Right inverse of $f(x)= x^2$ that is not $\pm \sqrt x$

Question

I am looking for three different right inverses of $f(x)=x^2$. So far I've only been able to come up with the positive and the negative square roots, $\pm \sqrt x$. Furthermore, the right inverse maps from the positive real numbers to the real numbers.

Answer 1

The complete set of right inverse maps is given as follows:

For any $A\subseteq R^+$, be $$g_A(x) = \begin{cases} \sqrt{x} & x\in A\\ -\sqrt{x} & x\notin A \end{cases}$$

Then $g_A(x)$ is a right inverse to $f(x)=x^2$, and further we have $g_A=g_B \iff A=B$. There are no further right inverses

Proof:

• $g_A$ is a right inverse:

  $$f\circ g_A(x) = \begin{cases} \left(\sqrt{x}\right)^2 = x & x\in A\\ \left(-\sqrt{x}\right)^2 = x & x\notin A \end{cases}$$

• $g_A = g_B \iff A=B$:

  Obviously $A=B\implies g_A=g_B$. Therefore assume $A\ne B$. Therefore there exists a $x$ that is in only one of the sets. Assume w.l.o.g $x\in A$, $x\notin B$. Then $g_A(x)=\sqrt{x}$ and $g_B(x)=-\sqrt{x}$. Since $x\ne 0$ (because $0\notin\mathbb R^+$), $\sqrt{x}\ne-\sqrt{x}$ and thus $g_A\ne g_B$ since they differ in $x$.

• There are no further right inverses.

  Be $h$ a right inverse of $f$. Then for any $x\in\mathbb R^+_0$ we have $x=f\circ h(x)=(h(x))^2$. But given $x$, the only solutions for $a$ of the equation $x=a^2$ are $a=\sqrt{x}$ and $a=-\sqrt{x}$. Therefore for any $x$, we have either $h(x)=\sqrt{x}$ or $h(x)=-\sqrt{x}$. Be $A=\{x\in\mathbb R^+|h(x)=\sqrt{x}\}$. Then $h(x)=g_A(x)$. $\square$

So you've got $2^{2^{\aleph_0}}$ different functions to choose from. Choosing an arbitrary set of three of them is left as exercise to the reader.

Answer 2

You might get a combination of the two: $\sqrt x$ when $x<1$, and $-\sqrt x$ when $x\ge 1$, or any other combination (rationals vs irrationals, etc).

Answer 3

You have identified the continuous right inverses. But think also about discontinuous right inverses such as this, which is defined separately for $x$ rational and irrational:

For $x$ nonnegative rational, $g(x) = \sqrt{x}$

For $x$ positive irrational, $g(x) = -\sqrt{x}$

Answer 4

Assuming that $f\colon \Bbb R\to \Bbb R_{\geq 0}$, we have that $f$ is surjective. We also know that if $g\colon\Bbb R_{\geq 0}\to \Bbb R$ is its right inverse, then $f\circ g = \operatorname{id}$ and thus $g(x)\in f^{-1}(\{x\}),\ \forall x\in\Bbb R_{\geq 0}$. Conversely, if $g(x)\in f^{-1}(\{x\}),\ \forall x\in\Bbb R_{\geq 0}$, then $g$ is right inverse of $f$. But, for all $x$ we have that $f^{-1}(\{x\}) = \{\sqrt x,-\sqrt x\}$ so any right inverse $g$ of $f$ is of the form

$$g(x) = \left\{ \begin{array}{r l} \sqrt x,& x\in A\\ -\sqrt x, &x\in \Bbb R_{\geq 0}\setminus A \end{array}\right.$$

for arbitrary $A\subseteq \Bbb R_{\geq 0}$.