Tangent space manifold Let M be a differentiable manifold of dimension m and also let $\{\xi_1,\dots,\xi_m\}\subset \text{T}_pM$ be an linearly independent set of the tangent bundle of M at a certain point $p\in M$.
I have to proof that there is a chart $(\varphi=(u_1,\dots,u_m), U)$ with $\varphi(p)=0$ and:$$ \xi_i =\left( \frac{d}{du_i}\right)_p$$
-My approach: Let $(\varphi, U)$ be a chart with $\varphi(p)=0$, there is an unique linear endomorphism $F$ of $\text{T}_pM$ that sends (since both sets are bases), $$\left( \frac{d}{du_i}\right)_p\longrightarrow \xi_i$$
from this point I supose you should be able to define a diffeomorphism $f:M\rightarrow M$ with $df(p)=F$ wich would end the argument by taking the chart $(\varphi \circ f,U)$. I really don't know how to define such a map or if it even exists.
 A: You've made an initial choice of a random chart $(\phi,U)$ such that $\phi(p)=0$. Let $V_1,\ldots,V_m \in T_0(\mathbb{R}^m)$ be the image of $\xi_1,\ldots,\xi_m$ under $d\phi_p$. The desired effect is that $V_1,\ldots,V_m$ is the standard basis for $T_0(\mathbb{R}^m)$, but it might not be.
Instead of doing a precomposition in $M$, do a postcomposition in $\mathbb{R}^m$.
Using the standard identification of $T_0(\mathbb{R}^m)$ with $\mathbb{R}^m$, let $F$ be the linear transformation of $\mathbb{R}^m$ taking $V_1,\ldots,V_m$ to the standard basis for $\mathbb{R}^m$. 
Finally, let $\phi' = F \circ \phi$. An application of the chain rule together with linearity of $F$ will give the proof that $\phi'$ is the desired chart.
Some more details: Here's the final calculations to put this all together. Applying the chain rule,
$$d\phi'_p = dF_{\phi(p)} \circ d\phi_p = dF_0 \circ d\phi_p
$$
Applying linearity of $F$ we have $dF_0=F$ and so
$$d\phi'_p = F \circ d\phi_p
$$
Letting $e_1,\ldots,e_m$ be the standard basis, $F$ was chosen so that $F(V_i)=e_i$, and so
$$d\phi'_p(\xi_i) = F(d\phi_p(\xi_i)) = F(V_i) = e_i
$$
which is what was wanted to be shown.
