Let $\mathcal{M}$ be a smooth $n$-dimensional submanifold of $\mathbb{R}^{n + d}$; $p \in \mathcal{M}$; $c : (-\varepsilon, \varepsilon) \rightarrow \mathbb{R}^{n + d}$ a differentiable curve with $c ((-\varepsilon, \varepsilon)) \subset \mathcal{M}$ and $c (0) = p$; $T_p \mathcal{M} = \operatorname{range} (D f_u)$ the tangent space at $p$, where $f$ is a parametrization of $\mathcal{M}$ with $f (u) = p$. Show that $T_p \mathcal{M}$ doesn't depend on the chosen parametrization $f$.
My attempt:
Let $f : U \rightarrow \mathbb{R}^{n + d}$ and $\tilde{f} : \widetilde{U} \rightarrow \mathbb{R}^{n + d}$ both parametrize $\mathcal{M}$ around $f (u) = p = \tilde{f} (\widetilde{u})$. We have $\operatorname{range} (D f_u) = T_p \mathcal{M} = \operatorname{range} (D \tilde{f}_{\tilde{u}})$ and $\operatorname{rank} (D f_u) = \operatorname{rank} (D \tilde{f}_{\tilde{u}}) = n$. Hence, there is $\alpha \in \mathbb{R}^n$ such that $v = D f_u \alpha$ iff there is $\widetilde{\alpha} \in \mathbb{R}^n$ such that $v = D \tilde{f}_{\tilde{u}} \widetilde{\alpha}$ for $v \in T_p \mathcal{M}$, and the characterization of $T_p \mathcal{M}$ is thus independent of the parametrization.
This seems too simple indeed and I'm not sure whether I'm really proving the claim. Is there anything that I'm missing?
The definition of the tangent space at $p$ is $$T_pM:=\{ v\in \mathbb R^n\mid \text{there exists $\varepsilon\gt 0$ and a diff. curve $c:(-\varepsilon,\varepsilon)\to\mathcal M$, s. th. $c(0)=p$ and $c'(0)=v$} \}$$
Thats why you should look how $v=c'(0)$ behaves under the change of parametrization $\phi:=\tilde{f}^{-1}\circ f:U\to\tilde{U}$. Use $u:=f^{-1}\circ c:(-\varepsilon,\varepsilon)\to U$.