The function $G: x \mapsto 2^{x^2}$ maps $\mathbb{R}$ onto $\{ x \in \mathbb{R} : x \geq 1 \}$ 
Let $X = \mathbb{R}$ and $Y = \{x \in \mathbb{R} :x ≥ 1\}$, and define $G : X → Y$ by $$G(x) = e^{x^2}.$$
Prove that $G$ is onto.

Is this going along the right path and if so how do get the function to equal $y$?

$G: \mathbb{R} \to\mathbb{N}_1$. Let $y$$\in $$\mathbb{N_1}$.
claim: $\sqrt{\ln y}$ maps to $y$.
Does $\sqrt{\ln y}$ belong to $\mathbb{N_1}$? Yes because $y \in \mathbb{N_1}$, $G( \sqrt{\ln y})=e^{(\sqrt{\ln y})^2}$.

 A: For any y $\in$ Y we have to show that there exists an x in X such that G(x) = y.
Now,
$$G(x)=y$$
$$\implies e^{x^2} = y$$
$$\implies x^2 = \ln y $$
$$\implies x = \pm \sqrt {\ln y}$$
Since, y $\in$ Y, y $\ge 1$ and hence $\ln y\ge 0$ and $\pm \sqrt {\ln y}$ is well defined and is in X.Thus for any real y in Y there are two reals x in X, such that G(x) = y. Thus, $G:X \rightarrow Y$ is onto. 
A: Pick an arbitrary element from within the range of the function, and show that the preimage of the element is non-empty.
A: For any $y\in Y$, there is $x= \sqrt{\ln y} \in X$ such that $G(x)=y$.
A: Hint: $g(x)=e^{x^2}$ is symmetric about the y-axis. Also $g$ is continuous and strictly increasing in $\mathbb{R}^+$ with $g(0)=1$
A: A function is surjective (onto) iff every element in the codomain is mapped to by at least one element in the domain.
So to determine if: $\forall y\in [1, \infty), \exists x\in \mathbb R: y=e^{x^2}$, we ask, do the roots $x=\pm \sqrt[2]{\ln y}$ have a real value for all $y\in [1,\infty)$?
Alternatively, we observe the value of the function at $x=0$ and the behaviour as $0>x\to-\infty$ and $0<x\to\infty$.
A: You should write $G$ as a composite function of onto functions. Then as an exercise, prove again that a composite function of onto functions is onto.
