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Let $M$ be a smooth manifold of dimension $m$. Let's consider $(U,\varphi)$ a coordinate neighbourhood, such that $\varphi =(x^1,...,x^m)$. And let $(y^1,...,y^m)$ smooth maps between an open neighbourhood $V$ of a point $p \in U$ of $M$ and $\mathbb{R}$.

Then I have to prove the following:

$(y^1,..,y^m)$ defines a system of coordinates for some neighbourhood of $p$ if and only if det$((\frac{\partial y^j}{\partial x^i}))\neq0$.

I think I could prove the direct implication. Because if $(y^1,..,y^m)$ defines a system of coordinates for some neighbourhood of $p$ then $\{(\partial y^1)_p,..,(\partial y^m)_p)\}$ is a basis for $T_p(M)$ and so it is $\{(\partial x^1)_p,..,(\partial x^m)_p)\}$. Hence, $(\frac{\partial y^j}{\partial x^i})$ is an endomorphism of $T_p(M)$ which carries one basis of it into another, hence is injective. Thus, det$((\frac{\partial y^j}{\partial x^i}))\neq0$.

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I am trying to understand your question and notations. You have $M$ a manifold, $x\in M$ a point. $(U,\phi)$ a chart at $x$ such that $\phi(y)=(x^1(y),...,x^n(y))$.

You consider $n$ smooth functions $y^1,...,y^n$ and you denote by $({\partial y^j\over{\partial x^j}})$ the Jacobian at $\phi(u)$ of the map, $f=(y^1,...,y^n)\circ\phi^{-1}$. The Local inversion theorem implies that if this Jacobian is not zero, there exists a neighborhood $V\subset\phi(U)$ of $u$ such that the restriction of $f$ to $V$ is a local diffeomorphism. This implies that the restriction of $(y^1,...,y^n)$ to $\phi^{-1}(V)$ is a local diffeomorphism since it is equal to $f_{\mid V}\circ \phi_{\mid \phi^{-1}(V)}$ and a chart since it is injective.