A description/proof of that for map between arbitrary subsets of Euclidean \(C^\infty\) manifolds, map is \(C^k\) at point if restriction on subspace open neighborhood of point domain is \(C^k\) at point
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of map between arbitrary subsets of Euclidean \(C^\infty\) manifolds \(C^k\) at point, where \(k\) excludes \(0\) and includes \(\infty\).
- The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that for any map between any arbitrary subsets of any Euclidean \(C^\infty\) manifolds, the map is \(C^k\) at any point if the restriction on any subspace open neighborhood of the point domain is \(C^k\) at the point, where \(k\) includes \(\infty\).
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Description
For any Euclidean \(C^\infty\) manifolds, \(\mathbb{R}^{d_1}, \mathbb{R}^{d_2}\), any subsets, \(S_1 \subseteq \mathbb{R}^{d_1}, S_2 \subseteq \mathbb{R}^{d_2}\), any point, \(p \in S_1\), any natural number (including 0) or \(\infty\) \(k\), and any map, \(f: S_1 \to S_2\), \(f\) is \(C^k\) at \(p\) if there is any open neighborhood, \(U \subseteq S_1\), of \(p\) on \(S_1\) such that \(f \vert_{U}: U \to S_2\) is \(C^k\) at \(p\).
2: Proof
Let us suppose that \(k = 0\).
For any open neighborhood, \(U_{f (p)} \subseteq S_2\), of \(f (p)\), there is an open neighborhood, \(U_p \subseteq U\), of \(p\) such that \(f \vert_{U} (U_p) \subseteq U_{f (p)}\). \(U_p \subseteq S_1\) is an open neighborhood of \(p\) on \(S_1\), by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space, and \(f (U_p) = f \vert_{U} (U_p) \subseteq U_{f (p)}\).
Let us suppose that \(1 \le k\) including \(\infty\).
There are an open neighborhood, \(U'_p \subseteq \mathbb{R}^{d_1}\), of \(p\) and a map, \(f': U'_p \to \mathbb{R}^{d_2}\), \(C^k\) at \(p\) such that \(f' \vert_{U'_p \cap U} = f \vert_U \vert_{U'_p \cap U}\). \(U = U' \cap S_1\) for an open \(U' \subseteq \mathbb{R}^{d_1}\). \(U'_p \cap U' \subseteq \mathbb{R}^{d_1}\) is an open neighborhood of \(p\) and \(f' \vert_{U'_p \cap U'}: U'_p \cap U' \to \mathbb{R}^{d_2}\) is \(C^k\) at \(p\) and satisfies \(f' \vert_{U'_p \cap U'} \vert_{U'_p \cap U' \cap S_1} = f' \vert_{U'_p \cap U} = f \vert_U \vert_{U'_p \cap U} = f \vert_{U'_p \cap U' \cap S_1}\).
