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 $f(x) = (y^{\frac{1}{3}})^3 = y$

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.
