Let $M$ be a $n$-dimensional differentiable manifold, $p\in M$, $x:U\longrightarrow V$ a chart with $p\in U$. One can identify the tangent space $T_pM$, as equivalence classes of curves, which have same derivative under a chart. Lemma 1.3.5 states there is a isomorphism
$$dx|_p:T_p M\longrightarrow \mathbb{R}^n,\quad \overset{\cdot}c(0)\mapsto\frac{d}{dt}(x\circ c)|_{t=0}$$
However, if I can find a parametrisation, I learned that the tangent space is the span of the columns of the Jacobi matrix of that parametrisation:
Example: Let $M=S^2$. Then one has the parametrisation
$$\alpha:B_1(0)\subset\mathbb{R}^2\longrightarrow \{x\in S^2: x_3>0\}\subset S^2,\quad (x,y)\mapsto(x,y,\sqrt{1-x^2-y^2})$$
So one gets
$$T_{(x,y,\sqrt{1-x^2-y^2})}S^2=<D\alpha|_{(x,y)}e_1,D\alpha|_{(x,y)}e_2>_\mathbb{R}$$
which is intuitively clear, considering a geometric picture.
Question: How can I identify the abstract tangential vectors with the derivatives under a parametrisation? I want to find the exact isomorphism of those two $\mathbb{R}$-vector spaces.
Of course I can consider the curve into $S^2$
$$c_1(t):=\alpha(\alpha^{-1}(p)+te_1)$$
Then one has
$$d\alpha^{-1}|_p(\overset{\cdot}{c_1}(0))=\frac{d}{dt}(\alpha^{-1}\circ c_1)|_{t=0}=e_1$$
On the other hand, since $\alpha$ has domain and values in $\mathbb{R}^k$ for some $k$ one has by using the chain rule
$$\frac{d}{dt}(\alpha^{-1}\circ c_1)|_{t=0}=D\alpha^{-1}|_{ c_1(0)}\cdot D\alpha|_{ \alpha^{-1}(p)}\cdot e_1=e_1$$
Can I now conclude that somehow
$$\big( D\alpha^{-1}|_{ c_1(0)}\big)^{-1}= D\alpha|_{ \alpha^{-1}(p)}?$$
I am having problems with this, since $D\alpha^{-1}|_{ c_1(0)}$ is a $2\times 3$-matrix and $D\alpha|_{ \alpha^{-1}(p)}$ is a $3\times 2$-matrix. So in what sense are those matrizes inverse?
Is it now possible to consider the map
$$\big( D\alpha^{-1}|_{ c_1(0)}\big)^{-1}\circ d\alpha^{-1}|_p: T_pS^2\longrightarrow<D\alpha|_{ \alpha^{-1}(p)}e_1,D\alpha|_{ \alpha^{-1}(p)}e_2>$$
which maps
$$\overset{\cdot}{c_1}(0)\longmapsto \frac{d}{dt}(\alpha^{-1}\circ c_1)|_{t=0}\longmapsto\big( D\alpha^{-1}|_{ c_1(0)}\big)^{-1}e_1$$
as being a vector space isomorphism which identifies the abstract tangential vectors with the tangential vectors achieving by the derivate of a parametrisation?
The best way to check that these tangent vectors are identical is to show that they define the same derivation, i.e., the same $1$-form on the space of smooth functions defined in an arbitrarily small neighborhood of a point $p$ on the manifold which satisfies the Leibniz rule at $p$. Then there is less chance of getting lost in the notation. This can be checked in local coordinates, of course.