Jonathan wrote:

> **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'\$$.

Nice again! The functions \$$g'\$$ doubles any integer; \$$h'\$$ doubles it and adds one, but \$$f'\$$ takes each even integer and the next odd integer to the same value, so \$$f' \circ g' = f' \circ h'\$$, and in fact \$$f' \circ g' = f' \circ h' = 1_{\mathbb{Z}}\$$.

> 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.

Right. Everyone who hasn't pondered this already should tackle these:

**Puzzle.** Exactly which functions have left inverses?

**Puzzle.** Exactly which functions have right inverses?

The answer in each case is a famous class of functions.