Continued from last week...
Solution.
Let’s start by taking a “step back” and look what we’re really assuming here. We’re given a map \(Y \xrightarrow{h} Z\) and asked to consider “all parallel pairs” \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\text{.}\)
Looking back over the reading for this session, the authors are incredibly strict about getting Danilo to say “For each object \(X\) in \(\mathcal{C}\) there is exactly one map...”. I think I was on the right track earlier when I started referring to the objects of the category by index.
Let’s assume our initial object of the category is \(S\) and terminal object is \(T\text{.}\) We’ll call \(S = C_0\) and \(T = C_1\text{.}\) Our triple of maps \(f,g,h\) necessitates the existence of three objects \(X,Y,Z\text{,}\) which may or may not be isomorphic to the objects we’ve already defined.
By the definition of initial object, there there is “exactly one map” from \(C_0 \xrightarrow{f_i} C_i\) for every \(C_i\) in \(\mathcal{C}\text{.}\) Clearly our category must have a unique map \(C_0 \xrightarrow{f_0} C_0 = 1_S\) by the “identity property” of the category \(\mathcal{C}\) and definition of initial object. The map \(C_0 \xrightarrow{f_1} C_1\) would be uniquely defined by domain and codomain also.
Similarly, by the definition of terminal object there’s uniquely defined maps \(C_i \xrightarrow{g_i} C_1\) for every object in the category. It’s therefore safe to conclude that \(f_1 = g_0\) need to both be the unique map \(S \rightarrow T\text{.}\) If \(f_1\) is invertible, it would imply every object is isomorphic to the initial object.
Consider an object \(C_2\) defined by the disjoint union of \(C_0+C_1\text{.}\) By the definition of coproduct, for any object \(C_i\) in the category \(\mathcal{C}\) there should be a unique map \(C_0+C_1 \xrightarrow{j} C_2\) such that the following diagram commutes:
By assigning our initial objects indexes such that \(C_0\) is initial and \(C_1\) is terminal, we’ve basically set us up to say that \(f_i\) is non-invertible for every \(i\) with the exception of \(f_0\text{.}\) Likewise, very \(g_i\) needs to be invertible for every \(i\) with the execption of \(g_0\text{.}\) Generalizing this notion, we could simply say that any object \(X\) in the category \(\mathcal{C}\) has a uniquely defined set of 3 maps \(C_0 \xrightarrow{f_x} X \xrightarrow{g_x} C_1 \xrightarrow{\bar{g_x}} X\) such that \(\bar{g_x} g_x = 1_X\text{.}\)
How convenient? The wild-card \(X\) is effectively a collection of maps isomorphic to \(\mathbf{3}\) and the last object we named was \(C_2\text{.}\) It’s almost like \(X\) wants to be \(C_3\text{,}\) but is it? Given some \(Y \neq X\text{,}\) there would need to some point in \(Y\) which doesn’t come from \(X\text{.}\) We could join that point and keep counting to construct a monoid \(\mathbb{N}\text{.}\)
I think that actually gives me an idea about how to answer to the first part of this. Maybe the unique map \(h\) is the retraction that returns the index of the object \(C_i\text{?}\)
Consider a map \(\mathbb{N} \xrightarrow{i} X\) that indexes every \(\mathbf{1} \xrightarrow{x_i} X\text{.}\) This map defines a unique inverse map \(X \xrightarrow{r_X} \mathbb{N}\) that returns the index given the element. Similar maps exist for the objects \(Y\) and \(Z\) , but there could be at most one map or \(Z \xrightarrow{r_Z} \mathbb{N}\) that will preserve the indexes from \(X\) for all \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y \xrightarrow{h} Z
\text{.}\) We could then compose \(i r_Z\) to get a unique map \(Z \rightarrow X\text{.}\)
I think that’s basically the requirements of our “directed graph”? Choosing \(X_h = \mathbb{N}\) to preserve the index of the elements of \(X\) gives us a unique retraction \(r\) that preserves the structure of every source and target map \(\mathbb{N} \xrightarrow{i} X \mathrel{\substack{s \\ \longrightarrow \\ \longrightarrow \\ t}} Y \xrightarrow{r} \mathbb{N}
\text{.}\) This establishes our “reflexive” property.
