Hello fellow mathematicians, I am currently learning about differential geometry and I have read the chapter about the tangential space. Now, there is a note following the definition of tangential vectors via charts. It demonstrates that for manifolds in the $\mathbb{R}^n$, the tangential vectors behave just as the directional derivatives but I do not understand the explanation.
Let $M\subset\mathbb{R}^n$ be open and let $\psi,\varphi:M\rightarrow\mathbb{R}^n$ be charts. Let $y\in\mathbb{R}^n$ be a tangential vector to $M$ in $x$. Then we define $u:=\varphi^\prime_{|x}(y)$, $:=\psi^\prime_{|x}(y)$. Thus, we have $(\psi\circ\varphi^{-1})^\prime_{|\varphi(x)}(u)=v$.
I have tried and tried to deduce the same result but I have always failed so far. Can anybody help me?
So, I have decided to switch to Euler notation, since I feel that it makes the differentiation clearer in this particular case.
First, we note that the Leibniz rule gives us $$D(\Psi\circ \varphi^{-1})_{\varphi(x)}=D\Psi_{(\varphi^{-1}\circ\varphi)(x)}\circ {D\varphi^{-1}}_{\varphi(x)}=D\Psi_x\circ {D\varphi^{-1}}_{\varphi(x)}.$$
Now we look at $u:={D\varphi}_x$ and we obtain: $${D\varphi^{-1}}_{\varphi(x)}\circ{D\varphi}_x={D(\varphi^{-1}\circ\varphi)}_x=D(id_{\mathbb{R}^n})_x=1,$$ using the Leibniz rule backwards.
Hence, all together we yield $D(\Psi\circ \varphi^{-1})_{\varphi(x)}\circ u=D\Psi_x\circ 1=v,\qquad$ q.e.d..