Jacobian matrix and differential coincide Let $f:\mathbb{R}^2\to \mathbb{R}^2$ be a differentiable map. If I write elements of $\mathbb{R}^2$ as $(x,y)$ and the components of $f$ as $f=(f_1,f_2)$ then its Jacobian Matrix is given as:
$J=
  \left(\begin{array}{cc}
  \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\\frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y}\end{array}\right)$.
However, I can consider $f$ as a continous map between the 2-dimensional real manifolds $\mathbb{R}^2$. Its differential is then given as
$$df|p:T_p\mathbb{R}^2\to T_p\mathbb{R}^2$$
where a tangent vector $[\gamma]$ is mapped to the tangent vector $[f\circ\gamma]$.
As $T_p\mathbb{R}^2\cong \mathbb{R}^2$ this differential map can be described by an 2x2-matrix, which should be exactly the jacobian Matrix $J$ as above.
My Question: I wonder how one can that the matrix representation of $df|p$ actually IS the Jacobian of $f$ and would be thankful for an explicit calculation of this.
Thanks!
 A: Actually, this is mostly basic linear algebra. To get a matrix representation $A$ of a linear map $L: V\rightarrow W$, you have to choose basises on $V$ and $W$. Let $\{v_1,\dots,v_n\}$ and $\{w_1,\dots,w_m\}$ be those sets, respectively. Then the components of $A$ are given by the expansion
$$
Lv_i = \sum_j A_{ji} w_j.
$$
In your case $V=W=T_pM$ and $L=df_p$, the basis is given by the coordinate vector fields of the choosen chart, i.e. the identity. By definition, for every smooth function $h: \mathbb{R}^2\rightarrow \mathbb{R}$ it holds:
$$
df_p(\partial_{i})(h)=\partial_i(h\circ f)|_p.
$$
In particular, if we choose the projection on the $j$-th component $h=pr_j$ (which is the $j$-th component of our chart):
$$
df_p(\partial_{i})(pr_j)=\partial_i(f_j)|_p.
$$
On the other hand, we have the above expansion:
$$
df_p(\partial_{i})(pr_j)=\sum_k J_{ki} \partial_k(pr_j)|_p = \sum_k J_{ki} \delta_{kj} = J_{ji}.
$$
Hence,
$$
J_{ij}=\partial_j(f_i)|_p = \frac{\partial f_i}{\partial x_j}(p).
$$
