[Math] Proving $f(x)^3$ is bijective

discrete mathematicsfunctions

Problem: Given that $f: \mathbb{R} \rightarrow \mathbb{R}$ is bijective, prove that $f^3$ is bijective.

Proof 1a: $f(x)^3$ is injective

  1. Let $f(x)^3 = f(y)^3$
  2. $\sqrt[3]{f(x)^3} = \sqrt[3]{f(y)^3}$
  3. $f(x) = f(y)$
  4. Because $f$ is injective, $x = y$

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

  1. Let $f(a)^3 = b$ for arbitrary constants $a$ and $b$
  2. $\sqrt[3]{f(a)^3} = \sqrt[3]{b}$
  3. $f(a) = \sqrt[3]{b}$
  4. Let $b' = \sqrt[3]{b} \rightarrow f(a) = b'$
  5. $f(x)^3$ 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!

Best Answer

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.

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