The coordinate representation of a map between manifolds

Let $M$ be a topological $m$-manifold, let $N$ be a topological $n$-manifold, let $F:M\to N$ be any function, and let $(U,\phi )$ and $(V,\psi )$ be charts on $M$ and $N$ respectively (Manifold charts, coordinate domains, and coordinate functions). The coordinate representation (or local representative) of $F$ with respect to these charts is the function

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

which is defined on the image of $U\cap F^{-1}(V)$. When $F$ is continuous, the preimage $F^{-1}(V)$ is open in $M$, so $U\cap F^{-1}(V)$ is open in the open set $U$ and its image under the homeomorphism $\phi$ is an open subset of ${\mathbb{R}}^m$; in that case $\widehat{F}$ is a map between open subsets of Euclidean spaces, and only such representatives are ever tested for smoothness.

With coordinates $(x^1,\dots ,x^m)$ on $U$ and $(y^1,\dots ,y^n)$ on $V$, the representative expresses the image coordinates as functions of the source coordinates: $\widehat{F}(x^1,\dots ,x^m)=(y^1(F({\phi }^{-1}(x))),\dots ,y^n(F({\phi }^{-1}(x))))\mathrm{.}$

• The domain is written explicitly. The restriction to $\phi (U\cap F^{-1}(V))$ is exactly where the composite formula makes sense; on the remaining part of $\phi (U)$ the image of ${\phi }^{-1}$ need not even lie in $V$. This restriction is the precise content of "the representative with respect to these two charts".

• Continuity is what makes the domain open. Without continuity of $F$ the set $U\cap F^{-1}(V)$ can fail to be open, and the representative is not a map between open subsets of Euclidean spaces; the definition of smoothness below therefore applies the representative only to continuous maps.