Continued from Session 23 Part 2
Solution.
I think I want to start here by organizing the information I have so far (or at least what I think I have).
| Category | Initial Object |
| \(\mathcal{C}\) | \(I\) |
| \(\mathcal{S}\) | \(\emptyset\) |
| \(\mathcal{S}^\circlearrowright\) | \(\mathbb{N}^{()+1}\text{?}\) |
| \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\) | ? |
If I’m right about \(\mathbb{N}^{()+1}\) being the initial object in \(\mathcal{S}^\circlearrowright\text{,}\) maybe the initial object in \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\) will somehow pair each arrow with its opposite. The authors have previously discussed an object \(\Omega\) which acts like the inverse to \(\mathbb{N}^{()+1}\text{.}\) This would take a graph like \(\boxed{\bullet \rightarrow \bullet \circlearrowleft}\) and pair it with the graph \(\boxed{\bullet \leftarrow \bullet \circlearrowleft}\text{.}\) This second graph is important because while it can exist in \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\text{,}\) it does not exist in \(\mathcal{S}^\circlearrowright\) because an endomap can’t have two arrows originating at the same point.
In the category \(S\) we already saw that our definition of product means that \(\emptyset \rightarrow \emptyset+\mathbf{1}\) and \(\mathbf{1} \rightarrow \emptyset+\mathbf{1}\) are two uniquely defined maps. If we think of the sum as “put together with no overlap and no interaction”, combining the empty set with anything doesn’t change it giving us a unique map defined by \(\emptyset+\mathbf{1} \rightarrow \mathbf{1}\text{.}\) Combining something with a new object necessarily would have to change the set, so we’d need to have \(\mathbf{1} + \mathbf{1} = \mathbf{2}\text{.}\) This effectively establishes a bijection between \(\mathbb{N}\) and sets of the respective sizes: \(\{\emptyset,\mathbf{1},\mathbf{2},...\}\text{.}\)
In the category \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\) our terminal objects aren’t points but loops. However, there should be an isomorphism between the set of \(n\) points and a cycle of that length. If we think of \(\mathbb{N}\) as a “cycle of length 0”, then \(C_0 = \mathbb{N}\) allows use to form an isomorphism \(\mathbb{N} \leftrightarrows \{\emptyset,\mathbf{1},\mathbf{2},...\}
\leftrightarrows
\{\mathbb{N},C_1,C_2,...\}
\text{.}\)
While experimenting with Lean 4 the other week, I noticed that the category theory library contained separate definitions for “small” and “large” categories. The documentation suggested that the difference was that a “large category” could contain itself as an object. I’m starting to wonder if that property is related to role of \(\mathbb{N}\) in this bijection \(\mathbb{N}
\leftrightarrows
\{\mathbb{N},C_1,C_2,...\}
\text{.}\)
In \(\mathcal{S}^\circlearrowright\text{,}\) we observed that there were \(n\) possible structure preserving maps from \(C_n \rightarrow C_n\text{.}\) Maybe we can use the notion of “initial object” to narrow that down to one. Given any cycle \(C_n\text{,}\) we have a unique map \(I \rightarrow C_n\) that we could use to identify which element is “point 0”.
Consider the cycle of two. There are precisely two \(\mathcal{S}^\circlearrowright\)-maps \(C_2 \rightarrow C_2\text{.}\) If we combine the cycle with the initial object, we get a unique map \(I \rightarrow C_2+I\) and through composition with the unique map \(I \rightarrow C_2\) can construct a unique map \(C_2 \rightarrow C_2+I\text{.}\) However, this operation \(+I\) can’t preserve the structure of \(\mathbb{N}^{()+1}\) because there is no map \(C_2 \rightarrow C_3\) in \(\mathcal{S}^\circlearrowright\text{.}\) It would follow that any composition \(\mathbb{N} \rightarrow I \rightarrow C_n\) would have to be unique.
In an attempt to put this all together, I’m thinking that an my objects in \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\) should have a general structure that looks like this:
If the “cycle” has a length \(n\) and the tail has length \(m\) then there’s precisely one map from tail of length \(m+n\) to this object. This is consistent with my earlier hypothesis that our initial object in \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\) is actually just \(\mathbb{N}\) itself.
