L&S Cat Theory: Attempted solutions to "Conceptual Mathematics" by Lawvere and Schanuel

This problem goes back to a difference in definitions of “map” between myself and the authors that I first observed in Article I. My model of “map” allowed for the output of the map to be “null” while the authors’ didn’t. Setting aside the merits of the authors’ definition for the moment, my definition of “gender map” has the advantage that it allows for people to be non-binary in the resulting model.

In contrast, L&S have defined “map” in such a way that every element in the domain must have an arrow pointing to an element in the codomain. This makes the bookkeeping of maps easier because there is a one-to-one correspondence between dots and arrows, but it does not allow for the expression of non-binary people in the same way as my model. Instead, a map \(\mathbf{P} \rightarrow \mathbf{G}\) forces each person into one of the two genders.

Essentially, by assuming that “if you’re a person then you have a gender” the authors have directly implied the contrapositive that “if you don’t have a gender then you’re not a person”. I think I might understand why they did it, but it doesn’t change how messed up it is to deny personhood to an entire group like that. If the goal of this is to convince me that the author’s definition of map is better than the one I started with, this exercise isn’t doing a very good job at it. It’s starting to look more like a “vacuous statement”. Given that non-binary people exist, the only logical conclusion is that the map \(\mathbf{P} \rightarrow \mathbf{G}\) doesn’t exist. Any conditional statement beginning with “if \(\mathbf{P} \rightarrow \mathbf{G}\) is a map from the set of persons to genders” is therefore automatically true because the hypothesis is always false.

Gender has been a recurring theme throughout the text so far. We saw it in Session 5 when talking about determination problems, in “Composition of opposed maps” at the end of Part 2, in Article 3 Exercise 17, Session 12 Exercise 3, Session 15 Exercise 6, and now again in Session 21 Exercise 3. Looking over my notes, we’ve gradually moved from looking at gender being defined as a map in \(\mathcal{S}\) to a map in \(\mathcal{S}^{\circlearrowright, \circlearrowleft}\) that preserves the structure of “father” and “mother” maps. The idea here was that gender is preserved under the operations “father” and “mother” because it becomes a constant map after composition.

The authors referred to this as “objectification of the subjective”, implying that we can infer a person’s gender from examining the behavior of these “mother” and “father” maps. In Session 15, I joked that this model excluded Stormtroopers as people because as clones of a male they lack a well defined “mother” map, but the truth is that this joke was merely a reflection of the fact that this model of gender fails to work for me specifically because I don’t consider myself to have a “father map”.

Okay, maybe the truth is a little more complicated. I have something like a “father map”, but it’s a map from \(\mathbf{P} \rightarrow \mathbf{P} \times \mathbf{P}\) rather than \(\mathbf{P} \rightarrow \mathbf{P}\) so it doesn’t preserve the structure of this “gender map” in quite the same way.

Growing up, there was a family legend suggesting that one of my ancestors was Native American. In 2020, I decided to take a DNA test to determine whether or not there was any truth to it and wrote about the experience on my blog

¹

suburbanlion.com/blog/2020/11/14/land-acknowledgement/

. What I omitted from the original post was the fact that my half-sister took the same test so we could compare results, and I inadvertently found out that I was born through a process of in vitro fertilization using donor sperm. My parents chose to keep this secret from me until such a time that I was going to find it out anyway, and I’m still not exactly sure how to feel about it.

It’s kind of interesting that back in Session 1 I chose to look at a product of \(\text{NATURE} \times \text{NURTURE}\) because this parallels the properties of my “father map”. I have a “nature father” that provided genetic material and a “nurture father” that was actually involved in my life. Under my old definition of “father map” it would be easy for me to think of the output being undefined, but the definitions used in the text exclude this as a possibility for the specified domain and codomain. The fact that one input corresponds to two outputs means that my “father map” does not meet the criteria for an endomap on \(\mathbf{P}\text{.}\) The authors are using “gender” as an example of a universal property, but it’s not a universal property if there are people it doesn’t apply to.

I’ve commented before how the sections including classroom dialogues have been fun to read, but I’m not sure if this “father map” problem is a question I’d be able to bring up in class. While looking back at the Composition of opposed maps section for clues to this problem, the idea of being put in Alysia’s shoes is terrifying. Would I have the guts to say “I don’t have a father map” in a class full of strangers? Probably not. It’s far more likely that I would lie about the personal situation in favor of a simpler mathematical model.

I think what makes it even more frustrating is the fact that the logical half of my brain is screaming “IT DOESN’T EVEN MATTER!!” this whole time. With the way that L&S have explicitly defined the product, there should still be exactly one map such that all triangles commute. The universal mapping property is worded in such a way that it shouldn’t matter whether I have two dads or no dads — there’s still an unique isomorphism between the models. It’s just easier to see without the additional complexity.

Looking back over Session 12 again, the authors make it very clear that the model of “gender” in Exercise 3 is completely detached from reality. They explicitly state that “[y]ou will notice that this category contains many objects and maps that cannot reasonably be interpreted as a set of people with ‘father’ and ‘mother’ maps”. There’s this underlying notion that even if this model of gender is “wrong”, it can still be “useful”. I get that basic idea in principle, but this doesn’t change how dehumanizing it feels to be excluded. How much harm are we willing to impose for the sake of utility? IVF was a relatively new procedure when the first edition of this text was printed, but more recent evidence suggests that people like myself are a rapidly growing demographic

²

www.ncbi.nlm.nih.gov/pmc/articles/PMC6765402/

. At some point mathematicians will need to stop assuming everyone comes from a traditional family structure.

I don’t think I’m exaggerating when I say this is a “traumatic experience”. I probably wouldn’t have even recognized it had it not been for my experiences learning about trauma sensitive instruction

³

www.apa.org/ed/schools/primer/trauma

. I have to constantly remind myself that I need to treat myself with grace and kindness in this situation.

Having read Eugenia Chen’s X+Y before starting this venture through L&S, I kind of expected gender to make an appearance. What I didn’t expect was to start questioning my own — again. I wrote before

⁴

suburbanlion.com/blog/2020/03/19/choosing-whiteness/

about the ways in which I consciously chose to identify as male, but now there’s this alternate definition where I’m not so sure about the label. I have a gender map in \(\mathcal{S}\) but not in \(\mathcal{S}^{\circlearrowright, \circlearrowleft}\text{.}\) It’s almost as if category theory has just handed me an objective definition by which I am non-binary, should I choose to accept it, and I feel a little frightened about the prospects.

Perhaps the big idea here is that I need to do a better job of distinguishing between maps in \(\mathcal{S}\) and maps in \(\mathcal{S}^{\circlearrowright, \circlearrowleft}\text{.}\) In Session 12, maybe they were dropping a hint when they describe a “person” who is “its own ‘mother’ and its own ‘father’”. Such a ‘person’ would satisfy the properties of a terminal object \(T^{\circlearrowright, \circlearrowleft}\) in the category \(\mathcal{S}^{\circlearrowright, \circlearrowleft}\text{.}\) This would at least give me an object to observe the system through. I could enumerate all possible maps in \(\mathcal{S}\) and sort them into two groups based on whether or not they preserve the structure in \(\mathcal{S}^{\circlearrowright, \circlearrowleft}\text{.}\)

In order for \(\mathbf{G} \times \mathbf{C}\) to be a product, it must satisfy the following conditions:

First, we need to show \(\mathbf{G} \times \mathbf{C}\) is an object in \(\mathcal{S}^{\circlearrowleft, \circlearrowright}\text{,}\) and we need a pair of maps \(\mathbf{G} \times \mathbf{C} \xrightarrow{p_1} \mathbf{G}\) and \(\mathbf{G} \times \mathbf{C} \xrightarrow{p_2} \mathbf{C}\text{.}\) Second, we need to show that for every object \(T\) and every pair of maps \(T \xrightarrow{q_1} \mathbf{G}\) and \(T \xrightarrow{q_2} \mathbf{C}\text{,}\) there is exactly one map \(T \xrightarrow{q} \mathbf{G} \times \mathbf{C}\) for which \(q_1 = p_1 \circ q\) and \(q_2 = p_2 \circ q\text{.}\)

Ignoring the wildcard object \(T\) for now, we can take all this information and construct an external diagram of the sitation as follows:

The first task was essentially the goal of Session 12 Exercise 3. Last time I exhaustively listed the options, but I think there’s a more direct approach where we can construct a map from the product of endomaps to the endomaps of products.

Given any \(p\) in \(\mathbf{P}^{\circlearrowleft m_P, \circlearrowright f_P}\text{,}\) and structure preserving maps \(\mathbf{P}^{\circlearrowleft m_P, \circlearrowright f_P} \xrightarrow{q_1} \mathbf{G}^{\circlearrowleft m_G, \circlearrowright f_G}\) and \(\mathbf{P}^{\circlearrowleft m_P, \circlearrowright f_P} \xrightarrow{q_2} \mathbf{C}^{\circlearrowleft m_C, \circlearrowright f_C} \text{,}\) we can define a map \(\mathbf{P}^{\circlearrowleft m_P, \circlearrowright f_P} \xrightarrow{q} (\mathbf{G} \times \mathbf{C})^{\circlearrowleft m_{G \times C}, \circlearrowright f_{G \times C}}\) such that \(m_{G \times C} \circ q \circ p = \langle m_G \circ q_1 \circ p, m_C \circ q_2 \circ p \rangle = \langle q_1 \circ m_P \circ p, q_2 \circ f_P \circ p \rangle\) and \(f_{G \times C} \circ q \circ p = \langle f_G \circ q_1 \circ p, f_C \circ q_2 \circ p \rangle = \langle q_1 \circ f_P \circ p, q_2 \circ f_P \circ p \rangle\text{.}\)

Since \(\mathbf{P}^{\circlearrowleft m_P, \circlearrowright f_P} \xrightarrow{q_1} \mathbf{G}^{\circlearrowleft m_G, \circlearrowright f_G}\) needs to preserve the structure of \(\mathcal{S}^{\circlearrowleft \circlearrowright}\text{,}\) we must have the property that \(q_1 m_P = m_G q_1\) and \(q_1 f_P = f_G q_1\text{.}\) Likewise, \(\mathbf{P}^{\circlearrowleft m_P, \circlearrowright f_P} \xrightarrow{q_2} \mathbf{C}^{\circlearrowleft m_C, \circlearrowright f_C}\) must preserve \(q_2 \circ m_P = m_C \circ q_2\) and \(q_2 \circ f_P = f_C \circ q_2\text{.}\) By substituting into our definition of \(q\text{,}\) we get the following:

Since this is true for any \(p\text{,}\) we can equate the maps \(m_{G \times C} \circ q = q \circ m_P\) and \(f_{G \times C} \circ q = q \circ f_P\text{.}\) This demonstrates that \(\mathbf{P} \rightarrow \mathbf{G} \times \mathbf{C}\) is a valid map in the category.

Next, let’s have a look at our projection maps \(p_1,p_2\) defined by \(p_1 \langle x_1, x_2 \rangle = x_1\) and \(p_2 \langle x_1, x_2 \rangle = x_2\text{.}\) For any person \(p\text{,}\) we’d defined \(q \circ p = \langle q_1 p, q_2 p\rangle\text{.}\) Applying our projection maps \(p_1, p_2\) on the left gives us \(p_1 \circ q \circ p = q_1 p\) and \(p_2 \circ q \circ p = q_2 p\text{.}\) This gives us the properties that \(q_1 = p_1 \circ q\) and \(q_2 = p_2 \circ q\text{.}\) I should probably point out that the necessary structure preservation properties \(m_G \circ p_1 = p_1 \circ m_{G \times C}\text{,}\) \(f_G \circ p_1 = p_1 \circ f_{G \times C}\text{,}\) \(m_C \circ p_2 = p_2 \circ m_{G \times C}\text{,}\) and \(f_C \circ p_2 = p_2 \circ f_{G \times C}\text{,}\) all follow directly from our choice of \(m_{G \times C} = m_G \times m_C\) and \(f_{G \times C} = f_G \times f_C\text{.}\)

I think it might also be helpful to reproduce the diagrams from Session 12 in the style used by this Session:

I think that one of the key ideas here is to identify a terminal object in the category \(\mathcal{S}^{\circlearrowleft \circlearrowright}\text{.}\) I’m going to make the claim that the following object has the desired properties:

I think one of the important features of this object is that it can’t be mapped back to any of elements in \(\mathbf{G} \times \mathbf{C}\text{.}\) There’s are 4 maps from \(\mathbf{1} \rightarrow \mathbf{G} \times \mathbf{C}\text{,}\) but none of them preserve the structure of \(\mathbf{T}^{\circlearrowleft m_T,\circlearrowright f_T}\) because \(m_{G \times C},f_{G \times C}\) are different maps while \(m_T = f_T\) for the only element in \(\mathbf{T}\text{.}\) There’s a unique set of isomorphism in \(\mathcal{S}\) given by \(\mathbf{1} \mathrel{\substack{\longrightarrow \\ \longleftarrow}} \mathbf{T}\text{,}\) \(\mathbf{1} \mathrel{\substack{\longrightarrow \\ \longleftarrow}} \mathbf{G}\text{,}\) \(\mathbf{1} \mathrel{\substack{\longrightarrow \\ \longleftarrow}} \mathbf{C}\text{,}\) \(\mathbf{1} \mathrel{\substack{\longrightarrow \\ \longleftarrow}} \mathbf{G} \times \mathbf{C}\text{,}\) and \(\mathbf{1} \mathrel{\substack{\longrightarrow \\ \longleftarrow}} \mathbf{P}\text{.}\) However, none of the composite maps \(\mathcal{S}\)-maps \(\mathbf{T} \rightarrow \mathbf{G}\text{,}\) \(\mathbf{T} \rightarrow \mathbf{C}\text{,}\) \(\mathbf{T} \rightarrow \mathbf{G} \times \mathbf{C}\) preserve the structure required of maps in \(\mathcal{S}^{\circlearrowleft, \circlearrowright}\) because there’s two unique maps in the codomain and only one in the domain. This seems like a reasonable way to say our product uniquely commutes, which is what we needed to complete the proof.

I’ve been trying to avoid using the sum here, but I think this relation is nicely summarized by the following diagram:

Once we start looking at the maps from \(\mathbf{4} \rightarrow \mathbf{4}\) which are invertable, the number of ways of breaking them into pairs is limited. One of the things I noticed while exploring was that if I defined a “parent of opposite gender” map \(\mathbf{P} \xrightarrow{\alpha} \mathbf{P}\text{,}\) composing this map with \(q\) to get \(\mathbf{G} \times \mathbf{C}\) into a cycle of length 4 under the respective operation \(\alpha'\text{.}\)

The structure of these relations reminded me of the presentations of dynamical systems we say is Session 15. In particular, it got me thinking about the way I was trying to standardize my presentations using an isomorphism \(\mathbb{N} \leftrightarrow \mathbb{N} \times \mathbb{N}\text{.}\) In particular, there were two different ways of enumerating my table of \(\langle \text{generator}, \text{iterations} \rangle\) pairs:

These two methods both enumerate all possible values, but the do so in a slightly different order. When the ordered pairs all commute, such that \(\langle x_1, x_2 \rangle = \langle x_2, x_1 \rangle\text{,}\) then the two representations converge to a single solution. Perhaps this is why the phrase “all triangles commute” does such heavy lifting.

I can’t help but wonder if the point of this whole \(\mathcal{S}^{\circlearrowleft,\circlearrowright}\) category is to generalize from my zig-zag ordering of elements into two separate operators — such that \(m\) increments the row and \(f\) increments the column. Even two observers have different enumeration maps, each map will have a unique pair of \(m,f\) maps such that they both agree every value in the table.

I don’t think I’ve fully answered this exercise quite yet, but at least I’ll have some better notes for next time this model shows up.