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Exercise 50 - Chapter 1

edited April 2018 in Exercises

We can make any set \(X\) into a poset where we say \(x \leq y\) if and only if \(x = y\). A poset of this form is called the discrete poset.

Show that monotone maps between discrete posets are just functions.

Comments

  • 1.

    Monotone maps are functions by definition. I'm not sure what this exercise was really asking.

    Comment Source:Monotone maps are functions by definition. I'm not sure what this exercise was really asking.
  • 2.
    edited April 2018

    I think the exercise is getting at the issue that there is no monotonic relation that needs to be preserved by the mapping, hence it is "just" a function.

    Comment Source:I think the exercise is getting at the issue that there is no monotonic relation that needs to be preserved by the mapping, hence it is "just" a function.
  • 3.
    edited April 2018

    Let \(P, Q\) be discrete preorders, and let \(f : P \to Q\) be a function. Let \(x, y \in P\), and suppose \(x \le y\); then \(x = y\), since \(P\) is a discrete preorder. So \(f(x) = f(y)\), hence \(f(x) \le f(y)\). Therefore, \(f\) is a monotone map.

    Notice that we didn't actually use the fact that \(Q\) is a discrete preorder. Thus, any function out of a discrete preorder (to an arbitrary preorder) is a monotone map.

    Comment Source:Let \\(P, Q\\) be discrete preorders, and let \\(f : P \to Q\\) be a function. Let \\(x, y \in P\\), and suppose \\(x \le y\\); then \\(x = y\\), since \\(P\\) is a discrete preorder. So \\(f(x) = f(y)\\), hence \\(f(x) \le f(y)\\). Therefore, \\(f\\) is a monotone map. Notice that we didn't actually use the fact that \\(Q\\) is a discrete preorder. Thus, any function out of a discrete preorder (to an arbitrary preorder) is a monotone map.
  • 4.
    edited April 2018

    When a mathematician says "monotone maps between discrete posets are just functions", they mean that every function between discrete posets is a monotone map.

    This is much stronger than saying "monotone maps between discrete posets are functions", which is also true, but boring. The word just changes everything.

    But Jonathan is right: we can say even more. Remember, for any set \(X\):

    1. Setting \(x \leq y\) if and only if \(x = y\) gives you a preorder on \(X\) called the discrete preorder.

    2. Setting \(x \leq y \) for all \(x,y \in X\) gives you a preorder on \(X\) called the codiscrete preorder.

    Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map!

    Comment Source:When a mathematician says "monotone maps between discrete posets are _just_ functions", they mean that _every_ function between discrete posets is a monotone map. This is much stronger than saying "monotone maps between discrete posets are functions", which is also true, but boring. The word _just_ changes everything. But Jonathan is right: we can say even more. Remember, for any set \\(X\\): 1. Setting \\(x \leq y\\) if and only if \\(x = y\\) gives you a preorder on \\(X\\) called the **discrete** preorder. 2. Setting \\(x \leq y \\) for all \\(x,y \in X\\) gives you a preorder on \\(X\\) called the **codiscrete** preorder. Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map!
  • 5.

    John wrote:

    Setting \(x \leq y \) for all \(x,y \in X\) gives you a preorder on \(X\) called the codiscrete preorder.

    Why is this preorder called "codiscrete"? In what sense is it dual to the discrete preorder?

    Usually to obtain the dual of a category we reverse its arrows; but for the discrete preorder, reversing the inequalities should still give a discrete preorder.

    Comment Source:[John](https://forum.azimuthproject.org/profile/17/John%20Baez) wrote: > Setting \\(x \leq y \\) for all \\(x,y \in X\\) gives you a preorder on \\(X\\) called the **codiscrete** preorder. Why is this preorder called "codiscrete"? In what sense is it dual to the discrete preorder? Usually to obtain the dual of a category we reverse its arrows; but for the discrete preorder, reversing the inequalities should still give a discrete preorder.
  • 6.

    I assume it has something to do with this:

    Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map!

    I'd love to hear what the reason is well!

    Comment Source:I assume it has something to do with this: > Any function from a discrete preorder to any preorder is a monotone map. Similarly, any function from any preorder to a codiscrete preorder is a monotone map! I'd love to hear what the reason is well!
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