Jonathan wrote:

> **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\$$.

Yes, that's nice! These functions \$$g\$$ and \$$h\$$ take different values on odd integers, but they agree on even integers - they just halve any even integer - so they both provide a left inverse to the function \$$f\$$, which double integers.

You could in fact have chosen \$$g\$$ to map the odd integers to whatever you wanted!