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

Okay my first step here is to reproduce our diagram:

While creating the diagram, there’s some obvious structural patterns. In order to draw this, I started by placing the six points in \(P = \{k,m,n,p,q,r\}\text{.}\) The other objects are the arrows, namely \(X = \{a,b,c,d,e\}\text{.}\) Our maps \(X \xrightarrow{s} P\) and \(X \xrightarrow{t} P\) represent the source and target of each arrow. For me, it seemed natural to structure this information into a table:

Based on this table, the questions become fairly straight forward. The input \(b \in X\) has the same output points for \(s\) and \(t\text{,}\) specifically \(s(b) = t(b) = m\text{.}\) There is no element \(x \in X\) with \(t(x) = k\text{.}\)

I’m thinking these sub-questions are important because they show that these arrow maps have neither "retraction" nor "sections". Unless we’re very specific about the qualities of our maps, our source and target maps will likely be non-invertable. I find arrows \(a\) and \(c\) particularly interesting because they are two different arrows with the same source \(s(a) = s(c) = k\) and same target \(t(a) = t(c) = m\text{.}\) I feel like I’m fighting a natural tendancy to assume that "having the same source and target" makes them "the same arrow" but that equivalency may not necessarily hold true in a broader category.

Let’s begin with a recap of what we know about \(f\text{:}\)

We’re told that \(f\) "preserves source and target relations", which is defined by having a pair of maps \(X \xrightarrow{f_A} Y\) and \(P \xrightarrow{f_D} Q\) satisfying \(f_D s = s' f_A\) and \(f_D t = t' f_A\text{.}\)

Consider another map \(g\) with the same property:

In order to preserve source and target relations, there would need to exist \(Y \xrightarrow{g_A} Z\) and \(Q \xrightarrow{g_D} R\) with \(g_D s' = s'' g_A\) and \(g_D t' = t''' g_A\text{.}\)

We can form a composition of maps \(g_D \circ f_D\) and \(g_A \circ f_A\) as shown below:

In order to show that the composition preserves source and target relations, we’d need to demonstrate that both \((g_D \circ f_D) s = s'' (g_A \circ f_A)\) and \((g_D \circ f_D) t = t'' (g_A \circ f_A)\text{.}\)

By the associative property, \((g_D \circ f_D) s = g_D \circ (f_D \circ s)\text{.}\) Using the substition \(f_D \circ s = s' f_D\) gives us \(g_D \circ s' \circ f_A\text{,}\) and allows us to the perform a subsitution using \(g_D \circ s' = s'' g_A\) to show \(g_D \circ s' \circ f_A = s'' \circ g_A \circ f_A\text{.}\) It follows that \((g_D \circ f_D) s = s'' (g_A \circ f_A)\text{.}\)

We can apply a similar process on \((g_D \circ f_D) t\text{.}\) The associative property gives us \(g_D \circ (f_D \circ t)\text{,}\) which allows us to use the substition \(f_D \circ t = t' f_D\) to get \(g_D \circ t' \circ f_A\text{.}\) Next we use our property that \(g_D \circ t' = t'' g_A\) to perform another subsitution resulting in \(g_D \circ t' \circ f_A = t'' \circ g_A \circ f_A\text{.}\) It follows that \((g_D \circ f_D) t = t'' (g_A \circ f_A)\text{.}\)

Having established the source and target preservation properties, we can conclude that \(g \circ f \) is an well defined \(\mathcal{S}^{\downarrow^{\bullet}_{\bullet} \downarrow}\)-map:

We want to show that \(I(g \circ f) = I(g) \circ I(f)\text{.}\) Or, "the composition of embeddings is the same as the embedding of the composition".

Our embedding of \(I(f)\) can be illustrated by the following diagram:

In order to preserve structure, the \(\mathcal{S}^{\downarrow^{\bullet}_{\bullet} \downarrow}\)-map is required to preserve the relations \(f \circ 1_X = 1_Y \circ f\) and \(f \circ \alpha = \beta \circ f\text{.}\) The former is a trivial result of the identity map, while the later is a property \(f\) inherits from being an object in \(\mathcal{S}^{\circlearrowright}\text{.}\)

Similarly, our embedding of \(I(g)\) can be illustrated as follows:

Like before, this \(\boxed{Y^{\circlearrowright \beta}} \xrightarrow{g} \boxed{Z^{\circlearrowright \gamma}}\) preserves the properties \(g \circ 1_Y = 1_Z \circ g\) and \(g \circ \beta = \gamma \circ g\text{.}\)

The embedding of our composition also should have similar structure:

This needs to satisfy \(g \circ f \circ 1_X = 1_Z \circ g \circ f\) and \(g \circ f \circ \alpha = \gamma \circ g \circ f\text{.}\) The former is again a trivial result of the identity map. The latter is implied by the fact that since \(\boxed{X^{\circlearrowright \alpha}} \xrightarrow{f} \boxed{Y^{\circlearrowright \beta}}\) and \(\boxed{Y^{\circlearrowright \beta}} \xrightarrow{g} \boxed{Z^{\circlearrowright \gamma}}\) are both objects in \(\mathcal{S}^{\circlearrowright}\) that we must also have \(g \circ f \in \mathcal{S}^{\circlearrowright}\text{.}\) This \(\boxed{X^{\circlearrowright \alpha}} \xrightarrow{g \circ f} \boxed{Z^{\circlearrowright \gamma}}\) establishes the relation \(g \circ f \circ \alpha = \gamma \circ g \circ f\text{.}\)

Finally, let’s compose these embeddings together and verify everything still holds:

It’s easy to see how the embedding illustrations above are composed together together to produce an embedding of the compositions. The identity property ensures that \(g \circ f \circ 1_X = g \circ 1_Y \circ f = 1_Z \circ g \circ f\text{.}\) The structure inherited from the endomaps ensures that \(g \circ f \circ \alpha = g \circ \beta \circ f = \gamma g \circ f\text{.}\)

Having shown that these properties are preserved, I think we’ve sufficiently demonstrated that \(I(g \circ f) = I(g) \circ I(f)\text{.}\)

We’re given an \(\mathcal{S}^{\downarrow^{\bullet}_{\bullet} \downarrow}\)-morphism that comes via the insertion \(I\) that we used previously:

The source preservation property of \(f\) implies that \(f_D \circ 1_X = 1_Y \circ f_A\text{,}\) but since \(1_X\) and \(1_Y\) are identity maps we can just say \(f_D = f_A = f\text{.}\) Since the target preservation property says \(f_D \circ \alpha = \beta \circ f_A\text{,}\) we can substitute \(f\) to get \(f \circ \alpha = \beta \circ f\text{.}\) This property establishes that \(f\) exists and is a valid \(\mathcal{S}^{\circlearrowright}\)-morphism: \(\boxed{X^{\circlearrowright \alpha}} \xrightarrow{f} \boxed{Y^{\circlearrowright \beta}}\) .

To be honest, I wasn’t quite sure how to start this one at first so I just started exploring small endomaps. I explored too many to review them all, but in the process I learned some things.

I think I had conceptually confused \(f_D\) and \(f_A\) at first. I was under the impression that \(f_A\) "produced an arrow" and \(f_D\) "produced a dot", but a closer reading of the book stated that \(f_A\) "operates on arrows" and \(f_D\) "operates on dots". Eventually I made a mental shift to the Spanish \(antes\) (before) and \(despues\) (after) because it helped me remember the relationship with the endomaps.

I found tables to be a nice way of organizing all the information about my "arrows" and "dots". Having my information organized reminded me of Exercise 10. Looking back over my notes, I had noticed earlier that the diagram included a pair of two different arrows with the same source and destination. I took this as a hint.

I also found myself repeatedly going back to the map in Example 9 as a reference. There was something about the way the endomap removes an element that seemed like it might be important. I labeled the arrows in addition to the points to help keep track of what I was working with:

I had a hunch that \(\alpha^3=\alpha\) might help establish that \(f_D \circ \alpha = \beta \circ f_A\text{.}\) My idea revolved around the idea that if I could somehow make \(f_D\text{,}\) \(f_A\text{,}\) and \(\beta\) by composing \(\alpha\) with itself multiple times that I could set up a situation where they balance out.

I think I also was confused about the inclusion \(J\) and the property that the "source and target structures are the same map". I wasn’t sure if I needed to enforce this restriction on that all arrows in my endomaps were loops or not. I started to wonder how much of this ambiguity was intentional, and if looser of a definition of "loop" might be in order.

A flash of insight came as I was trying to pick out a cat photo for the week and see Alphonse nuzzling Konshu:

I started to think about Moon Knight and the relation between the depiction of Konshu in the comic books versus the mythological depiction of the same character. I wondered how my mental model of Konshu might vary depending on which version I encountered first. To appease my "fussy professor", I’m going to define my set \(X\) to be the set of "Konshu Models" \(\{\text{Comic Konshu},\text{Myth Konshu},\text{My Konshu}\}\) and endomap \(\alpha\) to be the "influences" acting on those models.

I reasoned that the Moon Knight from the comics used the mythology around Konshu to develop the character, but as more people read the comic they start to project qualities from the comic character back to the deity. Whether those projections align with the mythology will impact the design of later comics, forming a sort of feedback loop. Prior to seeing the Netflix show, my model of Konshu isn’t really impacted by either — I’m just off in my own little world.

Now let’s denote \(f\) to be a map describing how my metal model might change after exposure to new ideas. Maybe seeing the Netflix show prompts me to read a bunch of comics or mythology. In either case, it forms a feedback loop between my model of the comic Konshu and my model of the myth Konshu. Let’s call this our set \(Y\) and endomap \(\beta\text{:}\)

Now let’s consider or maps \(f_A\) and \(f_D\text{.}\) Let’s say that \(f_A\) is the case where I learn about Konshu mythos before watching the Netflix show and \(f_D\) to be the case where I’m exposed to the mythology after. For the sake of convenience, I’m going to relabel the points in the \(X\) as \(\{a,b,c\}\) and the points in \(Y\) as \(\{0,1\}\text{:}\)

Since \(f_A(a) = 0\) and \(f_D(a) = 1\text{,}\) clearly \(f_D \neq f_A\text{.}\) However, we can also show that \(f_D \circ \alpha = \beta \circ f_A\) by evaluting them over the entire codomain:

In the table above, notice that the columns for \(\beta \circ f_A\) and \(f_D \circ \alpha\) are equivalent for all three points. Thus, we have \(f_D \circ \alpha = \beta \circ f_A\) with \(f_A \neq f_D\text{.}\)

I’m thinking that the moral of the story here is these endomaps contain more than self-loops, but our insertion \(J\) fails to enforce that condition. If the endomap contains only self loops it has an insertion that makes a commutative diagram, but just because we have a commutative diagram like this doesn’t mean it came from that insertion.