Say that \(Y \xrightarrow{h} Z\) is cosolution...
Solution.
We’re defining a cosolution \(Y \xrightarrow{h} Z\) such that \(h f = h g\) for a pair of parallel maps \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\text{.}\) We want to show that if this cosolution is “universal”, any other cosolution \(Y \xrightarrow{h'} Z'\) can be uniquely expressed as \(h' = q h\text{.}\)
Well, I think the first step would be figuring out what this \(q\) is. Based on the equation \(h' = q h\text{,}\) we can infer that the domain and codomain are \(Z \xrightarrow{q} Z'\text{.}\)
The choice of \(q\) also seems like a successor reference to the \(p\) in the definition of equalizer. My intuition suggests that the “universality” of our cosolution implies that the map \(q\) is the unique isomorphism such that \(q^{-1} q = 1_Z\) and \(q q^{-1} = 1_{Z'}\text{.}\)
We want to show that \(h\) is an “epimorphism”. This was defined back in Article II as a map \(Y \xrightarrow{h} Z\) such that for any object \(T\) and pair of maps \(Z \mathrel{\substack{t_1 \\ \longrightarrow \\ \longrightarrow \\ t_2}} T\text{,}\) if \(t_1 h = t_2 h\) then \(t_1 = t_2\text{.}\)
Having laid out that information, what if we assume \(h\) is not an “epimorphism”? The statement \(t_1 h = t_2 h \implies t_1 = t_2\) is only false when we have two maps \(Z \mathrel{\substack{t_1 \\ \longrightarrow \\ \longrightarrow \\ t_2}} T\) satisfying \(t_1 h = t_2 h\) but with \(t_1 \neq t_2\text{.}\) However, if \(T\) is a terminal object then there’s precisely one map \(Z \rightarrow T\) so no such \(t_1, t_2\) can exist. It follows that our assumption was false and \(h\) is an epimorphism.
That seems so simple I’m not even sure if I trust it. I feel like I need to more explicitly connect the map \(Z \xrightarrow{q} Z'\) to this somehow. It seems completely reasonable that if \(Z\) and \(Z'\) are both terminal objects they’d have to be isomorphic, but I’m not really sure I’ve established either of them in such a way. Nor did I reference the original maps \(f,g\text{.}\) I was also expecting a need to invoke the sum \(X+Y\text{.}\)
Before I wrap up this exercise, let’s also try to address the parenthetical. I think it would be good to attempt a formal definition of co-equalizer by swapping domain and codomains around in the definition of equalizer:
Definition 5.46.2.
\(Y \xrightarrow{q} E\) is a co-equalizer of \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\) if \(q f = q g\) and for each \(Y \xrightarrow{\bar{y}} T\) for which \(\bar{y} f = \bar{y} g\text{,}\) there is exactly one \(E \xrightarrow{\bar{h}} T\) for which \(\bar{y} = \bar{h} q\text{.}\)
I’m not really sure if that’s correct. In naively swapping all the arrows like I did, I ended up naming maps \(\bar{h},\bar{y}\) that are uniquely defined by domain and codomain. The idea of using a “for each” statement over such uniquely defined maps seems like a contradiction in terms. Maybe I need to turn my equation around and express it in terms of a unique \(T \xrightarrow{h} E\) satisfying \(q y = h\) for all \(T \xrightarrow{y} Y\text{?}\)
