923: For C∞ Immersion Between C∞ Manifolds with Boundary, Its Global Differential Is C∞ Immersion

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description/proof of that for $C^{\mathrm{\infty }}$ immersion between $C^{\mathrm{\infty }}$ manifolds with boundary, its global differential is $C^{\mathrm{\infty }}$ immersion

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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: Structured Description
• 2: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ immersion.
• The reader knows a definition of global differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary.
• The reader knows a definition of double of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the rank theorem for $C^{\mathrm{\infty }}$ immersion.
• The reader admits the proposition that for any maps between any arbitrary subsets of any $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at corresponding points, where $k$ includes $\mathrm{\infty }$, the composition is $C^k$ at the point.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ immersion between any $C^{\mathrm{\infty }}$ manifolds with boundary, its global differential is a $C^{\mathrm{\infty }}$ immersion.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$M_1$: $\in \{\text{ the }{d}_1\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M_2$: $\in \{\text{ the }{d}_2\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ immersions }\}$
$df$: $:T{M}_1\to T{M}_2$, $=\text{ the global differential }$
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Statements:
$df\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ immersions }\}$.
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2: Proof

Whole Strategy: do it in the 2 parts: the 1st part: $M_2$ has the empty boundary; the 2nd part: $M_2$ has a nonempty boundary; Step 1: suppose that $M_2$ has the empty boundary; Step 2: around each $m\in M_1$, take a chart, $(U_m\subseteq M_1,{\varphi }_m)$, and a chart, $(U_{f(m)}\subseteq M_2,{\varphi }_{f(m)})$, such that $\hat{f}:={\varphi }_{f(m)}\circ f\circ {{\varphi }_m}^{-1}$ is $:(x^1,...,x^{d_1})\mapsto (x^1,...,x^{d_1},0,...,0)$; Step 3: let ${\pi }_1:T{M}_1\to M_1$ and ${\pi }_2:T{M}_2\to M_2$ be the projections; take the induced charts, $({{\pi }_1}^{-1}(U_m)\subseteq T{M}_1,\widetilde{{\varphi }_m})$ and $({{\pi }_2}^{-1}(U_{f(m)})\subseteq T{M}_2,\widetilde{{\varphi }_{f(m)}})$, and see that $\widehat{df}:=\widetilde{{\varphi }_{f(m)}}\circ df\circ {\widetilde{{\varphi }_m}}^{-1}$ is $:(v^1,...,v^k,x^1,...,x^{d_1})\mapsto (v^1,...,v^k,0,...,0,x^1,...,x^{d_1},0,...,0)$; Step 4: see that $df$ is a $C^{\mathrm{\infty }}$ immersion; Step 5: suppose that $M_2$ has a nonempty boundary; Step 6: take the double of $M_2$, $D(M_2)$, let $\widetilde{M_2}\subseteq D(M_2)$ be a regular domain diffeomorphic to $M_2$, with a diffeomorphism, $g:M_2\to \widetilde{M_2}$, let $\tilde{\iota }:\widetilde{M_2}\to D(M_2)$ be the inclusion, take $h:=\tilde{\iota }\circ g\circ f$, apply the 1st part conclusion to see that $dh$ is a $C^{\mathrm{\infty }}$ immersion, and see that $df$ is a $C^{\mathrm{\infty }}$ immersion.

Step 1:

Le us suppose that $M_2$ has the empty boundary.

Step 2:

Around each $m\in M_1$, let us take a chart, $(U_m\subseteq M_1,{\varphi }_m)$, and a chart, $(U_{f(m)}\subseteq M_2,{\varphi }_{f(m)})$, such that $f(U_m)\subseteq U_{f(m)}$ and $\hat{f}:={\varphi }_{f(m)}\circ f\circ {{\varphi }_m}^{-1}:{\varphi }_m(U_m)\to {\varphi }_{f(m)}(U_{f(m)})$, the coordinates function of $f$, is $:(x^1,...,x^{d_1})\mapsto (x^1,...,x^{d_1},0,...,0)$, which is possible by the rank theorem for $C^{\mathrm{\infty }}$ immersion (which requires $M_2$ to be without boundary, which is the reason why we have made the supposition of Step 1).

Step 3:

Let ${\pi }_1:T{M}_1\to M_1$ and ${\pi }_2:T{M}_2\to M_2$ be the projections.

Let us take the induced charts, $({{\pi }_1}^{-1}(U_m)\subseteq T{M}_1,\widetilde{{\varphi }_m})$ and $({{\pi }_2}^{-1}(U_{f(m)})\subseteq T{M}_2,\widetilde{{\varphi }_{f(m)}})$.

It is a well-known fact that $\widehat{df}:=\widetilde{{\varphi }_{f(m)}}\circ df\circ {\widetilde{{\varphi }_m}}^{-1}:\widetilde{{\varphi }_m}({{\pi }_1}^{-1}(U_m))\to \widetilde{{\varphi }_{f(m)}}({{\pi }_2}^{-1}(U_{f(m)}))$ is $:(v^1,...,v^{d_1},x^1,...,x^{d_1})\mapsto (v^1,...,v^{d_1},0,...,0,x^1,...,x^{d_1},0,...,0)$: the components, $(v^1,...,v^{d_1})$, are mapped to $({\partial }_j{\hat{f}}^1{v}^j,...,{\partial }_j{\hat{f}}^{d_2}{v}^j)$, but ${\hat{f}}^j=x^j$ for each $1\le j\le d_1$ and ${\hat{f}}^j=0$ for each $d_1+1\le j\le d_2$.

So, $df$ is $C^{\mathrm{\infty }}$.

Step 4:

Let ${\pi }_1^{\prime }:TT{M}_1\to T{M}_1$ and ${\pi }_2^{\prime }:TT{M}_2\to T{M}_2$ be the projections.

There are the induced charts, $({{\pi }_1^{\prime }}^{-1}({{\pi }_1}^{-1}(U_m))\subseteq TT{M}_1,\widetilde{\stackrel{~}{{\varphi }_m}})$ and $({{\pi }_2^{\prime }}^{-1}({{\pi }_2}^{-1}(U_{f(m)}))\subseteq TT{M}_2,\widetilde{\stackrel{~}{{\varphi }_{f(m)}}})$.

It is a well-known fact that $\widehat{ddf}:=\widetilde{\stackrel{~}{{\varphi }_{f(m)}}}\circ ddf\circ {\widetilde{\stackrel{~}{{\varphi }_m}}}^{-1}:\widetilde{\stackrel{~}{{\varphi }_m}}({{\pi }_1^{\prime }}^{-1}({{\pi }_1}^{-1}(U_m)))\to \widetilde{\stackrel{~}{{\varphi }_{f(m)}}}({{\pi }_2^{\prime }}^{-1}({{\pi }_2}^{-1}(U_{f(m)})))$ is $:(w^1,...,w^{2d_1},v^1,...,v^{d_1},x^1,...,x^{d_1})\mapsto (w^1,...,w^{d_1},0,...,0,w^{d_1+1},...,w^{2d_1},0,...,0,v^1,...,v^{d_1},0,...,0,x^1,...,x^{d_1},0,...,0)$: the components, $(w^1,...,w^{2d_1})$, are mapped to $({\partial }_j{\widehat{df}}^1{w}^j,...,{\partial }_j{\widehat{df}}^{2d_2}{w}^j)$ where ${\partial }_j$ s for $1\le j\le d_1$ are by $v^j$ s and ${\partial }_j$ s for $d_1+1\le j\le 2d_1$ are by $x^{j-d_1}$ s, but ${\widehat{df}}^j=v^j$ for each $1\le j\le d_1$, ${\widehat{df}}^j=0$ for each $d_1+1\le j\le d_2$, ${\widehat{df}}^j=x^{j-d_2}$ for each $d_2+1\le j\le d_2+d_1$, and ${\widehat{df}}^j=0$ for each $d_2+d_1+1\le j\le 2d_2$.

That is obviously injective.

So, $df$ is a $C^{\mathrm{\infty }}$ immersion.

Step 5:

Let us suppose that $M_2$ has a nonempty boundary.

Step 6:

Let us take the double of $M_2$, $D(M_2)$.

$D(M_2)$ is a $C^{\mathrm{\infty }}$ manifold without boundary that has a regular domain, $\widetilde{M_2}$, that is diffeomorphic to $M_2$.

Let a diffeomorphism be $g:M_2\to \widetilde{M_2}$.

Let $\tilde{\iota }:\widetilde{M_2}\to D(M_2)$ be the inclusion, which is a $C^{\mathrm{\infty }}$ immersion (in fact, a $C^{\mathrm{\infty }}$ embedding), because $\widetilde{M_2}$ is an embedded submanifold (in fact, a regular domain) of $D(M_2)$.

Let us think of $h:M_1\to D(M_2)=\tilde{\iota }\circ g\circ f:M_1\to M_2\to \widetilde{M_2}\to D(M_2)$.

$h$ is $C^{\mathrm{\infty }}$, by the proposition that for any maps between any arbitrary subsets of any $C^{\mathrm{\infty }}$ manifolds with boundary $C^k$ at corresponding points, where $k$ includes $\mathrm{\infty }$, the composition is $C^k$ at the point.

$h$ is a $C^{\mathrm{\infty }}$ immersion, because $dh=d\tilde{\iota }\circ dg\circ df$ and $df$, $dg$, and $d\tilde{\iota }$ are injective.

By the 1st part conclusion, $dh=d\tilde{\iota }\circ dg\circ df$ is a $C^{\mathrm{\infty }}$ immersion, which means that $ddh=dd\tilde{\iota }\circ ddg\circ ddf$ is injective on each fiber.

Then, $ddf$ is injective on each fiber, because otherwise, the 2 vectors that were mapped to the same vector under $ddf$ would not be mapped to any different vectors under $ddh$.

So, $df$ is a $C^{\mathrm{\infty }}$ immersion.

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References

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