Prove that the matrix representing the tangent map $f_{*q_0} $ is the jacobian matrix of f 
Proposition: Let $\Omega \subseteq \mathbb{R}^n$ an open set, $f: \Omega \rightarrow \mathbb{R}^m$ with $f \in C^1(\Omega)$, now $$(f_{*q_0}(\omega))(h) =\omega(h \circ f) \tag{1}$$
where $\omega \in T_{q_0} \Omega$ , $h:\mathbb{R}^m \rightarrow \mathbb{R}$ and $ f_{*q_0}: T_{q_0}\Omega \rightarrow T_{f(q_0)} \mathbb{R}^m$ is the tangent map of $f$ at $q_0$.


Problem: Using the previous proposition , prove that the matrix representing the tangent map $f_{*q_0} $ is the jacobian matrix.
Suggestion: Compute $f_{*q_0}(\frac{\partial}{\partial x_i})(y_j)$, where $(x_1,\dots,x_n)$ is a coordinate system over $\Omega$ and $(y_1,\dots,y_m)$ is a coordinate system over $\mathbb{R}^m$.

I think that the suggestion is wrong for two reasons:

*

*The "$|_{q_0}$"is missing: It  should be $f_{*q_0}(\frac{\partial}{\partial x_i}|_{q_0})(y_j)$ because a basis of the space $T_{q_0}\Omega$ is given by $\big\{ \frac{\partial}{\partial x_i}|_{q_0}: i=1,\dots,n \big\}$ as far as I know.


*In order to apply the proposition I need a function $h:\mathbb{R}^m \rightarrow \mathbb{R}$ and I don't think $y_j$ has that form because it is just a coordinate, how can a coordinate be a  function of $m$ variables? what would  its arguments be? Therefore the composition $y_j\circ f$ would not make sense
Can someone shed some light?
Somehow I have to find that $f_{*q_0}(\frac{\partial}{\partial x_i}|_{q_0})= \frac{\partial f_k}{\partial x_i}|_{q_0}\frac{\partial}{\partial y_k}|_{f(q_0)}$ and from that conclude that the matrix $A$ that represents the tangent mapping is $A_{ki}=\frac{\partial f_k}{\partial x_i}|_{q_0}$.
 A: For your two questions:


*

*The "$|_{q_0}$"is missing: It  should be $f_{*q_0}(\frac{\partial}{\partial x_i}|_{q_0})(y_j)$ because a basis of the space $T_{q_0}\Omega$ is given by $\big\{ \frac{\partial}{\partial x_i}|_{q_0}: i=1,\dots,n \big\}$ as far as I know.


Usually this is omitted from notation, and it is understood from the context. So yes, technically speaking it should be $\left. \frac{\partial}{\partial x_i} \right|_{q_0}$, but you should understand that writing $\frac{\partial}{\partial x_i}$ can often mean either the vector field or at a specific point, depending on context.



*In order to apply the proposition I need a function $h:\mathbb{R}^m \rightarrow \mathbb{R}$ and I don't think $y_j$ has that form because it is just a coordinate, how can a coordinate be a  function of $m$ variables? what would  its arguments be? Therefore the composition $y_j\circ f$ would not make sense


Coordinates make perfect sense as functions. The coordinate "$y_j$" really means the function
$$ h(y_1,y_2,\dots,y_m) = y_j $$
That is a perfectly valid function of $m$ variables: it just spits out the $j^\mathrm{th}$ input. A function $f \colon \Bbb{R}^n \to \Bbb{R}^m$ is given by $m$ functions of $n$ variables:
$$ f(x_1,\dots,x_n) = \Big( f_1(x_1,\dots,x_n),\dots,f_m(x_1,\dots,x_n) \Big) $$
The composition is then just $y_j \circ f = f_j$.
