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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?

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    $\begingroup$ "Don't forget the ${}+C$." $\endgroup$ – Eric Towers Jan 17 '16 at 22:05
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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 $r\in [m,m+1[$.

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  • $\begingroup$ Could you please give two more examples? $\endgroup$ – FreeMind Jan 17 '16 at 13:02
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    $\begingroup$ @FreeMind: Asking for "two more examples" is an odd request that strongly suggests someone asked you for three examples. Could you please clarify? $\endgroup$ – Andrew D. Hwang Jan 17 '16 at 14:37
  • $\begingroup$ @FreeMind Do you understand how Arnaud D. created the above two examples? If not, I would ask him for clarification. $\endgroup$ – PyRulez Jan 17 '16 at 15:39
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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 $n\in\mathbb N$.

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    $\begingroup$ Indeed, with that method you can easily show that the set of right inverses can have any cardinality, by just replacing $\mathbb N$ with a set of the desired cardinality. $\endgroup$ – celtschk Jan 17 '16 at 14:17
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    $\begingroup$ +1 I have to give kudos for having the simplest example possible! $\endgroup$ – Cort Ammon Jan 17 '16 at 18:45
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    $\begingroup$ To be precise, this $f$ does not have to be like that. Instead of the second bullet, you may simply have infinitely many points with non-singleton preimages (e.g. the projection onto ${\bf N}$ of $\{1,2\}\times {\bf N}$). In this case, there will be at least continuum many right inverses (assuming axiom of choice ;) ). (But you must have one of the two.) $\endgroup$ – tomasz Jan 18 '16 at 12:01
  • $\begingroup$ @tomasz thanks for that. I have edited the answer to include this. $\endgroup$ – Mathmo123 Jan 18 '16 at 12:17
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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 $Y$.

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