Is the set $f=\{(x^3,x):x\in\mathbb R\}$ a function from $\mathbb R$ to $\mathbb R$? I need to find out if $f$ is a function or not. That is, whether or not the first coordinate of the ordered pair occurs only once in $f$. If yes, then it is a function.
My answer is yes, $f$ is a function because $f=x^{1/3}$ has a different value for every $x$.
Is this enough or do I need more explanation?
 A: If we consider the equation $x^3=r$ then for any $r\in \Bbb R$ this equation has a unique real solution(the other two are complex which are conjugates to each other). So the correspondence $x^3\rightarrow x$ is a function.
Note that the definition of a function comes from the Cartesian product. If $f:A\mapsto B$ is a function then $f\subseteq A×B$. So 
$(1)$ $f=\{(x^3,x):x\in\mathbb R\}$ and 
$(2)$ $f:A\rightarrow B$ such that $f(x^3)=x$ , both of them are function and same function.
A: Yes. Just also note that the function $x\mapsto x^3$ is surjective from $\mathbb R$ to itself so that you verify that you have domain $\mathbb R$.
A: Let $y = g(x):= x^3$,  $x \in \mathbb{R}$.
Domain$_g = \mathbb{R}; $ Range$_g = \mathbb{R}$.
$g$ is injective and surjective, I.e. bijective. 
Injective: $g$ is strictly monotonic.
Surjective: $g$ is continuous on $[a,b]$ , 
for any $a \in \mathbb{R}.$
$\Rightarrow :$
An inverse function $g^{-1}$ exists, continuous and strictly monotonic.
$g^{-1}(y) = y^{1/3},$ $y \in \mathbb{R}$.
Any similarity to your function $f?$
