is $f(t)=(1+t^2)\left(\operatorname{cos(t)},\operatorname{sin(t)},1\right)$ a submanifold?

Question

How can i show that the curve $f:\mathbb{(-\pi,2\pi)}\to\mathbb{R}^3$ defined by \begin{equation} f(t)=(1+t^2)\left(\operatorname{cos(t)},\operatorname{sin(t)},1\right) \end{equation} is injective ? Does a general way to show it exist for general curves? I can see that de diferential is injective but i can't view that f is injective.

Answer 1

Note that $|f(t)|=\sqrt{2}(1+t^2)$, so for $f(t)=f(s)$ we need \begin{align*} 1+t^2&=1+s^2,\\ (\cos t,\sin t, 1)&=(\cos s,\sin s, 1). \end{align*} The first equation holds iff $$ s=\pm t, $$ and the second holds iff $$ s=t + 2k\pi,\quad k\in {\mathbb Z}. $$

But the domain is $(-\pi,2\pi)$, so $s=-t$ iff $s, t\in (-\pi,\pi)$, which is the only symmetric part for negation.

Then $s=t + 2k\pi$ means $k$ can only be 0, since for $s,t\in (-\pi,\pi)$, $|s-t|<2\pi$.

Therefore $s=t$ is the only solution.