2025-12-14

1504: Composition of \(C^\infty\) Immersions Is \(C^\infty\) Immersion

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description/proof of that composition of \(C^\infty\) immersions is \(C^\infty\) immersion

Topics


About: \(C^\infty\) manifold

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that the composition of any \(C^\infty\) immersions is a \(C^\infty\) immersion.

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:
\(\{M'_1, ..., M'_{n + 1}\}\): \(\subseteq \{\text{ the } C^\infty \text{ manifolds with boundary }\}\)
\(\{M_2, ..., M_{n + 1}\}\): \(M_j \in \{\text{ the immersed submanifolds with boundary of } M'_j\}\)
\(\{f_1, ..., f_n\}\): \(f_j: M'_j \to M_{j + 1}\), \(\in \{\text{ the } C^\infty \text{ immersions }\}\)
\(f_n \circ ... \circ f_1\): \(M'_1 \to M_{n + 1}, m \mapsto f_n (f_{n - 1} (... (f_1 (m))))\)
//

Statements:
\(f_n \circ ... \circ f_1 \in \{\text{ the } C^\infty \text{ immersions }\}\)
//


2: Note


The concept of 'composition' allows \(M_j \subset M'_j\).

The codomain of each \(f_j\) is an immersed submanifold with boundary (instead of just a subset), so, is a \(C^\infty\) manifold with boundary, because we have defined '\(C^\infty\) immersion' as from any \(C^\infty\) manifold with boundary into any \(C^\infty\) manifold with boundary.

Of course, \(M_j\) can be \(M'_j\).


3: Proof


Whole Strategy: Step 1: take each inclusion, \(\iota_j: M_j \to M'_j\); Step 2: see that \(f_n \circ ... \circ f_1 = f_n \circ \iota_n ... \iota_2 \circ f_1\); Step 3: see that \(d (f_n \circ ... \circ f_1) = d f_n \circ d \iota_n \circ ... d \iota_2 \circ d f_1\); Step 4: conclude the proposition.

Step 1:

For each \(j \in \{2, ..., n + 1\}\), let \(\iota_j: M_j \to M'_j\) be the inclusion, which is a \(C^\infty\) immersion, by the definition of immersed submanifold with boundary.

Step 2:

\(f_n \circ ... \circ f_1 = f_n \circ \iota_n ... \iota_2 \circ f_1\).

Step 3:

\(d (f_n \circ ... \circ f_1) = d (f_n \circ \iota_n ... \iota_2 \circ f_1) = d f_n \circ d \iota_n ... d \iota_2 \circ d f_1\), as is well known: the transformation in Step 2 was needed for this: "\(d f_n \circ ... \circ d f_1\)" would be invalid, because \(d f_1 (v) \in T_{f_1 (m)}M_2\), which could not be passed into \(d f_2: T_{f_1 (m)}M'_2 \to T_{f_2 \circ f_1 (m)}M_3\), because \(d f_1 (v) \notin T_{f_1 (m)}M'_2\).

Step 4:

\(f_n \circ \iota_n ... \iota_2 \circ f_1\) is \(C^\infty\), by the proposition that for any maps between any arbitrary subsets of any \(C^\infty\) manifolds with boundary \(C^k\) at corresponding points, where \(k\) includes \(\infty\), the composition is \(C^k\) at the point.

Each \(d f_j\) is injective, because \(f_j\) is a \(C^\infty\) immersion.

Each \(d \iota_j\) is injective, because \(\iota_j\) is a \(C^\infty\) immersion.

So, \(d f_n \circ d \iota_n ... d \iota_2 \circ d f_1\) is injective, by the proposition that any finite composition of injections is an injection.

So, \(f_n \circ ... \circ f_1 = f_n \circ \iota_n ... \iota_2 \circ f_1\) is a \(C^\infty\) immersion.


References


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