Both parts of the distributive laws...
Solution.
First, we know that the initial object \(\mathbf{0}\) is not an empty set. In \(\mathbf{1} / \mathcal{S}\text{,}\) we need to pair the object \(\mathbf{0}\) with a map \(\mathbf{1} \xrightarrow{\mathbf{0}_0} \mathbf{0}\) to the “distinguished point”. Let’s depict this object visually as \(\mathbf{0} = \boxed{\star}\text{.}\)
Similary, the terminal object \(\mathbf{1}\) is also a set with a single distinguished point \(\boxed{\star}\text{.}\) Since there’s only one object in \(\mathbf{1}\text{,}\) the map to this distinquished point would technically the identity map \(\mathbf{1} \xrightarrow{1_\mathbf{1}} \mathbf{1}\text{.}\)
Since \(\mathbf{0} = \mathbf{1}\text{,}\) the products defined by \(\mathbf{0} \times \mathbf{1}\text{,}\) \(\mathbf{1} \times \mathbf{0}\text{,}\) \(\mathbf{0} \times \mathbf{0}\text{,}\) and \(\mathbf{1} \times \mathbf{1}\) would all need to be isomorphic to \(\mathbf{0} = \mathbf{1}\) as depicted by the following diagram:
Since we’ve defined sums in this category by “gluing together at the base point”, it also means that the sums \(\mathbf{0} + \mathbf{1}\text{,}\) \(\mathbf{1} + \mathbf{0}\text{,}\) \(\mathbf{0} + \mathbf{0}\text{,}\) and \(\mathbf{1} + \mathbf{1}\) would also all need to be isomorphic to \(\mathbf{0} = \mathbf{1}\text{:}\)
So when happens when there’s some object \(X\) in my category which is neither initial nor terminal? It seems reasonable that simpliest such object would be a set with both a disinguished and non-distinuguished point such as \(A = \boxed{\star ~ ~ \bullet}\text{.}\) Since we have a unique map \(\mathbf{0} \rightarrow A\) by the definition of initial object, and a unique map \(A \rightarrow \mathbf{1}\) by the definition of terminal object, the associative property of the category says that we should be able to compose these maps around the unique identity map \(1_A\text{.}\) The composition \(\mathbf{0} \rightarrow A \xrightarrow{1_A}
A \rightarrow \mathbf{1}\) should look something like this:
Suppose we have some other endomap \(A \xrightarrow{e} A\) such that \(e \neq 1_{A}\text{.}\) Since every map \(X \xrightarrow{f} Y\) in \(\mathbf{1} / \mathcal{S}\) must preserve the structure of the distinguished point such that \(f x_0 = y_0\) for the maps \(\mathbf{1} \xrightarrow{x_0} X\) and \(\mathbf{1} \xrightarrow{y_0} Y\text{.}\) Since in our case \(X = A = Y\text{,}\) that means our \(e\) must satisfy \(e a_0 = a_0\) for the distinguished map \(\mathbf{1} \xrightarrow{a_0} A\text{.}\) Thus, the condition that \(e \neq 1_A\) implies that the only other point in \(A\text{,}\) let’s call it \(a_1\text{,}\) must be mapped to something other than itself, which necessarily means that \(e a_1 = a_0\text{.}\) This map \(e\) would need to be the standard idempotent \(\boxed{
\circlearrowright \star \leftarrow \bullet}\)
So, what happens when we take the product \(\mathbf{0} \times A\text{?}\) Pictorally it should resemble the following:
The big question here is whether or not the map \(\mathbf{0} \xrightarrow{p} \mathbf{0} \times A\) is an “isomorphism”. By definition, \(p\) is an isomorphism if and only if there exists a map \(\mathbf{0} \times A \xrightarrow{p^{-1}} \mathbf{0}\) satisfying \(p^{-1} p = 1_\mathbf{0}\) and \(p p^{-1} = 1_\mathbf{0} \times A \text{.}\)
Clearly, the projection map \(\mathbf{0} \times A \xrightarrow{p_1} \mathbf{0}\) satisfies the properties of a “section”. The composition \(p_1 p\) has domain and codomain \(\mathbf{0} \rightarrow \mathbf{0}\) and would therefore need to be the unique map \(1_\mathbf{0}\text{.}\) If a retraction \(p^{-1}\) exists, it must be equivalent to the section. Does the composition defined by \(\mathbf{0} \times A \xrightarrow{p p_1} \mathbf{0} \times A\) result in the identity map \(1_{\mathbf{0} \times A}\text{?}\) My gut tells me no, but its not exactly clear why.
There must exist a pair of maps \(\mathbf{1} \mathrel{\substack{
\langle \star, \star \rangle \\
\longrightarrow \\ \longrightarrow \\
\langle \star, \bullet \rangle
}} \mathbf{0} \times A\) to the two points in the product. Post composing by \(p_2\) would then give us a pair of maps to \(A\) defined by \(\mathbf{1} \mathrel{\substack{
p_2 \langle \star, \star \rangle \\
\longrightarrow \\ \longrightarrow \\
p_2 \langle \star, \bullet \rangle
}} A\text{.}\) However, as a maps in \(\mathbf{1} / \mathcal{S}\text{,}\) these maps must satisfy both send the distinguished point in \(\mathbf{1}\) to the distinguished point in \(A\text{.}\) This feels like a contradiction in terms. Perhaps I’m confusing myself by referring to the distinuguished objects in \(\mathbf{0}\) and \(\mathbf{1}\) by the same symbol?
Let’s draw it out again, but this time add the “sum” \(\mathbf{0}+\mathbf{0} \times A +A\) into the picture as well:
What I find interesting about this is the fact that \(A \times A\) would need to have 3 non-distinguished points instead of just two:
This is interesting because \(A+A\) gets two non-distinguished points. Maybe I wasn’t far off when I originally considered the map \(\mathbf{0} \rightarrow \mathbf{0} \times A\) to behave like “monoid”? What if the solution I’m really looking for here is the object \(A = \mathbb{N}\text{?}\)
Consider this. For any map \(X \xrightarrow{f} Y\) in \(\mathbf{1} / \mathcal{S}\text{,}\) we can think of \(X=C_i\) and \(Y=C_j\) for some “preferred” indexing of the objects in the category. If there is further some indexing of objects in \(X\) such that a map \(\mathbb{N} \xrightarrow{p} X\) enumerates the elements of \(x\text{,}\) we can pair every point in every object with a pair of indexes in \(\mathbb{N} \times \mathbb{N}\) such that one projection map points to the index of the object \(C_i\) in the category and the other returns the index \(p\) of a particular map \(\mathbf{1} \xrightarrow{x_p} X\text{.}\)
What makes \(\mathbb{N}\) special here is that theres a unique isomorphism \(\mathbb{N} \leftrightarrows \mathbb{N} \times \mathbb{N}\) in \(\mathcal{S}\) that we previously saw in Section 5.36. I also remember the authors establishing that it does \(not\) preserve the structure of \(\mathbb{N}^{\circlearrowright \sigma}\text{.}\)
Likewise, I think the same is true of the unique isomorphism \(\mathbb{N} \leftrightarrows \mathbb{N}+\mathbb{N}\text{.}\) Maybe we can those projections and injections to construct an isomporphism \(\mathbb{N} \times \mathbb{N} \leftrightarrows \mathbb{N}+\mathbb{N}\) that matches up with our “identity matrix” some how?
I also find it curious that the fundamental theorem of arithmetic hasn’t come into play yet. It seems like that further gives me maps \(\mathbb{N} \rightarrow \mathbb{N}^m\) for arbitrarily large \(m\) by using the uniqueness of the prime-factorization of numbers. It seems like would behave similarly to \(\mathbf{1}/\mathcal{S}\) in that the objects \(0,1\) would require special treatment.
