# Biextensional Equivalence

post by Scott Garrabrant · 2020-10-28T14:07:38.289Z · LW · GW · 13 comments## Contents

1. Isomorphism 2. Homotopy Equivalence Homotopic Morphisms Homotopy Equivalence 3. Biextensional Equivalence Biextensionality Biextensional Collapse Biextensional Equivalence Example Relationship to Additive Operations 4. Some Small Cartesian Frames Footnotes None 13 comments

This is the third post in the Cartesian frames sequence. Read the first post here [LW · GW].

This post will introduce the standard equivalence relations we'll be using for Cartesian frames. Our primary interest will be in homotopy equivalence, which will allow us to classify frames according to their agents' and environments' effect on possible worlds.

## 1. Isomorphism

Before defining homotopy equivalence, I want to define isomorphism between Cartesian frames.

**Definition:** A morphism is an *isomorphism* if both and are bijective. If there is an isomorphism between and , we say .

**Claim:** is an equivalence relation.

**Proof:** Reflexivity is trivial because the identity is an isomorphism. For symmetry, we have that if is an isomorphism from to , then is an isomorphism from to . Transitivity follows from the fact that bijection is transitive.

**Claim:** if and only if there is a pair of morphisms and that compose to the identity morphism in both orders.

**Proof:** If , we have with both and bijective, and we can take and .

Conversely, if is the identity morphism on , then is the identity on , so must be injective. Similarly, is the identity on , so must be surjective. Surjectivity of and injectivity of follow similarly from the fact that is the identity on .

Isomorphism is pretty intuitive. It is basically saying that it doesn't matter what the possible agents and possible environments are, other than how they interact with the evaluation function.

We will basically always be working up to at least isomorphism. For example, in the last post ("Additive Operations on Cartesian Frames [LW · GW]"), we noted that and are commutative and associative up to isomorphism.

## 2. Homotopy Equivalence

**2.1. Homotopic Morphisms**

Our initial definition of homotopy equivalence will be devoid of interpretation, but the meaning will become clear later.

We say that two morphisms from to are homotopic if you can take the first function from the first morphism and the second function from the second morphism, and the resulting object is still a morphism.

**Definition:** Two morphisms with the same source and target are called *homotopic* if is also a morphism.

Note that the mere existence of two morphisms from to doesn't entail that those morphisms are homotopic. Consider the frame given by

.

There are two morphisms from to itself: the identity morphism , and a morphism that flips 's rows and columns, sending to , to , to , and to . These two morphisms are not homotopic, because neither nor is a morphism.

Being homotopic is an equivalence relation on morphisms. (As such, in the above example it would have been superfluous to demonstrate that both and aren't morphisms.)

**Claim: **Homotopic is an equivalence relation.

**Proof:** Let .

Reflexivity is trivial. For symmetry, we want to show that if , , and are all morphisms, then so is . Indeed, for all and ,

For transitivity, we want to show that if , , , , and are all morphisms, then so is . Indeed, for all and ,

Being homotopic is also respected by composition.

**Claim: **If is homotopic to , and is homotopic to , then is homotopic to .

**Proof: **Let , let , and let . We want to show that is a morphism. Indeed, since and are morphisms,

Next, we define when two Cartesian frames are homotopy equivalent in the standard way.

**2.2. Homotopy Equivalence**

Homotopy equivalence relies on the existence of morphisms between and that we can compose in either order and end up with something that is homotopic to the identity morphism.

**Definition:** is *homotopy equivalent* to , written , if there exists a pair of morphisms and such that is homotopic to the identity on and is homotopic to the identity on .

**Claim:** is an equivalence relation.

**Proof:** Reflexivity is trivial, by taking to be the identity. Symmetry is also trivial by swapping and .

For transitivity, assume that for , we have and such that and are homotopic to the identity. It suffices to show that and are both homotopic to the identity. In both cases, since composition respects what is homotopic, we have that the inner pair of morphisms cancels, and then the outer pair of morphisms cancels.

Note that homotopy equivalence is weaker than isomorphism.

**Claim:** If , then .

**Proof:** Trivial.

We now have the homotopy equivalence relation on Cartesian frames, but no real philosophical interpretation. To understand the meaning of , we will first need to define biextensional collapse.

## 3. Biextensional Equivalence

**3.1. Biextensionality**

**Definition:** A Cartesian frame is called *biextensional* if whenever are such that , for all , we have , and whenever are such that , for all , we have .

This is basically saying that a Cartesian frame is biextensional if all of its possible agents and possible environments are distinct when viewed as functions from environment to world and functions from agent to world respectively. The agent doesn't have two options that invariably produce the same outcomes as each other, nor does the environment.

Viewed as a matrix, is biextensional if all of its rows and columns are distinct.

We have the following lemma that hints at the relationship between biextensionality and homotopy equivalence.

**Lemma:** Let and be biextensional Cartesian frames. Then, if and only if .

**Proof:** The "if" direction is trivial. For the "only if" direction, let and be two biextensional Cartesian frames, and let . Thus, there is a pair of morphisms and such that is homotopic to the identity on , and is homotopic to the identity on . Thus and are both morphisms, where is the identity on the set . This means that for all and , which since is biextensional implies that is the identity on . Similarly, since is a morphism, we have that is the identity on . Thus is a bijection.

By the symmetry of homotopic, we also have that and , which similarly gives us that is the identity on and is the identity of , so is a bijection. Thus, .

Thus, we now understand how to interpret homotopy equivalence for biextensional Cartesian frames: it is equivalent to isomorphism.

To understand homotopy equivalence in general, we will first show how to collapse any Cartesian frame into a biextensional one.

**3.2. Biextensional Collapse**

Given a Cartesian frame , we can define an equivalence relation on that says two possible agents are equivalent if they implement the same function from to ; and we can similarly say that two elements of are equivalent if they implement the same function from to .

**Definition:** Given a Cartesian frame , for , we say if for all . For , we say that if for all .

**Claim:** is an equivalence relation on and on .

**Proof:** Trivial.

**Definition: **Given a Cartesian frame , for , let denote the equivalence class of up to . Let denote the set of equivalence classes of in . Similarly, for , let denote the equivalence class of up to , and let denote the set of equivalence classes of in .

**Definition:** Given a Cartesian frame , the *biextensional collapse* of , denoted , is the Cartesian frame , where .

**Claim:** is well-defined.

**Proof:** We need to show that is well defined, meaning we need to show that for all and , we have that . This is immediate from the definition of .

Viewed as a matrix, is basically formed from by deleting any duplicate rows and any duplicate columns. It doesn't matter whether you delete duplicate rows or duplicate columns first. After doing both, you will end up with a matrix with no duplicates.

**Claim: ** is biextensional for all Cartesian frames .

**Proof:** Let . We want to show that for all , there exists an such that . Indeed, since , we have that , so there exists an such that , which gives us that . Similarly, for all , there exists an such that .

**Claim:** is biextensional if and only if .

**Proof: **If is biextensional, then all equivalence classes up to on both and are singletons. Thus, the morphism given by and is well-defined, and both and are bijective, so .

Conversely, if then is isomorphic to a biextensional Cartesian frame, and since biextensionality is clearly preserved by isomorphism, is also biextensional.

**3.3. Biextensional Equivalence**

We can now (finally) use biextensional collapse to give an intuitive meaning to homotopy equivalence.

**Claim:** if and only if .

**Proof:** It suffices to show that for all Cartesian frames . Then, we will have that if , then . Since homotopy equivalence is the same as isomorphism on biextensional Cartesian frames, this gives . And conversely, if then , so .

Let . We want to show that . We do this by constructing a pair of morphisms , and . We will define by , and by . For , and , we can send each equivalence class to any one member of that class. The choice does not matter.

Now, we want to show that is homotopic to the identity on , and that is homotopic to the identity on . The first case is trivial, since and are the identity on and respectively. and need not be the identity on and , but and for all and . To show that is homotopic to the identity on , we just need to show that is a morphism, where is the identity on . However, this just means that for all and , which follows from the fact that .

We now have that two Cartesian frames are homotopy equivalent if and only if their biextensional collapses are isomorphic. Thus, when and are homotopy equivalent, we will also call them biextensionally equivalent.

**Definition:** We say and are *biextensionally equivalent* if .

When working up to biextensional equivalence, we are basically saying that we are ignoring any multiplicity in the space of possible worlds and possible environments.

**Claim:** Each biextensional equivalence class contains a unique biextensional Cartesian frame.

**Proof: **Each biextensional equivalence class has at least one element, , and is in the same equivalence class as and is biextensional, so there must be at least one biextensional Cartesian frame in the class. If there were two biextensional Cartesian frames, they would have to be isomorphic, because isomorphic is equivalent to biextensional equivalence on biextensional Cartesian frames.

From my perspective, the value of this equivalence relation is that it lets us be less realist about possible agents and possible environments, and instead just care about differences between possible worlds.

This fits well with our general approach in this sequence. Cartesian frames are particular ways of looking at the world and mentally carving it up into an agent component and an environment component, but we allow many different carvings, and we do not give any one carving privileged status as the "true" carving. Thus, we put less weight on our conception of the agent and environment, and more weight on the worlds themselves.

Giving less realism to possible agents/environments also fits with the fact that "worlds" may include details about the agent and environment, "possible agents" may specify features of the agent beyond its "actions," and so on.

Imagine an agent with two unrelated choices: which color to think about (green , or red ) and whether to go for a walk or stay home ( or ). This yields the possible agents . The environment either is safe or has bears: . If we represent this scenario with the Cartesian frame

,

then the possible worlds and differ only in which *thought *the agent is thinking; likewise and , etc.

We could have instead described a frame

,

in which case we would not be treating the agent's thoughts as a relevant difference between possible worlds.^{1} [LW(p) · GW(p)] But we have the *option* of fully representing "agent-internal" properties using possible worlds, just the same as "environment-internal" properties. As such, we don't need to separately reify possible agents or possible environments.

**3.4. Example**

One reason there are two definitions here is because the homotopy definition is easier to work with categorically, while the biextensionality definition is easier to work with directly with matrices.

Let and be the Cartesian frames given by:

and .

Note that when working up to isomorphism, there is no need to label the rows or columns.

We can then see that because

.

To verify the equivalence using the the homotopy definition would be far more tedious.

**3.5. Relationship to Additive Operations**

Since we will often want to work with Cartesian frames up to biextensional equivalence, it will be helpful to know that all of our additive operations respect biextensional equivalence.

**Claim:** If and , then , , and .

**Proof:** It is clear from the definition of biextensional collapse that commutes with . Thus since , we have , so .

For the rest, it suffices to show that if , then . Then, since is symmetric up to isomorphism, we have

and using the fact that and are De Morgan dual, we have

We will use the homotopy equivalence definition. Let and let . Let and compose to something homotopic to the identity in both orders. We want to construct a and , that similarly compose to something homotopic to the identity in both orders. We will take to be given by if , and if . Similarly, we will take to be given by .

Without loss of generality, it suffices to show that is a morphism and that is homotopic to the identity on . The fact that is a morphism and is homotopic to the identity will follow symmetrically. Let .

To show that is a morphism, observe that for all and , we have

Similarly, for all and , we have

To show that is homotopic to the identity on , we just need that for all and all , we have . Indeed, if , then , and if , then

Image is also clearly preserved by biextensional equivalence.

**Claim:** If , then .

**Proof:** Trivial from the biextensional collapse definition.

## 4. Some Small Cartesian Frames

We will now classify all biextensional Cartesian frames (and thus biextensional equivalence classes of Cartesian frames) in which the agent's size is at most one and/or the environment's size is at most one.

**Definition:** is the Cartesian frame with empty agent, empty environment, and empty evaluation function.

If you have an empty Cartesian frame—one with no image, no elements of —then it must be biextensionally equivalent to either , , or .

**Claim:** If and , then . If and , then . If , then .

**Proof:** If and , then all environments are equivalent up to so has one possible environment and no possible agents, so , so . Similarly, if and , all agents are equivalent up to , so and . If , then is already equal to .

**Claim:** The only three biextensional Cartesian frames with are , , and .

**Proof: **A Cartesian frame has empty image if and only if it has empty agent or empty environment.** **All three of 0, , and are clearly biextensional, and any other Cartesian frame with empty image is biextensionally equivalent to one of them, and so cannot be biextensional.

We now understand all biextensional Cartesian frames with empty agent or empty environment. Let's look at the case where either the agent or environment is a singleton.

is the biextensional Cartesian frame you get when the agent has only one option, and the frame's image is some set of possible worlds . Since will be in bijective correspondence with and the labels on don't matter, we will identify with .

**Definition:** Given , is the Cartesian frame , where for all . is the Cartesian frame .

We can think of as the perspective of a bystander who has no control, and is just observing which world the environment brings about.

is the transpose of , where the environment has only one option and the agent's options are . You can think of as a powerful agent facing no obstacles, beyond being constrained to : it gets to choose exactly what world we're in.

**Definition:** Given , is the Cartesian frame , where for all . is the Cartesian frame .

The names and will make more sense later, when we define multiplicative operations on Cartesian frames.^{2} [LW(p) · GW(p)]

We can think of as a powerless, all-knowing agent, and as with a promise from the environment that the world will be in . Similarly, we can think of as an all-powerful agent, and as with a commitment to do .

The class of frames where the agent has only one option, , contains at one extreme (where ) and at the other extreme (where ). Meanwhile, the class of frames where the environment has only one option, , contains at one extreme (where ) and at the other (where ).

**Claim:** , , , , .

**Proof: **Trivial.

**Claim:** If , then , where . If , then , where .

**Proof:** If , then equivalence classes of environments are given by where they send . There will be one such equivalence class for each , and it will send to . Thus , so . The case is the same with agent and environment swapped.

Now that we have built up language for talking about Cartesian frames categorically, we are ready to revisit controllables and observables and interpret them through the lens of category theory. This will be the focus of our next post.

Note: This Sunday at 2pm, I'll be giving a talk on Cartesian frames on Gather.Town, with the first two posts [? · GW] (or my last talk) as prerequisites. See Rob's post [LW · GW] for details.

## Footnotes

1. Similarly, we could have decided that we don't care about certain things about the environment. For example, if we only care whether there are bears in possible worlds where the agent went for a walk and might therefore encounter them, then we could construct a frame

.

2. Indeed, this section on small Cartesian frames would make more sense as part of our discussion of multiplicative operations on Cartesian frames; our motivation for discussing these objects will be provided there. I'm introducing these objects early because they will be useful in a few contexts before we get to multiplicative operations. ↩ [LW · GW]

## 13 comments

Comments sorted by top scores.

Are the two frames labeled and in section 3.3 (with an agent that thinks about either red or green and walks or stays home, and an environment that's safe or has bears) equivalent? (I would guess so, given the section that example was in, but didn't see it stated explicitly.)

They're not equivalent. If two frames are 'homotopy equivalent' / 'biextensionally equivalent' (two names for the same thing, in Cartesian frames), it means that you can change one frame into the other (ignoring the labels of possible agents and environments, i.e., just looking at the possible worlds) by doing some combination of 'make a copy of a row', 'make a copy of a column', 'delete a row that's a copy of another row', and/or 'delete a column that's a copy of another column'.

The entries of and are totally different (, while , before we even get into asking how those entries are organized in the matrices), so they can't be biextensionally equivalent.

There *is* an important relationship between and , which Scott will discuss later in the sequence. But the reason they're brought up in this post is to make a more high-level point "here's a reason we want to reify agents and environments less than worlds, which is part of why we're interested in biextensional equivalence," not to provide an example of biextensional equivalence.

An example of frames that are biextensionally equivalent to :

... or any frame that enlarges one of those four frames by adding extra copies of any of the rows and/or columns.

This is helpful. Thanks!

Scott's post explaining the relationship between and exists as of now: Functors and Coarse Worlds [LW · GW].

Really interesting!

I think there might be a minor typo in Section 2.2:

For transitivity, assume that for

I think this should be based on the indexing in the rest of the paragraph.

in as equivalence relation

I presume this is supposed to be "is an"?

Does the biextensional collapse satisfy a universal property? There doesn't seem to be an obvious map either or (in each case one of the arrows is going the wrong way), but maybe there's some other way to make it universal?

I think the right way to think about biextensional collapse categorically is as a reflector.

But then shouldn't there be a natural biextensional equivalence ? Suppose , and denote . Then the map is clear enough, it's simply the quotient map. But there's not a unique map - any section of the quotient map will do, and it doesn't seem we can make this choice naturally.

I think maybe the subcategory of just "agent-extensional" frames is reflective, and then the subcategory of "environment-extensional" frames is *co*reflective.
And there's a canonical (i.e natural) zig-zag

You might be right, I am not sure.

It looks to me like it satisfies the definition on wikipedia, which does not require that the morphism is unique, only that it exists.