I have the following problem:
Let $\textbf{x}$: $(-\pi,\pi)\times(-\pi,\pi)\rightarrow R^3$ such that
$$\textbf{x}(\theta,t)=((2+\cos(t))\cos(\theta),(2+\cos(t))\sin(\theta),\sin(t))$$
I want to prove that $\textbf{x}$ is a injective function. I tried by the definition of injective function, but I didn't get it. That problem are in the frame of differential geometry
My attemp:
Let $(\theta_{1},t_{1}),(\theta_{2},t_{2})\in (-\pi,\pi)$ such that $\textbf{x}(\theta_{1},t_{1})=\textbf{x}(\theta_{2},t_{2})$. Then
- $$(2+\cos(t_{1}))\cos(\theta_{1})=(2+\cos(t_{2}))\cos(\theta_{2})$$
- $$(2+\cos(t_{1}))\sin(\theta_{1})=(2+\cos(t_{2}))\sin(\theta_{2})$$
- $$\sin(t_{1})=\sin(t_{2})$$
We can say, from the last equality:
$t_{1} = \pi - t_{2}$ , if $t_{1},t_{2}\geq 0$
$t_{1} = -\pi - t_{2}$ , if $t_{1},t_{2}\leq 0$
$t_{1}=t_{2}$
We want to prove that if any of the first two happens, then (1) or (2) is not true.
As you said, you have your system of equations. You can create a new one by using $(1)^2+(2)^2 $, which yields :
$$(2+cos(t_1))^2=(2+cos(t_2))^2 $$ And since everything is positive, you have : $$cos(t_1)=cos(t_2), sin(t1)=sin(t_2), (3) $$ So you have $t_1=t_2$, then, you can plug it in the first 2 equations and simplify by $(2+cos(t_{1,2}))$ (since they are the same), and you also get : $$sin(\theta_1)=sin(\theta_2), cos(\theta_1)=cos(\theta_2) $$ Then you can conclude that :
$$x(t_1,\theta_1)=x(t_2,\theta_2) \Rightarrow (t_1,\theta_1)=(t_2,\theta_2)$$
So your function is injective