865: For Set, To-Be-Atlas Determines Topology and Atlas

<The previous article in this series | The table of contents of this series | The next article in this series>

description/proof of that for set, to-be-atlas determines topology and atlas

---

Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Note
• 3: Proof

---

Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any topological space, the intersection of any basis and any subspace is a basis for the subspace.
• The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
• The reader admits some criteria for any collection of open sets to be a basis.
• The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
• The reader admits the proposition that for any injective map, the map image of the intersection of any sets is the intersection of the map images of the sets.

---

Target Context

• The reader will have a description and a proof of the proposition that for any set, any to-be-atlas determines the canonical topology and the atlas.

---

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:
$S$: $\in \{\text{ the sets }\}$
$B$: $\in \{\text{ the possibly infinite but countable index sets }\}$
$C$: $=\{U_{\beta }\subseteq S|\beta \in B\}$, $\in \{\text{ the covers of }S\}$
$A$: $=\{(U_{\beta }\subseteq S,{\varphi }_{\beta })|\beta \in B\}$, ${\varphi }_{\beta }:U_{\beta }\to {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d$ such that ${\varphi }_{\beta }\in \{\text{ the injections }\}$, ${\varphi }_{\beta }(U_{\beta })\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d\in \{\text{ the open subsets of }{\mathbb{R}}^d\text{ or }{\mathbb{H}}^d\}$, and $\mathrm{\forall }\beta ,{\beta }^{\prime }\in B({\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d\in \{\text{ the open subsets of }{\varphi }_{\beta }(U_{\beta })\}\wedge {\varphi }_{{\beta }^{\prime }}\circ {{\varphi }_{\beta }}^{-1}{|}_{{\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})}:{\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\to {\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }})\in \{\text{ the diffeomorphisms }\})$
$\{D_{\beta }|\beta \in B\}$: $D_{\beta }\in \{\text{ the bases of }{\varphi }_{\beta }(U_{\beta })\}\}$
$D:=\{{{\varphi }_{\beta }}^{-1}(U)\subseteq U_{\beta }|\beta \in B,U\in D_{\beta }\}$
//

Statements:
$D\in \{\text{ the topological bases for }S\}$
$\wedge$
(
$\mathrm{\forall }s,s^{\prime }\in S$
(
$1)\mathrm{\exists }{U}_{\beta }\in C(s,s^{\prime }\in U_{\beta })$
$\vee$
$2)\mathrm{\exists }{U}_{\beta },U_{{\beta }^{\prime }}\in C\text{ such that }{U}_{\beta }\cap U_{{\beta }^{\prime }}=\mathrm{\varnothing }(s\in U_{\beta }\wedge s^{\prime }\in U_{{\beta }^{\prime }})$
$\vee$
$3)\mathrm{\exists }{U}_{\beta },U_{{\beta }^{\prime }}\in C((s\in U_{\beta }\setminus \overline{U_{{\beta }^{\prime }}}\wedge s^{\prime }\in U_{{\beta }^{\prime }})\vee (s\in U_{\beta }\wedge s^{\prime }\in U_{{\beta }^{\prime }}\setminus \overline{U_{\beta }}))$
)
$⟹$
$A\in \{\text{ the atlases of the }{C}^{\mathrm{\infty }}\text{ manifold with boundary }S\text{ with the topology generated by }D\}$
)
//

---

2: Note

$C$ is not defined to be an open cover, because $S$ does not have any topology yet, but becomes an open cover with respect to the topology given later.

$A$ is called "to-be-atlas", because without $S$ having any topology yet, the homeomorphism of ${\varphi }_{\beta }$ cannot be talked about.

Formally, first, a topology is given to $S$ and then, an atlas is given to the topological space. But practically, sometimes, first, we determine a to-be-atlas without any topology specified yet, and then, the canonical topology is determined to make the to-be-atlas to be really an atlas, which is the purpose of this proposition.

The condition, 1), 2), or 3) is for confirming that the topological space is Hausdorff; 1) and 2) can be just checked, but 3) can be checked only after the topology is generated, because taking the closures does not make sense before that. If $C$ can be easily defined to satisfy 1) or 2), you can go for it. The reason why 3) is added as an option is that only 1) and 2) may be too harsh a requirement: for example, for $S=\mathbb{R}$, $C=\{(-\mathrm{\infty },2),(1,\mathrm{\infty })\}$ with the identity maps does not satisfy 1) or 2) for $s\le 1,2\le s^{\prime }$: for example, $s=0,s^{\prime }=3$ does not satisfy 1) or 2), but satisfies 3): $0\in (-\mathrm{\infty },2)\setminus [1,\mathrm{\infty }),3\in (1,\mathrm{\infty })$; in fact, $s=1,s^{\prime }=2$ does not satisfy 3), but we can add $(0.5,2.5)$ into $C$, and 1) will be satisfied as $1,2\in (0.5,2.5)$, while 1) or 2) is still not satisfied for $s=0,s^{\prime }=3$.

In fact, if you have checked the Hausdorff-ness by any other way, it will be enough.

---

3: Proof

Whole Strategy: Step 1: see that $D$ is indeed a basis, by a criterion for any collection of open sets to be a basis, and see that the topological space is 2nd-countable and Hausdorff; Step 2: see that $A$ is indeed an atlas of the $C^{\mathrm{\infty }}$ manifold with boundary, $S$, with the topology generated by $D$.

Step 1:

Let us see that $D$ is indeed a basis, by Description 2 for any collection of open sets to be a basis.

1) $S=\cup D$?

For each $\beta \in B$, ${\varphi }_{\beta }(U_{\beta })=\cup D_{\beta }$, because $D_{\beta }$ is a basis of ${\varphi }_{\beta }(U_{\beta })$, and so, $U_{\beta }={{\varphi }_{\beta }}^{-1}({\varphi }_{\beta }(U_{\beta }))={{\varphi }_{\beta }}^{-1}(\cup D_{\beta })=\cup \{{{\varphi }_{\beta }}^{-1}(U)|U\in D_{\beta }\}$, by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.

As $S=\cup C={\cup }_{\beta \in B}{U}_{\beta }$, $S={\cup }_{\beta \in B}(\cup \{{{\varphi }_{\beta }}^{-1}(U)|U\in D_{\beta }\})=\cup D$.

2) for each ${{\varphi }_{\beta }}^{-1}(U),{{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\in D$, and each point, $s\in {{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })$, is there a set, ${{\varphi }_{{\beta }^{″}}}^{-1}(U^{″})\in D$, such that $s\in {{\varphi }_{{\beta }^{″}}}^{-1}(U^{″})\subseteq {{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })$?

When $\beta ={\beta }^{\prime }$, ${{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{\beta }}^{-1}(U^{\prime })={{\varphi }_{\beta }}^{-1}(U\cap U^{\prime })$, by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets. But as $D_{\beta }$ is a basis, there is a $U^{″}\in D_{\beta }$ such that ${\varphi }_{\beta }(s)\in U^{″}\subseteq U\cap U^{\prime }$, and ${{\varphi }_{\beta }}^{-1}(U^{″})\in D$ satisfies that $s\in {{\varphi }_{\beta }}^{-1}(U^{″})\subseteq {{\varphi }_{\beta }}^{-1}(U\cap U^{\prime })={{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{\beta }}^{-1}(U^{\prime })$.

Let us suppose that $\beta \ne {\beta }^{\prime }$.

We are going to see that there is a $U^{″}\subseteq D_{\beta }$, such that $s\in {{\varphi }_{\beta }}^{-1}(U^{″})\subseteq {{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })$.

${{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })={{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\cap (U_{\beta }\cap U_{{\beta }^{\prime }})=({{\varphi }_{\beta }}^{-1}(U)\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))\cap ({{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))$.

So, ${\varphi }_{\beta }({{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime }))={\varphi }_{\beta }(({{\varphi }_{\beta }}^{-1}(U)\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))\cap ({{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\cap (U_{\beta }\cap U_{{\beta }^{\prime }})))={\varphi }_{\beta }({{\varphi }_{\beta }}^{-1}(U)\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))\cap {\varphi }_{\beta }({{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))={\varphi }_{\beta }({{\varphi }_{\beta }}^{-1}(U))\cap {\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\cap {\varphi }_{\beta }({{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))$, by the proposition that for any injective map, the map image of the intersection of any sets is the intersection of the map images of the sets, $=U\cap {\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\cap {\varphi }_{\beta }\circ {{\varphi }_{{\beta }^{\prime }}}^{-1}\circ {\varphi }_{{\beta }^{\prime }}({{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })\cap (U_{\beta }\cap U_{{\beta }^{\prime }}))=U\cap {\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\cap {\varphi }_{\beta }\circ {{\varphi }_{{\beta }^{\prime }}}^{-1}({\varphi }_{{\beta }^{\prime }}({{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime }))\cap {\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }}))$, by the proposition that for any injective map, the map image of the intersection of any sets is the intersection of the map images of the sets, $=U\cap {\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\cap {\varphi }_{\beta }\circ {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime }\cap {\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }})):=U^{‴}$.

$U$ is open on ${\varphi }_{\beta }(U_{\beta })$; ${\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})$ is open on ${\varphi }_{\beta }(U_{\beta })$ by the supposition; ${\varphi }_{\beta }\circ {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime }\cap {\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }}))$ is open on ${\varphi }_{\beta }(U_{\beta })$, because $U^{\prime }\cap {\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }})$ is open on ${\varphi }_{{\beta }^{\prime }}(U_{{\beta }^{\prime }})$ (because $U^{\prime }$ and ${\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }})$ are so, by the supposition) and ${\varphi }_{\beta }\circ {{\varphi }_{{\beta }^{\prime }}}^{-1}$ is diffeomorphic. So, $U^{‴}$ is open on ${\varphi }_{\beta }(U_{\beta })$.

As $D_{\beta }$ is a basis, there is a $U^{″}\in D_{\beta }$ such that ${\varphi }_{\beta }(s)\in U^{″}\subseteq U^{‴}$.

Then, ${{\varphi }_{\beta }}^{-1}(U^{″})\in D$ and $s\in {{\varphi }_{\beta }}^{-1}(U^{″})\subseteq {{\varphi }_{\beta }}^{-1}(U^{‴})={{\varphi }_{\beta }}^{-1}(U)\cap {{\varphi }_{{\beta }^{\prime }}}^{-1}(U^{\prime })$.

So, $D$ is indeed a basis for $S$ and generates the topology for $S$.

$D$ is countable, because $B$ is countable and for each $\beta \in B$, $D_{\beta }$ is countable.

$D$ is Hausdorff, because for each $s,s^{\prime }\in S$, when 1) $s,s^{\prime }\in U_{\beta }$, there are some open neighborhoods, $U_s,U_{s^{\prime }}\subseteq U_{\beta }$ such that $U_s\cap U_{s^{\prime }}=\mathrm{\varnothing }$, because as ${\varphi }_{\beta }(U_{\beta })$ is Hausdorff, there are some open neighborhoods of ${\varphi }_{\beta }(s)$ and ${\varphi }_{\beta }(s^{\prime })$, $U_{{\varphi }_{\beta }(s)},U_{{\varphi }_{\beta }(s^{\prime })}\in D_{\beta }$, such that $U_{{\varphi }_{\beta }(s)}\cap U_{{\varphi }_{\beta }(s^{\prime })}=\mathrm{\varnothing }$, and $U_s:={{\varphi }_{\beta }}^{-1}(U_{{\varphi }_{\beta }(s)}),U_{s^{\prime }}:={{\varphi }_{\beta }}^{-1}(U_{{\varphi }_{\beta }(s^{\prime })})$ will do; when 2) $U_{\beta }\cap U_{{\beta }^{\prime }}=\mathrm{\varnothing }$ and $s\in U_{\beta }$ and $s^{\prime }\in U_{{\beta }^{\prime }}$, $U_s:=U_{\beta },U_{s^{\prime }}:=U_{{\beta }^{\prime }}$ will do; when 3) $s\in U_{\beta }\setminus \overline{U_{{\beta }^{\prime }}}\wedge s^{\prime }\in U_{{\beta }^{\prime }}$, there is an open neighborhood of $s$, $U_s\subseteq U_{\beta }\setminus \overline{U_{{\beta }^{\prime }}}$, and with $U_{s^{\prime }}:=U_{{\beta }^{\prime }}$, $U_s\cap U_{s^{\prime }}\subseteq U_{\beta }\setminus \overline{U_{{\beta }^{\prime }}}\cap U_{{\beta }^{\prime }}\subseteq U_{\beta }\setminus U_{{\beta }^{\prime }}\cap U_{{\beta }^{\prime }}=\mathrm{\varnothing }$; for $s\in U_{\beta }\wedge s^{\prime }\in U_{{\beta }^{\prime }}\setminus \overline{U_{\beta }}$, likewise.

Step 2:

Let us see that $A$ is indeed an atlas of the $C^{\mathrm{\infty }}$ manifold with boundary, $S$, with the topology generated by $D$.

Each $(U_{\beta }\subseteq S,{\varphi }_{\beta })$ is indeed a chart, because $U_{\beta }$ is an open subset of $S$, ${\varphi }_{\beta }(U_{\beta })$ is an open subset of ${\mathbb{R}}^d$ or ${\mathbb{H}}^d$, ${\varphi }_{\beta }:U_{\beta }\to {\varphi }_{\beta }(U_{\beta })\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d$ is homeomorphic: each open subset of ${\varphi }_{\beta }(U_{\beta })$ is the union of some elements of $D_{\alpha }$, whose preimage under ${\varphi }_{\beta }$ is the union of the preimages of the elements, which (the elements) are the corresponding elements of $D$, which (the union) is open on $U_{\beta }$; each open subset of $U_{\beta }$ is the union of some elements of $D$, whose image under ${\varphi }_{\beta }$ is the union of the images of the elements, which (the elements) are the corresponding elements of $D_{\beta }$, which (the union) is open on ${\varphi }_{\beta }(U_{\beta })$.

Each transition, ${\varphi }_{{\beta }^{\prime }}\circ {{\varphi }_{\beta }}^{-1}{|}_{{\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})}:{\varphi }_{\beta }(U_{\beta }\cap U_{{\beta }^{\prime }})\to {\varphi }_{{\beta }^{\prime }}(U_{\beta }\cap U_{{\beta }^{\prime }})$ is diffeomorphic by the supposition.

The charts cover $S$.

So, $A$ is indeed an atlas of the $C^{\mathrm{\infty }}$ manifold with boundary, $S$, with the topology generated by $D$.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>