866: For Set and 2 Topology-Atlas Pairs, iff There Is Common Chart Domains Open Cover and Each Transition Is Diffeomorphism, Pairs Are Same

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description/proof of that for set and 2 topology-atlas pairs, iff there is common chart domains open cover and each transition is diffeomorphism, pairs are same

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any set and any 2 topologies for the set, iff there is a common open cover and each open subset of each element of the cover in one topology is open in the other topology and vice versa, the topologies are the same.

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Target Context

• The reader will have a description and a proof of the proposition that for any set and any 2 topology-atlas pairs, if and only if there is a common chart domains open cover and the transition for each common chart is a diffeomorphism, the pairs are the same.

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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:
$S$: $\in \{\text{ the sets }\}$
$O_1$: $\in \{\text{ the topologies for }S\}$
$O_2$: $\in \{\text{ the topologies for }S\}$
$A_1$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ -manifold-with-boundary atlases for }S\}$
$A_2$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ -manifold-with-boundary atlases for }S\}$
$(O_1,A_1)$:
$(O_2,A_2)$:
//

Statements:
(
$\mathrm{\exists }\{U_{\beta }|\beta \in B\}\in \{\text{ the chart domains open covers in }(O_1,A_1)\}\cap \{\text{ the chart domains open covers in }(O_2,A_2)\}$
$\wedge$
$\mathrm{\forall }\beta \in B$
(
${\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}:{\varphi }_{1,\beta }(U_{\beta })\to {\varphi }_{2,\beta }(U_{\beta })\in \{\text{ the diffeomorphisms }\}$, where $(U_{\beta }\subseteq S,{\varphi }_{1,\beta })$ and $(U_{\beta }\subseteq S,{\varphi }_{2,\beta })$ are some charts of $(O_1,A_1)$ and $(O_2,A_2)$
)
)
$⟺$
$(O_1,A_1)=(O_2,A_2)$
//

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2: Note

The immediate purpose of this proposition is to confirm that 2 possible pairs constructed in a certain way with some freedom are the same, which means that the construction is unique regardless of the freedom.

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3: Proof

Whole Strategy: Step 1: suppose that there is a common chart domains open cover and the transition for each common chart is a diffeomorphism, take a chart of $(O_1,A_1)$, $(U_{\beta }\subseteq S,{\varphi }_{1,\beta })$, and a chart of $(O_2,A_2)$, $(U_{\beta }\subseteq S,{\varphi }_{2,\beta })$, and take ${\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}:{\varphi }_{1,\beta }(U_{\beta })\to {\varphi }_{2,\beta }(U_{\beta })$, diffeomorphic; Step 2: take any open subset of $U_{\beta }$ in $O_1$, $U_1\subseteq U_{\beta }$, and see that $U_1={{\varphi }_{2,\beta }}^{-1}\circ {\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}({\varphi }_{1,\beta }(U_1))$ is an open subset of $U_{\beta }$ in $O_2$; Step 3: conclude that $O_1=O_2$; Step 4: conclude that $A_1=A_2$; Step 5: suppose that $(O_1,A_1)=(O_2,A_2)$, conclude that there is a common chart domains open cover and the transition for each common chart is a diffeomorphism.

Step 1:

Let us suppose that there is a common chart domains open cover and the transition for each common chart is a diffeomorphism.

Let us take a chart of $(O_1,A_1)$, $(U_{\beta }\subseteq S,{\varphi }_{1,\beta })$, and a chart of $(O_2,A_2)$, $(U_{\beta }\subseteq S,{\varphi }_{2,\beta })$.

${\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}:{\varphi }_{1,\beta }(U_{\beta })\to {\varphi }_{2,\beta }(U_{\beta })$ is diffeomorphic, because that is what the supposition means.

Step 2:

Let us take any open subset of $U_{\beta }$ in $O_1$, $U_1\subseteq U_{\beta }$.

$U_1={{\varphi }_{2,\beta }}^{-1}\circ {\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}({\varphi }_{1,\beta }(U_1))$.

${\varphi }_{1,\beta }(U_1)$ is open on ${\varphi }_{1,\beta }(U_{\beta })$, because ${\varphi }_{1,\beta }$ is homeomorphic. As ${\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}$ is diffeomorphic, it is homeomorphic, so, ${\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}({\varphi }_{1,\beta }(U_1))$ is open on ${\varphi }_{2,\beta }(U_{\beta })$. Then, $U_1={{\varphi }_{2,\beta }}^{-1}\circ {\varphi }_{2,\beta }\circ {{\varphi }_{1,\beta }}^{-1}({\varphi }_{1,\beta }(U_1))$ is open on $U_{\beta }$ in $O_2$, because ${\varphi }_{2,\beta }$ is homeomorphic.

Step 3:

By the symmetry, any open subset of $U_{\beta }$ in $O_2$ is open in $O_1$.

By the proposition that for any set and any 2 topologies for the set, iff there is a common open cover and each open subset of each element of the cover in one topology is open in the other topology and vice versa, the topologies are the same, $O_1=O_2$.

Step 4:

$(U_{\beta }\subseteq S,{\varphi }_{2,\beta })$ is $C^{\mathrm{\infty }}$ compatible with $(U_{\beta }\subseteq S,{\varphi }_{1,\beta })$, because the transition is $C^{\mathrm{\infty }}$, so, $(U_{\beta }\subseteq S,{\varphi }_{2,\beta })\in A_1$.

As $\{(U_{\beta }\subseteq S,{\varphi }_{2,\beta })|\beta \in B\}$ is an atlas, which both $A_1$ and $A_2$ contain, $A_1=A_2$: the maximal atlas that contains any atlas is unique.

Step 5:

Let us suppose that $(O_1,A_1)=(O_2,A_2)$.

There is a common chart domains open cover.

The transition for each common chart is a diffeomorphism.

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References

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