Exercise 47 - Chapter 1

Fredrick Eisele

Given preorders \(A\) and \(B\), let their product \(A \times B\) be the cartesian product of sets \(A\) and \(B\) made into a preorder as follows:

$$ (a,b) \le (a',b') \textrm{ if and only if } a \le a' \textrm{ and } b \le b' .$$ Draw the product of the two preorders drawn below:

Jared Summers

Jonathan Castello

This is pretty fascinating to me. You can imagine acquiring the product by extruding \(A\) from \(B\). But it also works just as well by imagining extruding \(B\) from \(A\)!

It's like we're replacing every element in one preorder with a copy of the other preorder, and distributing edges of the original preorder through the copies.

Product constructions aren't particularly new to me, but... There's something incredibly pleasant about a geometric explanation.

Jonathan Castello

Here's another image showing what I mean by "replacing every element". The outer edge "distributes" through the larger cells onto every analogous pair.

Fredrick Eisele

Nice illustrations @JonathanCastello . https://forum.azimuthproject.org/discussion/comment/17639/#Comment_17639 in particular shows commutivity of paths.