1123: Associativity for 3 Items Allows Any Association

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

description/proof of that associativity for 3 items allows any association

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Topics

About: structure

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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 structure.

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Target Context

• The reader will have a description and a proof of the proposition that for any structure, the associativity for any 3 items allows any association.

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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.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$S$: $\in \{\text{ the structures }\}$
$\ast$: $\in \{\text{ the operations of }S\}$
//

Statements:
$\mathrm{\forall }{s}_1,s_2,s_3\in S((s_1\ast s_2)\ast s_3=s_1\ast (s_2\ast s_3))$
$⟹$
$s_1\ast ...\ast s_n:=(...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{n-1})\ast s_n$ can be associated in any way
//

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2: Note

Associativity is generally defined with respect to 3 items, which is being understood to allow any associativity. Let us confirm that that is indeed the case.

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3: Proof

Whole Strategy: Step 1: for each $1\le j\le n-1$, associate $s_j$ and $s_{j+1}$ 1st; Step 2: conclude the proposition.

Step 1:

$s_1\ast ...\ast s_n:=(...(((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast s_j)\ast s_{j+1})\ast ...\ast s_{n-1})\ast s_n$.

Letting $a:=(...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})$, it is $(...((a\ast s_j)\ast s_{j+1})\ast ...\ast s_{n-1})\ast s_n$.

Applying the associativity for 3 items to $(a\ast s_j)\ast s_{j+1}$, $=(...(a\ast (s_j\ast s_{j+1}))\ast ...\ast s_{n-1})\ast s_n=(...((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast (s_j\ast s_{j+1}))\ast ...\ast s_{n-1})\ast s_n$.

So, any $s_j$ and $s_{j+1}$ can be associated.

Step 2:

Letting $b:=s_j\ast s_{j+1}$, it is $(...((...((s_1\ast s_2)\ast s_3)\ast ...\ast s_{j-1})\ast b)\ast ...\ast s_{n-1})\ast s_n$.

By Step 1, its any neighboring 2 items can be associated.

That is what it means by $s_1\ast ...\ast s_n$ "can be associated in any way".

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References

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

1122: For Map, Cardinality of Range Is Equal to or Smaller Than Cardinality of Domain

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

description/proof of that for map, cardinality of range is equal to or smaller than cardinality of domain

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Topics

About: set

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The table of contents of this article

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

---

Starting Context

• The reader knows a definition of map.
• The reader knows a definition of cardinality of set.
• The reader knows a definition of relation.

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Target Context

• The reader will have a description and a proof of the proposition that for any map, the cardinality of the range is equal to or smaller than the cardinality of the domain.

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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.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$S_1$: $\in \{\text{ the sets }\}$
$S_2$: $\in \{\text{ the sets }\}$
$f$: $:S_1\to S_2$
//

Statements:
$Card(f(S_1))\le Card(S_1)$
//

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2: Proof

Whole Strategy: Step 1: think of the relation, $R:=\{(s_2,s_1)\in S_2\times S_1|s_2=f(s_1)\}$; Step 2: apply the axiom of choice to have a function, $F\subseteq R$, such that $Dom(F)=Dom(R)$.

Step 1:

Let us think of the relation, $R:=\{(s_2,s_1)\in S_2\times S_1|s_2=f(s_1)\}$.

The domain of $R$ is $Dom(R)=f(S_1)$.

$R$ is not necessarily any function, because for an $s_2$, there may be some multiple $s_1$ s.

Step 2:

But by the axiom of choice, there is a function, $F\subseteq R$, such that $Dom(F)=Dom(R)$.

$F$ is a map from $Dom(F)$ into $S_1$.

$F$ is injective, because for any $s_2,s_2^{\prime }\in Dom(F)$ such that $s_2\ne s_2^{\prime }$, $F(s_2)\ne F(s_2^{\prime })$, because if $F(s_2)=F(s_2^{\prime })$, $s_2=f(F(s_2))=f(F(s_2^{\prime }))=s_2^{\prime }$, a contradiction.

So, $Card(f(S_1))=Card(Dom(R))=Card(Dom(F))\le Card(S_1)$.

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References

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

1121: For C∞ Manifold with Boundary, Interior Point Has r′-r-Open-Balls Charts Pair and Boundary Point Has r′-r-Open-Half-Balls Charts Pair

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

description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary, interior point has $r^{\prime }$-$r$-open-balls charts pair and boundary point has $r^{\prime }$-$r$-open-half-balls charts pair

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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

---

Starting Context

• The reader knows a definition of $r^{\prime }$-$r$-open-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $r^{\prime }$-$r$-open-half-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r$-open-ball chart and each boundary point has an $r$-open-half-ball chart for any positive $r$.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$.

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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.

---

Main Body

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$r^{\prime }$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$
$r$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$ such that $r<r^{\prime }$
//

Statements:
(
$m\in \{\text{ the interior points of }M\}$
$⟹$
$\mathrm{\exists }((B_{m,r^{\prime }}\subseteq M,{\varphi }_m),(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}}))\in \{\text{ the }{r}^{\prime }\text{ - }r\text{ -open-balls charts pairs around }m\text{ on }M\}$
)
$\wedge$
(
$m\in \{\text{ the boundary points of }M\}$
$⟹$
$\mathrm{\exists }((H_{m,r^{\prime }}\subseteq M,{\varphi }_m),(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}}))\in \{\text{ the }{r}^{\prime }\text{ - }r\text{ -open-half-balls charts pairs around }m\text{ on }M\}$
)
//

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2: Proof

Whole Strategy: Step 1: suppose that $m$ is any interior point, and take any $r^{\prime }$-open-ball chart around $m$, $(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$; Step 2: take $B_{{\varphi }_m(m),r}\subset B_{{\varphi }_m(m),r^{\prime }}$ and define $B_{m,r}:={\varphi }_m^{-1}(B_{{\varphi }_m(m),r})$; Step 3: see that $(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is an $r$-open-ball chart around $m$; Step 4: suppose that $m$ is any boundary point, and take any $r^{\prime }$-open-half-ball chart around $m$, $(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$; Step 5: take $H_{{\varphi }_m(m),r}\subset H_{{\varphi }_m(m),r^{\prime }}$ and define $H_{m,r}:={\varphi }_m^{-1}(H_{{\varphi }_m(m),r})$; Step 6: see that $(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is an $r$-open-half-ball chart around $m$.

Step 1:

Let us suppose that $m$ is any interior point.

Let us take any $r^{\prime }$-open-ball chart around $m$, $(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$, which is possible, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r$-open-ball chart and each boundary point has an $r$-open-half-ball chart for any positive $r$.

Step 2:

Let us take $B_{{\varphi }_m(m),r}\subset B_{{\varphi }_m(m),r^{\prime }}$.

Let us define $B_{m,r}:={\varphi }_m^{-1}(B_{{\varphi }_m(m),r})\subset B_{m,r^{\prime }}$.

$B_{m,r}$ is an open neighborhood of $m$ on $B_{m,r}$ and on $M$.

Step 3:

$(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is a chart, because ${\varphi }_m{|}_{B_{m,r}}:B_{m,r}\to B_{{\varphi }_m(m),r}$ is a homeomorphism and $(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is $C^{\mathrm{\infty }}$ compatible with larger $(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$.

So, $(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is an $r$-open-ball chart around $m$.

So, $((B_{m,r^{\prime }}\subseteq M,{\varphi }_m),(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}}))$ is an $r^{\prime }$-$r$-open-balls charts pair around $m$.

Step 4:

Let us suppose that $m$ is any boundary point.

Let us take any $r^{\prime }$-open-half-ball chart around $m$, $(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$, which is possible, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r$-open-ball chart and each boundary point has an $r$-open-half-ball chart for any positive $r$.

Step 5:

Let us take $H_{{\varphi }_m(m),r}\subset H_{{\varphi }_m(m),r^{\prime }}$.

Let us define $H_{m,r}:={\varphi }_m^{-1}(H_{{\varphi }_m(m),r})\subset H_{m,r^{\prime }}$.

$H_{m,r}$ is an open neighborhood of $m$ on $H_{m,r}$ and on $M$.

Step 6:

$(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is a chart, because ${\varphi }_m{|}_{H_{m,r}}:H_{m,r}\to H_{{\varphi }_m(m),r}$ is a homeomorphism and $(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is $C^{\mathrm{\infty }}$ compatible with larger $(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$.

So, $(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is an $r$-open-half-ball chart around $m$.

So, $((H_{m,r^{\prime }}\subseteq M,{\varphi }_m),(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}}))$ is an $r^{\prime }$-$r$-open-half-balls charts pair around $m$.

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References

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

1120: r′-r-Open-Half-Balls Charts Pair Around Point on C∞ Manifold with Boundary

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

definition of $r^{\prime }$-$r$-open-half-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary

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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

---

Starting Context

• The reader knows a definition of $r$-open-half-ball chart around point on $C^{\mathrm{\infty }}$ manifold with boundary.

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Target Context

• The reader will have a definition of $r^{\prime }$-$r$-open-half-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.

---

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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$r^{\prime }$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$
$r$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$ such that $r<r^{\prime }$
$(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$: $=\text{ the }{r}^{\prime }\text{ -open-half-ball chart }$
$(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$: $=\text{ the }r\text{ -open-half-ball chart }$ such that $H_{m,r}\subset H_{m,r^{\prime }}$
$\ast ((H_{m,r^{\prime }}\subseteq M,{\varphi }_m),(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}}))$:
//

Conditions:
//

$(H_{m,r^{\prime }}\subseteq M,{\varphi }_m)$ is called "outer (open-half-ball) chart".

$(H_{m,r}\subseteq M,{\varphi }_m{|}_{H_{m,r}})$ is called "inner (open-half-ball) chart".

---

2: Note

There is no $r^{\prime }$-$r$-open-half-balls charts pair around $m$ when $m$ is an interior point.

There is always an $r^{\prime }$-$r$-open-half-balls charts pair around $m$ for any positive $r^{\prime }$ and $r$ when $m$ is a boundary point, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$.

The reason why an $r^{\prime }$-$r$-open-half-balls charts pair is sometimes useful is that $\overline{H_{m,r}}\subset H_{m,r^{\prime }}$ while $\overline{H_{m,r}}$ is the closure on $H_{m,r^{\prime }}$ and on $M$ and is a compact subspace of $H_{m,r^{\prime }}$ and $M$.

Let us see that fact.

On $H_{m,r^{\prime }}$, $\overline{H_{m,r}}={\varphi }_m^{-1}(\overline{H_{{\varphi }_m(m),r}})$, by the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain equals the preimage of the closure of the subset if the map is open especially if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.

From $\overline{H_{{\varphi }_m(m),r}}\subset H_{{\varphi }_m(m),r^{\prime }}$, ${\varphi }_m^{-1}(\overline{H_{{\varphi }_m(m),r}})\subset {\varphi }_m^{-1}(H_{{\varphi }_m(m),r^{\prime }})$, but the left hand side is $\overline{H_{m,r}}$ and the right hand side is $H_{m,r^{\prime }}$.

As $\overline{H_{{\varphi }_m(m),r}}$ is compact on $H_{{\varphi }_m(m),r^{\prime }}$, $\overline{H_{m,r}}$ is compact on $H_{m,r^{\prime }}$.

$\overline{H_{m,r}}$ is compact on $M$, by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.

$\overline{H_{m,r}}$ is closed on $M$, by the proposition that each compact subset of any Hausdorff topological space is closed.

If the closure of $H_{m,r}$ on $M$ was not $\overline{H_{m,r}}$ but $C^{\prime }\subseteq M$, $C^{\prime }\subset \overline{H_{m,r}}$ and so, $C^{\prime }\cap H_{m,r^{\prime }}\subset \overline{H_{m,r}}$, but $C^{\prime }\cap H_{m,r^{\prime }}$ would be a closed subset of $H_{m,r^{\prime }}$ and $H_{m,r}\subseteq C^{\prime }\cap H_{m,r^{\prime }}$, and so, $\overline{H_{m,r}}$ such that $C^{\prime }\cap H_{m,r^{\prime }}\subset \overline{H_{m,r}}$ would not be any closure on $H_{m,r^{\prime }}$, a contradiction, so, $\overline{H_{m,r}}$ is the closure on $M$.

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References

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

1119: r′-r-Open-Balls Charts Pair Around Point on C∞ Manifold with Boundary

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

definition of $r^{\prime }$-$r$-open-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary

---

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

---

Starting Context

• The reader knows a definition of $r$-open-ball chart around point on $C^{\mathrm{\infty }}$ manifold with boundary.

---

Target Context

• The reader will have a definition of $r^{\prime }$-$r$-open-balls charts pair around point on $C^{\mathrm{\infty }}$ manifold with boundary.

---

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:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$m$: $\in M$
$r^{\prime }$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$
$r$: $\in \{r^{″}\in \mathbb{R}|0<r^{″}\}$ such that $r<r^{\prime }$
$(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$: $=\text{ the }{r}^{\prime }\text{ -open-ball chart }$
$(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$: $=\text{ the }r\text{ -open-ball chart }$ such that $B_{m,r}\subset B_{m,r^{\prime }}$
$\ast ((B_{m,r^{\prime }}\subseteq M,{\varphi }_m),(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}}))$:
//

Conditions:
//

$(B_{m,r^{\prime }}\subseteq M,{\varphi }_m)$ is called "outer (open-ball) chart".

$(B_{m,r}\subseteq M,{\varphi }_m{|}_{B_{m,r}})$ is called "inner (open-ball) chart".

---

2: Note

There is no $r^{\prime }$-$r$-open-balls charts pair around $m$ when $m$ is a boundary point.

There is always an $r^{\prime }$-$r$-open-balls charts pair around $m$ for any positive $r^{\prime }$ and $r$ when $m$ is an interior point, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, each interior point has an $r^{\prime }$-$r$-open-balls charts pair and each boundary point has an $r^{\prime }$-$r$-open-half-balls charts pair for any positive $r^{\prime }$ and $r$.

The reason why an $r^{\prime }$-$r$-open-balls charts pair is sometimes useful is that $\overline{B_{m,r}}\subset B_{m,r^{\prime }}$ while $\overline{B_{m,r}}$ is the closure on $B_{m,r^{\prime }}$ and on $M$ and is a compact subspace of $B_{m,r^{\prime }}$ and $M$.

Let us see that fact.

On $B_{m,r^{\prime }}$, $\overline{B_{m,r}}={\varphi }_m^{-1}(\overline{B_{{\varphi }_m(m),r}})$, by the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain equals the preimage of the closure of the subset if the map is open especially if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.

From $\overline{B_{{\varphi }_m(m),r}}\subset B_{{\varphi }_m(m),r^{\prime }}$, ${\varphi }_m^{-1}(\overline{B_{{\varphi }_m(m),r}})\subset {\varphi }_m^{-1}(B_{{\varphi }_m(m),r^{\prime }})$, but the left hand side is $\overline{B_{m,r}}$ and the right hand side is $B_{m,r^{\prime }}$.

As $\overline{B_{{\varphi }_m(m),r}}$ is compact on $B_{{\varphi }_m(m),r^{\prime }}$, $\overline{B_{m,r}}$ is compact on $B_{m,r^{\prime }}$.

$\overline{B_{m,r}}$ is compact on $M$, by the proposition that for any topological space, any compact subset of any subspace is compact on the base space.

$\overline{B_{m,r}}$ is closed on $M$, by the proposition that each compact subset of any Hausdorff topological space is closed.

If the closure of $B_{m,r}$ on $M$ was not $\overline{B_{m,r}}$ but $C^{\prime }\subseteq M$, $C^{\prime }\subset \overline{B_{m,r}}$ and so, $C^{\prime }\cap B_{m,r^{\prime }}\subset \overline{B_{m,r}}$, but $C^{\prime }\cap B_{m,r^{\prime }}$ would be a closed subset of $B_{m,r^{\prime }}$ and $B_{m,r}\subseteq C^{\prime }\cap B_{m,r^{\prime }}$, and so, $\overline{B_{m,r}}$ such that $C^{\prime }\cap B_{m,r^{\prime }}\subset \overline{B_{m,r}}$ would not be any closure on $B_{m,r^{\prime }}$, a contradiction, so, $\overline{B_{m,r}}$ is the closure on $M$.

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References

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

1118: Closure of Continuous Map Preimage of Subset Equals Preimage of Closure of Subset if Map Is Open Especially if Map Is Surjective and Open Subset of Domain Is Preimage of Open Subset of Codomain

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

description/proof of that closure of continuous map preimage of subset equals preimage of closure of subset if map is open especially if map is surjective and open subset of domain is preimage of open subset of codomain

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of closure of subset of topological space.
• The reader knows a definition of continuous map.
• The reader admits the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.
• 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 any map between any topological spaces is open if but not only if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.
• The reader admits the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain equals the preimage of the closure of the subset if the map is open especially if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.

---

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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S$: $\subseteq T_2$
//

Statements:
(
$f\in \{\text{ the open maps }\}$
$⟹$
$\overline{f^{-1}(S)}=f^{-1}(\bar{S})$
)
$\wedge$
(
(
$f\in \{\text{ the surjections }\}$
$\wedge$
$\mathrm{\forall }{U}_1\in \{\text{ the open subsets of }{T}_1\}(\mathrm{\exists }{U}_2\in \{\text{ the open subsets of }{T}_2\}(U_1=f^{-1}(U_2)))$
)
$⟹$
$\overline{f^{-1}(S)}=f^{-1}(\bar{S})$
)
//

---

2: Proof

Whole Strategy: Step 1: see that $\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$; Step 2: suppose that $f$ is open, and see that $f^{-1}(\bar{S})\subseteq \overline{f^{-1}(S)}$; Step 3: suppose that $f$ is surjective and any open subset of the domain is the preimage of an open subset of the codomain, and see that $f$ is open.

Step 1:

$\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$, by the proposition that for any continuous map between topological spaces, the closure of the map preimage of any subset is contained in but not necessarily equal to the preimage of the closure of the subset.

So, it is about $f^{-1}(\bar{S})\subseteq \overline{f^{-1}(S)}$.

Step 2:

Let us suppose that $f$ is open.

For any $t\in f^{-1}(\bar{S})$, $t\in \overline{f^{-1}(S)}$?

For each open neighborhood of $t$, $U_t\subseteq T_1$, $U_t\cap f^{-1}(S)\ne \mathrm{\varnothing }$?

As $f(t)\in f(U_t)$ and $f(U_t)$ is open on $T_2$, $f(U_t)\subseteq T_2$ is an open neighborhood of $f(t)$.

As $f(t)\in \bar{S}$, $f(U_t)\cap S\ne \mathrm{\varnothing }$. There is a point, $t^{\prime }\in f(U_t)\cap S$. There is a point, $t^{″}\in U_t$, such that $t^{\prime }=f(t^{″})$. $f(t^{″})\in S$, so, $t^{″}\in f^{-1}(S)$. So, $t^{″}\in U_t\cap f^{-1}(S)$, so, $U_t\cap f^{-1}(S)\ne \mathrm{\varnothing }$.

So, $t\in \overline{f^{-1}(S)}$, by the proposition that the closure of any subset is the union of the subset and the accumulation points set of the subset.

Step 3:

Let us suppose that $f$ is surjective and any open subset of the domain is the preimage of an open subset of the codomain.

$f$ is open, by the proposition that any map between any topological spaces is open if but not only if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.

So, the conclusion follows by Step 2.

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References

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

1117: Map Between Topological Spaces Is Open if but Not Only if Map Is Surjective and Open Subset of Domain Is Preimage of Open Subset of Codomain

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

description/proof of that map between topological spaces is open if but not only if map is surjective and open subset of domain is preimage of open subset of codomain

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Topics

About: topological space

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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
• 4: Note

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Starting Context

• The reader knows a definition of open map.
• The reader knows a definition of Euclidean topology.
• The reader admits the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset.

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Target Context

• The reader will have a description and a proof of the proposition that any map between any topological spaces is open if but not only if the map is surjective and any open subset of the domain is the preimage of an open subset of the codomain.

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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

---

1: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the maps }\}$
//

Statements:
(
$f\in \{\text{ the surjections }\}$
$\wedge$
$\mathrm{\forall }{U}_1\in \{\text{ the open subsets of }{T}_1\}(\mathrm{\exists }{U}_2\in \{\text{ the open subsets of }{T}_2\}(U_1=f^{-1}(U_2)))$
)
$⟹$
$f\in \{\text{ the open maps }\}$
//

The reverse does not necessarily hold.

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2: Note

$f$ does not need to be continuous.

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3: Proof

Whole Strategy: Step 1: suppose that $f$ is surjective and $U_1=f^{-1}(U_2)$, and see that $f$ is open; Step 2: see an example in which the reverse does not hold.

Step 1:

Let us suppose that $f$ is surjective and any open subset, $U_1\subseteq T_1$, is $U_1=f^{-1}(U_2)$ for an open subset, $U_2\subseteq T_2$.

$f(U_1)=f\circ f^{-1}(U_2)$, but $f\circ f^{-1}(U_2)=U_2$, by the proposition that for any map between any sets, the composition of the map after the preimage of any subset of the codomain is identical if the map is surjective with respect to the argument subset. So, $f(U_1)=U_2$, open on $T_2$.

Step 2:

Let us see an example in which the reverse does not hold.

Let $T_1$ be $\mathbb{R}$ with the Euclidean topology, $T_2$ be $\{0\}$ with the inevitable topology, $f$ be the constant map. $f$ is open. $f$ is surjective, but the open subset, $(0,1)$, is not the preimage of any open subset of $T_2$.

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4: Note

The surjectivity is required. As a counterexample, let $T_1$ be $(0,1]\subseteq \mathbb{R}$ with the subspace topology of the Euclidean topological space, $T_2$ be $\mathbb{R}$ with the Euclidean topology, $f$ be the identity map. For any open subset, $U_1\subseteq T_1$, $U_1=U_1^{\prime }\cap (0,1]$ where $U_1^{\prime }\subseteq \mathbb{R}$ is open, by the definition of subspace topology, and $U_1=f^{-1}(U_1^{\prime })$. But while $(1/2,1]\subseteq T_1$ is open, $f((1/2,1])=(1/2,1]$ is not open on $T_2$. That is because $f$ is not surjective.

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References

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

1116: For Open Surjective Continuous Map Between Topological Spaces, Image of Basis of Domain Is Basis of Codomain

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

description/proof of that for open surjective continuous map between topological spaces, image of basis of domain is basis of codomain

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Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of continuous map.
• The reader knows a definition of open map.
• The reader knows a definition of surjection.
• The reader knows a definition of basis of topological space.
• The reader admits the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.

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Target Context

• The reader will have a description and a proof of the proposition that for any open surjective continuous map between any topological spaces, the image of any basis of the domain is a basis of the codomain.

---

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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the open surjective continuous maps }\}$
$J$: $\in \{\text{ the possibly uncountable index sets }\}$
$B$: $\in \{\text{ the bases of }{T}_1\}$, $=\{B_j|j\in J\}$
$B^{\prime }$: $=f(B):=\{f(B_j)|B_j\in B\}$
//

Statements:
$B^{\prime }\in \{\text{ the bases of }{T}_2\}$
//

---

2: Proof

Whole Strategy: Step 1: see that each $f(B_j)$ is open on $T_2$; Step 2: see that for each point, $t\in T_2$, and each open neighborhood of $t$, $U_t$, there is an $f(B_j)$ such that $t\in f(B_j)\subseteq U_t$.

Step 1:

Each $f(B_j)$ is open on $T_2$ because $f$ is open.

Step 2:

For any point, $t\in T_2$, and any open neighborhood of $t$, $U_t\subseteq T_2$, is there an $f(B_j)\in B^{\prime }$, such that $t\in f(B_j)\subseteq U_t$?

$f^{-1}(U_t)$ is open on $T_1$ because $f$ is continuous, and is not empty because $f$ is surjective.

There is a point, $t^{\prime }\in f^{-1}(U_t)$, such that $f(t^{\prime })=t$, because $f$ is surjective, and $f^{-1}(U_t)$ is an open neighborhood of $t^{\prime }$.

There is a $B_j$ such that $t^{\prime }\in B_j\subseteq f^{-1}(U_t)$, because $B$ is a basis.

$t=f(t^{\prime })\in f(B_j)$ and $f(B_j)\subseteq f\circ f^{-1}(U_t)\subseteq U_t$, by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.

So, yes.

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3: Note 1

So, if $T_1$ is 2nd-countable, $T_2$ is 2nd-countable, if $f$ satisfies the requirements of this proposition.

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4: Note 2

The requirements for $f$ are necessary for this proposition because if $f(B_j)$ is not open, $B^{\prime }$ cannot be any basis; if $f$ is not surjective, $B^{\prime }$ cannot cover the area into which $f$ does not map; if $f$ is not continuous, the openness of a subset of $T_2$ is not related with the topology of $T_1$, so, there is no guarantee that there is a $B_j$ such that $f(B_j)$ is contained in the subset.

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5: Note 3

As a quotient map is not necessarily open, this proposition cannot be applied to general quotient maps, so, a quotient space of a 2nd-countable topological space is not guaranteed to be 2nd-countable by this proposition, and in fact, there are some non-2nd-countable quotient spaces of 2nd-countable topological spaces.

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References

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

1115: Open Map

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

definition of open map

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Topics

About: topological space

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description

---

Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of map.

---

Target Context

• The reader will have a definition of open map.

---

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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$\ast f$: $:T_1\to T_2$
//

Conditions:
$\mathrm{\forall }U\in \{\text{ the open subsets of }{T}_1\}(f(U)\in \{\text{ the open subsets of }{T}_2\})$
//

---

References

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

1114: Closure of Continuous Map Preimage of Subset Is Contained in but Not Necessarily Equal to Preimage of Closure of Subset

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

description/proof of that closure of continuous map preimage of subset is contained in but not necessarily equal to preimage of closure of subset

---

Topics

About: topological space

---

The table of contents of this article

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

---

Starting Context

• The reader knows a definition of closure of subset.
• The reader knows a definition of continuous map.
• The reader knows a definition of Euclidean topology.
• The reader admits the proposition the preimage of any closed subset under any continuous map is a closed subset.

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Target Context

• The reader will have a description and a proof of the proposition that for any continuous map between any topological spaces, the closure of the map preimage of any subset of the codomain is contained in but not necessarily equal to the preimage of the closure of the subset.

---

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:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$f$: $:T_1\to T_2$, $\in \{\text{ the continuous maps }\}$
$S$: $\subseteq T_2$
//

Statements:
$\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$
//

---

2: Proof

Whole Strategy: Step 1: see that $f^{-1}(S)\subseteq f^{-1}(\bar{S})$ and $f^{-1}(\bar{S})\subseteq T_1$ is closed, and think of the definition of closure to see that $\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$; Step 2: see an example that $\overline{f^{-1}(S)}\subset f^{-1}(\bar{S})$.

Step 1:

$f^{-1}(S)\subseteq f^{-1}(\bar{S})$, obviously.

$f^{-1}(\bar{S})$ is closed, by the proposition the preimage of any closed subset under any continuous map is a closed subset.

As $\overline{f^{-1}(S)}$ is the smallest closed subset that contains $f^{-1}(S)$ by the definition of closure of subset, $\overline{f^{-1}(S)}\subseteq f^{-1}(\bar{S})$, because the right hand side is one of such closed subsets.

Step 2:

As an example that the equality does not hold, let $T_1$ be $\mathbb{R}$ with the discrete topology, $T_2$ be $\mathbb{R}$ with the Euclidean topology, $f$ be the identity map, $S$ be $(0,1)$. $f$ is continuous, because the preimage of any open subset is open. $\overline{f^{-1}(S)}=(0,1)$, because $[1,\mathrm{\infty })$ and $(-\mathrm{\infty },0]$ are open on $T_1$, while $f^{-1}(\bar{S})=[0,1]$, so, $\overline{f^{-1}(S)}\subset f^{-1}(\bar{S})$.

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References

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

1113: Antisymmetrization of Tensor Product of Tensors Is Antisymmetrizations Applied Sequentially

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

description/proof of that antisymmetrization of tensor product of tensors is antisymmetrizations applied sequentially

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Topics

About: vectors space

---

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 tensor product of tensors.
• The reader knows a definition of antisymmetrization of tensor with respect to some arguments.

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Target Context

• The reader will have a description and a proof of the proposition that the antisymmetrization of the tensor product of any tensors is the antisymmetrizations applied sequentially.

---

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:
$F$: $\in \{\text{ the fields }\}$
$\{V,W\}$: $\subseteq \{\text{ the }F\text{ vectors spaces }\}$
$t_1$: $\in L(V,...,V:W)$, where $V$ appears $k_1$ times
$t_2$: $\in L(V,...,V:W)$, where $V$ appears $k_2$ times
//

Statements:
$Asym(t_1\otimes t_2)=Asym(Asym(t_1)\otimes t_2)=Asym(t_1\otimes Asym(t_2))$
//

---

2: Note

By applying this proposition sequentially, $Asym(t_1\otimes ...\otimes t_n)$ can be expressed in many ways.

For example, $Asym(t_1\otimes t_2\otimes t_3)=Asym((t_1\otimes t_2)\otimes t_3)=Asym(Asym(t_1\otimes t_2)\otimes t_3)=Asym(Asym(Asym(t_1)\otimes t_2)\otimes t_3)$.

There can be a more general proposition on partial antisymmetrization, but as it seems to become cumbersome and our immediate necessity requires only full antisymmetrizations, this proposition deals with only full antisymmetrizations.

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3: Proof

Whole Strategy: Step 1: let $(v_1,...,v_{k_1+k_2})\in V\times ...\times V$ be any; Step 2: let $Asym(t_1\otimes t_2)$ operate on $(v_1,...,v_{k_1+k_2})$ and expand the result; Step 3: let $Asym(Asym(t_1)\otimes t_2)$ operate on $(v_1,...,v_{k_1+k_2})$ and expand the result; Step 4: let $Asym(t_1\otimes Asym(t_2))$ operate on $(v_1,...,v_{k_1+k_2})$ and expand the result.

Step 1:

Let $(v_1,...,v_{k_1+k_2})\in V\times ...\times V$ be any.

If some 2 tensors operate on it with the same result, the 2 tensors will be the same.

Step 2:

Let $Asym(t_1\otimes t_2)$ operate on $(v_1,...,v_{k_1+k_2})$.

$Asym(t_1\otimes t_2)((v_1,...,v_{k_1+k_2}))=1/(k_1+k_2)!\sum _{\sigma \in S^{k_1+k_2}}sgn\sigma t_1\otimes t_2(v_{{\sigma }_1},...,v_{{\sigma }_{k_1+k_2}})$.

$\sigma$ can be expressed as ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$, where ${\sigma }^{\prime }$ is the permutation that permutates $({\sigma }_1,...,{\sigma }_{k_1})$ and $({\sigma }_{k_1+1},...,{\sigma }_{k_1+k_2})$ into the increasing orders after $\sigma$, and ${\sigma }_1$ or ${\sigma }_2$ returns $({\sigma }_1^{\prime },...,{\sigma }_{k_1}^{\prime })$ or $({\sigma }_{k_1+1}^{\prime },...,{\sigma }_{k_1+k_2}^{\prime })$ back to $({\sigma }_1,...,{\sigma }_{k_1})$ or $({\sigma }_{k_1+1},...,{\sigma }_{k_1+k_2})$, respectively.

The set of ${\sigma }_1$ s is practically the symmetric group, $S^{k_1}$; the set of ${\sigma }_2$ s is practically the symmetric group, $S^{k_2}$.

So, ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$ means that any $S^{k_1+k_2}$ element is 1st, permutating $(1,...,k_1+k_2)$ such that $({\sigma }_1^{\prime },...,{\sigma }_{k_1}^{\prime })$ and $({\sigma }_{k_1+1}^{\prime },...,{\sigma }_{k_1+k_2}^{\prime })$ are in the increasing orders, and then, permuting $({\sigma }_1^{\prime },...,{\sigma }_{k_1}^{\prime })$ and $({\sigma }_{k_1+1}^{\prime },...,{\sigma }_{k_1+k_2}^{\prime })$.

$sgn\sigma =sgn{\sigma }_{1}sgn{\sigma }_{2}sgn{\sigma }^{\prime }$.

So, $Asym(t_1\otimes t_2)((v_1,...,v_{k_1+k_2}))=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1\sum _{{\sigma }_2}sgn{\sigma }_2{t}_1\otimes t_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})k_2!/k_2!\sum _{{\sigma }_2}sgn{\sigma }_2{t}_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})=(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})$, because ${\sigma }_2$ does not change the concerned components; likewise, $(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

So, $Asym(t_1\otimes t_2)((v_1,...,v_{k_1+k_2}))=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{k}_1!/k_1!\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})k_2!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }{k}_1!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})k_2!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

Step 3:

Let $Asym(Asym(t_1)\otimes t_2)$ operate on $(v_1,...,v_{k_1+k_2})$.

$Asym(Asym(t_1)\otimes t_2)(v_1,...,v_{k_1+k_2})=1/(k_1+k_2)!\sum _{\sigma \in S^{k_1+k_2}}sgn\sigma Asym(t_1)\otimes t_2(v_{{\sigma }_1},...,v_{{\sigma }_{k_1+k_2}})$.

As before, $\sigma$ can be expressed as ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$.

So, $Asym(Asym(t_1)\otimes t_2)(v_1,...,v_{k_1+k_2})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_1)\otimes t_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}Asym(t_1)(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})(k_2)!/(k_2)!\sum _{{\sigma }_2}sgn{\sigma }_2{t}_2(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}Asym(t_1)(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})(k_2)!/(k_2)!\sum _{{\sigma }_2}sgn{\sigma }_2{t}_2(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}Asym(t_1)(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

$Asym(t_1)(v_{{{\sigma }_1\circ {\sigma }^{\prime }}_1},...,v_{{{\sigma }_1\circ {\sigma }^{\prime }}_{k_1}})=sgn{\sigma }_{1}Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})$, because $Asym(t_1)$ is antisymmetric.

So, $=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_{1}sgn{\sigma }_{1}Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

That is the same with the result of Step 2.

So, $Asym(t_1\otimes t_2)=Asym(Asym(t_1)\otimes t_2)$.

Step 4:

Let $Asym(t_1\otimes Asym(t_2))$ operate on $(v_1,...,v_{k_1+k_2})$.

$Asym(t_1\otimes Asym(t_2))(v_1,...,v_{k_1+k_2})=1/(k_1+k_2)!\sum _{\sigma \in S^{k_1+k_2}}sgn\sigma t_1\otimes Asym(t_2)(v_{{\sigma }_1},...,v_{{\sigma }_{k_1+k_2}})$.

As before, $\sigma$ can be expressed as ${\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }$.

So, $Asym(t_1\otimes Asym(t_2))(v_1,...,v_{k_1+k_2})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1\sum _{{\sigma }_2}sgn{\sigma }_2{t}_1\otimes Asym(t_2)(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_2)(v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_1\circ {\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!/(k_1)!\sum _{{\sigma }_1}sgn{\sigma }_1{t}_1(v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_1},...,v_{({\sigma }_1\circ {\sigma }^{\prime }{)}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_2)(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}Asym(t_2)(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})$.

$Asym(t_2)(v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+1}},...,v_{({\sigma }_2\circ {\sigma }^{\prime }{)}_{k_1+k_2}})=sgn{\sigma }_{2}Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$, because $Asym(t_2)$ is antisymmetric.

So, $=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})\sum _{{\sigma }_2}sgn{\sigma }_{2}sgn{\sigma }_{2}Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})\sum _{{\sigma }_2}Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})=1/(k_1+k_2)!\sum _{{\sigma }^{\prime }}sgn{\sigma }^{\prime }(k_1)!Asym(t_1)(v_{{{\sigma }^{\prime }}_1},...,v_{{{\sigma }^{\prime }}_{k_1}})(k_2)!Asym(t_2)(v_{{{\sigma }^{\prime }}_{k_1+1}},...,v_{{{\sigma }^{\prime }}_{k_1+k_2}})$.

That is the same with the result of Step 2.

So, $Asym(t_1\otimes t_2)=Asym(t_1\otimes Asym(t_2))$.

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References

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