477: For Map Between Arbitrary Subsets of C∞ Manifolds with Boundary, Map Is Ck at Point if Restriction on Subspace Open Neighborhood of Point Domain Is Ck at Point

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A description/proof of that for map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, map is $C^k$ at point if restriction on subspace open neighborhood of point domain is $C^k$ at point

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Topics

About: $C^{\mathrm{\infty }}$ manifold

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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Starting Context

• The reader knows a definition of map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at point, where $k$ excludes $0$ and includes $\mathrm{\infty }$.
• 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.

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Target Context

• The reader will have a description and a proof of the proposition that for any map between any arbitrary subsets of any $C^{\mathrm{\infty }}$ manifolds with boundary, 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 $\mathrm{\infty }$.

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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.

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Main Body

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1: Description

For any $C^{\mathrm{\infty }}$ manifolds with (possibly empty) boundary, $M_1,M_2$, any subsets, $S_1\subseteq M_1,S_2\subseteq M_2$, any point, $p\in S_1$, any natural number (including 0) or $\mathrm{\infty }$ $k$, 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 $f{|}_U:U\to S_2$ is $C^k$ at $p$.

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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 $f{|}_U(U_p)\subseteq U_{f(p)}$. $U_p\subseteq S_1$ 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 $f(U_p)=f{|}_U(U_p)\subseteq U_{f(p)}$.

Let us suppose that $1\le k$ including $\mathrm{\infty }$.

There are a chart, $(U_p^{\prime }\subseteq M_1,{\varphi }_p^{\prime })$, around $p$ and a chart, $(U_{f(p)}\subseteq M_2,{\varphi }_{f(p)})$, around $f(p)$ such that $f{|}_U(U_p^{\prime }\cap U)\subseteq U_{f(p)}$ and ${\varphi }_{f(p)}\circ f{|}_U\circ {{\varphi }_p^{\prime }}^{-1}{|}_{{\varphi }_p^{\prime }(U_p^{\prime }\cap U)}$ is $C^k$ at ${\varphi }_p^{\prime }(p)$.

$U=U^{\prime }\cap S_1$ for an open subset, $U^{\prime }\subseteq M_1$. $(U_p^{\prime }\cap U^{\prime }\subseteq M_1,{\varphi }_p^{\prime }{|}_{U_p^{\prime }\cap U^{\prime }})$ is a chart. $f(U_p^{\prime }\cap U^{\prime }\cap S_1)=f(U_p^{\prime }\cap U)\subseteq U_{f(p)}$ and ${\varphi }_{f(p)}\circ f\circ {{\varphi }_p^{\prime }{|}_{U_p^{\prime }\cap U^{\prime }}}^{-1}{|}_{{\varphi }_p^{\prime }{|}_{U_p^{\prime }\cap U^{\prime }}(U_p^{\prime }\cap U^{\prime }\cap S_1)}={\varphi }_{f(p)}\circ f{|}_U\circ {{\varphi }_p^{\prime }}^{-1}{|}_{{\varphi }_p^{\prime }(U_p^{\prime }\cap U)}$ is $C^k$ at ${\varphi }_p^{\prime }(p)$.

So, the pair, $(U_p^{\prime }\cap U^{\prime }\subseteq M_1,{\varphi }_p^{\prime }{|}_{U_p^{\prime }\cap U^{\prime }})$ and $(U_{f(p)}\subseteq M_2,{\varphi }_{f(p)})$, can be used for $f$.

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References

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