Exercise 50 - Chapter 1

Vishal Lama

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.

Andrew Ardill

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

Fredrick Eisele

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.

Jonathan Castello

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.

John Baez

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!

Dan Oneata

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.

Jonathan Castello

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!