This is the third post in the Cartesian frames sequence. Read the first post here.

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"), 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 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

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.




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.

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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.

Fixed. Thanks.

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.

 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 coreflective. 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.