Solution.
Some of my key take-aways from the reading this week were that:
- not every \(\mathbf{1}/\mathcal{S}\) is invertible (as I had falsely assumed earlier)
- “the empty set is not an object of this category”
- values which are “undefined” get sent to the distiguished point
I think there’s a close connection between this category \(\mathbf{1}/\mathcal{S}\) as a form of “trinary logic”. In Section 5.13, I examined this idea with respect to the existence of a “gender map”. We can make a “binary” question into a “trinary” one by adding an option for “undefined”.
I’m also wondering if this is connected to what I was doing in Section 5.47. We can interpret a pair of maps \(X \mathrel{\substack{\longrightarrow \\ \longrightarrow}}
Y\) as inducing an endomap \(Y^{\circlearrowright \beta}\) and it’s inverse \(\beta^{-1}\text{.}\) However, just because the endomap is invertable doesn’t mean the original maps are. By adding in a “distinguished point”, I’m thinking that will introduce a way to tell if \(X\) and \(Y\) are the same object or not, because it gives me a place to send the point \(y_9\) in my example that did not come from any \(\mathbf{1} \xrightarrow{x_i} X\text{.}\)
I think the diagrams from Exercise 20 are still relevant, but I want to have one more as a reference:
I’m also thinking I’ll need an updated “universal sum” definition, so let’s attempt that also:
Definition 5.50.3.
A sum of an indexed family is an object \(Q\) together with maps \(C_i \xrightarrow{q_i} Q\) (one for each \(i\)) having the universal property:
Given any object \(Y\) and maps \(C_i \xrightarrow{g_i} Y\text{,}\) there is exactly one map \(Q \xrightarrow{g} Y\) such that all triangles below commute, or \(g q_i = g_i\) for all \(i\text{.}\)
