**Puzzle 141.** Define \\(f : \mathbb{Z} \to \mathbb{Z}\\) by \\(f(x) = 2x\\). Then both \\(g(x) = \lfloor \frac{x}{2} \rfloor\\) and \\(h(x) = \lfloor \frac{x + 1}{2}\rfloor\\) are left inverses of \\(f\\).

**Puzzle 142.** Define \\(f' : \mathbb{Z} \to \mathbb{Z}\\) by \\(f'(x) = \lfloor \frac{x}{2} \rfloor\\). Then both \\(g'(x) = 2x\\) and \\(h'(x) = 2x + 1\\) are right inverses of \\(f'\\).

Notice that \\(f\\) is injective but not surjective, and \\(f'\\) is surjective but not injective. In \\(\textbf{Set}\\), functions with inverses are the same as bijections, and we want to avoid those here.

There also seems to be a relationship suggested here with Galois connections, since several of these functions showed up back when we were discussing those.

**Puzzle 142.** Define \\(f' : \mathbb{Z} \to \mathbb{Z}\\) by \\(f'(x) = \lfloor \frac{x}{2} \rfloor\\). Then both \\(g'(x) = 2x\\) and \\(h'(x) = 2x + 1\\) are right inverses of \\(f'\\).

Notice that \\(f\\) is injective but not surjective, and \\(f'\\) is surjective but not injective. In \\(\textbf{Set}\\), functions with inverses are the same as bijections, and we want to avoid those here.

There also seems to be a relationship suggested here with Galois connections, since several of these functions showed up back when we were discussing those.