**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.