Proving $f(x)^3$ is bijective Problem: Given that $f: \mathbb{R} \rightarrow \mathbb{R}$ is bijective, prove that $f^3$ is bijective.
Proof 1a: $f(x)^3$ is injective


*

*Let $f(x)^3 = f(y)^3$

*$\sqrt[3]{f(x)^3} = \sqrt[3]{f(y)^3}$

*$f(x) = f(y)$

*Because $f$ is injective, $x = y$


Proof 1b: $f(x)^3$ is surjective


*

*Let $f(a)^3 = b$ for arbitrary constants $a$ and $b$

*$\sqrt[3]{f(a)^3} = \sqrt[3]{b}$

*$f(a) = \sqrt[3]{b}$

*Let $b' = \sqrt[3]{b} \rightarrow f(a) = b'$

*$f(x)^3$ must be surjective because $\forall b' \in \mathbb{R}$, $\exists a$ such that $f(a) = b'$


I'm wondering if this proof is correct. If some steps have logical errors, I would appreciate if anyone had suggestions to correct them. Thank you!
 A: Things can be simplified a bit if you know that the composition of two bijective functions is bijective.
Then you only have to prove that $x\mapsto x^3$ is bijective (which is easy, since its inverse is familiar to us), and you can immediately conclude that the composition with $f$ is bijective.
A: Your proof for injectivity is correct, but you should change some wording in your proof for surjectivity. To prove a function $g(x)$ is surjective, you must show that for every $y$ in the codomain, there exists $x$ such that $g(x)=y$. So instead of starting with "Let $f(a)^3=b$ for arbitrary constants $a$ and $b$," you want to start by saying "let $b \in \mathbb{R}$." Then proceed to show there exists $a$ such that $f(a)^3=b$. The equations you wrote down are the "scratch work" that you do to figure out what that $a$ should be. But the actual proof should instead provided a candidate for $a$ and then show it works. I.e., "let $a=f^{-1}(\sqrt[3]{b})$ (which is well-defined because $f$ is bijective, so has an inverse); then $f(a)^3=f(f^{-1}(\sqrt[3]{b}))^3=(\sqrt[3]{b})^3=b$."
