This is the ninth post in the 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.

1. Definitions of Additive and Multiplicative Subagent

1.1. Sub-Sum and Sub-Tensor Definitions

Definition: $C$ is an additive subagent of $D$, written $C{◃}_{+}D$, if there exists a $C^{\prime }$ and a $D^{\prime }\simeq D$ with $D^{\prime }\in C⊞C^{\prime }$.

Definition: $C$ is a multiplicative subagent of $D$, written $C{◃}_{\times }D$, if there exists a $C^{\prime }$ and $D^{\prime }\simeq D$ with $D^{\prime }\in C⊠C^{\prime }$.

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 $D^{\prime }\in C⊞C^{\prime }$. Similarly, $C^{\prime }$ is called a sister to $C$ in $D$ if $D\simeq D^{\prime }$ for some $D^{\prime }\in C⊠C^{\prime }$.

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 $C{◃}_{+}D$ if $C$ has a brother in $D$ and $C{◃}_{\times }D$ 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 $C{◃}_{+}D$ if there exist three sets $X$, $Y$, and $Z$, with $X\subseteq Y$, and a function $f:Y\times Z\to W$ 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 ${◃}_{+}$.

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 $D^{\prime }\in C⊞C^{\prime }$. Then, if we let $X=A$, let $Y=B^{\prime }=A\bigsqcup A^{\prime }$, 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 $f:Y\times Z\to W$.  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 $C_1=(Y\setminus X,Z,{\diamond }_1)$ 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

if $y\in X$, and is given by

otherwise. Consider the diagonal subset $S\subseteq Z\times Z$ given by $S=\left\{\right(z,z)|z\in Z\}$. 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 ${\bullet }^{\prime \prime }:Y\times S\to W$ given by $y{\bullet }^{\prime \prime }(z,z)=y\bullet z$. Thus $(Y,S,{\bullet }^{\prime \prime })\cong (Y,Z,\bullet )\text{,}$ 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 $\left(Y\setminus X\right)\times S$, we get ${\bullet }_0$ and ${\bullet }_1$ respectively, given by $x{\bullet }_0=x{\diamond }_{0}z$ and $x{\bullet }_1(z,z)=x{\diamond }_{1}z$. Thus $(X,S,{\bullet }_0)\cong (X,Z,{\diamond }_0)$ and $(Y\setminus X,S,{\bullet }_1)\cong (Y\setminus X,Z,{\diamond }_1)$, with the isomorphisms coming from the identities on $Y$ and $X\setminus Y$, and the bijection between $S$ and $Z$.

Thus $(Y,S,{\bullet }^{\prime \prime })\in C_0⊞C_1$, 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 $C{◃}_{\times }D$ if there exist three sets $X$, $Y$, and $Z$, and a function $f:X\times Y\times Z\to W$ 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 ${◃}_{\times }$.

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 $D^{\prime }\in C⊠C^{\prime }$. Then, if we let $X=A$, let $Y=A^{\prime }$, let $Z=F^{\prime }\subseteq \text{hom}(C,{C^{\prime }}^{\ast })$, and let

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 $f:X\times Y\times Z\to W$. 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 $C_0\otimes C_1=(X\times Y,\text{hom}(C_0,C_1^{\ast }),{\bullet }^{\prime })$, where ${\bullet }^{\prime }$ is given by

For every $z\in Z$, there is a morphism $(g_z,h_z):C_0\to C_1^{\ast }$, where $g_z:X\to X\times Z$ is given by $g_z\left(x\right)=(x,z)$, and $g_z:Y\to Y\times Z$ is given by $g_z\left(x\right)=(x,z)$. This is clearly a morphism. Consider the subset $S\subseteq \text{hom}(C_0,C_1^{\ast })$ given by $S=\left\{\right(g_z,h_z)|z\in Z\}$. 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 ${\bullet }^{\prime \prime }:(X\times Y)\times S\to W$ 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 $(X\times Y,S,{\bullet }^{\prime \prime })\in C_0⊠C_1$, 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

Indeed,  $x{\bullet }_0(y,(g_h,z_h\left)\right)=x{\diamond }_0(y,z)$ and $y{\bullet }_1(x,(g_h,z_h\left)\right)=y{\diamond }_1(x,z)$, 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 $(X\times Y,S,{\bullet }^{\prime \prime })\in C_0⊞C_1$, 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, $C\cong \text{null}$, and either $D\simeq \text{null}$ or $D\simeq 0$, depending on whether $Z$ is empty. It is easy to verify that $\text{null}⊠\text{null}=\{0,\text{null}\}$, since $\text{null}\otimes \text{null}\cong 0$, taking the two subsets of the singleton environment in $0$ yields $0$ and $\text{null}$ as candidate sub-tensors, and both are valid sub-tensors, since either way, the conditions reduce to $\text{null}\simeq \text{null}$. $\square$

Next, we have some definitions that more directly relate to our original definitions of subagent.

1.4. Currying Definitions

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

Claim: This definition is equivalent to all of the above definitions of ${◃}_{+}$.

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 $p:Y\times Z\to W$ 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 $(g_0,h_0):D\to (Y,Z,\bullet )$ and $(g_1,h_1):(Y,Z,\bullet )\to D$ 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  $x\cdot e=g_1\left(x\right).$ Observe that $D^{\circ }\left(M\right)=(X,\{e\}\times F,{\star }^{\prime })$, where ${\star }^{\prime }$ is given by 

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

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

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

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

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

Let $X\subseteq Y$ be given by $X=\text{Image}\left(M\right)$. Since $|\text{Env}\left(M\right)|=1$, we have $M\simeq {\perp }_X$. Thus, $C\simeq D^{\circ }\left(M\right)\simeq D^{\circ }\left({\perp }_X\right)$. Unpacking the definition of $D^{\circ }\left({\perp }_X\right)$, we get $D^{\circ }\left({\perp }_X\right)=(X,\{e\}\times Z,\cdot )$, 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 $C{◃}_{\times }D$ if there exists a Cartesian frame $M$ over $\text{Agent}\left(D\right)$ with $\text{Image}\left(M\right)=\text{Agent}\left(D\right)$, such that $C\simeq D^{\circ }\left(M\right)$.

Claim: This definition is equivalent to all of the above definitions of ${◃}_{\times }$.

Proof: We show equivalence to the externalizing definition. 

First, assume there exist three sets $X$, $Y$, and $Z$, and a function $p:X\times Y\times Z\to W$ 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 $(g_0,h_0):D\to (X\times Y,Z,\bullet )$ and $(g_1,h_1):(X\times Y,Z,\bullet )\to D$ compose to something homotopic to the identity in both orders.

We define $B^{\prime }=B\bigsqcup \left\{a\right\}$, 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 $g_0\left(b\right)=(x,y)$, and $x\cdot (y,b)=g_1(x,y)$ otherwise. Clearly, $\text{Image}\left(M\right)=B$, since for any $b\in B$, if we let $(x,y)=g_0\left(b\right)$, 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 $g_0\left(b\right)=(x,y)$, then

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

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

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

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

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

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

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

Thus $C\simeq D^{\circ }\left(M\right)\cong (X,Y\times Z,\diamond )$, 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 $(g_0,h_0):D\to (X\times Y,Z,\bullet )$ and $(g_1,h_1):D\to (X\times Y,Z,\bullet )$ that compose to something homotopic to the identity in both orders. Let $h_0$ and $h_1$ be the identity on $Z$. Let $g_1:X\times Y\to B$ be given by $g_1(x,y)=x\cdot y$. Since $g_1$ is surjective, it has a right inverse. Let $g_0:B\to X\times Y$ be any choice of right inverse of $g_1$, so $g_1\left(g_0\right(b\left)\right)=b$ for all $b\in B$. 

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

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

$(g_0,h_0)$ 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 $\text{Agent}\left(C\right)$ and whose possible worlds are $\text{Agent}\left(D\right)$. When $C$ is a subagent of $D$, (up to biextensional equivalence) there exists a function from $\text{Agent}\left(C\right)$, paired with $\text{Env}\left(M\right)$, to $\text{Agent}\left(D\right)$.

Just as we did in "Subagents of Cartesian Frames" §1.2 (Currying Definition), we can think of this function as a (possibly) nondeterministic function from $\text{Agent}\left(C\right)$ to $\text{Agent}\left(D\right)$, where $\text{Env}\left(M\right)$ represents the nondeterminism. In the case of additive subagents, $\text{Env}\left(M\right)$ is a singleton, meaning that the function from $\text{Agent}\left(C\right)$ to  $\text{Agent}\left(D\right)$ 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), 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 ${◃}_{+}$ and ${◃}_{\times }$ 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. However, we will also have to weaken the definition slightly in order to only require the morphisms to be homotopic.

Definition: We say $C{◃}_{+}D$ if there exists a single morphism ${\varphi }_0:C\to D$ such that for every morphism $\varphi :C\to \perp$ there exists a morphism ${\varphi }_1:D\to \perp$ such that $\varphi$ is homotopic to ${\varphi }_1\circ {\varphi }_0$ .

Claim: This definition is equivalent to all the above definitions of ${◃}_{+}$.

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 $(g_0,h_0):C\to D$ be such that for all $(g,h):C\to \perp$, there exists a $(g^{\prime },h^{\prime }):D\to \perp$ such that $(g,h)$ is homotopic to $(g^{\prime },h^{\prime })\circ (g_0,h_0)$. Let $\perp =(W,\{i\},\star )$.

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

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

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

Defining $h_2:E\to Z$ will be a bit more complicated.  Given an $e\in E$, let $(g_e,h_e)$ be the morphism from $C$ to $\perp$, given by $h_e\left(i\right)=e$ and $g_e\left(a\right)=a\cdot e$. Let $(g_e^{\prime },h_e^{\prime }):D\to \perp$ 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 $h_2\left(e\right)=h_e^{\prime }\left(i\right)$.

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

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,

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,