It looks like you're new here. If you want to get involved, click one of these buttons!

- All Categories 2.2K
- Applied Category Theory Course 352
- Applied Category Theory Seminar 4
- Exercises 149
- Discussion Groups 49
- How to Use MathJax 15
- Chat 479
- Azimuth Code Project 108
- News and Information 145
- Azimuth Blog 149
- Azimuth Forum 29
- Azimuth Project 189
- - Strategy 108
- - Conventions and Policies 21
- - Questions 43
- Azimuth Wiki 711
- - Latest Changes 701
- - - Action 14
- - - Biodiversity 8
- - - Books 2
- - - Carbon 9
- - - Computational methods 38
- - - Climate 53
- - - Earth science 23
- - - Ecology 43
- - - Energy 29
- - - Experiments 30
- - - Geoengineering 0
- - - Mathematical methods 69
- - - Meta 9
- - - Methodology 16
- - - Natural resources 7
- - - Oceans 4
- - - Organizations 34
- - - People 6
- - - Publishing 4
- - - Reports 3
- - - Software 21
- - - Statistical methods 2
- - - Sustainability 4
- - - Things to do 2
- - - Visualisation 1
- General 39

Options

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

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

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

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.

`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.`

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.

`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.`

When a mathematician says "monotone maps between discrete posets are

justfunctions", they mean thateveryfunction 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

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

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

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

codiscretepreorder.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!

`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!`

John wrote:

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.

`[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.`

I assume it has something to do with this:

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

`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!`