I have some understanding issues with the following theorem.
Let $M \subset \mathbb{R}^N$ be a d-dimensional smooth manifold. $p \in M$ with $(\phi, U)$ a chart around $p$. Then $T_pM = (d\phi |_p)^{-1}(\mathbb{R}^d_0) $.
Now I wanted to try this theorem on $M = S^1=\{(x,y)^T \in \mathbb{R^2} | x^2+y^2=1\}$
So first I have to find a chart $(\phi, U) $.
My Chart is:
$\phi : S^1 \setminus \{ (1,0)^T\} \to (-\pi, \pi) \times \{0\} $
$ \phi(u,v):=(\arccos(u), 0)^T $
with $U:=(-\pi, \pi) \times \{0\} $
now $$d\phi |_p = \begin{pmatrix} -\frac{1}{\sqrt{1-p_1^2}} & 0 \\ 0 & 0 \end{pmatrix} $$ But this matrix is not invertible because the rank is 1. Can someone tell me where my mistake is?
The image of $\phi$ is rather unusual. $\phi(U)$ should be an open set in a copy of $\mathbb{R}^n$ where $n$ is the dimension of your manifold. $(-\pi,\pi)\times\{0\}$ is not an open set in $\mathbb{R}^2$. Instead, $\phi$ should map to $\mathbb{R}$ (forgetting about the $\times\{0\}$) and then the differential of $\phi$ should make more sense.
Also, as written, you are writing $d\phi|_p$ as a map from the tangent space of $\mathbb{R}^2$ (which is one-dimensional) instead of the tangent space of $S^1$, which is a one-dimensional subspace of the tangent space of $\mathbb{R}^2$.