For a given object \(G\text{...}\)
Solution.
Before we get too into this, let’s start by examining the examples we’re asked to consider. Having already solved Session 12 Exercise 3 and Article III Exercise 17, let’s review what we learned from each.
Both of these questions deal with a "gender" construct, but it’s defined a little differently in each.
In Session 12 Exercise 3, we defined gender as a map \(g\) from the set of people, \(P\text{,}\) to a set of genders, \(G\text{.}\) In this situation, we essentially get a pair of "mother maps" and "father maps" depending on whether they operate in "people space" (\(m_P,f_P\)) or "gender space" (\(m_G,f_G\)).
In Article 3 Exercise 17, we don’t actually describe a set of genders directly, we just subdivide people into sets of males, \(M\text{,}\) and females, \(F\text{,}\) and a set of maps between combinations of those two sets. In this situation, we have pairs of mother and father maps again, but this time these are all effectively in "people space" and are differentiated by domain and codomain.
My previously created external diagrams for these exercises are reproduced below.
So how are these representations connected?
My first observation is that \(P = M \cup F\text{.}\) In this universe where everyone is either male or female, we can think of the "set of all people" \(P\) as a union the "set of all males" \(M\) with the "set of all females". We’re also assuming implicitly that no one belongs to more than one gender so we also know \(M \cap F = \emptyset\text{.}\)
Effectively, the map \(g: P \rightarrow G\) provides a "sorting" of the set of people into these two groups. For every \(p \in P\text{,}\) \(g p = \text{male} \iff p \in M\) and \(g p = \text{female} \iff p \in F\text{.}\)
In Session 12 Exercise 3, we needed two relations to preserve structure. We needed \(g \circ m_P = m_G \circ g\) to assert "the gender of my mother is the gender of my mother’s mother" and \(g \circ f_P = m_G \circ g\) to assert "the gender of my father is the gender of my father’s father".
In Article 3 Exercise 17, we needed a total of 4 relations to preserve structure between these "gender-like" structures: \(f_M \circ \phi_0 = \phi_1 \circ f_M\text{,}\) \(f_F \circ \mu_0 = \mu_1 \circ f_F\text{,}\) \(f_F \circ \mu_0' = \mu_1' \circ f_M\text{,}\) and \(f_M \circ \phi_0' = \phi_1' \circ f_F\text{.}\) However, some of these maps are essentially the same if we define them over \(P\) instead of \(M\) and \(F\) using a generalized "father map" \(\phi_P: P \rightarrow P\) and "mother map" \(\mu_P: P \rightarrow P\text{.}\)
Specifically, for any \(p \in P\) we can define \(\phi(p)\) such that \(\phi_P(p) = \phi_0(p)\) for \(p\) such that \(g p = \text{male}\) and \(\phi_P(p) = \phi_0'(p)\) for \(p\) such that \(g p = \text{female}\text{.}\) Likewise, we can define \(\mu_P(p)\) such that \(\mu_P(p) = \mu_0(p)\) for \(p\) such that \(g p = \text{female}\) and \(\mu_P(p) = \mu_0'(p)\) for \(p\) such that \(g p = \text{male}\text{.}\)
Furthermore, the distinction between the pairs \(\phi_0,\phi_1\) and \(\mu_0,\mu_1\) is basically an illusion because \(M_0\) and \(M_1\) are both really just \(M\) and \(F_0,F_1\) are really \(F\text{.}\) The only reason we needed this distinction to begin with was to enforce the condition that "no person is their own parent". Equivalently, \(\forall p \in P: \phi p \neq p\) and \(\forall p \in P: \mu p \neq p\text{.}\) These are both maps with no fixed point.
My hunch is that the authors are trying to reinforce the idea that the existence of fixed points provides a form of parity on maps in a way that behaves similarly to this gender object. Let’s return to the commutative diagram we were given at the start of the exercise:
I think this is actually close to one of the ideas I was playing around with in Session 15 where I was using a non-structure preserving map from \(X \rightarrow \mathbb{N}\) as an itermediate step to finding the structure preserving maps from \(X^{\circlearrowright \alpha} \rightarrow \mathbb{N}^{\circlearrowright \sigma}\text{.}\)
In this diagram, it looks like \(f\) is really an endomap on \(X\) with \(X'\) denoting the "image" of \(X\) under the map \(f\text{.}\) This basically allows us to frame a question about whether or not the the structure of \(X^{\circlearrowright f}\) is preserved by the map \(s: X \rightarrow G\text{.}\)
For each pairing of an element \(x \in X\) and map \(f: X \rightarrow X\text{,}\) we can "ask" if \(f x = x\text{.}\) For example, assign \(s_f x = 0\) if \(f x = x\) and \(s_f x = 1\) if \(f x \neq x\text{.}\) In the same way that we split \(P = M \cup F\text{,}\) we could then split \(X\) into the union of \(X_{f,0} \cup X_{f,1}\) where \(x \in X_{f,0} \iff s_f x = 0\) and \(x \in X_{f,1} \iff s x = 1\text{.}\)
I think this brings me to the point in Session 15 where I was confused about how to handle the "null map" and "identity map". For the "identity map", every point is fixed so \(s_{1_X} x = 1\) for every \(x \in X\text{.}\) In contrast, the "null map" doesn’t have any points so how could it possibly have a fixed point? It seems reasonable to assume \(s_{\emptyset} = 0\text{.}\) At the same time, the fact that the "null map" returns the empty set after being given the empty set also makes \(s_{\emptyset} = 1\) seem reasonable as well — but that’s if and only if we treat the empty set as a object itself.
Let’s take another step back to that commutative diagram. We essentially have two paths we could follow to get from \(X\) to \(G\text{.}\) We can either apply \(s\) alone or through the composition \(s' \circ f\text{.}\) This allows use to ask the question: does \(s = s' \circ f\text{?}\)
If the answer is "yes" then \(\forall x \in X: s(x) = (s' \circ f)(x)\text{.}\) If the answer is "no" then we must have some counterexample. Namely, \(\exists x \in X: s(x) \neq (s' \circ f)(x)\text{.}\) However, we’ve defined \(s\) in such a way that either \(s(x) = 0\) or \(s(x) = 1\text{.}\) Since there’s only two choices, that means that \((s' \circ f)(x) = (\alpha \circ s)(x)\) where \(\alpha\) is our "antipodal map" given by \(\alpha 0 = 1\) and \(\alpha 1 = 0\text{.}\) Since \(s\) and \(s'\) are essentially the same map, these two situtations are equivalent to the relations \(s \circ f = s\) and \(s \circ f = \alpha \circ s\text{.}\)
What happens if we know \(s \circ f = f \circ s\text{?}\) The relation \(s \circ f = s = f \circ s\) would seem to imply \(f = 1_X\text{.}\) The other relation gives us \(s \circ f = \alpha \circ s = f \circ s\text{.}\) Given that \(\alpha^2 = 1_X\text{,}\) it would follow that \(\alpha \circ s \circ f = \alpha \circ \alpha \circ s = s = \alpha \circ f \circ s\text{.}\) This basically guarantees that either \(s \circ f = s = f \circ s\) or \(\alpha \circ s \circ f = s = \alpha \circ f \circ s\text{.}\)
I’m not really sure where I’m going with this anymore, but I’m reminded of two diagrams I drew while exploring Session 15. It seemed that there were two "natural" maps from \(X^{\circlearrowright \alpha} \rightarrow \mathbb{N}\) while I’ll present side-by-side for comparison:
I’m beginning to wonder if one of the things I was missing here was that the object \(\mathbb{N}^{\circlearrowright \sigma}\) is effectively the "antipodal map" of the object defined as \(\Omega\) with the structure \(\boxed{\circlearrowright 0 \leftarrow 1
\leftarrow 2 \leftarrow 3 \leftarrow ... \infty \circlearrowright}\text{.}\)
The reason this seems relevant now is that when I iterate the map by applying \(\alpha\) a second time, the numbers in both of these diagrams either "shrink" or "stabilize" — like they do the object \(\Omega\text{.}\) If I define \(X' = \text{Image}_\alpha(X)\text{,}\) then these natural maps become the following:
In fact, after an additional application of \(\alpha\) both of these graphs stabilize completely into a pair of cycles:
Basically, this splitting of "people into genders" is analagous to the splitting of "endomaps into clusters". Maybe that means that my missing maps in Session 15 are somehow related to the two possible "common sections" that assign \(0\) to the possible generators. Namely, the map which assigns \(\{a,b,c,d\} \rightarrow 0\) and the map which assigns \(\{a,b,c,\alpha d\} \rightarrow 0\text{.}\)
In an effort to wrap this up, what exactly is this category \(\mathcal{C}/G\text{?}\)
Well, we know \(G\) is an object in the category \(\mathcal{C}\text{.}\) The objects of \(\mathcal{C}/G\) are described as "objects of \(\mathcal{C}\) equipped with a given \(\mathcal{C}\)-sorting \(X \xrightarrow{s} G\)", but what does that mean?
I’m thinking that if every object of \(\mathcal{C}\) is some arbitrary graph \(G_0, G_1, ..., G_n\text{,}\) we can think of the objects in \(\mathcal{C}/G\) as being an "ordered pairs" \(< G_i, s_i>\) such that \(s_i \circ G_i = G\text{.}\) Since one of our objects \(G_i\) must be the same as \(G\text{,}\) the existence of a corresponding object in \(\mathcal{C}/G\) is guaranteed by the identity map in \(\mathcal{C}\) since \(1_\mathcal{C} \circ G = G\text{.}\) Maybe it would make sense to define this ensured element as \(< G_0, s > = < G, 1_\mathcal{C}>\) where \(s\) is a "common section" formed by merging the individual sections over all the possible domains \(G_i\text{.}\)
If those are our objects of \(\mathcal{C}/G\text{,}\) what about our maps? We’re told they are "commutative triangles in \(\mathcal{C}\)". I’m thinking this means a map \(f\) in \(\mathcal{C}/G\) would need to "preserve structure" by satisfying the relation \(s G_i = s f G_i\) for every \(< G_i, s >\) pair in \(\mathcal{C}/G\text{.}\)
In order to confirm that this \(\mathcal{C}/G\) is a valid category, we still need to verify the identity and associative laws. Our identity map needs to take an object in \(\mathcal{C}/G\) and return the same object. It seems like we should get that naturally from our identity map on \(\mathcal{C}\text{,}\) leaving us with only the associative property to worry about.
Suppose \(f_0,f_1\) are both maps in \(\mathcal{C}/G\text{.}\) If \(s(G_i) = (s \circ f_1)(G_i)\) and \(s(G_i) = (s \circ f_2)(G_i)\) for every \(< G_i, s >\) pair in \(\mathcal{C}/G\text{,}\) then consider the composition of \(f_1 \circ f_0\text{.}\)
For any \(G_i\text{,}\) the expression \((s \circ (f_1 \circ f_0))(G_i)\) could be rewritten as \(((s \circ f_1) \circ f_0)(G_i)\) using the associative law from \(\mathcal{C}\text{.}\) Since we know \(s \circ f_1 = s\) and \(s \circ f_0 = s\text{,}\) we can substiute the former to get \((s \circ f_0)(G_i)\) and then subsitute the latter to get \(s(G_i)\text{.}\) It follows that \((s \circ (f_1 \circ f_0))(G_i) = s(G_i)\) for all \(G_i\text{,}\) which establishes that \(s \circ (f_1 \circ f_0) = s\) and confirms that the composition of maps preserves the desired structure.
I have a feeling that a far simpler proof of this probably exists, but we’ll call that a wrap for this week.
