Given a basis $\{e_1,\ldots , e_n \}$ of $T_pX$; there exists a chart $(U;u_1,\ldots , u_n)$ such that $\frac{\partial}{\partial u_i} |_p =e_i$ I am stuck with a well-known differential geometry property I cannot show. 
Let $X$ be a manifold and $p \in X$. Given a basis $\{e_1,\ldots , e_n \}$ of $T_pX$; there exists a chart $(U;u_1,\ldots , u_n)$ centered at $p$ such that $\frac{\partial}{\partial u_i} |_p =e_i$.
 A: Let $\epsilon > 0$ be such that $\exp_p: B_\epsilon(0) \to X$ is a diffeomorphism onto its image. Define $\varphi\colon U \to X$ by $$\varphi(u_1,\ldots,u_n) = \exp_p(u_1e_1+\cdots+u_ne_n),$$where $U = \{ 
(u_1,\ldots,u_n) \in \Bbb R^n \mid u_1e_1+\cdots+u_ne_n \in B_\epsilon(0)\}$. Then $(\varphi^{-1}, \varphi[U])$ is the chart we want, since $$\frac{\partial}{\partial u_i}\bigg|_p = \frac{\partial \varphi}{\partial u_i}(0,\ldots,0) = {\rm d}(\exp_p)_0(e_i) = e_i,$$as wanted.

Let's explain a couple of things:


*

*We have a map $\exp_p$ defined in some open set in $T_pX$ as $\exp_p(v) = \gamma(1)$, where $\gamma$ is the unique geodesic in $X$ such that $\gamma(0) = p$ and $\gamma'(0) = v$.

*$B_\epsilon(0)$ denotes the open ball of radius $\epsilon$ in the vector space $T_pX$. It follows that there is $\epsilon > 0$ such that $\exp_p\big|_{B_\epsilon(0)}$ is a diffeomorphism onto its image in $X$, by the inverse function theorem (since ${\rm d}(\exp_p)_0 = {\rm Id}_{T_pX})$.
A: Let $\varphi:\mathbb{R}^n\to X$ be a local parametrization with $\varphi(0)=p.$ Write $$v_i:=d\varphi^{-1}(e_i)\in T_0\mathbb{R}^n=\mathbb{R}^n,\quad i=1,\ldots, n.$$ Let $\frac{\partial}{\partial x^k},\;k=1,\ldots,n,$ denote the standard basis of $T_0\mathbb{R}^n=\mathbb{R}^n,$ and let $A=(a_i^k)$ be the matrix given by $$v_i=a_i^k\frac{\partial}{\partial x^k}.$$ As can be confirmed by a simple computation, the parametrization $\psi=\varphi\circ A$ has the desired property. (Here $A$ is thought of as a linear map $\mathbb{R}^n\to\mathbb{R}^n$).
A: Take any coordinate system around $p$. We can write
$$(e_1, \ldots, e_n) = (\frac{\partial}{\partial y_1},\ldots , \frac{\partial}{\partial y_n})\cdot (a_{ij})$$ at $p$.
Let $(x_1, \ldots, x_n)$ another coordinate system around the point. Then 
$$(\frac{\partial}{\partial x_1},\ldots , \frac{\partial}{\partial x_n})=(\frac{\partial}{\partial y_1},\ldots , \frac{\partial}{\partial y_n})\cdot (\frac{ \partial y_i}{\partial x_j})$$
If we want the equality $(e_1, \ldots, e_n) = (\frac{\partial}{\partial x_1},\ldots , \frac{\partial}{\partial x_n})$ at $p$, we must have 
$$(a_{ij}) = (\frac{ \partial y_i}{\partial x_j})_p$$
In fact we can have this equality around $p$ by taking $y_i = \sum a_{ij} x_j$ for all $i$, or, equivalently, $x_i = \sum b_{ij} y_j$ for all $i$, where $(b_{ij})$ is the inverse of $(a_{ij})$.
Example: say in some coordinate system of $\mathbb{R}$, we have $e= 2\frac{\partial }{\partial y}$. Then we can write $e = \frac{\partial}{\partial{\frac{y}{2}}}$
