Need two functions always be composed to prove they are inverses? Normally, if I claimed that $f: A \rightarrow B$ and $g: B \rightarrow A$ were inverses of each other, I would check for the following results: $f \circ g(b) = b$ and $g \circ f(a) = a$.
Suppose I knew, however, that $f$ is bijective. And suppose I knew that $g$ is a working pre-image mapping. Then, can I claim that $g$ is the inverse of $f$, even if I haven't composed $g$ with $f$? A concrete example:

Suppose we proved that $f: \mathbb{R} \rightarrow \mathbb{R}$ where $f(x) = x^{3}$ is a bijection. As part of proving surjectivity of $f$, we found for any $y \in \mathbb{R}$ an $x$ such that $f(x) = y$. In particular, we decided to let $x = y^{1/3}$.
Can I safely claim that $h(y) = y^{1/3}$ is $f^{-1}$ without needing to compose $h$ with $f$?
I would think so, since if $f$ is a bijection, then its inverse is also a bijection and well-defined. Thus, am I right that composition is not the only means to proving two functions are inverses?
 A: Yes, if you know $f$ is bijective, you can skip one of the verifications. This follows because inverses are unique. 
If $f:A \to B$ is a bijection, there is a unique function $f^{-1}:B \to A$ such that $f\circ f^{-1} = \mathrm{id}_B$ and $f^{-1} \circ f = \mathrm{id}_A$.
If you show that $g \circ f = \mathrm{id}_A$ then $g=g \circ \mathrm{id}_B = g \circ (f\circ f^{-1}) = (g \circ f) \circ f^{-1} = \mathrm{id}_A \circ f^{-1} = f^{-1}$. Therefore, since $g=f^{-1}$, the other composition comes for free.
A: Your set-up: we found for any $y∈\Bbb R$ an x such that $f(x)=y$. In particular, we decided to let $x=y^{1/3}$. ... $h(y)=y^{1/3}$.
Then this is an inverse function.  Since $f(x)=y$ and $h(y)=y^{1/3}=x$, then these two statements imply $h(f(x))=x$ for all $x$.
So the only thing you are really "avoiding" checking is that $f(h(y)) = y$.  The way you can see that is:
$h\circ$ $f(h(y)) = (h\circ f)(h(y)) = h(y)$, since we know that $h\circ f=\iota$
So $h\circ$ $f(h(y)) = h\circ y$ for all $y$; and since $h$ is a bijection, then this implies $f(h(y)) = y$ for all $y$.  So yes, you get it for free when proving that $f$ a bijection.
