Continued from last week...
Solution.
So I think I made things unnecessarily difficult last time with how I chose to enumerate the objects of the category. I’m starting to think it would be simpler if I define \(C_0 = S\) and \(C_1 = T\text{.}\) If our category has a product, then there’s a unique object \(C_2 = S \times T\) such that the diagram below would commute.
Since \(S\) is initial there’s a unique map \(S \xrightarrow{p} S \times T\text{.}\) That means the compositions with our projection maps would need to be the unique maps \(p_1 p = S \rightarrow S\) and \(p_2 p = S \rightarrow T\text{.}\) It seems to follow that the product of an initial object with any object would also need to be initial itself.
Now let’s allow \(X \xrightarrow{f} Y\) to be an arbitrary map in the category. If we define \(\Gamma = X \xrightarrow{\langle ?,f \rangle} X \times Y\) to map each \(1 \xrightarrow{x_i} X\) in \(X\) to the respective pair \(\langle x_i, f x_i \rangle\text{,}\) then clearly this definition implies both both \(p_X \Gamma = 1_X\) and \(p_Y \Gamma = f\text{.}\)
Since \(p_X \Gamma = 1_X\text{,}\) that means \(\Gamma\) is a section for \(p_X\text{.}\) Maybe we just need to prove its unique?
Suppose some other map \(X \xrightarrow{g} X \times Y\) satisfied \(p_X g = 1_X\) and \(p_Y g = f\text{.}\) The condition that \(\Gamma \neq g\) implies the existence of a point \(T \xrightarrow{x} X\) such that \(\Gamma x \neq g x\text{.}\) Maybe we can try precomposing by \(S \rightarrow T\text{?}\) That would give us two separate maps \(S \rightarrow Y\) which is supposed to be uniquely defined. This contradiction means our assumption is false and \(\Gamma\) is unique.
I’m wondering if the main idea here is that even if \(X \xrightarrow{f} Y\) is not invertible, the map defined here as \(X \xrightarrow{\Gamma} X \times Y\) must be. For \(f^{-1}\) to not exist, there must be some \(T \xrightarrow{y} Y\) with two different maps \(T \xrightarrow{x_1} X\) and \(T \xrightarrow{x_2} X\) satisfying \(y = f x_1 = f x_2\text{.}\) Even if \(f\) isn’t one-to-one, we come up with a map that is by treating the points \(\langle x_1, y \rangle\) and \(\langle x_2, y \rangle\) as two distinct values.
