Consider the function $\gamma : \mathbb R → \mathbb R^2 $ defined by $\begin{pmatrix} \sin(t)\\ \sin(2t) \end{pmatrix} $ $M$ = {$z= \gamma (t)$for some $t $ $\epsilon$ $ (0,1)$} How do I show that this is a manifold without using the parameterization method.
I understand how to prove it is a manifold using the parameterization but how would I go backwards to use the implicit function theorem?
You may apply the preimage theorem. Since $\sin(2t)=2\sin t \cos t$ and $\cos t>0$ when $t\in (0,1)$, we have $\sin(2t)=2\sin t \sqrt{1-\sin^2 t}$. Plug in $x=\sin t, y=\sin (2t)$, we see that the manifold is determined by equation $2x\sqrt{1-x^2}-y=0$.
Consider map $f=2x\sqrt{1-x^2}-y:(0, \sin 1)\times \mathbb R \to \mathbb R$. Since $\partial f/\partial y=-1$, $0$ is a regular value of $f$ and hence $f^{-1}0=M$ is a manifold.