Question # Is f(s)=x^3 a bijection from RR to RR?

Transformations of functions
ANSWERED Is $$\displaystyle{f{{\left({s}\right)}}}={x}^{{3}}$$ a bijection from $$\displaystyle\mathbb{R}$$ to $$\displaystyle\mathbb{R}$$? 2021-09-14
$$\displaystyle{f}:{A}\to{B}$$ is one-to-one if $$\displaystyle∀{a},{b}\in{A},{f{{\left({a}\right)}}}={f{{\left({b}\right)}}}\to{a}={b}$$
Assume that $$\displaystyle∀{x},{y}\in\mathbb{R},{f{{\left({x}\right)}}}={f{{\left({y}\right)}}}.$$
f(x)=f(y)
$$\displaystyle{x}^{{3}}={y}^{{3}}$$
x=y
Thus,
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}$$ is one-to-one
"For the function to be onto:
f(x)=y
let $$\displaystyle{y}\in\mathbb{R}$$
Choose x to be $$\displaystyle{y}^{{\frac{{1}}{{3}}}}.$$
Then,
$$\displaystyle{x}\in\mathbb{R}$$
and
$$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}$$
$$\displaystyle={\left({y}^{{\frac{{1}}{{3}}}}\right)}^{{3}}$$
=y
Thus $$\displaystyle{f{{\left({x}\right)}}}={x}^{{3}}$$ is onto
Since it's both one-to-one and onto, it's a bijection function from $$\displaystyle\mathbb{R}$$ to $$\displaystyle\mathbb{R}$$."