Showing Surjectivitity of $f(x) = x^3$ I want to show that the function $f: \mathbb{R} \to \mathbb{R},\; f(x) = x^3$ is surjective.

First Question: If a function has an inverse, it is bijective yes?
Second Question: Is my process below, for showing surjectivity of $y = x^3$, incorrect because I am assuming that the function has an inverse? i.e., by assuming it has an inverse I am assuming it is bijective and using that to show it is surjective.

Surjective statement: For a function $f: A \to B$, the function is surjective if $\forall_{ y \in B}, \; \exists_{x \in A}\;|\;f(x) = y$
So then $\forall_{y \in \mathbb{R}}, \; \exists_{x \in \mathbb{R}}\; | \; f(x) = y $
If such an real $x$ exists, then we have $x^3 = y$ and $x = y^{\frac{1}{3}}$. $x$ is a real number (how do I show this?). It follows that $f(x) = (y^{\frac{1}{3}})^3 = y$
 A: (1) Yes, by definition, the inverse of a map $f: X \to Y$ is a map $f^{-1}: Y \to X$ such that $f^{-1} \circ f = id_X$ and $f \circ f^{-1} = id_Y$, so in particular $f$ must be bijective.
Note that one commonly one will say that a function $f$ defined by a rule (but without specifying the codomain) has an inverse, but in the sense of the above definition this requires that we take the codomain of $f$ to be its image.
For example, we often say that $f(x) := e^x$ is invertible, but if we take the domain of $f$ to be $\mathbb{R}$, in order for $f$ to have an inverse, we must regard $f$ as a function $\mathbb{R} \to \mathbb{R}_+$, where the codomain here is the set of positive numbers. If we instead treat it as a map $\mathbb{R} \to \mathbb{R}$ then it is not invertible, despite that it has the same rule as an invertible function.
The fact that two functions with the same rule and same domain can differ in whether they are invertible is part of the reason why a function, strictly speaking, includes as part of its definition its codomain in addition to its rule and domain.
(2) If you already know what the function we denote $x^{1/3}$ is, then your proof is sufficient to show that $x\mapsto x^3$ is bijective. On the other hand, if you're simply using $x^{1/3}$ as a symbol for the inverse of $x\mapsto x^3$ without any more information, then your answer is circular.
Much thanks to @DanZimm, who pointed out a confusion in a previous answer.
A: You don't have to assume your function has an inverse. You can show that it has in inverse (which you did) and from this it follows that it is bijective. For a function to be bijective it has to be both injective and surjective. So by showing that a function has an inverse, you also show it is surjective.
A: check f'(x) and the function will be strictly increasing and unbounded. so it will be surjective
A: $$\lim_{x\to\pm\infty }f(x)=\pm\infty $$
then $f(\mathbb R)=\mathbb R$, and so $f$ is surjective.
