I'm trying to show that the tangent space at $x$, $T_{x}X$, of a smooth manifold $X \subset \mathbb{R}^{N}$ is well defined.
The definition I'm using is:
- Take a parametrisation $\phi: U \to X$ such that $\phi(0)=x$, where $U \subset \mathbb{R}^{k}$ is open.
- Viewing $\phi$ as a map into $\mathbb{R}^{N}$, compute $d\phi_{0}: \mathbb{R}^{k} \to \mathbb{R}^{N}$
- Define $T_{x}X=d\phi_{0}(\mathbb{R}^{k})$
Now, I want to show that $T_{x}X$ is independent of the choice of parametrisation. The proof I'm following begins by defining $\psi: V \to X$ as a another parametrisation, and then:
by shrinking both of $U$ and $V$, we can choose $\phi(U)=\psi(V)$.
I think this is true, by taking $U'=U\cap \phi^{-1}(\psi(V))$, and $V'=V \cap \psi^{-1}(\phi(U))$. Then, while this changed the parametrisations we used, we still have $d\phi_{0}=d(\phi|_{U'})_{0}$, so the tangent space is unchanged. Is that right?
In principle, yes, that's correct, but by changing both neigbourhoods the way you did you may end up with a different one again. Just make one so small it's image is contained in the image of the other one and then use the other map to reduce the size of the second domain. Since you are looking at local homeomorphisms openness will be preserved.
You need a neighbourhood so that the differential is defined in the point you are interested in, but in that point the differential of $\phi\circ \psi^{-1}$ will always be the same regardless of the size of the neighbourhood.