Exercise 58 - Chapter 1

Fredrick Eisele

Check the two claims made in Proposition 1.53.

Proposition 1.53. For any preorder \( (P, \le_P ) \), the identity function is monotone.

If \( (Q, \le_Q ) \) and \( (R, \le_R) \) are preorders and \( f : P \rightarrow Q \) and \( g : Q \rightarrow R \) are monotone, then \( ( f .g) : P \rightarrow R \) is also monotone.

Deepak Suresh

Let \(x_p\) and \(y_p\) be elements of \(P\) such that \(x_p\leq y_p\).

Since \(f\) is monotone, it maps to elements \(x_q\) and \(y_q\), where \(x_q\leq y_q\)

Monotone function \(g\) maps \(x_q\) and \(y_q\) to \(x_r\) and \(y_r\) such that \(x_r\leq y_r\)

Thus, \( ( f .g) : P \rightarrow R \) maps \(x_p\) and \(y_p\) to \(x_r\) and \(y_r\), since \(x_p\leq y_p\) and \(x_r\leq y_r\), \( ( f .g) \) is monotone

John Baez

Looks good, Deepak! Like 99.5% of mathematicians, I usually use the notation \(g \circ f\) or just \(gf\) for the composite of functions \(f : P \to Q\) and \(g : Q \to R\), but Fong and Spivak use \( (f.g) \), which has certain advantages.