Function with infinitely many right inverses?

Question

I was thinking about a function with infinitely many right inverses but I could not come up with anything.
Does there exist a function with infinitely many right inverses?

Answer 1

The function $$f:\mathbb{Z}\to \{0,1\}:x\mapsto \left\{\begin{matrix} 0 \text{ if }x \text{ is even}\\ 1\text{ if }x\text{ is odd}\end{matrix}\right.$$has infinitely many right inverses, since it suffices to map $0$ to an even number and $1$ to an odd one.

Another example is $\sin$, for which any function of the form $$t\in [-1,1]\mapsto \arcsin(t)+2k\pi $$with $k\in \mathbb{Z}$ is a right inverse. This also works for other trigonometric functions, of course.

A third example would be the floor function $f:\mathbb{R} \to \mathbb{Z} : x\mapsto \lfloor x\rfloor $; here it suffices to map any $m\in \mathbb{Z}$ to any $r\in [m,m+1[$.

Answer 2

You need to find a function $f:S\to T$ which is:

• surjective (so that a right inverse exists)
• so that there are infinitely many right inverses, either
  • for which there is some $y\in T$ with $f^{-1}(y)$ an infinite set
  • or for which there are infinitely many $y\in T$ with $\#f^{-1}(y) \ge 2$

Any function with these two properties will have infinitely many right inverses. A simple example is the trivial function $$f:\mathbb N\to \{0\}$$ for which the map $$i_n:\{0\}\to\mathbb N\\0\mapsto n$$is a right inverse for every $n\in\mathbb N$.

Answer 3

For another example let $X$ be any infinite set, and let

$$f:X\times X\to X:\langle x,y\rangle\mapsto y\;.$$

For each $x\in X$ let

$$g_x:X\to X\times X:y\mapsto\langle x,y\rangle\;.$$

Then $g_x$ is a right inverse for $f$ for each $x\in X$. This is one of the most natural constructions. It can be generalized by letting $Y$ be any non-empty set, replacing each instance of $X\times X$ by $X\times Y$, and replacing the range of $f$ and the domain of the functions $g_x$ by $Y$.