496: For Map Between Arbitrary Subsets of C∞ Manifolds with Boundary Ck at Point, Restriction or Expansion on Codomain That Contains Range 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 $C^k$ at point, restriction or expansion on codomain that contains range is $C^k$ at point

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Topics

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

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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 $\mathrm{\infty }$.

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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 $C^k$ at any point, where $k$ includes $\mathrm{\infty }$, the restriction or expansion on any codomain that contains the range is $C^k$ at the point.

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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 $\mathrm{\infty }$ $k$, any map, $f:S_1\to S_2$, such that $f$ is $C^k$ at $p$, and any subset, $S_2^{\prime }\subseteq M_2$, such that $f(S_1)\subseteq S_2^{\prime }$, the codomain restriction or expansion, $f^{\prime }:S_1\to S_2^{\prime }$ is $C^k$ at $p$.

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2: Proof

Let us suppose that $k=0$.

Let $U_{f^{\prime }(p)}\subseteq S_2^{\prime }$ be any open neighborhood of $f^{\prime }(p)$. $U_{f^{\prime }(p)}=U_{f^{\prime }(p)}^{\prime }\cap S_2^{\prime }$, where $U_{f^{\prime }(p)}^{\prime }\subseteq M_2$ is an open neighborhood of $f^{\prime }(p)$ on $M_2$. Let $U_{f^{\prime }(p)}^{″}=U_{f^{\prime }(p)}^{\prime }\cap S_2\subseteq S_2$ be the open neighborhood of $f^{\prime }(p)$ on $S_2$. There is an open neighborhood, $U_p\subseteq S_1$, of $p$ such that $f^{\prime }(U_p)=f(U_p)\subseteq U_{f^{\prime }(p)}^{″}$. Then, $f^{\prime }(U_p)\subseteq U_{f^{\prime }(p)}$, because $f^{\prime }(U_p)\subseteq U_{f^{\prime }(p)}^{\prime }\cap S_2^{\prime }=U_{f^{\prime }(p)}$.

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 $f(U_p^{\prime }\cap S_1)\subseteq U_{f(p)}$ and ${\varphi }_{f(p)}\circ f\circ {{\varphi }_p^{\prime }}^{-1}{|}_{{\varphi }_p^{\prime }(U_p^{\prime }\cap S_1)}:{\varphi }_p^{\prime }(U_p^{\prime }\cap S_1)\to {\varphi }_{f(p)}(U_{f(p)})$ is $C^k$ at ${\varphi }_p^{\prime }(p)$.

But the same charts pair can be used for $f^{\prime }$, because $f^{\prime }(U_p^{\prime }\cap S_1)\subseteq U_{f(p)}$ and ${\varphi }_{f(p)}\circ f^{\prime }\circ {{\varphi }_p^{\prime }}^{-1}{|}_{{\varphi }_p^{\prime }(U_p^{\prime }\cap S_1)}:{\varphi }_p^{\prime }(U_p^{\prime }\cap S_1)\to {\varphi }_{f(p)}(U_{f(p)})$ is $C^k$ at ${\varphi }_p^{\prime }(p)$.

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References

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