Tangent space is independent of the coordiante function.
Let $X$ be a smooth n-manifold and $\phi: U\to X$ be a local parametrization at $x$ ($\phi(0)=x$). Define tangent space as $T_x(X)=d\phi_0(\mathbb R^n)$
We claim tangent space is independent of the local parametrization so we take another local parametrization $\psi: V\to X$ and we shrink $U,V$ such that $\phi(U)=\psi(V)$ namely we take a diffeomorphism $h=\psi^{-1}\circ \phi:U\to V$
Such that $\phi=\psi\circ h$ and differentiating it yields:$$d\phi_0=d\psi_0\circ dh_0$$
Question: Why last equation would imply image of $d\psi$ consists image of $d\phi$?
Best Answer
To spell it out, if $x\in\operatorname{im}(d\phi_0)$, then $x=d\phi_0(y)$ for some $y$, and so $x=(d\psi_0\circ dh_0)(y)=(d\psi_0)(dh_0(y))$ is in $\operatorname{im}(d\psi_0)$. That is, $\operatorname{im}(d\phi_0)\subset\operatorname{im}(d\psi_0)$.
Also, you may want to check out J. Milnor's Topology from the differentiable viewpoint for what I find to be a clear exposition.