For any map \(X \xrightarrow{f} Y\text{...}\)
Solution.
I don’t know if this will become relevant, but I’ve got this overwhelming urge to complete the diagram above this problem:
My intuition suggests that the existence of a retraction like \(r\) in the diagram depends whether or not \(Y \subset X\text{.}\) I think that’s conceptually related to the “dual” of this exercise.
I think there’s also an implication that the result I’m looking for follows from the “universal property of products”. This was defined over a (possibly empty) set of indices \(I\) where each \(i\) in \(I\) defines a (possibly duplicated) object \(C_i\) in the category \(\mathcal{C}\text{.}\)
Definition 5.43.2.
A product is an object \(P\) together with maps \(P \xrightarrow{p_1} C_i\) (one for each \(i\)), having the universal property:
Given any object \(X\) and any maps \(X \xrightarrow{f_i} C_i\) (one for each \(i\)), there is exactly one map \(X \xrightarrow{p} P\) such that \(p_i p = f_i\) for each \(i\) in \(I\text{.}\)
In this hypothetical category \(\mathcal{C}\) with initial object \(S\) and terminal object \(T\text{,}\) there should be unique maps \(\alpha,\beta\) where \(S \xrightarrow{\alpha} X \xrightarrow{f} Y \xrightarrow{\beta} T\) satisfy the associative property. Let’s start enumerating these objects \(C_i\) of our category: \(C_0 = S\text{,}\) \(C_1 = X\text{,}\) \(C_2 = Y\text{,}\) \(C_3 = T\text{.}\)
By the universalized definition of product, for any \(X\) there should be a unique object \(P\) and map \(X \xrightarrow{f} P\text{,}\) and maps \(P \xrightarrow{p_i} C_i\) for each \(i\) such that \(p_i f = f_i\) for every \(X \xrightarrow{f_i} C_I\text{.}\) Since this product is also an object in the category, let’s call it \(C_4 = S \times X \times Y \times T\text{.}\) What if we just need to iterate over all of these \(C_i\) to handle the cases where \(X = Y\) and \(X \neq Y\) at the same time?
Let’s start with \(X = C_0 = S\text{.}\) Any map \(X \xrightarrow{f_0} C_0\) would have to the unique map \(S \xrightarrow{1_S} S\text{.}\) Likewise, any \(X \xrightarrow{f_1} C_1\) would be the unique map \(S \rightarrow X\text{,}\) \(X \xrightarrow{f_2} C_2\) would be the unique map \(S \rightarrow Y\text{,}\) \(X \xrightarrow{f_3} C_3\) would be the unique map \(S \rightarrow T\text{,}\) \(X \xrightarrow{f_4} C_4\) would be the unique map \(S \rightarrow X \times Y\text{.}\)
Let’s move on \(X = C_1 = X\text{.}\) If any map \(X \xrightarrow{f_0} C_0\) existed, that would imply that \(X\) is itself an intial object. We could then used the uniqueness of the inverse map \(C_1 \rightarrow C_0\) to uniquely define the rest of the maps we’re looking for through precomposition with the maps we saw previously. In fact, I think we can go one step further and assume that no \(C_i \rightarrow C_0\) exists for any \(i > 0\text{.}\)
Assuming that no map \(X \rightarrow C_0\) exists, then it seems reasonable to conclude that the only way \(P \xrightarrow{p_1} X\) can satisfy \(p_1 p = f_1\) for all \(X \xrightarrow{f_1} X\) is if \(p_1 p = 1_X\) for all \(X\text{.}\) I think this partially explains the wildcard notation of \(?\) in the definition of \(\Gamma = \langle ?, f \rangle\text{.}\)
The existence of some \(X \xrightarrow{f} Y\) necessarily means we have maps \(T \xrightarrow{x_i} X \xrightarrow{f} Y \xrightarrow{\bar{y_i}} T\text{.}\) I’m thinking we could have take the product \(X\rightarrow X \times X\) that maps \(x_i\) to the pair \(\langle x_i,x_i\rangle\text{.}\) and then somehow compare this with the projection map \(X \rightarrow X \times Y\) that pairs each \(x_i\) with the respective \(y_i = f x_i\text{.}\) I’m thinking we can do this by “stepping out” somehow through a shared map \(X \rightarrow X \times (X+Y)\text{.}\)
