Parallel maps continued...
Solution.
I think this time I want to approach this more visually. After all, the question explicitly talks about the graph reprepresentation. Let’s start with another simple example of parallel maps \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\text{.}\)
I tried to make sure these maps were neither onto nor one-to-one, and tried to include a variety of relationships between mapped pairs. I’ll keep the notation of “red dashed” for \(f\) and “blue dotted” for \(g\) as I go forward.
I think my “simpliest non-trivial equalizer” to this parallel pair of maps is the unique map defined by \(\{x_1\} \rightarrow \{y_1\}\text{.}\) Clearly \(f x_1 = g x_1\) in my diagram, and the map \(T \rightarrow \{x_1\}\) is uniquely defined. This equalizer would be illustrated by the solid arrow in the following diagram:
I use the extra qualifiers to exlude the maps from \(\{\}\text{,}\) because that’s our initial object when we’re working in \(\mathcal{S}\text{.}\) Maybe it would be appropriate to call these maps \(\{\} \xrightarrow{p_0} X\) and \(\{x_1\} \xrightarrow{p_1} X\text{,}\) in an effort to enumerate all possible equalizers.
The next equalizer I see is the unique map \(\{x_3\} \xrightarrow{p_2} X\text{.}\)
Once again, it’s both true that \(f x_3 = g x_3 = y_4\) in the diagram and the map \(T \rightarrow \{x_3\}\) would need to be uniquely defined by domain and codomain.
We can find yet another example like this given by \(\{x_4\} \xrightarrow{p_3} \{y_4\}\text{:}\)
As before, we have both \(f x_4 = g x_4\) and \(T \rightarrow \{x_4\}\) unique. In fact, these maps are the only three maps \(T \xrightarrow{x_i} X\) for which \(f x_i = g x_i\text{.}\)
I think where this gets complicated is that we can combine these maps through set union to produce new equalizers. Consider what happens the map \(\{x_1,x_3\} \xrightarrow{p_4} X\) such that there is a unique \(\{x_1,x_3\} \xrightarrow{\bar{p_4}} X\) such that \(\bar{p_4} p_4 = 1_{\{x_1,x_3\}}\text{.}\) When we compose this with \(f,g\) we get the following:
Again, we have the case that \(f x_1 = g x_1\) and \(f x_3 = g x_3\text{.}\) For each of those two values, there are unique maps \(\{x_1,x_3\} \rightarrow \{x_1\}\) and \(\{x_1,x_3\} \rightarrow \{x_3\}\) that produce those two respective points in \(X\text{.}\)
For the same reason, we should also have a equalizer \(\{x_1,x_4\} \xrightarrow{p_5} X\text{:}\)
Given that we seem to be okay to combine \(p_1\) with \(p_2\) or \(p_3\text{,}\) what happens if we combine maps \(p_2\) and \(p_3\) together. Consider the following composite \(\{x_3,x_4\} \xrightarrow{p_6} X\text{:}\)
We still have the case that \(f x_i = g x_i\) for \(x_i\) in \(\{x_3,x_4\}\text{,}\) but does this map satisfy the “exactly one” property?
At first, I didn’t think it would, but I’m beginning to be more convinced that it does. There are unique maps \(\{x_3,x_4\} \rightarrow \{x_3\}\) and \(\{x_3,x_4\} \rightarrow \{x_4\}\) that could be paired one-to-one with the two unique maps \(T \rightarrow \{x_3\}\) and \(T \rightarrow \{x_4\}\text{.}\)
So let’s take this one last step and combine all three into one last equalizer \(\{x_1,x_3,x_4\} \xrightarrow{p_7} X\text{:}\)
So, I think this gives me a set of 8 equalizers all together: \(\{\} \xrightarrow{p_0} X\text{,}\) \(\{x_1\} \xrightarrow{p_1} X\text{,}\) \(\{x_3\} \xrightarrow{p_2} X\text{,}\) \(\{x_4\} \xrightarrow{p_3} X\text{,}\) \(\{x_1,x_3\} \xrightarrow{p_4} X\text{,}\) \(\{x_1,x_4\} \xrightarrow{p_5} X\text{,}\) \(\{x_3,x_4\} \xrightarrow{p_6} X\text{,}\) and \(\{x_1,x_3,x_4\} \xrightarrow{p_7} X\text{.}\) That last one strikes me as being “special” in that it combines all the others. It also strikes me as notable that there are exactly as many equalizers as points in \(X\text{,}\) but that might just be a coincidence.
This is nice and all, but I think I’m straying from the question. I should really be concerned with the interpretation of this pair \(f,g\) as defining a endomap on \(X+Y\) where I treat \(f\) as the source and \(g\) as the target. Let’s see what happens to my last diagram when I add the arrows in \(Y\text{:}\)
Note that I could have just easily swap interpretations to make \(g\) the source and \(f\) the target.
In both cases, it would seem that the equalizer has the property that it maps points in \(X\) to explicit self-loops of this endomap \(Y^\beta\text{.}\) With those fixed points, it doesn’t matter which of \(f,g\) is the source and target because they’re both the same point anyway. Maybe that’s all this exercise was looking for in the end?
