description/proof of that injective \(C^\infty\) immersion is C^\infty embedding iff restriction of map on range codomain is open
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of injection.
- The reader knows a definition of \(C^\infty\) immersion.
- The reader knows a definition of \(C^\infty\) embedding.
- The reader admits the proposition that any \(C^\infty\) immersion is locally a \(C^\infty\) embedding.
- The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace..
- The reader admits the proposition that any expansion of any continuous map on the codomain is continuous.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any injective \(C^\infty\) immersion is a \(C^\infty\) embedding if and only if the restriction of the map on the range codomain is open.
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:
\(M_1\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(M_2\): \(\in \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(f\): \(: M_1 \to M_2\), \(\in \{\text{ the injective } C^\infty \text{ immersions }\}\)
\(f'\): \(: M_1 \to f (M_1), m_1 \mapsto f (m_1)\), where \(f (M_1) \subseteq M_2\) has the subspace topology
//
Statements:
\(f \in \{\text{ the } C^\infty \text{ embeddings }\}\)
\(\iff\)
\(f' \in \{\text{ the open maps }\}\)
//
2: Proof
Whole Strategy: Step 1: see that \(f\) is locally a \(C^\infty\) embedding, and for each \(m_1 \in M_1\), take an open neighborhood of \(m_1\), \(U_{m_1}\), such that \(f'_{m_1} := f \vert_{U_{m_1}}: U_{m_1} \to f (U_{m_1})\) is a homeomorphism; Step 2: suppose that \(f'\) is open; Step 3: see that \(f'^{-1}\) is continuous; Step 4; suppose that \(f\) is a \(C^\infty\) embedding; Step 5: see that \(f'\) is open.
Step 1:
As \(f\) is injective, \(f'\) is a bijection, so, there is the inverse, \(f'^{-1}: f (M_1) \to M_1\).
As \(f\) is a \(C^\infty\) immersion, \(f\) is locally an \(C^\infty\) embedding, by the proposition that any \(C^\infty\) immersion is locally a \(C^\infty\) embedding, which means that for any point, \(m_1 \in M_1\), there is an open neighborhood of \(m_1\), \(U_{m_1}\), such that \(f'_{m_1} := f \vert_{U_{m_1}}: U_{m_1} \to f (U_{m_1})\) is a homeomorphism with the inverse, \({f'_{m_1}}^{-1}: f (U_{m_1}) \to U_{m_1}\): while \(f (U_{m_1})\) is a topological subspace of \(M_2\), it is a topological subspace of \(f (M_1)\), by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
Also \({f'_{m_1}}^{-1}: f (U_{m_1}) \to M_1\) is continuous, by the proposition that any expansion of any continuous map on the codomain is continuous, which is a domain restriction of \(f'^{-1}\), while \(\{f (U_{m_1})\}\) is a (not necessarily open) cover of \(f (M_1)\).
Step 2:
Let us suppose that \(f'\) is open.
Step 3:
\(\{f (U_{m_1})\}\) is an open cover of \(f (M_1)\).
As the domain restriction of \(f'^{-1}\) to each element of the open cover is continuous, \(f'^{-1}\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous, so, \(f'\) is a homeomorphism, so, \(f\) is a \(C^\infty\) embedding.
Step 4:
Let us suppose that \(f\) is a \(C^\infty\) embedding.
Step 5:
\(f'\) is a homeomorphism. \(f'^{-1}\) is continuous.
For any open subset, \(U \subseteq M_1\), \({f'^{-1}}^{-1} (U) = f' (U)\) is open on \(f (M_1)\), so, \(f'\) is open.
