Answer
$G\left( {u,v} \right)$ is not one-to-one.
However, $G\left( {u,v} \right)$ is one-to-one, for $u \ge 0$.
(a) the image of the $u$-axis is the positive $x$-axis including the origin.
the image of the $v$-axis is the $y$-axis.
(b) the image of ${\cal R}$:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1, - 1 \le y \le 1} \right\}$
(c) the image of the line segment joining $\left( {0,0} \right)$ and $\left( {1,1} \right)$ is a curve parametrized by $t \mapsto \left( {{t^2},t} \right)$ for $0 \le t \le 1$ in the $xy$-plane.
(d) the image of the triangle with vertices $\left( {0,0} \right)$, $\left( {0,1} \right)$, and $\left( {1,1} \right)$ is a region bounded by the curves: $t \mapsto \left( {0,1} \right)t$, $t \mapsto \left( {{t^2},1} \right)$, and $t \mapsto \left( {{t^2},t} \right)$, for $0 \le t \le 1$.
Work Step by Step
We have the mapping $G\left( {u,v} \right) = \left( {{u^2},v} \right)$.
For any value of $v$, we choose ${u_1} = 1$ and ${u_2} = - 1$. The images are $G\left( {1,v} \right) = \left( {1,v} \right)$ and $G\left( { - 1,v} \right) = \left( {1,v} \right)$. Since $\left( {{u_1},v} \right) \ne \left( {{u_2},v} \right)$ but $G\left( {{u_1},v} \right) = G\left( {{u_2},v} \right)$, we conclude that $G\left( {u,v} \right)$ is not one-to-one.
However, we can choose the domain $u,v \in \mathbb{R}$ and $u \ge 0$ such that $G\left( {u,v} \right)$ is one-to-one.
(a)
1. the $u$-axis is the set of points where $v=0$. So, its image is $G\left( {u,0} \right) = \left( {{u^2},0} \right)$.
$x = {u^2}$, ${\ \ \ \ \ }$ $y=0$
Since $y=0$ and $x \ge 0$, the image of the $u$-axis is the positive $x$-axis including the origin.
2. the $v$-axis is the set of points where $u=0$. So, its image is $G\left( {0,v} \right) = \left( {0,v} \right)$.
$x=0$, ${\ \ \ \ \ }$ $y=v$
Since $x=0$ and $v \in \mathbb{R}$, the image of the $v$-axis is the $y$-axis.
(b) We consider the rectangle $\mathcal{R} = \left[ { - 1,1} \right] \times \left[ { - 1,1} \right]$ consisting of the following line segments:
1. the line segment from $\left( { - 1, - 1} \right)$ to $\left( {1, - 1} \right)$
A point in this segment has the image given by $G\left( {u, - 1} \right) = \left( {{u^2}, - 1} \right)$. So,
$x = {u^2}$, ${\ \ \ \ \ }$ $y=-1$
Since $x = {u^2} \ge 0$ and $y=-1$, we conclude that the image of this line segment is the line segment $0 \le x \le 1$, $y=-1$.
2. the line segment from $\left( {1, - 1} \right)$ to $\left( {1,1} \right)$
A point in this segment has the image given by $G\left( {1,v} \right) = \left( {1,v} \right)$. So,
$x=1$, ${\ \ \ \ \ }$ $y=v$
We conclude that the image of this line segment is the line segment $x=1$, $ - 1 \le y \le 1$.
3. the line segment from $\left( {1,1} \right)$ to $\left( { - 1,1} \right)$
A point in this segment has the image given by $G\left( {u,1} \right) = \left( {{u^2},1} \right)$. So,
$x = {u^2}$, ${\ \ \ \ \ }$ $y=1$
Since $x = {u^2} \ge 0$ and $y=1$, we conclude that the image of this line segment is the line segment $0 \le x \le 1$, $y=1$.
4. the line segment from $\left( { - 1,1} \right)$ to $\left( { - 1, - 1} \right)$
A point in this segment has the image given by $G\left( { - 1,v} \right) = \left( {1,v} \right)$. So,
$x=1$, ${\ \ \ \ \ }$ $y=v$
We conclude that the image of this line segment is the line segment $x=1$, $ - 1 \le y \le 1$. Notice that this is the same line with the one in point (2).
Since $x$ is nonnegative, we obtain the image ${\cal D}$ in the $xy$-plane:
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1, - 1 \le y \le 1} \right\}$
(c) The line segment joining $\left( {0,0} \right)$ and $\left( {1,1} \right)$ in the $uv$-plane can be parametrized by
$\ell = t\left( {1,1} \right)$, ${\ \ \ \ \ }$ for $0 \le t \le 1$
Using the mapping $G\left( {u,v} \right) = \left( {{u^2},v} \right)$, we obtain the image
$G\left( {t,t} \right) = \left( {{t^2},t} \right)$
Thus, the image of the line segment joining $\left( {0,0} \right)$ and $\left( {1,1} \right)$ is a curve parametrized by $t \mapsto \left( {{t^2},t} \right)$ for $0 \le t \le 1$ in the $xy$-plane.
(d) The triangle with vertices $\left( {0,0} \right)$, $\left( {0,1} \right)$, and $\left( {1,1} \right)$ consists of three line segments:
1. line segment joining $\left( {0,0} \right)$ and $\left( {0,1} \right)$
It can be parametrized by ${\ell _1} = \left( {0,1} \right)t$ for $0 \le t \le 1$.
Using the mapping $G\left( {u,v} \right) = \left( {{u^2},v} \right)$, we obtain the image
$G\left( {0,t} \right) = \left( {0,t} \right)$
Thus, the image of the line segment joining $\left( {0,0} \right)$ and $\left( {0,1} \right)$ is also a line parametrized by $t \mapsto \left( {0,1} \right)t$ for $0 \le t \le 1$ in the $xy$-plane.
2. line segment joining $\left( {0,1} \right)$ and $\left( {1,1} \right)$
It can be parametrized by ${\ell _2} = \left( {0,1} \right) + \left( {1,0} \right)t = \left( {t,1} \right)$ for $0 \le t \le 1$ in the $xy$-plane.
Using the map $G\left( {u,v} \right) = \left( {{u^2},v} \right)$, we obtain the image
$G\left( {t,1} \right) = \left( {{t^2},1} \right)$
Thus, the image of line segment joining $\left( {0,1} \right)$ and $\left( {1,1} \right)$ is a line parametrized by $t \mapsto \left( {{t^2},1} \right)$ for $0 \le t \le 1$ in the $xy$-plane.
3. the line segment joining $\left( {0,0} \right)$ and $\left( {1,1} \right)$
This is the line segment in part (c). From the result in part (c), we obtain the image is a curve parametrized by $t \mapsto \left( {{t^2},t} \right)$ for $0 \le t \le 1$ in the $xy$-plane.
Thus, the image of the triangle with vertices $\left( {0,0} \right)$, $\left( {0,1} \right)$, and $\left( {1,1} \right)$ is a region bounded by the curves: $t \mapsto \left( {0,1} \right)t$, $t \mapsto \left( {{t^2},1} \right)$, and $t \mapsto \left( {{t^2},t} \right)$, for $0 \le t \le 1$.
