Smoothly compatible charts and the smoothness of Euclidean transition maps

Let $m\ge 1$, let $q\in {\mathbb{N}}_0$, let $W\subseteq {\mathbb{R}}^m$ be open and let $f:W\to {\mathbb{R}}^q$. The map $f$ is smooth (of class $C^{\mathrm{\infty }}$) when every component $f_l:W\to \mathbb{R}$, $l<q$, is of class $C^k$ for every $k\in \mathbb{N}$, where scalar $C^k$ means that every iterated coordinate derivative through order $k$ exists and is continuous on $W$ ($C^k$ maps and multi-index derivative notation in Euclidean space). For $m=0$ the domain is the one-point space or the empty set and every map from it is declared smooth; for $q=0$ there are no components and every map into the one-point space is smooth.

Let $M$ be a topological $n$-manifold and let $(U,\phi )$ and $(V,\psi )$ be charts on $M$ (Manifold charts, coordinate domains, and coordinate functions). The charts are smoothly compatible when $U\cap V=\mathrm{\varnothing }$, or when $n\ge 1$ and both transition maps

$\psi \circ {\phi }^{-1}:\phi (U\cap V)⟶\psi (U\cap V),\phi \circ {\psi }^{-1}:\psi (U\cap V)⟶\phi (U\cap V)$

are smooth in the sense above. Both directions of the transition are part of the definition; in dimension zero overlapping charts have the same one-point image, the only transition is the identity, and overlapping charts are declared compatible.

• Smoothness is a property of pairs of charts, not of a chart alone. One chart has no smoothness condition: any homeomorphism onto an open set is a chart. Smoothness enters only when two charts must agree on their overlap.

• Both transition directions are required by definition. A bijective map whose one direction is smooth need not have a smooth inverse (for example, the two charts $(\mathbb{R},\mathrm{id})$ and $(\mathbb{R},x\mapsto x^3)$ have reverse transition $x\mapsto x^{1\mathrm{/}3}$, which is not differentiable at $0$); requiring both directions outright makes compatibility genuinely symmetric, as Smooth chart compatibility is symmetric and reflexive records.