$C^r$ and smooth maps between smooth manifolds

Let $M$ and $N$ be smooth manifolds, let $F:M\to N$ be a map, let $p\in M$ and let $r\in {\mathbb{N}}_0\cup \{\mathrm{\infty }\}$, where ${\mathbb{N}}_0=\{0,1,2,\dots \}$. For $m,q\in {\mathbb{N}}_0$ and an open set $E\subseteq {\mathbb{R}}^m$, a map $g:E\to {\mathbb{R}}^q$ is of class $C^r$ as follows. When $m=0$, every such map is declared $C^r$. When $m\ge 1$, each component $g_l$ with $l<q$ must be $C^r$ in the sense of $C^k$ maps and multi-index derivative notation in Euclidean space; when $q=0$ this component condition is vacuous. Suppose $F$ is continuous at $p$. Then $F$ is of class $C^r$ at $p$ when there are smooth charts $(U,\phi )$ of $M$ at $p$ and $(V,\psi )$ of $N$ at $F(p)$ with $F(U)\subseteq V$ such that the coordinate representative

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

(The coordinate representation of a map between manifolds) is of class $C^r$ on a neighbourhood of $\phi (p)$ in this componentwise Euclidean sense. By Chart independence of $C^r$ smoothness ↗, this condition is independent of the chosen charts: if one such representative is $C^r$, then every one is, so "some charts" may be read as "any charts". A map is $C^r$ on an open set $W\subseteq M$ when it is continuous on $W$ and $C^r$ at every point of $W$. A map that is $C^r$ for every finite $r$ — equivalently $C^{\mathrm{\infty }}$ — is called smooth; the term $C^r$ map between smooth manifolds is reserved for the case where $F$ is continuous and the representative condition holds at every point of $M$.

• Continuity is part of the hypothesis, not a consequence, at this point. The representative is only a map between open Euclidean sets when $F$ is continuous, as The coordinate representation of a map between manifolds records; that Smooth maps are continuous later derives continuity from the $C^1$ representative condition does not change the definition.

• The choice of charts is discharged. The well-definedness obligation — that testing one chart pair agrees with testing every chart pair — is discharged by Chart independence of $C^r$ smoothness ↗, which is why that lemma is named in justified_by rather than deps.