Graph of Quadratic describes Parabola/Corollary 2

Theorem
The locus of the equation of the square root function on the non-negative reals:
 * $\forall x \in \R_{\ge 0}: f \left({x}\right) = \sqrt x$

describes half of a parabola.

Proof
From Graph of Quadratic describes Parabola: Corollary 1, where:
 * $y = x^2$

is the equation of a parabola.

Let $f: \R \to \R$ be the real function defined as:
 * $f \left({x}\right) = x^2$

From Square of Real Number is Non-Negative, the image of $f$ is $\R_{\ge 0}$.

Also we have from Positive Real Number has Two Square Roots:
 * $\forall x \in \R: \left({-x}\right)^2 = x^2$

Thus it is necessary to apply a bijective restriction upon $f$.

Let $g: \R_{\ge 0} \to \R_{\ge 0}$ be the bijective restriction of $f$ to $\R_{\ge 0} \times \R_{\ge 0}$:


 * $\forall x \in \R_{\ge 0}: g \left({x}\right) = x^2$

From Inverse of Bijection is Bijection, $g^{-1}: \R_{\ge 0} \to \R_{\ge 0}$ is also a bijection.

By definition:
 * $\forall x \in \R_{\ge 0}: g^{-1} \left({x}\right) = + \sqrt x$

Then from Graph of Inverse Mapping, the graph of $g^{-1}$ is the same as the graph of $g$, reflected in the line $x = y$.

As the graph of $f$ is a parabola, the graph of $g$ is also a parabola, but because of the restriction to $\R_{\ge 0}$, just half of it.

Thus the graph of $g^{-1}$ is also half a parabola.