Proving that $f=\{(x^n,x): x \in \mathbb{R}\}$ is a function for positive odd integers $n$

Question: Prove or disprove that, for any positive odd integer n, $f=\{(x^n,x): x \in \mathbb{R}\}$ is a function.

I believe the set defines a function.

My attempt: Let $n$ be a positive odd integer.

Claim: $f$ defines a function from $\mathbb{R}$ to $\mathbb{R}$.

1. Note that if $x \in \mathbb{R}$ then, $x^n \in \mathbb{R}$. Thus $f \subset \mathbb{R^2}$.
2. Let $y \in \mathbb{R}$. Then, $(y,y^{1/n}) \in f$. Thus, we have shown the existence of an element $x$, such that $(y,x) \in f$.
3. Suppose that for some $y_1,y_2 \in \mathbb{R}$, $(y,y_1^{1/n}),(y,y_2^{1/n}) \in f$. Thus, $y_1^{1/n}=y_2^{1/n}$. Thus $y_1=y_2.$

From 2 and 3, we conclude that $\forall y \in \mathbb{R}, \exists ! x \in \mathbb{R}, (y,x) \in f$. Hence $f$ is a function. Since $n$ was arbitrary, $f$ is a function for any odd integer $n$.

Could someone tell me if it is correct? Please let me know if there are any points to improve on and if anything is incorrect. Thank you.

Solution 1:

Hint: If $n$ is odd, then $x^n=y^n$ implies $x=y$. Then the assignment $x^n\mapsto x$ is a function.

If $n$ is even, then $x^n=y^n$ implies $x=\pm y$. E.g., if $n=2$, then $4=2^2\mapsto 2$ and $4=(-2)^2\mapsto -2$ which does not yield a function.