This is the tenth post in the Cartesian frames sequence.

Here, we define a bunch of ways to construct new (additive/multiplicative) (sub/super)-(agents/environments) from a given Cartesian frame. Throughout this post, we will start with a single Cartesian frame over $W$, $C=(A,E,\cdot )$.

We will start by defining operations from taking subsets and partitions of $A$ and $E$. We will then define similar operations from taking subsets and partitions of $W$.

1. Definitions from Agents and Environments

1.1. Committing

Definition: Given a subset $B\subseteq A$, let ${\text{Commit}}^B\left(C\right)$ denote the Cartesian frame $(B,E,\star )$, with $\star$ given by $b\star e=b\cdot e$. Let ${\text{Commit}}^{\setminus B}\left(C\right)$ denote the Cartesian frame $(A\setminus B,E,\diamond )$, with $\diamond$ given by $a\diamond e=a\cdot e$.

${\text{Commit}}^B\left(C\right)$ represents the perspective you get when the agent of $C$ makes a commitment to choose an element of $B$, while ${\text{Commit}}^{\setminus B}\left(C\right)$ represents the perspective you get when the agent of $C$ makes a commitment to choose an element outside of $B$.

Claim: For all $B\subseteq A$, ${\text{Commit}}^B\left(C\right){◃}_{+}C$ and ${\text{Commit}}^{\setminus B}\left(C\right){◃}_{+}C$. Further, ${\text{Commit}}^{\setminus B}\left(C\right)$ and ${\text{Commit}}^B\left(C\right)$ are brothers in $C$.

Proof: That ${\text{Commit}}^B\left(C\right){◃}_{+}C$ and ${\text{Commit}}^{\setminus B}\left(C\right){◃}_{+}C$ is trivial from the committing definition of additive subagent.

Observe that ${\text{Commit}}^B\left(C\right)\oplus {\text{Commit}}^{\setminus B}\left(C\right)=(A,E\times E,\bullet )$, where $\bullet$ is given by $a\bullet (e,f)=a\cdot e$ if $a\in B$, and $a\bullet (e,f)=a\cdot f$ if $a\notin B$. Let $D\subset E\times E$ be the diagonal, $\left\{\right(e,e)|e\in E\}$. We clearly have that $(A,D,{\bullet }^{\prime })$ is in ${\text{Commit}}^B\left(C\right)⊞{\text{Commit}}^{\setminus B}\left(C\right)$, where ${\bullet }^{\prime }$ is the restriction of $\bullet$ to $A\times D$, and that $(A,D,{\bullet }^{\prime })\cong C$; so ${\text{Commit}}^{\setminus B}\left(C\right){◃}_{+}C$ and ${\text{Commit}}^B\left(C\right){◃}_{+}C$ are brothers in $C$. $\square$

Claim: ${\text{Commit}}^{\setminus B}\left(C\right)\cong {\text{Commit}}^{A\setminus B}\left(C\right)$

Proof: Trivial. $\square$

1.2. Assuming

Assuming is the dual operation to committing.

Definition: Given a subset $F\subseteq E$, let ${\text{Assume}}^F\left(C\right)$ denote the Cartesian frame $(A,F,\star )$, with $\star$ given by $a\star f=a\cdot f$. Let ${\text{Assume}}^{\setminus F}\left(C\right)$ denote the Cartesian frame $(A,E\setminus F,\diamond )$, with $\diamond$ given by $a\diamond e=a\cdot e$.

${\text{Assume}}^F\left(C\right)$ represents the perspective you get when you assume the environment is chosen from $F$, while ${\text{Assume}}^{\setminus F}\left(C\right)$ represents the perspective you get when you assume the environment is chosen from outside of $F$.

In "Introduction to Cartesian Frames" §3.2 (Examples of Controllables), I noted the counter-intuitive result that agents have no control in worlds where a meteor (or other event) could have prevented their existence:

$C_0=\begin{array}{cc}& \begin{array}{ccc}r& s& m\end{array}\\ \begin{array}{c}u\\ n\\ u\leftrightarrow r\\ u\leftrightarrow s\end{array}& \left(\begin{array}{ccc}ur& us& m\\ nr& ns& m\\ ur& ns& m\\ nr& us& m\end{array}\right)\end{array}$.

Here, we see that we can use ${\text{Assume}}^{\setminus F}\left(C\right)$ to recover the more intuitive idea of "control." The subagent modified by the assumption "there's no meteor" can have controllables, even though the original agent has no controllables:

${\text{Assume}}^{\setminus \left\{m\right\}}\left(C_0\right)=\begin{array}{cc}& \begin{array}{cc}r& s\end{array}\\ \begin{array}{c}u\\ n\\ u\leftrightarrow r\\ u\leftrightarrow s\end{array}& \left(\begin{array}{cc}ur& us\\ nr& ns\\ ur& ns\\ nr& us\end{array}\right)\end{array}$.

Claim: For all $F\subseteq E$, $\left({\text{Assume}}^F\right(C){)}^{\ast }\cong {\text{Commit}}^F(C^{\ast })$ and $\left({\text{Assume}}^{\setminus F}\right(C){)}^{\ast }\cong {\text{Commit}}^{\setminus F}(C^{\ast })$. Similarly, for all $B\subseteq A$, $\left({\text{Commit}}^B\right(C){)}^{\ast }\cong {\text{Assume}}^B(C^{\ast })$ and $\left({\text{Commit}}^{\setminus B}\right(C){)}^{\ast }\cong {\text{Assume}}^{\setminus B}(C^{\ast })$.

Proof: Trivial. $\square$

Claim: For all $F\subseteq E$, $C{◃}_{+}^{\ast }{\text{Assume}}^F\left(C\right)$ and $C{◃}_{+}^{\ast }{\text{Assume}}^{\setminus F}\left(C\right)$. 

Proof: Trivial. $\square$

1.3. Externalizing

Note that for the following definitions, when we say "$X$ is a partition of $Y$," we mean that $X$ is a set of nonempty subsets of $Y$, such that for each $y\in Y$, there exists a unique $x\in X$ such that $y\in x$. 

Definition: Given a partition $X$ of $Y$, let $Y/X$ denote the set of all functions $q$ from $X$ to $Y$ such that $q\left(x\right)\in x$ for all $x\in X$.

Definition: Given a partition $B$ of $A$, let ${\text{External}}^B\left(C\right)$ denote the Cartesian Frame $(A/B,B\times E,\star )$, where $\star$ is given by $q\star (b,e)=q\left(b\right)\cdot e$. Let ${\text{External}}^{/B}\left(C\right)$ denote the Cartesian Frame $(B,A/B\times E,\diamond )$, where $\diamond$ is given by $b\diamond (q,e)=q\left(b\right)\cdot e$.

We say "externalizing $B$" for ${\text{External}}^B$ and "externalizing mod $B$" for ${\text{External}}^{/B}$.

${\text{External}}^B\left(C\right)$ can be thought of the result of the agent of $C$ first factoring its choice into choosing an equivalence class in $B$, and choosing an element of each equivalence class, and then externalizing the part of itself that chooses an equivalence class. I.e., we are drawing a new Cartesian frame which treats the choice of equivalence class as part of the environment, rather than part of the agent.

Similarly, ${\text{External}}^{/B}\left(C\right)$ can be thought of the result of the agent of $C$ factoring as above, then externalizing the part of itself that chooses an element of each equivalence class.

Claim: For all partitions $B$ of $A$, ${\text{External}}^B\left(C\right){◃}_{\times }C$ and ${\text{External}}^{/B}\left(C\right){◃}_{\times }C$. Further, ${\text{External}}^B\left(C\right)$ and ${\text{External}}^{/B}\left(C\right)$ are sisters in $C$.

Proof: Let ${\text{External}}^B\left(C\right)=(A/B,B\times E,\star )$ and ${\text{External}}^{/B}\left(C\right)=(B,A/B\times E,\diamond )$.

First, observe that for every $e\in E$, there exists a morphism $(g_e,h_e):{\text{External}}^B\left(C\right)\to \left({\text{External}}^{/B}\right(C){)}^{\ast }$, with $g_e:\to A/B\times E$ given by $g_e\left(q\right)=(q,e)$, and $h_e:B\to B\times E$ given by $h_e\left(b\right)=(b,e)$. To see that this is a morphism, observe that for all $q\in A/B$ and $b\in B$,

Let $E^{\prime }\subseteq \text{hom}\left({\text{External}}^B\right(C),({\text{External}}^{/B}\left(C\right){)}^{\ast })$ be given by $E^{\prime }=\left\{\right(g_e,h_e)|e\in E\}$, and let $D=(A/B\times B,E^{\prime },\bullet )$, where

Our goal is to show that $D\in {\text{External}}^B\left(C\right)⊠{\text{External}}^{/B}\left(C\right)$, and that $D\simeq C$.

To see $D\simeq C$, we define $(g_0,h_0):D\to C$ and $(g_1,h_1):C\to D$ as follows.

First, $g_0:A/p\times B\to A$ is given by $g_0(q,b)=q\left(b\right)$. We first need to confirm that $g_0$ is surjective. Given any $a\in A$, we can let $b\in B$ be the set with $a\in b$ and construct a function $q\in A/B$ by saying $q\left(b\right)=a$, and for each $b^{\prime }\ne b$, choosing an $a^{\prime }\in b^{\prime }$, and saying $q\left(b^{\prime }\right)=a^{\prime }$. Observing that $g_0(q,b)=a$, we have that $g_0$ is surjective and thus has a right inverse.

We choose $g_1:A\to A/B\times B$ to be any right inverse to $g_0$. Similarly, we define $h_0:E\to E^{\prime }$ by $h_0\left(e\right)=(g_e,h_e)$, which is clearly surjective, and define $h_1:E^{\prime }\to E$ to be any right inverse to $h_0$. (Indeed, $h_0$ is bijective as long as $A$ is nonempty.)

Then, for all $(q,b)\in A/B\times B$ and $e\in E$, we have

so $(g_0,h_0)$ is a morphism. This also gives us that for all $a\in A$ and $e^{\prime }\in E^{\prime }$ we have

so $(g_1,h_1)$ is a morphism. We know $(g_0,h_0)\circ (g_1,h_1)$ is homotopic to the identity on $C$, since $g_0\circ g_1$ is the identity on $A$, and we know that $(g_1,h_1)\circ (g_0,h_0)$ is homotopic to the identity on $D$, since $h_0\circ h_1$ is the identity on $E^{\prime }$. Thus, $D\simeq C$.

To show that $D\in {\text{External}}^B\left(C\right)⊠{\text{External}}^{/B}\left(C\right)$, it suffices to show that

and

where ${\star }^{\prime }$ and ${\diamond }^{\prime }$ are given by

Indeed, we show that if $A$ is nonempty, $(A/B,B\times E,\star )\cong (A/B,B\times E^{\prime },{\star }^{\prime })$, and $(B,A/B\times E,\diamond )\cong (B,A/B\times E^{\prime },{\diamond }^{\prime })$.

If $A$ is nonempty, then the function from $E$ to $E^{\prime }$ given by $e\mapsto (g_e,h_e)$ is a bijection, since it is clearly surjective, and is injective because $e$ is uniquely defined by $g_e\left(a\right)=(a,e)$. This gives a bijection between $B\times E$ and $B\times E^{\prime }$, and we have that for all $q\in A/B$, $b\in B$, and $e\in E$,

Similarly, we have a bijection between $A/B\times E$ and $A/B\times E^{\prime }$, and for all $q\in A/B$, $b\in B$, and $e\in E$,

If $A$ is empty, then $B$ is empty, and $A/p$ is a singleton empty function, so $(A/B,B\times E,\star )\simeq (A/B,B\times E^{\prime },{\star }^{\prime })\simeq \top$, and we either have $(B,A/B\times E,\diamond )\simeq (B,A/B\times E^{\prime },{\diamond }^{\prime })\simeq 0$ or $(B,A/B\times E,\diamond )\simeq (B,A/B\times E^{\prime },{\diamond }^{\prime })\simeq \text{null}$, depending on whether or not $E$ is empty.

Thus, $D\in {\text{External}}^B\left(C\right)⊠{\text{External}}^{/B}\left(C\right)$, so ${\text{External}}^B\left(C\right)$ and ${\text{External}}^{/B}\left(C\right)$ are sisters in $C$. $\square$

1.4. Internalizing

Definition: Given a partition $F$ of $E$, let ${\text{Internal}}^F\left(C\right)$ denote the Cartesian Frame $(F\times A,E/F,\star )$, where $\star$ is given by $(f,a)\star q=a\cdot q\left(f\right)$. Let  ${\text{Internal}}^{/F}\left(C\right)$ denote the Cartesian Frame $(E/F\times A,F,\diamond )$, where $\diamond$ is given by $(q,a)\diamond f=a\cdot q\left(f\right)$. 

We say "internalizing $p$" for ${\text{Internal}}^p$ and "internalizing mod $p$" for ${\text{Internal}}^{/p}$.

Claim: For all partitions $F$ of $E$, $\left({\text{Internal}}^F\right(C){)}^{\ast }\cong {\text{External}}^F(C^{\ast })$ and $\left({\text{Internal}}^{/F}\right(C){)}^{\ast }\cong {\text{External}}^{/F}(C^{\ast })$. Similarly, for all partitions $B$ of $A$, $\left({\text{External}}^B\right(C){)}^{\ast }\cong {\text{Internal}}^B(C^{\ast })$ and $\left({\text{External}}^{/B}\right(C){)}^{\ast }\cong {\text{Internal}}^{/B}(C^{\ast })$. 

Proof: Trivial. $\square$

Claim: For all partitions $F$ of $E$, $C{◃}_{\times }{\text{Internal}}^F\left(C\right)$ and $C{◃}_{\times }{\text{Internal}}^{/F}\left(C\right)$.

Proof: This follows from the fact that $\left({\text{Internal}}^F\right(C){)}^{\ast }\cong {\text{External}}^F(C^{\ast }){◃}_{\times }{C}^{\ast }$ and $\left({\text{Internal}}^{/F}\right(C){)}^{\ast }\cong {\text{External}}^{/F}(C^{\ast }){◃}_{\times }{C}^{\ast }$, and the fact that multiplicative subagent is equivalent to multiplicative sub-environment. $\square$

2. Definitions from Worlds

The above definitions are dependent on subsets and partitions of a given $A$ and $E$, and thus do not represent a single operation that can be applied to an arbitrary Cartesian frame over $W$. We will now use the above eight definitions to define another eight operations that instead use subsets and partitions of $W$.

Once we have the following definitions in hand, our future references to committing, assuming, externalizing, and internalizing will use the definitions from worlds unless noted otherwise.

2.1. Committing

Definition: Given a set $S\subseteq W$, we define ${\text{Commit}}_S\left(C\right)={\text{Commit}}^B\left(C\right)$ and ${\text{Commit}}_{\setminus S}\left(C\right)={\text{Commit}}^{\setminus B}\left(C\right)$, where $B\subseteq A$ is given by $B=\{a\in A|\forall e\in E,a\cdot e\in S\}$.

Claim: For all $S\subseteq W$, ${\text{Commit}}_S\left(C\right){◃}_{+}C$ and ${\text{Commit}}_{\setminus S}\left(C\right){◃}_{+}C$. Further, ${\text{Commit}}_{\setminus S}\left(C\right)$ and ${\text{Commit}}_S\left(C\right)$ are brothers in $C$.

Proof: Trivial. $\square$

Unlike before, it is not necessarily the case that ${\text{Commit}}_{\setminus S}\left(C\right)\cong {\text{Commit}}_{W\setminus S}\left(C\right)$. This is because there might be rows that contains both elements of $S$ and elements of $W\setminus S$.

2.2. Assuming

Definition: Given $S\subseteq W$, we define ${\text{Assume}}_S\left(C\right)={\text{Assume}}^F\left(C\right)$ and ${\text{Assume}}_{\setminus S}\left(C\right)={\text{Assume}}^{\setminus F}\left(C\right)$, where $F\subseteq E$ is given by $F=\{e\in E|\forall a\in A,a\cdot e\in S\}.$

Claim: For all $S\subseteq W$, $\left({\text{Assume}}_S\right(C){)}^{\ast }\cong {\text{Commit}}_S(C^{\ast })$ and $\left({\text{Assume}}_{\setminus S}\right(C){)}^{\ast }\cong {\text{Commit}}_{\setminus S}(C^{\ast })$, $\left({\text{Commit}}_S\right(C){)}^{\ast }\cong {\text{Assume}}_S(C^{\ast })$ and $\left({\text{Commit}}_{\setminus S}\right(C){)}^{\ast }\cong {\text{Assume}}_{\setminus S}(C^{\ast })$.

Proof: Trivial. $\square$

Claim: For all $S\subseteq W$, $C{◃}_{+}^{\ast }{\text{Assume}}_S\left(C\right)$ and $C{◃}_{+}^{\ast }{\text{Assume}}_{\setminus S}\left(C\right)$.

Proof: Trivial. $\square$

2.3. Externalizing

Definition: Given a partition $V$ of $W$, let $v:W\to V$ send each element $w\in W$ to the part that contains it. We define ${\text{External}}_V\left(C\right)={\text{External}}^B\left(C\right)$ and ${\text{External}}_{/V}\left(C\right)={\text{External}}^{/B}\left(C\right)$, where $B=\left\{\right\{a^{\prime }\in A|\forall e\in E,v(a^{\prime }\cdot e)=v(a\cdot e)\}|a\in A\}$.

Claim: For all partitions $V$ of $W$, ${\text{External}}_V\left(C\right){◃}_{\times }C$ and ${\text{External}}_{/V}\left(C\right){◃}_{\times }C$. Further, ${\text{External}}_V\left(C\right)$ and ${\text{External}}_{/V}\left(C\right)$ are sisters in $C$.

Proof: Trivial. $\square$

2.4. Internalizing

Definition: Given a partition $V$ of $W$, let $v:W\to V$ send each element $w\in W$ to the part that contains it. We define ${\text{Internal}}_V\left(C\right)={\text{Internal}}^F\left(C\right)$ and ${\text{Internal}}_{/V}\left(C\right)={\text{Internal}}^{/F}\left(C\right)$, where $F=\left\{\right\{e^{\prime }\in E|\forall a\in a,v(a\cdot e^{\prime })=v(a\cdot e)\}|e\in E\}$.

Claim: For all partitions $V$ of $W$, $C{◃}_{\times }{\text{Internal}}_V\left(C\right)$ and $C{◃}_{\times }{\text{Internal}}_{/V}\left(C\right)$. 

Proof: Trivial. $\square$

Claim: For all partitions $V$ of $W$, $\left({\text{Internal}}_V\right(C){)}^{\ast }\cong {\text{External}}_V(C^{\ast })$, $\left({\text{Internal}}_{/V}\right(C){)}^{\ast }\cong {\text{External}}_{/V}(C^{\ast })$, $\left({\text{External}}_V\right(C){)}^{\ast }\cong {\text{Internal}}_V(C^{\ast })$, and $\left({\text{External}}_{/V}\right(C){)}^{\ast }\cong {\text{Internal}}_{/V}(C^{\ast })$. 

Proof: Trivial. $\square$

3. Basic Properties

3.1. Biextensional Equivalence

Committing and assuming are well-defined up to biextensional equivalence.

Claim: If $C_0\simeq C_1$ are Cartesian frames over $W$, then for any subset $S\subseteq W$, ${\text{Commit}}_S\left(C_0\right)\simeq {\text{Commit}}_S\left(C_1\right)$, ${\text{Commit}}_{\setminus S}\left(C_0\right)\simeq {\text{Commit}}_{\setminus S}\left(C_1\right)$,${\text{Assume}}_S\left(C_0\right)\simeq {\text{Assume}}_S\left(C_1\right)$, and ${\text{Assume}}_{\setminus S}\left(C_0\right)\simeq {\text{Assume}}_{\setminus S}\left(C_1\right)$.

Proof: Let $C_i=(A_i,E_i,{\cdot }_i)$, and let $(g_0,h_0):C_0\to C_1$ and $(g_1,h_1):C_1\to C_0$ compose to something homotopic to the identity in both orders. Let $B_i\subset A_i$ be defined by $B_i=\{a\in A_i|\forall e\in E_i,a{\cdot }_{i}e\in S\}$.

Observe that if $b\in B_0$, then for all $e\in E_1$, $g_0\left(b\right){\cdot }_{1}e=b{\cdot }_0{h}_0\left(e\right)\in S$, so $g_0\left(b\right)\in B_1$. Similarly, if $b\in B_1$, then $g_1\left(b\right)\in B_0$. Thus, if we let $g_i^{\prime }:B_i\to B_{1-i}$ be given by $g_i^{\prime }\left(b\right)=g_i\left(b\right)$, we get morphisms $(g_i^{\prime },h_i):{\text{Commit}}_S\left(C_i\right)\to {\text{Commit}}_S\left(C_{1-i}\right)$, which are clearly morphisms, since they are restrictions of our original morphisms $(g_i,h_i)$.

Since $(g_0,h_0)$ and $(g_1,h_1)$ compose to something homotopic to the identity in both orders, $({\text{id}}_{A_i},h_{1-i}\circ h_i):C_i\to C_i$ is a morphism, so $({\text{id}}_{B_i},h_{1-i}\circ h_i):{\text{Commit}}_S\left(C_i\right)\to {\text{Commit}}_S\left(C_i\right)$ is a morphism, so $(g_0^{\prime },h_0)$ and $(g_1^{\prime },h_1)$ compose to something homotopic to the identity in both orders. Thus ${\text{Commit}}_S\left(C_0\right)\simeq {\text{Commit}}_S\left(C_1\right)$.

Similarly, if $b\in A_0\setminus B_0$, then there exists an $e\in E_0$ such that $b{\cdot }_{0}e\notin S$. But then

$\notin S$, so $g_0\left(b\right)\in A_1\setminus B_1$. Similarly, if $b\in A_1\setminus B_1$, then $g_1\left(b\right)\in A_0\setminus B_0$. Thus, if we let $g_i^{\prime \prime }:A_i\setminus B_i\to A_{1-i}\setminus B_{1-i}$ be given by $g_i^{\prime \prime }\left(b\right)=g_i\left(b\right)$, we get morphisms $(g_i^{\prime \prime },h_i):{\text{Commit}}_{\setminus S}\left(C_i\right)\to {\text{Commit}}_{\setminus S}\left(C_{1-i}\right)$, which (similarly to above) compose to something homotopic to the identity in both orders. Thus, ${\text{Commit}}_{\setminus S}\left(C_0\right)\simeq {\text{Commit}}_{\setminus S}\left(C_1\right)$.

We know that ${\text{Assume}}_S\left(C_0\right)\simeq {\text{Assume}}_S\left(C_1\right)$ and ${\text{Assume}}_{\setminus S}\left(C_0\right)\simeq {\text{Assume}}_{\setminus S}\left(C_1\right)$, because

and

$\square$

Externalizing and internalizing are also well-defined up to biextensional equivalence.

Claim: If $C_0\simeq C_1$ are Cartesian frames over $W$, then for all partitions $V$ of $W$, ${\text{External}}_V\left(C_0\right)\simeq {\text{External}}_V\left(C_1\right)$, ${\text{External}}_{/V}\left(C_0\right)\simeq {\text{External}}_{/V}\left(C_1\right)$, ${\text{Internal}}_V\left(C_0\right)\simeq {\text{Internal}}_V\left(C_1\right)$, and ${\text{Internal}}_{/V}\left(C_0\right)\simeq {\text{Internal}}_{/V}\left(C_1\right)$.

Proof: Let $C_i=(A_i,E_i,{\cdot }_i)$, and let $(g_0,h_0):C_0\to C_1$ and $(g_1,h_1):C_1\to C_0$ compose to something homotopic to the identity in both orders. Let $V$ be a partition of W, and let $v:W\to V$ send each element $w\in W$ to the part that contains it. Let $B_i$ be the partition of $A_i$ defined by $B_i=\left\{\right\{a^{\prime }\in A_i|\forall e\in E_i,v(a^{\prime }\cdot e)=v(a\cdot e)\}|a\in A_i\}$. 

Let ${\beta }_i:A_i\to B_i$, send each element of $A_i$ to its part in $B_i$, so ${\beta }_i\left(a\right)=\{a^{\prime }\in A_i|\forall e\in E_i,v(a^{\prime }\cdot e)=v(a\cdot e\left)\right\}$. Since ${\beta }_i$ is surjective, it has a right inverse. Let ${\alpha }_i:B_i\to A_i$ be any choice of right inverse to ${\beta }_i$. This gives a pair of functions ${\iota }_i:B_i\to B_{1-i}$ given by ${\iota }_i={\beta }_{1-i}\circ g_i\circ {\alpha }_i$.

We start by showing that ${\iota }_0$ and ${\iota }_1$ are inverses, and thus bijections between $B_0$ and $B_1$. We do this by showing that ${\beta }_i\circ g_{1-i}\circ g_i={\beta }_i$, and that ${\iota }_i\circ {\beta }_i={\beta }_{1-i}\circ g_i$ , and thus we will have

which is the identity on $B_i$.

To see that ${\beta }_i\circ g_{1-i}\circ g_i={\beta }_i$, observe that for all $a\in A_i$, we have that for all $e\in E_i$, $v\left(a{\cdot }_{i}e\right)=v\left(g_{1-i}\right(g_i\left(a\right)\left){\cdot }_{i}e\right)$, so, ${\beta }_i\left(a\right)={\beta }_i\left(g_{1-i}\right(g_i\left(a\right)\left)\right)$. Thus, ${\beta }_i={\beta }_i\circ g_{1-i}\circ g_i$.

To see that ${\iota }_i\circ {\beta }_i={\beta }_{1-i}\circ g_i$, first observe that for all $a\in A_i$, we have that ${\beta }_i\left({\alpha }_i\right({\beta }_i\left(a\right)\left)\right)={\beta }_i\left(a\right)$, and thus, for all $e\in E_{1-i}$,