# smooth maps between smooth manifolds - Jacobian coordinate independence

I have question about comment in Lee's Introduction to Smooth Manifolds - page 51.

Given smooth map $F:M\to N$ between smooth manifolds $M$ and $N$ we say that the total derivative of $F$ at $p\in M$ (given chart $(U,\varphi)$ around $p$ and $(V,\psi)$ around $F(p)$) is given by $D(\psi\circ F\circ \varphi^{-1})(\varphi(p))$. The comment in the book is that total derivative is chart independent. If there is another chart $(U',\varphi')$ around $p$ then we should have

$$D(\psi\circ F\circ\varphi'^{-1})(\varphi'(p))=D(\psi\circ F\circ \varphi^{-1}\circ \varphi \circ \varphi'^{-1})(\varphi'(p))=$$ $$D(\psi\circ F\circ\varphi^{-1})(\varphi(p))\cdot D(\varphi \circ\varphi'^{-1})(\varphi (p)).$$

If total derivative is chart independent we should have $D(\varphi\circ \varphi'^{-1})(\varphi'(p))=\mathbb{Id}$, which doesn't have to be the case.

Whats wrong in my reasoning?

-
Sorry about that, changed! –  dmm Jul 3 '12 at 9:35
A smooth vector field $v$ on a manifold $M$ is chart independent: treating it as a derivation on smooth functions, $v(f)$ is a scalar independent of which vector field in which you compute it. But the expression of the vector field $v$ in a coordinate basis $v = \sum v^i \partial_i$ is clearly dependent on the choice of charts. In fact, any nontrivial element $k\in T_pM$, when written in a coordinate basis, will depend on the choice of charts. This is simply the fact that different coordinate charts give rise to possibly different coordinate bases of $T_pM$, and the only element $k$ in a vector space whose coordinate representation is the same in all bases is the 0 element.
As you can easily verify: if $k\in T_pM$, then $DF(k) \in T_{F(p)}N$ is independent of choice of charts, since the change of representations of $k$ as a vector under the chart $(U,\varphi)$ and under the chart $(U',\varphi')$ precisely cancels the factor $D(\varphi\circ\varphi'^{-1})$ you get.