# Coordinate chart of full rank is an “open mapping”

In our Riemannian geometry class we were asked to verify the following:

Let $M$ be a $k$-dimensional manifold in $\mathbb{R}^n$ and let $\varphi:U\subset\mathbb{R}^k\to M\subset\mathbb{R}^n$ be a $C^1$ coordinate chart with $U$ open. Then if $\varphi'(x)$ has full rank $\forall x\in U$ we have $\varphi(U)$ is open in $M$ (which I'm assuming means $\varphi(U)=M\cap V$ for some $V\subset\mathbb{R}^n$ , $V$ open?)

Any points of advice to get me started in the right direction? I not sure how I should approach proving something is open in the subset-topology sense.

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Take a look at the inverse function theorem. –  Bill Cook Nov 6 '11 at 22:02

Let $f\colon \mathbb{R}^n \supset U \times \mathbb{R}^{n-k} \to \mathbb{R}^n$ be the extension of $\varphi$ given by $$f(x) = f(x_1,\dots,x_n) = (\varphi_1(x),\dots,\varphi_k(x),x_{k+1},\dots,x_n),$$ where $\varphi = (\varphi_i)$. By the assumption the square matrix $(D_j\varphi_i(x))_{i,j}$ has full rank (for appropriate choice of $j$'s).
Now you only need to show that the Jacobi matrix of $f$ is invertible, which is easy (or obvious).