going to get an inverse?), then \(f(x) < f(y)\)" but this last part is false, exactly for the reason you mentioned. I assumed, I think, that the preorder is antisymmetric, so that \(x \leq y\) and \(y \leq x\) implies that \(x = y\), which is of course true for \(\mathbb{N}\), but not for preorders in general.

    Comment Source:[Vladislav Papayan #86](https://forum.azimuthproject.org/profile/1791/Vladislav%20Papayan), I see what you are saying, but I'm pretty sure what I wrote is true. The right adjoint is an iff statement of the form \\(A \iff B\\). I am saying that this is the same as proving that \\(A \implies B\\) and \\(B \implies A\\). So I use the *and* merely to break the **iff** into two **implies** statements. Regarding the second part on uniqueness: I am not sure though whether I was saying that the left or right adjoint is unique in general - only that it is in this case (which follows from features of the numbers). Having said that, I do think that I made that mistake when I was asking > A question regarding the initially wrongly stated puzzle 11, is it the case that if f is monotone and has an inverse g, and if either (i) f is also the inverse of g or (ii) f is surjective, then it is the case that g must be monotone? Hence, it is both the right and left adjoint (and so is f for g)? The problem is thus with injections or rather not mapping to all values in the image space (which I guess is called codomain or something). I think my logic there was something like this: "If \\(x \neq y\\), and \\(x \leq y\\), then \\(f(x) \leq f(y)\\). Then, since \\(f(x) \neq f(y)\\) (true, otherwise how are you going to get an inverse?), then \\(f(x) < f(y)\\)" but this last part is false, exactly for the reason you mentioned. I assumed, I think, that the preorder is antisymmetric, so that \\(x \leq y\\) and \\(y \leq x\\) implies that \\(x = y\\), which is of course true for \\(\mathbb{N}\\), but not for preorders in general.
  • 88.

    Hi Marc, yep, you are right wrt

    "So I use the and merely to break the iff into two implies statements."

    Comment Source:Hi Marc, yep, you are right wrt >"So I use the and merely to break the iff into two implies statements."
  • 89.

    Marc wrote:

    A question regarding the initially wrongly stated puzzle 11, is it the case that if f is monotone and has an inverse g, and if either (i) f is also the inverse of g or (ii) f is surjective, then it is the case that g must be monotone?

    Condition (i) is strange, because if \(f\) has an inverse \(g\), then \(f\) is always also the inverse of \(g\). Remember, we say \(f: X \to Y\) has an inverse \(g: Y \to X\) when \(f g : Y \to Y\) and \(gf : X \to X\) are both identity functions! So, this is equivalent to saying that \(g\) has an inverse \(f\).

    (And the inverse of a function is unique when it exists, so people don't usually say "an" inverse - they say "the" inverse!)

    If \(f g : Y \to Y\) is the identity function we say \(f\) is a left inverse of \(g\), or \(g\) is a right inverse of \(f\). Is that what you were trying to talk about, Marc?

    As for (ii), the answer is no.

    Puzzle. Find two posets \(X\) and \(Y\) and a monotone function \(f: X \to Y\) that is surjective and injective, whose inverse \(g: Y \to X\) is not monotone. (Hint: you can choose both \(X\) and \(Y)) to have just two elements.)

    Your questions are related as follows: a function has a left inverse iff it is injective, and it has a right inverse iff it is surjective.

    Comment Source:Marc wrote: > A question regarding the initially wrongly stated puzzle 11, is it the case that if f is monotone and has an inverse g, and if either (i) f is also the inverse of g or (ii) f is surjective, then it is the case that g must be monotone? Condition (i) is strange, because if \\(f\\) has an inverse \\(g\\), then \\(f\\) is _always_ also the inverse of \\(g\\). Remember, we say \\(f: X \to Y\\) has an **inverse** \\(g: Y \to X\\) when \\(f g : Y \to Y\\) and \\(gf : X \to X\\) are _both_ identity functions! So, this is equivalent to saying that \\(g\\) has an inverse \\(f\\). (And the inverse of a function is unique when it exists, so people don't usually say "an" inverse - they say _**"the"**_ inverse!) If \\(f g : Y \to Y\\) is the identity function we say \\(f\\) is a left inverse of \\(g\\), or \\(g\\) is a right inverse of \\(f\\). Is that what you were trying to talk about, Marc? As for (ii), the answer is _no_. **Puzzle.** Find two posets \\(X\\) and \\(Y\\) and a monotone function \\(f: X \to Y\\) that is surjective _and_ injective, whose inverse \\(g: Y \to X\\) is not monotone. (Hint: you can choose both \\(X\\) and \\(Y\)) to have just two elements.) Your questions are related as follows: a function has a left inverse iff it is injective, and it has a right inverse iff it is surjective.
  • 90.

    John, that's obviously true. I am not sure what I was thinking. Probably I said "an inverse" because I realized that there might be multiple elements in \(Y\) that are all less than or equal to each other, and I only wanted a unique such thing -- kind of groping towards anti-symmetry, but being confused, since anti-symmetry is about the partial order, whereas inverses have nothing to do with the partial order.

    Regarding your puzzle, I think the answer Anindya gave further up. But something like \(X, Y= {1, 2}\) with \(Y\) having the usual order, and \(X\) having the order that \(1 \leq_X 1\) and \(2 \leq_X 2\), but no other comparisons. Then the identity map \(i: X \to Y\) is monotone, but the inverse \(i': Y \to X\) is not, since in \(Y\) we have that \(1 \leq_Y 2\), but we do not have \(1 \leq_X 2\).

    Comment Source:John, that's obviously true. I am not sure what I was thinking. Probably I said "an inverse" because I realized that there might be multiple elements in \\(Y\\) that are all less than or equal to each other, and I only wanted a unique such thing -- kind of groping towards anti-symmetry, but being confused, since anti-symmetry is about the partial order, whereas inverses have nothing to do with the partial order. Regarding your puzzle, I think the answer Anindya gave further up. But something like \\(X, Y= \{1, 2\}\\) with \\(Y\\) having the usual order, and \\(X\\) having the order that \\(1 \leq_X 1\\) and \\(2 \leq_X 2\\), but no other comparisons. Then the identity map \\(i: X \to Y\\) is monotone, but the inverse \\(i': Y \to X\\) is not, since in \\(Y\\) we have that \\(1 \leq_Y 2\\), but we do not have \\(1 \leq_X 2\\).
  • 91.
    edited May 2018

    I'm late to the party here having started 7 weeks late. Here's something I don't think I've seen mentioned elsewhere.

    In puzzle 11, John pointed out that if two monotone maps are inverses, they form a Galois connection. It seems that the interesting aspects of a Galois connection arise when they are not inverses, so that non-surjectivity and non-injectivity are the relevant properties to look for.

    I wanted to test that idea by seeing how important non-surjectivity is.

    Claim. If the maps \((f,g)\) of a Galois connection on two posets \((A,B)\) are both surjective, the Galois connection is just a bijection.

    Proof. Choose \(a \in A\), and choose \(b \in g^*(a)\) (by assumption, the preimage of \(a\) is guaranteed to contain at least one \(b\)).

    Since \(a \leq g(b) = a\), the Galois condition yields \(f(a) \leq b\).

    Next choose \(a' \in f^*(b)\), again guaranteed to exist by assumption but not to be unique. Since \(f(a') = b \leq b\), the Galois condition yields \(a' \leq g(b) = a\).

    Since we now have \(a' \leq a\), monotonicity of f gives \(f(a') = b \leq f(a)\).

    At this point we have \(f(a) \leq b\) and \(b \leq f(a)\). Since we have limited ourselves to posets, that means that \(f(a) = b\). But \(b\) was arbitrarily chosen from the preimage, so it must be that the preimage \(g^*(a)\) only has one element \(b = f(a)\). Thus \(g\) is injective, with \(f(a) = b\) and \(g(b) = a\).

    A similar derivation shows that \(f\) is also injective. So we have two injective, surjective maps between \(A\) and \(B\). Thus any doubly surjective Galois connection is simply a bijection. \(\blacksquare\)

    Comment Source:I'm late to the party here having started 7 weeks late. Here's something I don't think I've seen mentioned elsewhere. In puzzle 11, John pointed out that if two monotone maps are inverses, they form a Galois connection. It seems that the interesting aspects of a Galois connection arise when they are not inverses, so that non-surjectivity and non-injectivity are the relevant properties to look for. I wanted to test that idea by seeing how important non-surjectivity is. **Claim.** If the maps \\((f,g)\\) of a Galois connection on two *posets* \\((A,B)\\) are both surjective, the Galois connection is just a bijection. *Proof.* Choose \\(a \in A\\), and choose \\(b \in g^*(a)\\) (by assumption, the preimage of \\(a\\) is guaranteed to contain at least one \\(b\\\)). Since \\(a \leq g(b) = a\\), the Galois condition yields \\(f(a) \leq b\\). Next choose \\(a' \in f^*(b)\\), again guaranteed to exist by assumption but not to be unique. Since \\(f(a') = b \leq b\\), the Galois condition yields \\(a' \leq g(b) = a\\). Since we now have \\(a' \leq a\\), monotonicity of f gives \\(f(a') = b \leq f(a)\\). At this point we have \\(f(a) \leq b\\) and \\(b \leq f(a)\\). Since we have limited ourselves to posets, that means that \\(f(a) = b\\). But \\(b\\) was arbitrarily chosen from the preimage, so it must be that the preimage \\(g^*(a)\\) only has one element \\(b = f(a)\\). Thus \\(g\\) is injective, with \\(f(a) = b\\) and \\(g(b) = a\\). A similar derivation shows that \\(f\\) is also injective. So we have two injective, surjective maps between \\(A\\) and \\(B\\). Thus any doubly surjective Galois connection is simply a bijection. \\(\blacksquare\\)
  • 92.
    edited August 2018

    Uhg, so the first example in the book of a Galois connection defines \( \lceil x \rceil \) and \( \lfloor x \rfloor \) in non-standard ways: "Write \( \lceil z \rceil \) for the smallest natural number above \( z \in R \) " which would make \( \lceil 3 \rceil = 4 \) . Under this interpretation, there is not a Galois connection between \( \lceil −/3 \rceil \) and \( (3 \times −) \) ( \( \lceil 6/3 \rceil \nleq 2 \) but \( 6 \leq 3(2) \) ). This had me hung up for quite a bit.

    Comment Source:Uhg, so the first example in the book of a Galois connection defines \\( \lceil x \rceil \\) and \\( \lfloor x \rfloor \\) in non-standard ways: "Write \\( \lceil z \rceil \\) for the smallest natural number above \\( z \in R \\) " which would make \\( \lceil 3 \rceil = 4 \\) . Under this interpretation, there is not a Galois connection between \\( \lceil −/3 \rceil \\) and \\( (3 \times −) \\) ( \\( \lceil 6/3 \rceil \nleq 2 \\) but \\( 6 \leq 3(2) \\) ). This had me hung up for quite a bit.
  • 93.
    edited November 2018

    Now that I'm back to review the lessons, I have a tiny question re John's wording in 7:

    ... If a function \(f : A \to B\) is a monotone map between preorders and it has an inverse, its inverse may not be a monotone map.

    Does "may not be" here mean "is forbidden to be" or "is not necessarily"? My impression from the context is the latter but maybe I'm mistaken.

    Comment Source:Now that I'm back to review the lessons, I have a tiny question re [John's wording in 7](https://forum.azimuthproject.org/discussion/comment/16205/#Comment_16205): >... If a function \\(f : A \to B\\) is a monotone map between preorders and it has an inverse, its inverse _may not be_ a monotone map. Does "may not be" here mean "is forbidden to be" or "is not necessarily"? My impression from the context is the latter but maybe I'm mistaken.
  • 94.

    @Fredrick in 58:

    The definition for adjunction is $$ f(p) \le q \iff q \le g(p) ...$$

    Did you mean to say \( f(p) \le q \iff p \le g(q) \)?

    Comment Source:@Fredrick in [58](https://forum.azimuthproject.org/discussion/comment/17001/#Comment_17001): >The definition for adjunction is >$$ f(p) \le q \iff q \le g(p) ...$$ Did you mean to say \\( f(p) \le q \iff p \le g(q) \\)?
  • 95.
    edited November 2018

    Does "may not be" here mean "is forbidden to be" or "is not necessarily"? My impression from the context is the latter but maybe I'm mistaken.

    The second: The Identity is monotone and so is its inverse (as it is its own inverse). The example above with two different preorders on the set {1,2} is the other case, where g is not monotone.

    Comment Source:> Does "may not be" here mean "is forbidden to be" or "is not necessarily"? My impression from the context is the latter but maybe I'm mistaken. The second: The Identity is monotone and so is its inverse (as it is its own inverse). The example above with two different preorders on the set {1,2} is the other case, where g is not monotone.
Sign In or Register to comment.