Coordinate Ball

Last updated Nov 1, 2022

Let $M$ be a Topological Manifold of Manifold Dimension $n$. Let $(U, \varphi)$ be a Coordinate Chart for $M$ and let $\hat{B} \subset \varphi(U)$ be an Open Ball in $\mathbb{R}^{n}$. Then $\varphi^{-1}(\hat{B})$ is a Coordinate Ball on $M$.

1. Becuase $\varphi$ is a Homeomorphism and A Function is a Homeomorphism iff it is a Bijective Continuous Open Map, $\varphi(U)$ is Open, so $\hat{B}$ exists (by definition of Metric Topology).
2. $\varphi^{-1}(\hat{B})$ is Open because $\varphi$ is a Continuous Function by definition of Homeomorphism.

• Euclidean Space