Integer Power Function is Bijective iff Index is Odd

Theorem
Let $n \in \Z_{\ge 0}$ be a positive integer.

Let $f_n: \R \to \R$ be the real function defined as:
 * $f_n \left({x}\right) = x^n$

Then $f_n$ is a bijection iff $n$ is odd.

Even Index
Suppose $n$ is even.

Let $x \ne 0$.

Then $1^n = \left({-1}\right)^n$ by Power of Ring Negative, so $f_n$ is not injective.

Alternatively, by Even Powers are Positive, $f_n$ is not surjective.

Since a bijection must be both injective and surjective, and $f_n$ is neither,

for even $n$, $f_n$ is not bijective by definition.

Odd Index
Now suppose $n$ is odd.

Since Odd Power Function is Strictly Increasing, $f_n$ is injective.

From Existence of Root we have that:
 * $\forall x \in \R_{\ge 0}: \exists y \in \R: y^n = x$

From Power of Ring Negative we have that $\left({-x}\right)^n = - \left({x^n}\right)$ and so:
 * $\forall x \in \R_{\le 0}: \exists y \in \R: y^n = x$

Thus:
 * $\forall x \in \R: \exists y \in \R: y^n = x$

and so $f_n$ is a surjection.

So when $n$ is odd, $f_n$ is both injective and surjective, and so by definition bijective.

Hence the result.