Consider the diagram of graphs...
Solution.
It look’s like the reason why I might have been getting confused was that my commutative diagram was incorrect. I think part of the confusion is that I’m confusing the objects \(X \xrightarrow{p} P,P \xrightarrow{p_1} B_1,P \xrightarrow{p_2} B_2\) in \(X\) with the maps to the terminal object given by \(P \rightarrow T,B_1 \rightarrow T,B_2 \rightarrow T\) so I’m not really sure if my maps \(m_1,m_2\) are the same as the projection maps \(p_1,p_2\) or not. Perhaps it will be easier if I name my three “dots” \(D \xrightarrow{d_0} P, D \xrightarrow{d_1} B_1, D \xrightarrow{d_2} B_2\) relative to the respective terminal maps \(P \xrightarrow{\bar{d_0}} T,B_1 \xrightarrow{\bar{d_1}} T,B_2 \xrightarrow{\bar{d_2}} T\text{.}\) These should form a unique commutive diagram as follows:
The maps \(m_1,m_2\) are defined by being the only two maps from \(A \rightarrow P\) that preserve source and target relations. The fact that these arrows share a common source can be expressed through the commutative diagram below:
Suppose we have two arbitrary arrows \(A \xrightarrow{x_1} X,A \xrightarrow{x_2} X\) with the same source. We can express that fact as \(x_1 s = x_2 s\) or saying the following diagram commutes:
If \(x_1 \neq x_2\text{,}\) the fact that these arrows share a source implies the targets of these arrows must be different. Having a map \(X \rightarrow P\) in \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\) gives us a pair of maps \(s',t'\) such that \(x s = s' x\) and \(x t = t' x\text{.}\) Knowing that \(x_1 t \neq x_2 t\) means the set \(\{x_1 t, x_2 t\}\) is isomorphic to the shape \(\mathbf{2}\text{.}\) Since the set \(\{B_1,B_2\}\) is also isomorphic to \(\mathbf{2}\text{,}\) there’s a guaranteed isomorphism between them. Since we have shown \(D \rightarrow P\) has the form of a product, we can these maps together as follows:
I suppose I should also note some how that these maps might “reduce” but the same basic structure should still hold.
