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.

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′ and a D′≃D with D′∈C⊞C′.

Definition:C is a multiplicative subagent of D, written C◃×D, if there exists a C′ and D′≃D with D′∈C⊠C′.

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′ is called a brother to C in D if D≃D′ for some D′∈C⊞C′. Similarly, C′ is called a sister to C in D if D≃D′ for some D′∈C⊠C′.

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◃×D if C has a sister in D.

Claim: This definition is equivalent to the ones above.

Proof: Trivial. □

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⊆Y, and a function f:Y×Z→W such that C≃(X,Z,⋄) and D≃(Y,Z,∙), where ⋄ and ∙ are given by x⋄z=f(x,z) and y∙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,⋅), and let D=(B,F,⋆). Let C′=(A′,E′,⋅′) be brother to C in D. Let D′=(B′,F′,⋆′) be such that D′≃D and D′∈C⊞C′. Then, if we let X=A, let Y=B′=A⊔A′, let Z=F′, and let f(y,z)=y⋆′z, we get D≃D′=(Y,Z,∙), where y∙z=f(y,z), and by the definition of sub-sum, C≃(X,Z,⋄), where x⋄z=f(x,z).

Conversely, let X, Y, and Z be arbitrary sets with X⊆Y, and let f:Y×Z→W. Let C≃C0=(X,Z,⋄0), and let D≃D′=(Y,Z,∙), where x⋄0z=f(x,z) and y∙z=f(y,z). We want to show that C has a brother in D. It suffices to show that C0 has a brother in D, since sub-sum is well-defined up to biextensional equivalence. Indeed, we will show that C1=(Y∖X,Z,⋄1) is brother to C0 in D, where ⋄1 is given by x⋄1z=f(x,z).

Observe that C0⊕C1=(Y,Z×Z,∙′), where ∙′ is given by

y∙′(z0,z1)=y⋄0z0=y∙z0

if y∈X, and is given by

y∙′(z0,z1)=y⋄1z1=y∙z1

otherwise. Consider the diagonal subset S⊆Z×Z given by S={(z,z)|z∈Z}. Observe that the map z↦(gz,hz) is a bijection from Z to S. Observe that if we restrict ∙′ to Y×S, we get ∙′′:Y×S→W given by y∙′′(z,z)=y∙z. Thus (Y,S,∙′′)≅(Y,Z,∙), with the isomorphism coming from the identity on Y, and the bijection between S and Z.

If we further restrict ∙′′ to X×S or (Y∖X)×S, we get ∙0 and ∙1 respectively, given by x∙0=x⋄0z and x∙1(z,z)=x⋄1z. Thus (X,S,∙0)≅(X,Z,⋄0) and (Y∖X,S,∙1)≅(Y∖X,Z,⋄1), with the isomorphisms coming from the identities on Y and X∖Y, and the bijection between S and Z.

Thus (Y,S,∙′′)∈C0⊞C1, and (Y,S,∙′′)≅D′≃D, so C1 is brother to C0 in D, so C has a brother in D. □

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◃×D if there exist three sets X, Y, and Z, and a function f:X×Y×Z→W such that C≃(X,Y×Z,⋄) and D≃(X×Y,Z,∙), where ⋄ and ∙ are given by x⋄(y,z)=f(x,y,z) and (x,y)∙z=f(x,y,z).

Claim: This definition is equivalent to the sub-tensor and sister definitions of ◃×.

Proof: First, assume that C has a sister in D. Let C=(A,E,⋅), and let D=(B,F,⋆). Let C′=(A′,E′,⋅′) be sister to C in D. Let D′=(B′,F′,⋆′) be such that D′≃D and D′∈C⊠C′. Then, if we let X=A, let Y=A′, let Z=F′⊆hom(C,C′∗), and let

f(x,y,(g,h))=x⋅h(y)=y⋅′g(x),

we get D≃D′=(X×Y,Z,∙), where (x,y)∙z=f(x,y,z), and by the definition of sub-tensor, C≃(X,Y×Z,⋄), where x⋄(y,z)=f(x,y,z).

Conversely, let X, Y, and Z be arbitrary sets, and let f:X×Y×Z→W. Let C≃C0=(X,Y×Z,⋄0), and let D≃D′=(X×Y,Z,∙), where x⋄0(y,z)=(x,y)∙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 C0 has a sister in D, since sub-tensor is well-defined up to biextensional equivalence. Indeed, we will show that C1=(Y,X×Z,⋄1) is sister to C0 in D, where ⋄1 is given by y⋄1(x,z)=f(x,y,z).

Observe that C0⊗C1=(X×Y,hom(C0,C∗1),∙′), where ∙′ is given by

(x,y)∙′(g,h)=x⋄0h(y)=y⋆1g(x).

For every z∈Z, there is a morphism (gz,hz):C0→C∗1, where gz:X→X×Z is given by gz(x)=(x,z), and gz:Y→Y×Z is given by gz(x)=(x,z). This is clearly a morphism. Consider the subset S⊆hom(C0,C∗1) given by S={(gz,hz)|z∈Z}. Observe that the map z↦(gz,hz) 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 ∙′ to (X×Y)×S, we get ∙′′:(X×Y)×S→W given by y∙′′(gz,hz)=y∙z. Thus, (X×Y,S,∙′′)≅(X×Y,Z,∙), with the isomorphism coming from the identity on X×Y, and the bijection between S and Z.

To show that (X×Y,S,∙′′)∈C0⊠C1, we need to show that C0≃(X,Y×S,∙0) and C1≃(Y,X×S,∙1), where ∙0 and ∙1 are given by

x∙0(y,(gh,zh))=y∙1(x,(gh,zh))=(x,y)∙′′(gz,hz).

Indeed, x∙0(y,(gh,zh))=x⋄0(y,z) and y∙1(x,(gh,zh))=y⋄1(x,z), so (X,Y×S,∙0)≅(X,Y×Z,⋄0)=C0 and (Y,X×S,∙1)≅(Y,X×Z,⋄1)=C1, with the isomorphisms coming from the identities on X and Y, and the bijection between S and Z.

Thus (X×Y,S,∙′′)∈C0⊞C1, and (Y,S,∙′′)≅D′≃D, so C1 is sister to C0 in D, so C has a sister in D.

Finally, in the case where X and Y are both empty, C≅null, and either D≃null or D≃0, depending on whether Z is empty. It is easy to verify that null⊠null={0,null}, since null⊗null≅0, taking the two subsets of the singleton environment in 0 yields 0 and null as candidate sub-tensors, and both are valid sub-tensors, since either way, the conditions reduce to null≃null. □

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 Agent(D) with |Env(M)|=1, such that C≃D∘(M).

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⊆Y, and a function p:Y×Z→W such that C≃(X,Z,⋄) and D≃(Y,Z,∙), where ⋄ and ∙ are given by x⋄z=p(x,z) and y∙z=p(y,z).

Let D=(B,F,⋆), and let (g0,h0):D→(Y,Z,∙) and (g1,h1):(Y,Z,∙)→D compose to something homotopic to the identity in both orders.

We define M, a Cartesian frame over B, by M=(X,{e},⋅), where ⋅ is given x⋅e=g1(x). Observe that D∘(M)=(X,{e}×F,⋆′), where ⋆′ is given by

x⋆′(e,f)=(x⋅e)⋆f=g1(x)⋆f=x∙h1(f).

To show that (X,Z,⋄)≃D∘(M), we construct morphisms (g2,h2):(X,Z,⋄)→D∘(M) and (g3,h3):D∘(M)→(X,Z,⋄) that compose to something homotopic to the identity in both orders. Let g2 and g3 both be the identity on X. Let h2:{e}×F→Z be given by h2(e,f)=h1(f), and let h3:Z→{e}×F be given by h3(z)=(e,h0(z)).

We know (g2,h2) is a morphism, since for all x∈X and (e,f)∈{e}×F, we have

Observe that (g2,h2) and (g3,h3) clearly compose to something homotopic to the identity in both orders, since g2∘g3 and g3∘g2 are the identity on X.

Thus, C≃(X,Z,⋄)≃D∘(M), and |Env(M)|=1.

Conversely, assume C≃D∘(M), with |Env(M)|=1. We define Y=Agent(D) and Z=Env(D). We define f:Y×Z→W by f(y,z)=y∙z, where ∙=Eval(D).

Let X⊆Y be given by X=Image(M). Since |Env(M)|=1, we have M≃⊥X. Thus, C≃D∘(M)≃D∘(⊥X). Unpacking the definition of D∘(⊥X), we get D∘(⊥X)=(X,{e}×Z,⋅), where ⋅ is given by x⋅(e,z)=f(x,z), which is isomorphic to (X,Z,⋄), where ⋄ is given by x⋄z=f(x,z). Thus C≃(X,Z,⋄) and D=(Y,Z,∙), as in the committing definition. □

Definition: We say C◃×D if there exists a Cartesian frame M over Agent(D) with Image(M)=Agent(D), such that C≃D∘(M).

Claim: This definition is equivalent to all of the above definitions of ◃×.

Proof: We show equivalence to the externalizing definition.

First, assume there exist three sets X, Y, and Z, and a function p:X×Y×Z→W such that C≃(X,Y×Z,⋄) and D≃(X×Y,Z,∙), where ⋄ and ∙ are given by x⋄(y,z)=(x,y)∙z=p(x,y,z).

Let D=(B,F,⋆), and let (g0,h0):D→(X×Y,Z,∙) and (g1,h1):(X×Y,Z,∙)→D compose to something homotopic to the identity in both orders.

We define B′=B⊔{a}, and we define M, a Cartesian frame over B, by M=(X,Y×B′,⋅), where ⋅ is given by x⋅(y,b)=b if b∈B and g0(b)=(x,y), and x⋅(y,b)=g1(x,y) otherwise. Clearly, Image(M)=B, since for any b∈B, if we let (x,y)=g0(b), we have x⋅(y,b)=b.

Observe that for all x∈X, y∈Y, b∈B′ and f∈F, if b∈B and g0(b)=(x,y), then

(x⋅(y,b))⋆f=b⋆f=g1(g0(b))⋆f=g1(x,y)⋆f,

and on the other hand, if b=a or g0(b)≠(x,y), we also have (x⋅(y,b))⋆f=g1(x,y)⋆f.

Thus, we have that D∘(M)=(X,Y×B′×F,⋆′), where ⋆′ is given by

x⋆′(y,b,f)=(x⋅(y,b))⋆f=g1(x,y)⋆f=(x,y)∙h1(f).

To show that (X,Y×Z,⋄)≃D∘(M), we construct morphisms (g2,h2):(X,Y×Z,⋄)→D∘(M) and (g3,h3):D∘(M)→(X,Y×Z,⋄) that compose to something homotopic to the identity in both orders. Let g2 and g3 both be the identity on X. Let h2:Y×B′×F→Y×Z be given by h2(y,b,f)=(y,h1(f)), and let h3:Y×Z→Y×B′×F be given by h3(y,z)=(y,a,h0(z)).

We know (g2,h2) is a morphism, since for all x∈X and (y,b,f)∈Y×B′×F,

Observe that (g2,h2) and (g3,h3) clearly compose to something homotopic to the identity in both orders, since g2∘g3 and g3∘g2 are the identity on X.

Thus, C≃(X,Z,⋄)≃D∘(M), where Image(M)=Agent(D).

Conversely, assume C≃D∘(M), with Image(M)=Agent(D). Let X=Agent(M), let Y=Env(M), and let Z=Env(D). Let f:X×Y×Z→W be given by f(x,y,z)=(x⋅y)⋆z, where ⋅=Eval(M) and ⋆=Eval(D).

Thus C≃D∘(M)≅(X,Y×Z,⋄), where ⋄ is given by x⋄(y,z)=(x⋅y)⋆z=f(x,y,z). All that remains to show is that D≃(X×Y,Z,∙), where (x,y)∙z=f(x,y,z). Let D=(B,Z,⋆).

We construct morphisms (g0,h0):D→(X×Y,Z,∙) and (g1,h1):D→(X×Y,Z,∙) that compose to something homotopic to the identity in both orders. Let h0 and h1 be the identity on Z. Let g1:X×Y→B be given by g1(x,y)=x⋅y. Since g1 is surjective, it has a right inverse. Let g0:B→X×Y be any choice of right inverse of g1, so g1(g0(b))=b for all b∈B.

We know (g1,h1) is a morphism, since for all (x,y)∈X×Y and z∈Z,

g1(x,y)⋆z=(x⋅y)⋆z=f(x,y,z)=(x,y)∙z=(x,y)∙h1(z).

To see that (g0,h0) is a morphism, given b∈B and z∈Z, let (x,y)=g0(b), and observe

(g0,h0) and (g1,h1) clearly compose to something homotopic to the identity in both orders, since h0∘h1 and h1∘h0 are the identity on Z. Thus D≃(X×Y,Z,∙), completing the proof. □

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

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 Agent(C) to Agent(D), where Env(M) represents the nondeterminism. In the case of additive subagents, Env(M) is a singleton, meaning that the function from Agent(C) to 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), 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 ◃× 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 ϕ0:C→D such that for every morphism ϕ:C→⊥ there exists a morphism ϕ1:D→⊥ such that ϕ is homotopic to ϕ1∘ϕ0 .

Claim: This definition is equivalent to all the above definitions of ◃+.

Proof: We show equivalence to the committing definition.

First, let C=(A,E,⋅) and D=(B,F,∙) be Cartesian frames over W, and let (g0,h0):C→D be such that for all (g,h):C→⊥, there exists a (g′,h′):D→⊥ such that (g,h) is homotopic to (g′,h′)∘(g0,h0). Let ⊥=(W,{i},⋆).

Let Y=B, let Z=F, and let X={g0(a)|a∈A}. Let f:Y×Z→W be given by f(y,z)=y∙z. We already have D=(Y,Z,∙), and our goal is to show that C≃(X,Z,⋄), where ⋄ is given by x⋄z=f(x,z).

We construct (g1,h1):C→(X,Z,⋄) and (g2,h2):(X,Z,⋄)→C that compose to something homotopic to the identity in both orders.

We define g1:A→X by g1(a)=g0(a). g1 is surjective, and so has a right inverse. We let g2:X→A be any right inverse to g1, so g1(g2(x))=x for all x∈X. We let h1:Z→E be given by h1(z)=h0(z).

Defining h2:E→Z will be a bit more complicated. Given an e∈E, let (ge,he) be the morphism from C to ⊥, given by he(i)=e and ge(a)=a⋅e. Let (g′e,h′e):D→⊥ be such that (ge,he) is homotopic to (g′e,h′e)∘(g0,h0). We define h2 by h2(e)=h′e(i).

We trivially have that (g1,h1) is a morphism, since for all a∈A and z∈Z,

g1(a)⋄z=g0(a)∙z=a⋅h0(z)=a⋅h1(z).

To see that (g2,h2) is a morphism, consider x∈X and e∈E, and define (ge,he) and (g′e,h′e) as above. Then,

We trivially have that (g1,h1)∘(g2,h2) is homotopic to the identity, since g1∘g2 is the identity on X. To see that (g2,h2)∘(g1,h1) is homotopic to the identity on C, observe that for all a∈A and e∈E, defining (ge,he) and (g′e,h′e) as above,

Thus C≃(X,Z,⋄), and C◃+D according to the committing definition.

Conversely, let X, Y, and Z be arbitrary sets with X⊆Y, let f:Y×Z→W, and let C≃(X,Z,⋄) and D≃(Y,Z,∙), where ⋄ and ∙ are given by x⋄z=f(x,z) and y∙z=f(y,z).

Let (g1,h1):C→(X,Z,⋄) and (g2,h2):(X,Z,⋄)→C compose to something homotopic to the identity in both orders, and let (g3,h3):D→(Y,Z,∙) and (g4,h4):(Y,Z,∙)→D compose to something homotopic to the identity in both orders. Let (g0,h0):(X,Z,⋄)→(Y,Z,∙) be given by g0 is the embedding of X in Y and h0 is the identity on Z. (g0,h0) is clearly a morphism.

We let ϕ:C→D=(g4,h4)∘(g0,h0)∘(g1,h1).

Given a (g,h):C→⊥, our goal is to construct a (g′,h′):D→⊥ such that (g,h) is homotopic to (g′,h′)∘ϕ.

Let ⊥=(W,{i},⋆), let C=(A,E,⋅0), and let D=(B,F,⋅1). Let h′:{i}→F be given by h′=h3∘h2∘h. Let g′:B→W be given by g′(b)=b⋅1h′(i). This is clearly a morphism, since for all b∈B and i∈{i},

g′(b)⋆i=g′(b)=b⋅h′(i).

To see that (g,h) is homotopic to (g′,h′)∘(g4,h4)∘(g0,h0)∘(g1,h1), we just need to check that (g,h1∘h0∘h4∘h′):C→⊥ is a morphism. Or, equivalently, that (g,h1∘h4∘h3∘h2∘h):C→⊥, since h0 is the identity, and h′=h3∘h2∘h.

Thus (g,h) is homotopic to (g′,h′)∘ϕ, completing the proof. □

Definition: We say C◃×D if for every morphism ϕ:C→⊥, there exist morphisms ϕ0:C→D and ϕ1:D→⊥ such that ϕ=ϕ1∘ϕ0, and for every morphism ψ:1→D, there exist morphisms ψ0:1→C and ψ1:C→D such that ψ=ψ1∘ψ0.

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◃∗D, if D∗◃C∗.

We can similarly define additive and multiplicative sub-environments.

Definition: We say C is an additive sub-environment of D, written C◃∗+D, if D∗◃+C∗. We say C is an multiplicative sub-environment of D, written C◃∗×D, if D∗◃×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◃×D if and only if C◃∗×D.

Proof: We prove this using the externalizing definition of ◃×.

If C◃×D, then for some X, Y, Z, and f:X×Y×Z→W, we have C≃(X,Y×Z,⋄) and D≃(X×Y,Z,∙), where ⋄ and ∙ are given by x⋄(y,z)=f(x,y,z) and (x,y)∙z=f(x,y,z).

Observe that D∗≃(Z,Y×X,⋅) and C∗≃(Z×Y,X,⋆), where ⋅ and ⋆ are given by z⋅(y,x)=f(x,y,z) and (z,y)⋆x=f(x,y,z). Taking X′=Z, Y′=Y, Z′=X, and f′(x,y,z)=f(z,y,x), this is exactly the externalizing definition of D∗◃×C∗, so C◃∗×D.

Conversely, if C◃∗×D, then D∗◃×C∗, so C≅{C∗}∗◃×{D∗}∗≅D. □

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

Definition: We say C◃×D if C◃D and C◃∗D. Equivalently, we say C◃×D if C◃D and D∗◃C∗.

Claim: This definition is equivalent to the categorical definition of ◃×.

Proof: The condition that for every morphism ϕ:C→⊥, there exist morphisms ϕ0:C→D and ϕ1:D→⊥ such that ϕ=ϕ1∘ϕ0, is exactly the categorical definition of C◃D.

The condition that for every morphism ψ:1→D, there exist morphisms ψ0:1→C and ψ1:C→D such that ψ=ψ1∘ψ0, is equivalent to saying that for every morphism ψ∗:D∗→⊥, there exist morphisms ψ∗0:C∗→⊥ and ψ∗1:D∗→C∗ such that ψ∗=ψ∗1∘ψ∗0. This is the categorical definition of D∗◃C∗. □

Claim: The categorical and sub-environment definitions of ◃× 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,⋅) and D=(B,F,⋆) are Cartesian frames over W with C◃D and C◃∗D.

We define X=A, Z=F, and Y=hom(C,D). We define p:X×Y

## Additive and Multiplicative Subagents

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 DefinitionsDefinition:C is an additive subagent of D, written C◃+D, if there exists a C′ and a D′≃D with D′∈C⊞C′.Definition:C is a multiplicative subagent of D, written C◃×D, if there exists a C′ and D′≃D with D′∈C⊠C′.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 DefinitionsDefinition:C′ is called a brother to C in D if D≃D′ for some D′∈C⊞C′. Similarly, C′ is called a sister to C in D if D≃D′ for some D′∈C⊠C′.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

notcarry 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◃×D if C has a sister in D.Claim:This definition is equivalent to the ones above.Proof:Trivial. □1.3. Committing and Externalizing DefinitionsNext, 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⊆Y, and a function f:Y×Z→W such that C≃(X,Z,⋄) and D≃(Y,Z,∙), where ⋄ and ∙ are given by x⋄z=f(x,z) and y∙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,⋅), and let D=(B,F,⋆). Let C′=(A′,E′,⋅′) be brother to C in D. Let D′=(B′,F′,⋆′) be such that D′≃D and D′∈C⊞C′. Then, if we let X=A, let Y=B′=A⊔A′, let Z=F′, and let f(y,z)=y⋆′z, we get D≃D′=(Y,Z,∙), where y∙z=f(y,z), and by the definition of sub-sum, C≃(X,Z,⋄), where x⋄z=f(x,z).Conversely, let X, Y, and Z be arbitrary sets with X⊆Y, and let f:Y×Z→W. Let C≃C0=(X,Z,⋄0), and let D≃D′=(Y,Z,∙), where x⋄0z=f(x,z) and y∙z=f(y,z). We want to show that C has a brother in D. It suffices to show that C0 has a brother in D, since sub-sum is well-defined up to biextensional equivalence. Indeed, we will show that C1=(Y∖X,Z,⋄1) is brother to C0 in D, where ⋄1 is given by x⋄1z=f(x,z).

Observe that C0⊕C1=(Y,Z×Z,∙′), where ∙′ is given by

y∙′(z0,z1)=y⋄0z0=y∙z0if y∈X, and is given by

y∙′(z0,z1)=y⋄1z1=y∙z1otherwise. Consider the diagonal subset S⊆Z×Z given by S={(z,z) | z∈Z}. Observe that the map z↦(gz,hz) is a bijection from Z to S. Observe that if we restrict ∙′ to Y×S, we get ∙′′:Y×S→W given by y∙′′(z,z)=y∙z. Thus (Y,S,∙′′)≅(Y,Z,∙), with the isomorphism coming from the identity on Y, and the bijection between S and Z.

If we further restrict ∙′′ to X×S or (Y∖X)×S, we get ∙0 and ∙1 respectively, given by x∙0=x⋄0z and x∙1(z,z)=x⋄1z. Thus (X,S,∙0)≅(X,Z,⋄0) and (Y∖X,S,∙1)≅(Y∖X,Z,⋄1), with the isomorphisms coming from the identities on Y and X∖Y, and the bijection between S and Z.

Thus (Y,S,∙′′)∈C0⊞C1, and (Y,S,∙′′)≅D′≃D, so C1 is brother to C0 in D, so C has a brother in D. □

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◃×D if there exist three sets X, Y, and Z, and a function f:X×Y×Z→W such that C≃(X,Y×Z,⋄) and D≃(X×Y,Z,∙), where ⋄ and ∙ are given by x⋄(y,z)=f(x,y,z) and (x,y)∙z=f(x,y,z).Claim:This definition is equivalent to the sub-tensor and sister definitions of ◃×.

f(x,y,(g,h))=x⋅h(y)=y⋅′g(x),Proof:First, assume that C has a sister in D. Let C=(A,E,⋅), and let D=(B,F,⋆). Let C′=(A′,E′,⋅′) be sister to C in D. Let D′=(B′,F′,⋆′) be such that D′≃D and D′∈C⊠C′. Then, if we let X=A, let Y=A′, let Z=F′⊆hom(C,C′∗), and letwe get D≃D′=(X×Y,Z,∙), where (x,y)∙z=f(x,y,z), and by the definition of sub-tensor, C≃(X,Y×Z,⋄), where x⋄(y,z)=f(x,y,z).

Conversely, let X, Y, and Z be arbitrary sets, and let f:X×Y×Z→W. Let C≃C0=(X,Y×Z,⋄0), and let D≃D′=(X×Y,Z,∙), where x⋄0(y,z)=(x,y)∙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 C0 has a sister in D, since sub-tensor is well-defined up to biextensional equivalence. Indeed, we will show that C1=(Y,X×Z,⋄1) is sister to C0 in D, where ⋄1 is given by y⋄1(x,z)=f(x,y,z).

Observe that C0⊗C1=(X×Y,hom(C0,C∗1),∙′), where ∙′ is given by

(x,y)∙′(g,h)=x⋄0h(y)=y⋆1g(x).For every z∈Z, there is a morphism (gz,hz):C0→C∗1, where gz:X→X×Z is given by gz(x)=(x,z), and gz:Y→Y×Z is given by gz(x)=(x,z). This is clearly a morphism. Consider the subset S⊆hom(C0,C∗1) given by S={(gz,hz) | z∈Z}. Observe that the map z↦(gz,hz) 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 ∙′ to (X×Y)×S, we get ∙′′:(X×Y)×S→W given by y∙′′(gz,hz)=y∙z. Thus, (X×Y,S,∙′′)≅(X×Y,Z,∙), with the isomorphism coming from the identity on X×Y, and the bijection between S and Z.

To show that (X×Y,S,∙′′)∈C0⊠C1, we need to show that C0≃(X,Y×S,∙0) and C1≃(Y,X×S,∙1), where ∙0 and ∙1 are given by

x∙0(y,(gh,zh))=y∙1(x,(gh,zh))=(x,y)∙′′(gz,hz).Indeed, x∙0(y,(gh,zh))=x⋄0(y,z) and y∙1(x,(gh,zh))=y⋄1(x,z), so (X,Y×S,∙0)≅(X,Y×Z,⋄0)=C0 and (Y,X×S,∙1)≅(Y,X×Z,⋄1)=C1, with the isomorphisms coming from the identities on X and Y, and the bijection between S and Z.

Thus (X×Y,S,∙′′)∈C0⊞C1, and (Y,S,∙′′)≅D′≃D, so C1 is sister to C0 in D, so C has a sister in D.

Finally, in the case where X and Y are both empty, C≅null, and either D≃null or D≃0, depending on whether Z is empty. It is easy to verify that null⊠null={0,null}, since null⊗null≅0, taking the two subsets of the singleton environment in 0 yields 0 and null as candidate sub-tensors, and both are valid sub-tensors, since either way, the conditions reduce to null≃null. □

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

1.4. Currying DefinitionsDefinition:We say C◃+D if there exists a Cartesian frame M over Agent(D) with |Env(M)|=1, such that C≃D∘(M).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⊆Y, and a function p:Y×Z→W such that C≃(X,Z,⋄) and D≃(Y,Z,∙), where ⋄ and ∙ are given by x⋄z=p(x,z) and y∙z=p(y,z).

Let D=(B,F,⋆), and let (g0,h0):D→(Y,Z,∙) and (g1,h1):(Y,Z,∙)→D compose to something homotopic to the identity in both orders.

We define M, a Cartesian frame over B, by M=(X,{e},⋅), where ⋅ is given x⋅e=g1(x). Observe that D∘(M)=(X,{e}×F,⋆′), where ⋆′ is given by

x⋆′(e,f)=(x⋅e)⋆f=g1(x)⋆f=x∙h1(f).To show that (X,Z,⋄)≃D∘(M), we construct morphisms (g2,h2):(X,Z,⋄)→D∘(M) and (g3,h3):D∘(M)→(X,Z,⋄) that compose to something homotopic to the identity in both orders. Let g2 and g3 both be the identity on X. Let h2:{e}×F→Z be given by h2(e,f)=h1(f), and let h3:Z→{e}×F be given by h3(z)=(e,h0(z)).

We know (g2,h2) is a morphism, since for all x∈X and (e,f)∈{e}×F, we have

g2(x)⋆′(e,f)=x⋆′(e,f)=x∙h1(f)=x⋄h1(f)=x⋄(h2(e,f)).We also have that (g3,h3) is a morphism, since for all x∈X and z∈Z, we have

g3(x)⋄z=x⋄z=x∙z=x∙h1(h0(z))=x⋆′(e,h0(z))=x⋆′h3(z).Observe that (g2,h2) and (g3,h3) clearly compose to something homotopic to the identity in both orders, since g2∘g3 and g3∘g2 are the identity on X.

Thus, C≃(X,Z,⋄)≃D∘(M), and |Env(M)|=1.

Conversely, assume C≃D∘(M), with |Env(M)|=1. We define Y=Agent(D) and Z=Env(D). We define f:Y×Z→W by f(y,z)=y∙z, where ∙=Eval(D).

Let X⊆Y be given by X=Image(M). Since |Env(M)|=1, we have M≃⊥X. Thus, C≃D∘(M)≃D∘(⊥X). Unpacking the definition of D∘(⊥X), we get D∘(⊥X)=(X,{e}×Z,⋅), where ⋅ is given by x⋅(e,z)=f(x,z), which is isomorphic to (X,Z,⋄), where ⋄ is given by x⋄z=f(x,z). Thus C≃(X,Z,⋄) and D=(Y,Z,∙), as in the committing definition. □

Definition:We say C◃×D if there exists a Cartesian frame M over Agent(D) with Image(M)=Agent(D), such that C≃D∘(M).Claim:This definition is equivalent to all of the above definitions of ◃×.Proof:We show equivalence to the externalizing definition.First, assume there exist three sets X, Y, and Z, and a function p:X×Y×Z→W such that C≃(X,Y×Z,⋄) and D≃(X×Y,Z,∙), where ⋄ and ∙ are given by x⋄(y,z)=(x,y)∙z=p(x,y,z).

Let D=(B,F,⋆), and let (g0,h0):D→(X×Y,Z,∙) and (g1,h1):(X×Y,Z,∙)→D compose to something homotopic to the identity in both orders.

We define B′=B⊔{a}, and we define M, a Cartesian frame over B, by M=(X,Y×B′,⋅), where ⋅ is given by x⋅(y,b)=b if b∈B and g0(b)=(x,y), and x⋅(y,b)=g1(x,y) otherwise. Clearly, Image(M)=B, since for any b∈B, if we let (x,y)=g0(b), we have x⋅(y,b)=b.

Observe that for all x∈X, y∈Y, b∈B′ and f∈F, if b∈B and g0(b)=(x,y), then

(x⋅(y,b))⋆f=b⋆f=g1(g0(b))⋆f=g1(x,y)⋆f,and on the other hand, if b=a or g0(b)≠(x,y), we also have (x⋅(y,b))⋆f=g1(x,y)⋆f.

Thus, we have that D∘(M)=(X,Y×B′×F,⋆′), where ⋆′ is given by

x⋆′(y,b,f)=(x⋅(y,b))⋆f=g1(x,y)⋆f=(x,y)∙h1(f).To show that (X,Y×Z,⋄)≃D∘(M), we construct morphisms (g2,h2):(X,Y×Z,⋄)→D∘(M) and (g3,h3):D∘(M)→(X,Y×Z,⋄) that compose to something homotopic to the identity in both orders. Let g2 and g3 both be the identity on X. Let h2:Y×B′×F→Y×Z be given by h2(y,b,f)=(y,h1(f)), and let h3:Y×Z→Y×B′×F be given by h3(y,z)=(y,a,h0(z)).

We know (g2,h2) is a morphism, since for all x∈X and (y,b,f)∈Y×B′×F,

g2(x)⋆′(y,b,f)=x⋆′(y,b,f)=(x,y)∙h1(f)=p(x,y,h1(f))=x⋄(y,h1(f))=x⋄(h2(y,b,f)).We also have that (g3,h3) is a morphism, since for all x∈X and (y,z)∈Y×Z, we have

g3(x)⋄(y,z)=x⋄z=x⋄(y,z)=p(x,y,z)=(x,y)∙z=(x,y)∙h1(h0(z))=(x,y)⋆′(y,a,h0(z))=x⋆′h3(y,z).Observe that (g2,h2) and (g3,h3) clearly compose to something homotopic to the identity in both orders, since g2∘g3 and g3∘g2 are the identity on X.

Thus, C≃(X,Z,⋄)≃D∘(M), where Image(M)=Agent(D).

Conversely, assume C≃D∘(M), with Image(M)=Agent(D). Let X=Agent(M), let Y=Env(M), and let Z=Env(D). Let f:X×Y×Z→W be given by f(x,y,z)=(x⋅y)⋆z, where ⋅=Eval(M) and ⋆=Eval(D).

Thus C≃D∘(M)≅(X,Y×Z,⋄), where ⋄ is given by x⋄(y,z)=(x⋅y)⋆z=f(x,y,z). All that remains to show is that D≃(X×Y,Z,∙), where (x,y)∙z=f(x,y,z). Let D=(B,Z,⋆).

We construct morphisms (g0,h0):D→(X×Y,Z,∙) and (g1,h1):D→(X×Y,Z,∙) that compose to something homotopic to the identity in both orders. Let h0 and h1 be the identity on Z. Let g1:X×Y→B be given by g1(x,y)=x⋅y. Since g1 is surjective, it has a right inverse. Let g0:B→X×Y be any choice of right inverse of g1, so g1(g0(b))=b for all b∈B.

We know (g1,h1) is a morphism, since for all (x,y)∈X×Y and z∈Z,

g1(x,y)⋆z=(x⋅y)⋆z=f(x,y,z)=(x,y)∙z=(x,y)∙h1(z).To see that (g0,h0) is a morphism, given b∈B and z∈Z, let (x,y)=g0(b), and observe

g0(b)∙z=(x,y)∙z=f(x,y,z)=(x⋅y)⋆z=g1(x,y)⋆z=g1(g0(b))⋆z=b⋆h0(z).(g0,h0) and (g1,h1) clearly compose to something homotopic to the identity in both orders, since h0∘h1 and h1∘h0 are the identity on Z. Thus D≃(X×Y,Z,∙), completing the proof. □

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

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 Agent(C) to Agent(D), where Env(M) represents the nondeterminism. In the case of additive subagents, Env(M) is a singleton, meaning that the function from Agent(C) to 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), 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 ◃× show why this is the case.

1.5. Categorical DefinitionsWe 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 ϕ0:C→D such that for every morphism ϕ:C→⊥ there exists a morphism ϕ1:D→⊥ such that ϕ is homotopic to ϕ1∘ϕ0 .Claim:This definition is equivalent to all the above definitions of ◃+.Proof:We show equivalence to the committing definition.First, let C=(A,E,⋅) and D=(B,F,∙) be Cartesian frames over W, and let (g0,h0):C→D be such that for all (g,h):C→⊥, there exists a (g′,h′):D→⊥ such that (g,h) is homotopic to (g′,h′)∘(g0,h0). Let ⊥=(W,{i},⋆).

Let Y=B, let Z=F, and let X={g0(a) | a∈A}. Let f:Y×Z→W be given by f(y,z)=y∙z. We already have D=(Y,Z,∙), and our goal is to show that C≃(X,Z,⋄), where ⋄ is given by x⋄z=f(x,z).

We construct (g1,h1):C→(X,Z,⋄) and (g2,h2):(X,Z,⋄)→C that compose to something homotopic to the identity in both orders.

We define g1:A→X by g1(a)=g0(a). g1 is surjective, and so has a right inverse. We let g2:X→A be any right inverse to g1, so g1(g2(x))=x for all x∈X. We let h1:Z→E be given by h1(z)=h0(z).

Defining h2:E→Z will be a bit more complicated. Given an e∈E, let (ge,he) be the morphism from C to ⊥, given by he(i)=e and ge(a)=a⋅e. Let (g′e,h′e):D→⊥ be such that (ge,he) is homotopic to (g′e,h′e)∘(g0,h0). We define h2 by h2(e)=h′e(i).

We trivially have that (g1,h1) is a morphism, since for all a∈A and z∈Z,

g1(a)⋄z=g0(a)∙z=a⋅h0(z)=a⋅h1(z).To see that (g2,h2) is a morphism, consider x∈X and e∈E, and define (ge,he) and (g′e,h′e) as above. Then,

x⋄h2(e)=g1(g2(x))⋄h′e(i)=g′e(g0(g2(x)))⋆i=g2(x)⋅he(i)=g2(x)⋅e.We trivially have that (g1,h1)∘(g2,h2) is homotopic to the identity, since g1∘g2 is the identity on X. To see that (g2,h2)∘(g1,h1) is homotopic to the identity on C, observe that for all a∈A and e∈E, defining (ge,he) and (g′e,h′e) as above,

g2(g1(a))⋅e=g1(a)⋄h2(e)=g0(a)⋄h′e(i)=g′e(g0(a))⋆i=a⋆he(i)=a⋅e.Thus C≃(X,Z,⋄), and C◃+D according to the committing definition.

Conversely, let X, Y, and Z be arbitrary sets with X⊆Y, let f:Y×Z→W, and let C≃(X,Z,⋄) and D≃(Y,Z,∙), where ⋄ and ∙ are given by x⋄z=f(x,z) and y∙z=f(y,z).

Let (g1,h1):C→(X,Z,⋄) and (g2,h2):(X,Z,⋄)→C compose to something homotopic to the identity in both orders, and let (g3,h3):D→(Y,Z,∙) and (g4,h4):(Y,Z,∙)→D compose to something homotopic to the identity in both orders. Let (g0,h0):(X,Z,⋄)→(Y,Z,∙) be given by g0 is the embedding of X in Y and h0 is the identity on Z. (g0,h0) is clearly a morphism.

We let ϕ:C→D=(g4,h4)∘(g0,h0)∘(g1,h1).

Given a (g,h):C→⊥, our goal is to construct a (g′,h′):D→⊥ such that (g,h) is homotopic to (g′,h′)∘ϕ.

Let ⊥=(W,{i},⋆), let C=(A,E,⋅0), and let D=(B,F,⋅1). Let h′:{i}→F be given by h′=h3∘h2∘h. Let g′:B→W be given by g′(b)=b⋅1h′(i). This is clearly a morphism, since for all b∈B and i∈{i},

g′(b)⋆i=g′(b)=b⋅h′(i).To see that (g,h) is homotopic to (g′,h′)∘(g4,h4)∘(g0,h0)∘(g1,h1), we just need to check that (g,h1∘h0∘h4∘h′):C→⊥ is a morphism. Or, equivalently, that (g,h1∘h4∘h3∘h2∘h):C→⊥, since h0 is the identity, and h′=h3∘h2∘h.

Indeed, for all a∈A and i∈{i},

g(a)⋆i=a⋅0h(a)=a⋅0h1(h2(h(a)))=g1(a)⋄h2(h(a))=g1(a)∙h2(h(a))=g1(a)∙h4(h3(h2(h(a))))=g1(a)⋄h4(h3(h2(h(a))))=a⋅0h1(h4(h3(h2(h(a))))).Thus (g,h) is homotopic to (g′,h′)∘ϕ, completing the proof. □

Definition:We say C◃×D if for every morphism ϕ:C→⊥, there exist morphisms ϕ0:C→D and ϕ1:D→⊥ such that ϕ=ϕ1∘ϕ0, and for every morphism ψ:1→D, there exist morphisms ψ0:1→C and ψ1:C→D such that ψ=ψ1∘ψ0.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 DefinitionFirst, 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◃∗D, if D∗◃C∗.We can similarly define additive and multiplicative sub-environments.

Definition:We say C is an additive sub-environment of D, written C◃∗+D, if D∗◃+C∗. We say C is an multiplicative sub-environment of D, written C◃∗×D, if D∗◃×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◃×D if and only if C◃∗×D.Proof:We prove this using the externalizing definition of ◃×.If C◃×D, then for some X, Y, Z, and f:X×Y×Z→W, we have C≃(X,Y×Z,⋄) and D≃(X×Y,Z,∙), where ⋄ and ∙ are given by x⋄(y,z)=f(x,y,z) and (x,y)∙z=f(x,y,z).

Observe that D∗≃(Z,Y×X,⋅) and C∗≃(Z×Y,X,⋆), where ⋅ and ⋆ are given by z⋅(y,x)=f(x,y,z) and (z,y)⋆x=f(x,y,z). Taking X′=Z, Y′=Y, Z′=X, and f′(x,y,z)=f(z,y,x), this is exactly the externalizing definition of D∗◃×C∗, so C◃∗×D.

Conversely, if C◃∗×D, then D∗◃×C∗, so C≅{C∗}∗◃×{D∗}∗≅D. □

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

Definition:We say C◃×D if C◃D and C◃∗D. Equivalently, we say C◃×D if C◃D and D∗◃C∗.Claim:This definition is equivalent to the categorical definition of ◃×.Proof:The condition that for every morphism ϕ:C→⊥, there exist morphisms ϕ0:C→D and ϕ1:D→⊥ such that ϕ=ϕ1∘ϕ0, is exactly the categorical definition of C◃D.The condition that for every morphism ψ:1→D, there exist morphisms ψ0:1→C and ψ1:C→D such that ψ=ψ1∘ψ0, is equivalent to saying that for every morphism ψ∗:D∗→⊥, there exist morphisms ψ∗0:C∗→⊥ and ψ∗1:D∗→C∗ such that ψ∗=ψ∗1∘ψ∗0. This is the categorical definition of D∗◃C∗. □

Claim:The categorical and sub-environment definitions of ◃× 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,⋅) and D=(B,F,⋆) are Cartesian frames over W with C◃D and C◃∗D.We define X=A, Z=F, and Y=hom(C,D). We define p:X×Y