# Additive and Multiplicative Subagents

This is the ninth post in the [Cartesian frames](https://www.lesswrong.com/posts/BSpdshJWGAW6TuNzZ/introduction-to-cartesian-frames) sequence. Here, we refine our notion of subagent into additive and multiplicative subagents. As usual, we will give many equivalent definitions. 

The additive subagent relation can be thought of as representing the relationship between an agent that has made a commitment, and the same agent before making that commitment. The multiplicative subagent relation can be thought of as representing the relationship between a football player and a football team.

Another way to think about the distinction is that additive subagents have fewer options, while multiplicative subagents have less refined options.

We will introduce these concepts with a definition using [sub-sums and sub-tensors](https://www.lesswrong.com/posts/LAHXvi4qwXogmdTHd/sub-sums-and-sub-tensors).

1\. Definitions of Additive and Multiplicative Subagent
-------------------------------------------------------

**1.1. Sub-Sum and Sub-Tensor Definitions**

**Definition:** $C$ is an additive subagent of $D$, written $C\triangleleft_+D$, if there exists a $C^\prime$ and a $D^\prime\simeq D$ with $D^\prime\in C\boxplus C^\prime$.

**Definition:** $C$ is a multiplicative subagent of $D$, written $C\triangleleft_\times D$, if there exists a $C^\prime$ and $D^\prime\simeq D$ with $D^\prime\in C\boxtimes C^\prime$.

These definitions are nice because they motivate the names "additive" and "multiplicative." Another benefit of these definitions is that they draw attention to the Cartesian frames given by C′. This feature is emphasized more in the below (clearly equivalent) definition.

**1.2. Brother and Sister Definitions**

**Definition:** $C^\prime$ is called a brother to $C$ in $D$ if $D\simeq D^\prime$ for some $D^\prime\in C\boxplus C^\prime$. Similarly, $C^\prime$ is called a sister to $C$ in $D$ if $D\simeq D^\prime$ for some $D^\prime\in C\boxtimes C^\prime$.

E.g., one "sister" of a football player will be the entire rest of the football team. One "brother" of a person that precommitted to carry an umbrella will be the counterfactual version of themselves that instead precommitted to *not* carry an umbrella.

This allows us to trivially restate the above definitions as:

**Definition:** We say $C\triangleleft_+ D$ if $C$ has a brother in $D$ and $C\triangleleft_\times D$ if $C$ has a sister in $D$.

**Claim:** This definition is equivalent to the ones above.

**Proof:** Trivial. $\square$

**1.3. Committing and Externalizing Definitions**

Next, we will give the committing definition of additive subagent and an externalizing definition of multiplicative subagent. These definitions are often the easiest to work with directly in examples.

We call the following definition the "committing" definition because we are viewing $C$ as the result of $D$ making a commitment (up to biextensional equivalence).

**Definition:** Given Cartesian frames $C$ and $D$ over $W$, we say $C\triangleleft_+ D$ if there exist three sets $X$, $Y$, and $Z$, with $X\subseteq Y$, and a function $f:Y\times Z\rightarrow W$ such that $C\simeq (X,Z,\diamond)$ and $D\simeq (Y,Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond z=f(x,z)$ and $$$$$y\bullet z=f(y,z)$.

**Claim:** This definition is equivalent to the sub-sum and brother definitions of $\triangleleft_+$.

**Proof:** First, assume that $C$ has a brother in $D$. Let $C=(A,E,\cdot)$, and let $D=(B,F,\star)$. Let $C^\prime=(A^\prime,E^\prime,\cdot^\prime)$ be brother to $C$ in $D$. Let $D^\prime=(B^\prime,F^\prime,\star^\prime)$ be such that  $D^\prime\simeq D$ and $D^\prime\in C\boxplus C^\prime$. Then, if we let $X=A$, let $Y=B^\prime =A\sqcup A^\prime$, let $Z=F^\prime$, and let $f(y,z)=y\star^\prime z$, we get $D\simeq D^\prime=(Y,Z,\bullet)$, where $y\bullet z=f(y,z)$, and by the definition of sub-sum, $C\simeq (X,Z,\diamond)$, where  $x\diamond z=f(x,z)$.

Conversely, let $X$, $Y$, and $Z$ be arbitrary sets with $X\subseteq Y$, and let $f:Y\times Z\rightarrow W$.  Let $C\simeq C_0=(X,Z,\diamond_0)$, and let $D\simeq D^\prime=(Y,Z,\bullet)$, where $x\diamond_0 z=f(x,z)$ and $$$$$y\bullet z=f(y,z)$. We want to show that $C$ has a brother in $D$. It suffices to show that $C_0$ $$has a brother in $D$, since sub-sum is well-defined up to biextensional equivalence. Indeed, we will show that $C_1=(Y\backslash X , Z, \diamond_1)$ is brother to $C_0$ in $D$, where $\diamond_1$ is given by $x\diamond_1 z=f(x,z)$. 

Observe that $C_0\oplus C_1=(Y,Z\times Z,\bullet^\prime)$, where $\bullet^\prime$ is given by

$$
\begin{equation} \begin{split} y\bullet^\prime (z_0,z_1) & =y\diamond_0 z_0 \ & =y\bullet z_0 \end{split} \end{equation}
$$

if $y\in X$, and is given by

$$
\begin{equation} \begin{split} y\bullet^\prime (z_0,z_1) & =y\diamond_1 z_1 \ & =y\bullet z_1 \end{split} \end{equation}
$$

otherwise. Consider the diagonal subset $S\subseteq Z\times Z$ given by $S=\{(z,z)  |  z\in Z\}$. Observe that the map $z\mapsto (g_z,h_z)$ is a bijection from $Z$ to $S$. Observe that if we restrict $\bullet^\prime$ to $Y\times S$, we get $\bullet^{\prime\prime}:Y\times S\rightarrow W$ given by $y\bullet^{\prime\prime} (z,z)=y\bullet z$. Thus $(Y,S,\bullet^{\prime\prime})\cong (Y,Z,\bullet)\mbox{,}$ with the isomorphism coming from the identity on $Y$, and the bijection between $S$ and $Z$.

If we further restrict $\bullet^{\prime\prime}$ to $X\times S$ or $(Y\backslash X)\times S$, we get $\bullet_0$ and $\bullet_1$ respectively, given by $x\bullet_0=x\diamond_0z$ and $x\bullet_1(z,z)=x\diamond_1 z$. Thus $(X,S,\bullet_0)\cong (X,Z,\diamond_0)$ and $(Y\backslash X,S,\bullet_1)\cong (Y\backslash X,Z,\diamond_1)$, with the isomorphisms coming from the identities on $Y$ and $X\backslash Y$, and the bijection between $S$ and $Z$.

Thus $(Y,S,\bullet^{\prime\prime})\in C_0\boxplus C_1$, and $(Y,S,\bullet^{\prime\prime})\cong D^\prime\simeq D$, so $C_1$ is brother to $C_0$ in $D$, so $C$ has a brother in $D$. $\square$

Next, we have the externalizing definition of multiplicative subagent. Here, we are viewing $C$ as the result of $D$ sending some of its decisions into the environment (up to biextensional equivalence).

**Definition:** Given Cartesian frames $C$ and $D$ over $W$, we say $C\triangleleft_\times D$ if there exist three sets $X$, $Y$, and $Z$, and a function $f:X\times Y\times Z\rightarrow W$ such that $C\simeq (X,Y\times Z,\diamond)$ and $D\simeq (X\times Y,Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond (y,z)=f(x,y,z)$ and $$$$$(x,y)\bullet z=f(x,y,z)$.

**Claim:** This definition is equivalent to the sub-tensor and sister definitions of $\triangleleft_\times$.

**Proof:** First, assume that $C$ has a sister in $D$. Let $C=(A,E,\cdot)$, and let $D=(B,F,\star)$. Let $C^\prime=(A^\prime,E^\prime,\cdot^\prime)$ be sister to $C$ in $D$. Let $D^\prime=(B^\prime,F^\prime,\star^\prime)$ be such that  $D^\prime\simeq D$ and $D^\prime\in C\boxtimes C^\prime$. Then, if we let $X=A$, let $Y=A^\prime$, let $Z=F^\prime\subseteq \mbox{hom}(C,{C^\prime}^*)$, and let

$$
\begin{equation} \begin{split} f(x,y,(g,h)) & =x\cdot h(y) \ & =y\cdot^\prime g(x)\mbox{,} \end{split} \end{equation}
$$

we get $D\simeq D^\prime=(X\times Y,Z,\bullet)$, where $(x,y)\bullet z=f(x,y,z)$, and by the definition of sub-tensor, $C\simeq (X,Y\times Z,\diamond)$, where  $x\diamond (y,z)=f(x,y,z)$.

Conversely, let $X$, $Y$, and $Z$ be arbitrary sets, and let $f:X\times Y\times Z\rightarrow W$. Let $C\simeq C_0=(X,Y\times Z,\diamond_0)$, and let $D\simeq D^\prime=(X\times Y,Z,\bullet)$, where $x\diamond_0 (y,z)=(x,y)\bullet z=f(x,y,z)$. We will assume for now that at least one of $X$ and $Y$ is nonempty, as the case where both are empty is degenerate. 

We want to show that $C$ has a sister in $D$. It suffices to show that $C_0$ $$has a sister in $D$, since sub-tensor is well-defined up to biextensional equivalence. Indeed, we will show that $C_1=(Y ,X\times Z, \diamond_1)$ is sister to $C_0$ in $D$, where $\diamond_1$ is given by $y\diamond_1 (x,z)=f(x,y,z)$. 

Observe that $C_0\otimes C_1=(X\times Y,\mbox{hom}(C_0,C_1^*),\bullet^\prime)$, where $\bullet^\prime$ is given by

$$
\begin{equation} \begin{split} (x,y)\bullet^\prime(g,h) & =x\diamond_0 h(y) \ & =y\star_1 g(x). \end{split} \end{equation}
$$

For every $z\in Z$, there is a morphism $(g_z,h_z):C_0\rightarrow C_1^*$, where $g_z:X\rightarrow X\times Z$ is given by $g_z(x)=(x,z)$, and $g_z:Y\rightarrow Y\times Z$ is given by $g_z(x)=(x,z)$. This is clearly a morphism. Consider the subset $S\subseteq \mbox{hom}(C_0,C_1^*)$ given by $S=\{(g_z,h_z)  |  z\in Z\}$. Observe that the map $z\mapsto (g_z,h_z)$ is a bijection from $Z$ to $S$. (We need that at least one of $X$ and $Y$ is nonempty here for injectivity.)

If we restrict $\bullet^\prime$ to $(X\times Y)\times S$, we get $\bullet^{\prime\prime}:(X\times Y)\times S\rightarrow W$ given by $y\bullet^{\prime\prime} (g_z,h_z)=y\bullet z$. Thus, $(X\times Y,S,\bullet^{\prime\prime})\cong (X\times Y,Z,\bullet)$, with the isomorphism coming from the identity on $X\times Y$, and the bijection between $S$ and $Z$.

To show that $(X\times Y,S,\bullet^{\prime\prime})\in C_0\boxtimes C_1$, we need to show that $C_0\simeq (X,Y\times S,\bullet_0)$ and $C_1\simeq (Y,X\times S,\bullet_1)$, where $\bullet_0$ and $\bullet_1$ are given by

$$
\begin{equation} \begin{split} x\bullet_0(y,(g_h,z_h)) &=y\bullet_1(x,(g_h,z_h)) \ &=(x,y)\bullet^{\prime\prime}(g_z,h_z). \end{split} \end{equation}
$$

Indeed,  $x\bullet_0(y,(g_h,z_h))=x\diamond_0(y,z)$ and $y\bullet_1(x,(g_h,z_h))=y\diamond_1(x,z)$, so $(X,Y\times S,\bullet_0)\cong (X,Y\times Z,\diamond_0)=C_0$ and $(Y,X\times S,\bullet_1)\cong (Y,X\times Z,\diamond_1)=C_1$, with the isomorphisms coming from the identities on $X$ and $Y$, and the bijection between $S$ and $Z$.

Thus $(X\times Y,S,\bullet^{\prime\prime})\in C_0\boxplus C_1$, and $(Y,S,\bullet^{\prime\prime})\cong D^\prime\simeq D$, so $C_1$ is sister to $C_0$ in $D$, so $C$ has a sister in $D$.

Finally, in the case where $X$ and $Y$ are both empty, $C\cong \mbox{null}$, and either $D\simeq \mbox{null}$ or $D\simeq 0$, depending on whether $Z$ is empty. It is easy to verify that $\mbox{null}\boxtimes\mbox{null}=\{0,\mbox{null}\}$, since $\mbox{null}\otimes\mbox{null}\cong 0$, taking the two subsets of the singleton environment in $0$ yields $0$ and $\mbox{null}$ as candidate sub-tensors, and both are valid sub-tensors, since either way, the conditions reduce to $\mbox{null}\simeq\mbox{null}$. $\square$

Next, we have some definitions that more directly relate to our original [definitions of subagent](https://www.lesswrong.com/posts/nwrkwTd6uKBesYYfx/subagents-of-cartesian-frames).

**1.4. Currying Definitions**

**Definition:** We say $C\triangleleft_+ D$ if there exists a Cartesian frame $M$ over $\mbox{Agent}(D)$ with $|\mbox{Env}(M)|=1$, such that $C\simeq D^\circ(M)$.

**Claim:** This definition is equivalent to all of the above definitions of $\triangleleft_+$.

**Proof:** We show equivalence to the committing definition. 

First, assume that there exist three sets $X$, $Y$, and $Z$, with $X\subseteq Y$, and a function $p:Y\times Z\rightarrow W$ such that $C\simeq (X,Z,\diamond)$ and $D\simeq (Y,Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond z=p(x,z)$ and $$$$$y\bullet z=p(y,z)$. 

Let $D=(B,F,\star)$, and let $(g_0,h_0):D\rightarrow(Y,Z,\bullet)$ and $(g_1,h_1):(Y,Z,\bullet)\rightarrow D$ compose to something homotopic to the identity in both orders. 

We define $M$, a Cartesian frame over $B$, by $M=(X,\{e\},\cdot)$, where $\cdot$ is given $$ $x\cdot e=g_1(x).$ Observe that $D^\circ(M)=(X,\{e\}\times F, \star^\prime)$, where $\star^\prime$ is given by 

$$
\begin{equation} \begin{split} x \star^\prime (e,f) &=(x\cdot e)\star f \ &=g_1(x)\star f \ &=x\bullet h_1(f). \end{split} \end{equation}
$$

To show that $(X,Z,\diamond)\simeq D^\circ(M)$, we construct morphisms $(g_2,h_2):(X,Z,\diamond)\rightarrow D^\circ(M)$ and $(g_3,h_3):D^\circ(M)\rightarrow(X,Z,\diamond)$ that compose$$$$ to something homotopic to the identity in both orders. Let $g_2$ and $g_3$ both be the identity on $X$. Let $h_2:\{e\}\times F\rightarrow Z$ be given by $h_2(e,f)=h_1(f)$, and let $h_3:Z\rightarrow\{e\}\times F$ be given by $h_3(z)=(e,h_0(z))$.

We know $(g_2,h_2)$ is a morphism, since for all $x\in X$ and $(e,f)\in \{e\}\times F$, we have 

$$
\begin{equation}\begin{split} g_2(x)\star^\prime (e,f) & =x\star^\prime(e,f)\ & =x\bullet h_1(f)\ & =x\diamond h_1(f)\ & =x\diamond(h_2(e,f)). \end{split} \end{equation}
$$

We also have that $(g_3,h_3)$ is a morphism, since for all $x\in X$ and $z\in Z$, we have

$$
\begin{equation} \begin{split} g_3(x)\diamond z & =x\diamond z \ & =x\bullet z \ & =x\bullet h_1(h_0(z)) \ & =x\star^\prime(e,h_0(z)) \ & =x\star^\prime h_3(z). \end{split} \end{equation}
$$

Observe that$$ $(g_2,h_2)$ and $(g_3,h_3)$ clearly compose to something homotopic to the identity in both $$orders, since $g_2\circ g_3$ and $g_3\circ g_2$ are the identity on $X$.

Thus, $C\simeq (X,Z,\diamond)\simeq D^\circ(M)$, and $|\mbox{Env}(M)|=1$.

Conversely, assume $C\simeq D^\circ(M)$, with $|\mbox{Env}(M)|=1$.  We define $Y=\mbox{Agent}(D)$ and $Z=\mbox{Env}(D)$. We define $f:Y\times Z\rightarrow W$ by $f(y,z)=y \bullet z$, where $\bullet=\mbox{Eval}(D)$.

Let $X\subseteq Y$ be given by $X=\mbox{Image}(M)$. Since $|\mbox{Env}(M)|=1$, we have $M\simeq \bot_{X}$. Thus, $C\simeq D^\circ(M)\simeq D^\circ(\bot_{X})$. Unpacking the definition of $D^\circ(\bot_X)$, we get $D^\circ(\bot_X)=(X, \{e\}\times Z,\cdot)$, where $\cdot$ is given by $x\cdot (e,z)=f(x,z)$, which is $$ isomorphic to $(X, Z,\diamond)$, where $\diamond$ is given by $x\diamond z=f(x,z)$. Thus $C\simeq (X,Z,\diamond)$ and $D=(Y,Z,\bullet)$, as in the committing definition. $\square$

**Definition:** We say $C\triangleleft_\times D$ if there exists a Cartesian frame $M$ over $\mbox{Agent}(D)$ with $\mbox{Image}(M)=\mbox{Agent}(D)$, such that $C\simeq D^\circ(M)$.

**Claim:** This definition is equivalent to all of the above definitions of $\triangleleft_\times$.

**Proof:** We show equivalence to the externalizing definition. 

First, assume there exist three sets $X$, $Y$, and $Z$, and a function $p:X\times Y\times Z\rightarrow W$ such that $C\simeq (X,Y\times Z,\diamond)$ and $D\simeq (X\times Y,Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond (y,z)=(x,y)\bullet z = p(x,y,z)$. 

Let $D=(B,F,\star)$, and let $(g_0,h_0):D\rightarrow(X\times Y,Z,\bullet)$ and $(g_1,h_1):(X\times Y,Z,\bullet)\rightarrow D$ compose to something homotopic to the identity in both orders.

We define $B^\prime=B\sqcup \{a\}$, and we define $M$, a Cartesian frame over $B$,$$ by $M=(X,Y\times B^\prime,\cdot)$, where $\cdot$ is given by $x\cdot (y,b)=b$ if $b\in B$ and $g_0(b)=(x,y)$, and $x\cdot (y,b)=g_1(x,y)$ otherwise. Clearly, $\mbox{Image}(M)=B$, since for any $b\in B$, if we let $(x,y)=g_0(b)$, we have $x\cdot (y,b)=b$.

Observe that for all $x\in X$, $y\in Y$, $b\in B^\prime$ and $f\in F$, if $b\in B$ and $g_0(b)=(x,y)$, then

$$
\begin{equation} \begin{split} (x\cdot (y,b))\star f & =b\star f\ & = g_1(g_0(b))\star f\ & =g_1(x,y)\star f, \end{split} \end{equation}
$$

and on the other hand, if $b=a$ or $g_0(b)\neq (x,y)$, we also have $(x\cdot (y,b))\star f=g_1(x,y)\star f$.

Thus, we have that $D^\circ(M)=(X,Y\times B^\prime\times  F, \star^\prime)$, where $\star^\prime$ is given by

$$
\begin{equation} \begin{split} x \star^\prime (y,b,f) & =(x\cdot (y,b))\star f\ & =g_1(x,y)\star f\ & =(x,y) \bullet h_1(f). \end{split} \end{equation}
$$

To show that $(X,Y\times Z,\diamond)\simeq D^\circ(M)$, we construct morphisms $(g_2,h_2):(X,Y\times Z,\diamond)\rightarrow D^\circ(M)$ and $(g_3,h_3):D^\circ(M)\rightarrow(X,Y\times Z,\diamond)$ that compose to something homotopic to the identity in both orders. Let $g_2$ and $g_3$ both be the identity on $X$. Let $h_2:Y\times B^\prime\times F\rightarrow Y\times Z$ be given by $h_2(y,b,f)=(y,h_1(f))$, and let $h_3:Y\times Z\rightarrow Y\times B^\prime\times F$ be given by $h_3(y,z)=(y,a,h_0(z))$.

We know $(g_2,h_2)$ is a morphism, since for all $x\in X$ and $(y,b,f)\in Y\times B^\prime\times F$,

$$
\begin{equation} \begin{split} g_2(x)\star^\prime (y,b,f) & =x\star^\prime(y,b,f) \ & =(x,y)\bullet h_1(f) \ & =p(x,y,h_1(f)) \ & =x\diamond(y,h_1(f)) \ & =x\diamond(h_2(y,b,f)). \end{split} \end{equation}
$$

We also have that $(g_3,h_3)$ is a morphism, since for all $x\in X$ and $(y,z)\in Y\times Z$, we have

$$
\begin{equation} \begin{split} g_3(x)\diamond(y,z) & =x\diamond z \ & =x\diamond(y,z) \ & =p(x,y,z) \ & =(x,y)\bullet z \ & =(x,y)\bullet h_1(h_0(z)) \ & =(x,y)\star^\prime(y,a,h_0(z)) \ & =x\star^\prime h_3(y,z). \end{split} \end{equation}
$$

Observe that $(g_2,h_2)$ and $(g_3,h_3)$ clearly compose to something homotopic to the identity in both orders, since $g_2\circ g_3$ and $g_3\circ g_2$ are the identity on $X$.

Thus, $C\simeq (X,Z,\diamond)\simeq D^\circ(M)$, where $\mbox{Image}(M)=\mbox{Agent}(D)$. 

Conversely, assume $C\simeq D^\circ(M)$, with  $\mbox{Image}(M)=\mbox{Agent}(D)$. Let $X=\mbox{Agent}(M)$, let $Y=\mbox{Env}(M)$, and let $Z=\mbox{Env}(D)$. Let $f:X\times Y\times Z\rightarrow W$ be given by $f(x,y,z)= (x \cdot y) \star z$, where $\cdot=\mbox{Eval}(M)$ and $\star=\mbox{Eval}(D)$.

Thus $C\simeq D^\circ(M)\cong (X,Y\times Z, \diamond)$, where $\diamond$ is given by $x\diamond(y,z)= (x \cdot y) \star z=f(x,y,z)$. All that remains to show is that $D\simeq(X\times Y,Z,\bullet)$, where $(x,y)\bullet z=f(x,y,z)$. Let $D=(B,Z,\star)$.

We construct morphisms $(g_0,h_0):D\rightarrow (X\times Y,Z,\bullet)$ and $(g_1,h_1):D\rightarrow (X\times Y,Z,\bullet)$ that compose to something homotopic to the identity in both orders. Let $h_0$ and $h_1$ be the identity on $Z$. Let $g_1:X\times Y\rightarrow B$ be given by $g_1(x,y)=x\cdot y$. Since $g_1$ is surjective, it has a right inverse. Let $g_0:B\rightarrow X\times Y$ be any choice of right inverse of $g_1$, so $g_1(g_0(b))=b$ for all $b\in B$. 

We know $(g_1,h_1)$ is a morphism, since for all $(x,y)\in X\times Y$ and $z\in Z$,

$$
\begin{equation} \begin{split} g_1(x,y)\star z & =(x\cdot y)\star z \ & =f(x,y,z) \ & =(x,y)\bullet z \ & =(x,y)\bullet h_1(z). \end{split} \end{equation}
$$

To see that $(g_0,h_0)$ is a morphism, given $b\in B$ and $z\in Z$, let $(x,y)=g_0(b)$, and observe

$$
\begin{equation} \begin{split} g_0(b)\bullet z & =(x,y)\bullet z \ & =f(x,y,z) \ & =(x\cdot y)\star z \ & =g_1(x,y)\star z \ & =g_1(g_0(b))\star z \ & =b\star h_0(z). \end{split} \end{equation}
$$

$(g_0,h_0)$ and $(g_1,h_1)$ clearly compose to something homotopic to the identity in both orders, since $h_0\circ h_1$ and $h_1\circ h_0$ are the identity on $Z$. Thus $D\simeq(X\times Y,Z,\bullet)$, completing the proof. $\square$

Consider two Cartesian frames $C$ and $D$, and let $M$ be a frame whose possible agents are $\mbox{Agent}(C)$ and whose possible worlds are $\mbox{Agent}(D)$. When $C$ is a subagent of $D$, (up to biextensional equivalence) there exists a function from $\mbox{Agent}(C)$, paired with $\mbox{Env}(M)$, to $\mbox{Agent}(D)$.

Just as we did in "Subagents of Cartesian Frames" §1.2 ([Currying Definition](https://www.lesswrong.com/posts/nwrkwTd6uKBesYYfx/subagents-of-cartesian-frames#1_2__Currying_Definition)), we can think of this function as a (possibly) nondeterministic function from $\mbox{Agent}(C)$ to $\mbox{Agent}(D)$, where $\mbox{Env}(M)$ represents the nondeterminism. In the case of additive subagents, $\mbox{Env}(M)$ is a singleton, meaning that the function from $\mbox{Agent}(C)$ to  $\mbox{Agent}(D)$ is actually deterministic. In the case of multiplicative subagents, the (possibly) nondeterministic function is surjective.

Recall that in "Sub-Sums and Sub-Tensors" §3.3 ([Sub-Sums and Sub-Tensors Are Superagents](https://www.lesswrong.com/posts/LAHXvi4qwXogmdTHd/sub-sums-and-sub-tensors#3_3__Sub_Sums_and_Sub_Tensors_Are_Superagents)), we constructed a frame with a singleton environment to prove that sub-sums are superagents, and we constructed a frame with a surjective evaluation function to prove that sub-tensors are superagents. The currying definitions of $\triangleleft_+$ and $\triangleleft_\times$ show why this is the case.

**1.5. Categorical Definitions**

We also have definitions based on the categorical definition of subagent. The categorical definition of additive subagent is almost just swapping the quantifiers from [our original categorical definition of subagent](https://www.lesswrong.com/posts/nwrkwTd6uKBesYYfx/subagents-of-cartesian-frames). However, we will also have to weaken the definition slightly in order to only require the morphisms to be homotopic.

**Definition:** We say $C\triangleleft_+ D$ if there exists a single morphism $\phi_0:C\rightarrow D$ such that for every morphism $\phi:C\rightarrow\bot$ there exists a morphism $\phi_1:D\rightarrow\bot$ such that $\phi$ is homotopic to $\phi_1\circ\phi_0$ .

**Claim:** This definition is equivalent to all the above definitions of $\triangleleft_+$.

**Proof:** We show equivalence to the committing definition. 

First, let $C=(A,E,\cdot)$ and $D=(B,F,\bullet)$ be Cartesian frames over $W$, and let $(g_0,h_0):C\rightarrow D$ be such that for all $(g,h):C\rightarrow \bot$, there exists a $(g^\prime,h^\prime):D\rightarrow  \bot$ such that $(g,h)$ is homotopic to $(g^\prime,h^\prime)\circ (g_0,h_0)$. Let $\bot=(W,\{i\},\star)$.

Let $Y=B$, let $Z=F$, and let $X=\{g_0(a)  |  a\in A\}$. Let $f:Y\times Z\rightarrow W$ be given by $f(y,z)=y\bullet z$. We already have $D=(Y,Z,\bullet)$, and our goal is to show that $C\simeq(X,Z,\diamond)\mbox{,}$ where $\diamond$ is given by $x\diamond z=f(x,z)$.

We construct $(g_1,h_1):C\rightarrow(X,Z,\diamond)$ and $(g_2,h_2):(X,Z,\diamond)\rightarrow C$ that compose to something homotopic to the identity in both orders.

We define $g_1:A\rightarrow X$ by $g_1(a)=g_0(a)$. $g_1$ is surjective, and so has a right inverse. We let $g_2:X\rightarrow A$ be any right inverse to $g_1$, so $g_1(g_2(x))=x$ for all $x\in X$. We let $h_1:Z\rightarrow E$ be given by $h_1(z)=h_0(z)$.

Defining $h_2:E\rightarrow Z$ will be a bit more complicated.  Given an $e\in E$, let $(g_e,h_e)$ be the morphism from $C$ to $\bot$, given by $h_e(i)=e$ and $g_e(a)=a\cdot e$. Let $(g_e^\prime,h_e^\prime):D\rightarrow \bot$ be such that $(g_e,h_e)$ is homotopic to $(g_e^\prime,h_e^\prime)\circ (g_0,h_0)$. We define $h_2$ by $h_2(e)=h_e^\prime(i)$.

We trivially have that $(g_1,h_1)$ is a morphism, since for all $a\in A$ and $z\in Z$,

$$
\begin{equation} \begin{split} g_1(a) \diamond z & =g_0(a)\bullet z \ & =a \cdot h_0(z) \ & =a\cdot h_1(z). \end{split} \end{equation}
$$

To see that $(g_2,h_2)$ is a morphism, consider $x\in X$ and $e\in E$, and define $(g_e,h_e)$ and $(g_e^\prime,h_e^\prime)$ as above. Then,

$$
\begin{equation} \begin{split} x\diamond h_2(e) & =g_1(g_2(x))\diamond h_e^\prime(i) \ & = g_e^\prime(g_0(g_2(x)))\star i \ & =g_2(x)\cdot h_e(i) \ & =g_2(x)\cdot e. \end{split} \end{equation}
$$

We trivially have that $(g_1,h_1)\circ (g_2,h_2)$ is homotopic to the identity, since $g_1\circ g_2$ is the identity on $X$. To see that $(g_2,h_2)\circ (g_1,h_1)$ is homotopic to the identity on $C$, observe that for all $a\in A$ and $e\in E$, defining $(g_e,h_e)$ and $(g_e^\prime,h_e^\prime)$ as above,

$$
\begin{equation} \begin{split} g_2(g_1(a))\cdot e & =g_1(a) \diamond h_2(e) \ & =g_0(a) \diamond h_e^\prime (i) \ & =g_e^\prime(g_0(a))\star i \ & =a\star h_e(i) \ & =a\cdot e. \end{split} \end{equation}
$$

Thus $C\simeq(X,Z,\diamond)$, and $C\triangleleft_+ D$ according to the committing definition.

Conversely, let $X$, $Y$, and $Z$ be arbitrary sets with $X\subseteq Y$, let $f:Y\times Z\rightarrow W$, and let $C\simeq (X,Z,\diamond)$ and $D\simeq (Y,Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond z=f(x,z)$ and $$$$$y\bullet z=f(y,z)$.

Let $(g_1,h_1):C\rightarrow(X,Z,\diamond)$ and $(g_2,h_2):(X,Z,\diamond)\rightarrow C$ compose to something homotopic to the identity in both orders, and let $(g_3,h_3):D\rightarrow(Y,Z,\bullet)$ and $(g_4,h_4):(Y,Z,\bullet)\rightarrow D$ compose to something homotopic to the identity in both orders. Let $(g_0,h_0):(X,Z,\diamond)\rightarrow (Y,Z,\bullet)$ be given by $g_0$ is the embedding of $X$ in $Y$ and $h_0$ is the identity on $Z$. $(g_0,h_0)$ is clearly a morphism.

We let $\phi:C\rightarrow D=(g_4,h_4)\circ(g_0,h_0)\circ(g_1,h_1)$.

Given a $(g,h):C\rightarrow\bot$, our goal is to construct a $(g^\prime,h^\prime):D\rightarrow\bot$ such that $(g,h)$ is homotopic to $(g^\prime,h^\prime)\circ \phi$. 

Let $\bot=(W,\{i\},\star)$, let $C=(A,E,\cdot_0)$, and let $D=(B,F,\cdot_1)$. Let $h^\prime:\{i\}\rightarrow F$ be given by $h^\prime=h_3\circ h_2 \circ h$. Let $g^\prime:B\rightarrow W$ be given by $g^\prime(b)=b\cdot_1  h^\prime(i)$. This is clearly a morphism, since for all $b\in B$ and $i\in\{i\}$,

$$
\begin{equation} \begin{split} g^\prime(b)\star i & =g^\prime(b) \ & = b\cdot h^\prime(i). \end{split} \end{equation}
$$

To see that $(g,h)$ is homotopic to $(g^\prime,h^\prime)\circ(g_4,h_4)\circ(g_0,h_0)\circ(g_1,h_1)$, we just need to check that $(g,h_1\circ h_0\circ h_4\circ h^\prime):C\rightarrow \bot$ is a morphism. Or, equivalently, that $(g,h_1\circ h_4\circ h_3\circ h_2\circ h):C\rightarrow\bot$, since $h_0$ is the identity, and $h^\prime=h_3\circ h_2 \circ h$. 

Indeed, for all $a\in A$ and $i\in\{i\}$,

$$
\begin{equation} \begin{split} g(a)\star i & = a\cdot_0 h(a) \ & =a\cdot_0 h_1(h_2(h(a))) \ & =g_1(a)\diamond h_2(h(a)) \ & =g_1(a)\bullet h_2(h(a)) \ & =g_1(a)\bullet h_4(h_3(h_2(h(a)))) \ & =g_1(a)\diamond h_4(h_3(h_2(h(a)))) \ & =a\cdot_0 h_1(h_4(h_3(h_2(h(a))))). \end{split} \end{equation}
$$

Thus $(g,h)$ is homotopic to $(g^\prime,h^\prime)\circ\phi$, completing the proof. $\square$

**Definition:** We say $C\triangleleft_\times D$ if for every morphism $\phi:C\rightarrow\bot$, there exist morphisms $\phi_0:C\rightarrow D$ and $\phi_1:D\rightarrow\bot$ such that $\phi=\phi_1\circ \phi_0$, and for every morphism $\psi:1\rightarrow D$, there exist morphisms $\psi_0:1\rightarrow C$ and $\psi_1:C\rightarrow D$ such that $\psi=\psi_1\circ \psi_0$.

Before showing that this definition is equivalent to all of the above definitions, we will give one final definition of multiplicative subagent.

**1.6. Sub-Environment Definition**

First, we define the concept of a sub-environment, which is dual to the concept of a sub-agent.

**Definition:** We say $C$ is a sub-environment of $D$, written $C\triangleleft^* D$, if $D^* \triangleleft C^*$.

We can similarly define additive and multiplicative sub-environments.

**Definition:** We say $C$ is an additive sub-environment of $D$, written $C\triangleleft^*_+ D$, if $D^* \triangleleft_+ C^*$. We say $C$ is an multiplicative sub-environment of $D$, written $C\triangleleft^*_\times D$, if $D^* \triangleleft_\times C^*$.

This definition of a multiplicative sub-environment is redundant, because the set of frames with multiplicative sub-agents is exactly the set of frames with multiplicative sub-environments, as shown below:

**Claim:** $C\triangleleft_\times D$ if and only if $C\triangleleft^*_\times D$.

**Proof:** We prove this using the externalizing definition of $\triangleleft_\times$.

If $C\triangleleft_\times D$, then for some $X$, $Y$, $Z$, and $f:X\times Y\times Z\rightarrow W$, we have $C\simeq(X,Y\times Z,\diamond)$ and $D\simeq(X\times Y, Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond (y,z)=f(x,y,z)$ and $$$$$(x,y)\bullet z=f(x,y,z)$.

Observe that $D^*\simeq(Z,Y\times X,\cdot)$ and $C^*\simeq (Z\times Y,X,\star)$, where $\cdot$ and $\star$ are given by $z\cdot (y,x)=f(x,y,z)$ and $$$$$(z,y)\star x=f(x,y,z)$. Taking $X^\prime=Z$, $Y^\prime=Y$, $Z^\prime=X$, and $f^\prime(x,y,z)=f(z,y,x)$, this is exactly the externalizing definition of $D^*\triangleleft_\times C^*$, so $C\triangleleft^*_\times D$.

Conversely, if $C\triangleleft^*_\times D$, then $D^*\triangleleft_\times C^*$, so $C\cong \{C^*\}^*\triangleleft_\times \{D^*\}^*\cong D$. $\square$

We now give the sub-environment definition of multiplicative subagent:

**Definition:** We say $C\triangleleft_\times D$ if $C\triangleleft D$ and $C\triangleleft^* D$. Equivalently, we say $C\triangleleft_\times D$ if $C\triangleleft D$ and $D^* \triangleleft C^*$.

**Claim:** This definition is equivalent to the categorical definition of $\triangleleft_\times$.

**Proof:** The condition that for every morphism $\phi:C\rightarrow\bot$, there exist morphisms $\phi_0:C\rightarrow D$ and $\phi_1:D\rightarrow\bot$ such that $\phi=\phi_1\circ \phi_0$, is exactly the categorical definition of $C\triangleleft D$.  

The condition that for every morphism $\psi:1\rightarrow D$, there exist morphisms $\psi_0:1\rightarrow C$ and $\psi_1:C\rightarrow D$ such that $\psi=\psi_1\circ \psi_0$, is equivalent to saying that for every morphism  $\psi^*:D^*\rightarrow \bot$, there exist morphisms $\psi_0^*:C^*\rightarrow \bot$ and $\psi_1^*:D^*\rightarrow C^*$ such that $\psi^*=\psi_1^*\circ \psi_0^*$. This is the categorical definition of $D^*\triangleleft C^*$. $\square$

**Claim:** The categorical and sub-environment definitions of $\triangleleft_\times$ are equivalent to the other four definitions of multiplicative subagent above: sub-tensor, sister, externalizing, and currying.

**Proof:** We show equivalence between the externalizing and sub-environment definitions. First, assume that $C=(A,E,\cdot)$ and $D=(B,F,\star)$ are Cartesian frames over $W$ with $C\triangleleft D$ and $C\triangleleft^* D$.

We define $X=A$, $Z=F$, and $Y=\mbox{hom}(C,D)$. We define $p:X\times Y\times Z\rightarrow W$ by

$$
\begin{equation} \begin{split} p(a,(g,h),f) & =g(a)\star f \ &=a\cdot h(f). \end{split} \end{equation}
$$

We want to show that $C\simeq (X,Y\times Z,\diamond)$, and $D\simeq(X\times Y,Z,\bullet)$, where $\diamond$ and $\bullet$ are given by $x\diamond (y,z)=(x,y)\bullet z=p(x,y,z)$. 

To see $C\simeq (X,Y\times Z,\diamond)$, we construct $(g_0,h_0):C\rightarrow (X,Y\times Z,\diamond)$ and $(g_1,h_1):(X,Y\times Z,\diamond)\rightarrow C$ that compose to something homotopic to the identity in both orders. Let $g_0$ and $g_1$ be the identity on $X$ and let $h_0:Y\times Z\rightarrow E$ be defined by $h_0((g,h),f)=h(f)$. By the covering definition of subagent, $h_0$ is surjective, and so has a right inverse. Let $h_1:E\rightarrow Y\times Z$ be any right inverse of $h_0$, so $h_0(h_1(e))=e$ for all $e\in E$.

We know $(g_0,h_0)$ is a morphism, because for all $a\in A$ and $((g,h),f)\in Y\times Z$,

$$
\begin{equation} \begin{split} g_0(a)\diamond ((g,h),f) & =a\diamond((g,h),f) \ & =p(a,(g,h),f) \ & =a\cdot h(f) \ & =a\cdot h_0((g,h),f). \end{split} \end{equation}
$$

We know $(g_1,h_1)$ is a morphism, since for $x\in X$ and $e\in E$, if $((g,h),f)=h_1(e)$,

$$
\begin{equation} \begin{split} g_1(x)\cdot e & =x\cdot h_0((g,h),f) \ & =x\cdot h(f) \ & =p(x,(g,h),f) \ & =x\diamond((g,h),f) \ & =x\diamond h_1(e). \end{split} \end{equation}
$$

$(g_0,h_0)$ and $(g_1,h_1)$ clearly compose to something homotopic to the identity in both orders, since $g_0\circ g_1$ and $g_1\circ g_0$ are the identity on $X$.

To see $D\simeq (X\times Y,Z,\bullet)$, we construct $(g_2,h_2):D\rightarrow (X\times Y,Z,\bullet)$ and $(g_3,h_3):(X\times Y, Z,\bullet)\rightarrow D$ that compose to something homotopic to the identity in both orders. Let $h_2$ and $h_3$ be the identity on $Z$ and let $g_3:X\times Y\rightarrow B$ be defined by $g_3(a,(g,h))=g(a)$. By the covering definition of subagent and the fact that $D^*\triangleleft C^*$, $g_3$ is surjective, and so has a right inverse. Let $g_2:B\rightarrow X\times Y$ be any right inverse of $g_3$, so $g_3(g_2(b))=b$ for all $b\in B$.

We know $(g_3,h_3)$ is a morphism, because for all $f\in F$ and $(a,(g,h))\in X\times Y$,

$$
\begin{equation} \begin{split} g_3(a,(g,h))\star f & =g(a)\star f \ & =p(a,(g,h),f) \ & =(a,(g,h))\bullet f \ & = (a,(g,h))\bullet h_3(f). \end{split} \end{equation}
$$

We know $(g_2,h_2)$ is a morphism, since for $z\in Z$ and $b\in B$, if $(a,(g,h))=g_2(b)$,

$$
\begin{equation} \begin{split} g_2(b)\bullet z & =(a,(g,h))\bullet z \ & =p(a,(g,h),z) \ & =g(a)\star z \ & =g_3(a,(g,h))\star z \ & =b\star h_2(z). \end{split} \end{equation}
$$

Observe that $(g_2,h_2)$ and $(g_3,h_3)$ clearly compose to something homotopic to the identity in both orders, since $h_2\circ h_3$ and $h_3\circ h_2$ are the identity on $Z$.

Thus,  $C\simeq (X,Y\times Z,\diamond)$, and $D\simeq(X\times Y,Z,\bullet)$.

Conversely, if $C\triangleleft_\times D$ according to the externalizing definition, then we also have $D^*\triangleleft_\times C^*$. However, by the currying definitions of multiplicative subagent and of subagent, multiplicative subagent is stronger than subagent, so $C\triangleleft D$ and $D^*\triangleleft C^*$. $\square$

2\. Basic Properties
--------------------

Now that we have enough definitions of additive and multiplicative subagent, we can cover some basic properties.

First: Additive and multiplicative subagents are subagents.

**Claim:** If $C\triangleleft_+ D$, then $C\triangleleft D$. Similarly, if $C\triangleleft_\times D$, then $C\triangleleft D$.

**Proof:** Clear from the currying definitions. $\square$

Additive and multiplicative subagent are also well-defined up to biextensional equivalence.

**Claim:** If $C\triangleleft_+ D$, $C^\prime\simeq C$, and $D^\prime\simeq D$, then $C^\prime\triangleleft_+ D^\prime$. Similarly, if $C\triangleleft_\times D$, $C^\prime\simeq C$, and $D^\prime\simeq D$, then $C^\prime\triangleleft_\times D^\prime$.

**Proof:** Clear from the committing and externalizing definitions. $\square$

**Claim:** Both $\triangleleft_+$ and $\triangleleft_\times$ are reflexive and transitive.

**Proof:** Reflexivity is clear from the categorical definitions. Transitivity of  $\triangleleft_\times$ is clear from the transitivity of $\triangleleft$ and the sub-environment definition. Transitivity of $\triangleleft_+$ can be seen using the categorical definition, by composing the morphisms and using the fact that being homotopic is preserved by composition. $\square$

3\. Decomposition Theorems
--------------------------

We have two decomposition theorems involving additive and multiplicative subagents.

**3.1. First Decomposition Theorem**

**Theorem:** $C_0\triangleleft C_1$ if and only if there exists a $C_2$ such that $C_0\triangleleft_\times C_2\triangleleft_+ C_1$.

**Proof:** We will use the currying definitions of subagent and multiplicative subagent, and the committing definition of additive subagent. Let $C_0=(A_0,E_0,\cdot_0)$ and $C_1=(A_1,E_1,\cdot_1)$. If $C_0\triangleleft C_1$, there exists some Cartesian frame $D$ over $A_1$ such that $C_0=C_1^\circ(D)$.

Let $C_2=(\mbox{Image}(D),E_1,\cdot_2)$, where $\cdot_2$ is given by $a\cdot_2 e=a\cdot_1 e$. $C_2$ is created by deleting some possible agents from $C_1$, so by the committing definition of additive subagent $C_2\triangleleft_+ C_1$.

Also, if we let $D^\prime$ be the Cartesian frame over $\mbox{Image}(D)$ which is identical to $D$, but on a restricted codomain, then we clearly have that $C_1^\circ(D)\cong C_2^\circ(D^\prime)$. Thus $C_0\simeq C_2^\circ(D^\prime)$ and $\mbox{Image}(D^\prime)=\mbox{Agent}(C_2)$, so $C_0\triangleleft_\times C_2$.

The converse is trivial, since subagent is weaker than additive and multiplicative subagent and is transitive. $\square$

Imagine that that a group of kids, Alice, Bob, Carol, etc., is deciding whether to start a game of baseball or football against another group. If they choose baseball, they form a team represented by the frame $C_B$, while if they choose football, they form a team represented by the frame $C_F$. We can model this by imagining $$that $C_0$ is the group's initial state, and $C_B$ and $C_F$ are precommitment-style subagents of $C_0$.

Suppose the group chooses football. $C_F$'s choices are a function of Alice-the-football-player's choices, Bob-the-football-player's choices, etc. (Importantly, Alice here has different options and a different environment than if the original group had chosen baseball. So we will need to represent Alice-the-football-player, $C_{AF}$, with a different frame than Alice-the-baseball-player, $C_{AB}$; and likewise for Bob and the other team members.)

It is easy to see in this case that the relationship between Alice-the-football-player's frame ($C_{AF}$) and the entire group's initial frame ($C_0$) can be decomposed into the additive relationship between $C_0$ and $C_F$ and the multiplicative relationship between $C_F$ and $C_{AF}$, in that order.

The first decomposition theorem tells us that *every* subagent relation, even ones that don't seem to involve a combination of "making a commitment" and "being a team," can be decomposed into a combination of those two things. I've provided an example above where this factorization feels natural, but other cases may be less natural.

Using the framing from our discussion of the currying definitions: this decomposition is always possible because we can always decompose a possibly-nondeterministic function $f$ into (1) a possibly-nondeterministic surjective function onto $f$'s image, and (2) a deterministic function embedding $f$'s image in $f$'s codomain.

**3.2. Second Decomposition Theorem**

**Theorem:** There exists a morphism from $C_0$ to $C_1$ if and only if there exists a $C_2$ such that $C_0\triangleleft_+^* C_2\triangleleft_+ C_1$. 

**Proof:** First, let $C_0=(A,E,\cdot)$, let $C_1=(B,F,\star)$, and let $(g,h):C_0\rightarrow C_1$. We let $C_2=(A,F,\diamond)$, where $a\diamond f=g(a)\star f=a\cdot h(f)$. 

First, we show $C_2\triangleleft_+C_1$, To do this, we let $B^\prime\subseteq B$ be the image of $g$, and let $C_2^\prime=(B^\prime,F,\star^\prime)$, where $\star^\prime$ is given by $b\star^\prime f = b\star f$. By the committing definition of additive subagent, it suffices to show that $C_2^\prime \simeq C_2$. 

We define $(g_0,h_0): C_2\rightarrow C_2^\prime$ and $(g_1,h_1): C_2^\prime\rightarrow C_2$ as follows. We let $h_0$ and $h_1$ be the identity on $F$. We let $g_0:A\rightarrow B^\prime$ be given by $g_0(a)=g(a)$. Observe that $g_0$ is surjective, and thus has a right inverse. Let $g_1$ be any right inverse to $g_0$, so $g_0(g_1(b))=b$ for all $b\in B^\prime$. 

We know $(g_0,h_0)$ is a morphism, since for all $a\in A$ and $f\in F$, we have

$$
\begin{equation} \begin{split} g_0(a)\star^\prime f &=g(a)\star f \ &=a \diamond f \ &=a \diamond h_0(f). \end{split} \end{equation}
$$

Similarly, we know $(g_1,h_1)$ is a morphism, since for all $b\in B^\prime$ and $f\in F$, we have

$$
\begin{equation} \begin{split} g_1(b)\diamond f & =g_1(b)\diamond h_0(f) \ & =g_0(g_1(b))\star^\prime f \ & =b\star^\prime f \ & =b\star^\prime h_1(f). \end{split} \end{equation}
$$

Clearly, $(g_0,h_0)\circ(g_1,h_1)$ and  $(g_1,h_1)\circ (g_0,h_0)$ are homotopic to the identity, since $h_0\circ h_1$ and $h_1\circ h_0$ are the identity on $F$. Thus, $C_2^\prime \simeq C_2$. 

The fact that $C_0\triangleleft_+^* C_2$, or equivalently $C_2^*\triangleleft_+ C_0^*$, is symmetric, since the relationship between $C_2^*$ and $C_0^*$ is the same as the relationship between $C_2$ and $C_1$.

Conversely, if $C_2\triangleleft_+ C_1$, there is a morphism from $C_2$ to $C_1$ by the categorical definition of additive subagent. Similarly, if $C_0\triangleleft_+^* C_2$, then $C_2^*\triangleleft_+ C_0^*$, so there is a morphism from  $C_2^*$ to $C_0^*$, and thus a morphism from $C_0$ to $C_2$. These compose to a morphism from $C_0$ to $C_1.$ $\square$

When we introduced morphisms and [described them as "interfaces,"](https://www.lesswrong.com/posts/ewkYgtZapQRtDPT2F/additive-operations-on-cartesian-frames#1_1__Morphisms_as_Interfaces) we noted that every morphism $(g,h):C_0→C_1$ implies the existence of an intermediate frame $C_2$ that represents $\mbox{Agent}(C_0)$ interacting with $\mbox{Env}(C_1)$. The second decomposition theorem formalizes this claim, and also notes that this intermediate frame is a super-environment of $C_0$ and a subagent of $C_1$.

In our next post, we will provide several methods for constructing additive and multiplicative subagents: "Committing, Assuming, Externalizing, and Internalizing."