Manifold charts, coordinate domains, and coordinate functions

Let $M$ be a topological $n$-manifold (Topological manifolds without boundary: Hausdorff, second-countable, and locally Euclidean spaces). A chart on $M$ is a pair $(U,\phi )$ in which:

• $U\subseteq M$ is open, called the coordinate domain;
• $\phi :U\to \widehat{U}$ is a homeomorphism onto an open subset $\widehat{U}\subseteq {\mathbb{R}}^n$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological), called the coordinate map; the open set $\widehat{U}=\phi (U)$ is the chart image.

For $n\ge 1$ and $i<n$, the $i$-th coordinate function is $x^i:={\pi }_i\circ \phi :U⟶\mathbb{R},x^i(p)=\phi (p{)}_i,$ so that $\phi =(x^0,\dots ,x^{n-1})$. A chart is often written $(U,(x^0,\dots ,x^{n-1}))$ and the coordinates are then used to name points of $U$. In dimension zero the coordinate map is the unique map to $\{0\}$ and there are no coordinate functions.

• The domain is open in the manifold, the image is open in Euclidean space. The two openness statements are separate hypotheses; in particular the domain $U$ need not itself be an open subset of ${\mathbb{R}}^n$, although the chart map makes it homeomorphic to the Euclidean open set $\widehat{U}$. The companion false statement A chart domain need not be a Euclidean open set records the failure to keep them apart.

• A chart is a homeomorphism by definition, so $\phi$ is continuous, bijective, and ${\phi }^{-1}:\widehat{U}\to U$ is continuous. Smoothness of either map is a later condition on pairs of charts, not a hypothesis here.