description/proof of that identity map from subset of Euclidean \(C^\infty\) manifold or closed upper half Euclidean \(C^\infty\) manifold with Boundary into subset of Euclidean \(C^\infty\) manifold or closed upper half Euclidean \(C^\infty\) manifold with boundary is \(C^\infty\)
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of Euclidean \(C^\infty\) manifold.
- The reader knows a definition of closed upper half Euclidean \(C^\infty\) manifold with boundary.
- The reader knows a definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
Target Context
- The reader will have a description and a proof of the proposition that the identity map from any subset of any Euclidean \(C^\infty\) manifold or any closed upper half Euclidean \(C^\infty\) manifold with Boundary into any subset of the same-dimensional Euclidean \(C^\infty\) manifold or the same-dimensional closed upper half Euclidean \(C^\infty\) manifold with boundary is \(C^\infty\).
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: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{R}^d\): \(= \text{ the Euclidean } C^\infty \text{ manifold }\)
\(\mathbb{H}^d\): \(= \text{ the closed upper half Euclidean } C^\infty \text{ manifold with boundary }\)
\(S_1\): \(\subseteq \mathbb{R}^d \text{ or } \mathbb{H}^d\)
\(S_2\): \(\subseteq \mathbb{R}^d \text{ or } \mathbb{H}^d\)
\(f\): \(: S_1 \to S_2, s \mapsto s\)
//
Statements:
\(f \in \{\text{ the } C^\infty \text{ maps }\}\)
//
\(S_1\) and \(S_2\) are such that \(f\) is possible.
2: Note
This proposition may seem obvious, but let us be ease at conscience by proving it once and for all, based on the definition of '\(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\)'.
The point is that a subset may be of \(\mathbb{H}^d\) instead of of \(\mathbb{R}^d\) and \(S_1\) or \(S_2\) may be an arbitrary subset instead of an open subset.
3: Proof
Whole Strategy: Step 1: suppose that \(S_1 \subseteq \mathbb{R}^d\) and \(S_2 \subseteq \mathbb{R}^d\), take the charts, \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\) and \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\), and see that \(id \circ f \circ {id}^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{R}^d)\) is \(C^\infty\); Step 2: suppose that \(S_1 \subseteq \mathbb{R}^d\) and \(S_2 \subseteq \mathbb{H}^d\), take the charts, \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\) and \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\), and see that \(id \circ f \circ {id}^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{H}^d)\) is \(C^\infty\); Step 3: suppose that \(S_1 \subseteq \mathbb{H}^d\) and \(S_2 \subseteq \mathbb{R}^d\), take the charts, \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\) and \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\), and see that \(id \circ f \circ {id}^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{R}^d)\) is \(C^\infty\); Step 4: suppose that \(S_1 \subseteq \mathbb{H}^d\) and \(S_2 \subseteq \mathbb{H}^d\), take the charts, \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\) and \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\), and see that \(id \circ f \circ {id}^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{H}^d)\) is \(C^\infty\).
While all the Steps are almost the same, let us do all of them diligently.
Step 1:
Let us suppose that \(S_1 \subseteq \mathbb{R}^d\) and \(S_2 \subseteq \mathbb{R}^d\).
Let \(s \in S_1\) be any.
Let us take the chart around \(s\), \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\), and the chart around \(f (s)\), \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\).
Let us see that \(id \circ f \circ id^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{R}^d)\) is \(C^\infty\) at \(id (s)\), which is the proposition is about by the definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
Let us take the open neighborhood of \(id (s)\), \(U_{id (s)} := \mathbb{R}^d \subseteq \mathbb{R}^d\) and the \(C^\infty\) extension of \(id \circ f \circ id^{-1} \vert_{id (S_1)}\), \(f': \mathbb{R}^d \to \mathbb{R}^d, r \mapsto r\).
\(f'\) is indeed \(C^\infty\), because it is the identity map from \(\mathbb{R}^d\) onto \(\mathbb{R}^d\).
\(f' \vert_{id (S_1)} = id \circ f \circ id^{-1} \vert_{id (S_1)}\), obviously.
So, \(f\) is \(C^\infty\) at each \(s \in S_1\), and so, \(f\) is \(C^\infty\) all over.
Step 2:
Let us suppose that \(S_1 \subseteq \mathbb{R}^d\) and \(S_2 \subseteq \mathbb{H}^d\).
Let \(s \in S_1\) be any.
Let us take the chart around \(s\), \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\), and the chart around \(f (s)\), \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\).
Let us see that \(id \circ f \circ id^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{H}^d)\) is \(C^\infty\) at \(id (s)\), which is the proposition is about by the definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
Let us take the open neighborhood of \(id (s)\), \(U_{id (s)} := \mathbb{R}^d \subseteq \mathbb{R}^d\) and the \(C^\infty\) extension of \(id \circ f \circ id^{-1} \vert_{id (S_1)}\), \(f': \mathbb{R}^d \to \mathbb{R}^d, r \mapsto r\).
\(f'\) is indeed \(C^\infty\), because it is the identity map from \(\mathbb{R}^d\) onto \(\mathbb{R}^d\).
\(f' \vert_{id (S_1)} = id \circ f \circ id^{-1} \vert_{id (S_1)}\), obviously.
So, \(f\) is \(C^\infty\) at each \(s \in S_1\), and so, \(f\) is \(C^\infty\) all over.
Step 3:
Let us suppose that \(S_1 \subseteq \mathbb{H}^d\) and \(S_2 \subseteq \mathbb{R}^d\).
Let \(s \in S_1\) be any.
Let us take the chart around \(s\), \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\), and the chart around \(f (s)\), \((\mathbb{R}^d \subseteq \mathbb{R}^d, id)\).
Let us see that \(id \circ f \circ id^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{R}^d)\) is \(C^\infty\) at \(id (s)\), which is the proposition is about by the definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
Let us take the open neighborhood of \(id (s)\), \(U_{id (s)} := \mathbb{R}^d \subseteq \mathbb{R}^d\) and the \(C^\infty\) extension of \(id \circ f \circ id^{-1} \vert_{id (S_1)}\), \(f': \mathbb{R}^d \to \mathbb{R}^d, r \mapsto r\).
\(f'\) is indeed \(C^\infty\), because it is the identity map from \(\mathbb{R}^d\) onto \(\mathbb{R}^d\).
\(f' \vert_{id (S_1)} = id \circ f \circ id^{-1} \vert_{id (S_1)}\), obviously.
So, \(f\) is \(C^\infty\) at each \(s \in S_1\), and so, \(f\) is \(C^\infty\) all over.
Step 4:
Let us suppose that \(S_1 \subseteq \mathbb{H}^d\) and \(S_2 \subseteq \mathbb{H}^d\).
Let \(s \in S_1\) be any.
Let us take the chart around \(s\), \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\), and the chart around \(f (s)\), \((\mathbb{H}^d \subseteq \mathbb{H}^d, id)\).
Let us see that \(id \circ f \circ id^{-1} \vert_{id (S_1)}: id (S_1) \to id (\mathbb{H}^d)\) is \(C^\infty\) at \(id (s)\), which is the proposition is about by the definition of \(C^k\) map between arbitrary subsets of \(C^\infty\) manifolds with boundary, where \(k\) includes \(\infty\).
Let us take the open neighborhood of \(id (s)\), \(U_{id (s)} := \mathbb{R}^d \subseteq \mathbb{R}^d\) and the \(C^\infty\) extension of \(id \circ f \circ id^{-1} \vert_{id (S_1)}\), \(f': \mathbb{R}^d \to \mathbb{R}^d, r \mapsto r\).
\(f'\) is indeed \(C^\infty\), because it is the identity map from \(\mathbb{R}^d\) onto \(\mathbb{R}^d\).
\(f' \vert_{id (S_1)} = id \circ f \circ id^{-1} \vert_{id (S_1)}\), obviously.
So, \(f\) is \(C^\infty\) at each \(s \in S_1\), and so, \(f\) is \(C^\infty\) all over.
