Any parallel pair \(X \mathrel{\substack{
f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\) of maps in sets...
Solution.
I’m going out on a limb with my metagaming here and hypothesize that the arrows named by the equalizer are related to the either the “section of the product” or “retraction from the sum”. The diagram below the problem and choice of letter \(p\) are just too convenient.
Let’s consider a simple example \(\mathbb{N} \mathrel{\substack{
f \\ \longrightarrow \\ \longrightarrow \\ g}} \mathbb{N}_6\) defined by \(f(x) = \text{mod}_3 x\) and \(g(x) = \text{mod}_6 x\text{.}\) The output of \(f\) would be the sequence \(0,1,2,0,1,2,0,1,2...\) and the output of \(g\) would be the sequence \(0,1,2,3,4,5,0,1,2,...\text{.}\)
I’m thinking that my equalizer should be the map \(\mathbb{N} \xrightarrow{p} \mathbb{N}\) that produces a sequences of indices where \(f,g\) have the same outputs. These two sequences match up for the first 3 of every 6 steps, producing the sequence the sequence \(0,1,2,6,7,8,12,13,14,...\text{.}\) This map can be defined by the formula \(p(x) = 6 \lfloor \frac{x}{3} \rfloor + \text{mod}_3 x\text{,}\) but it strikes me as interesting that we kind of need to “step out” through \(\mathbb{Q}\) in order to do division.
So how can I demonstrate that \(p\) is an equalizer? Well, it makes sense that \(E = \mathbb{N}\) because there are countably infinite solutions to \(f x = g x\text{.}\) It is also true that for each \(T \xrightarrow{x} \mathbb{N}\) such that \(f x = g x\) there is exactly one \(T \xrightarrow{e} \mathbb{N}\) such that \(x = p e\) because \(p\) is “strictly monotonic” and therefore one-to-one. In this case our \(e\) would be uniquely defined by its relation to the inverse of \(p\text{.}\) I could probably even work out \(p^{-1}\) for my particular example if I had to, but I think I’m already trivializing the proof I’m looking for by choosing an example where \(f g = f\) in the first place.
I think what’s going on here is that we can test for equality two different ways by checking if \(f-g=0\) or \(g-f=0\text{.}\) For any point where \(f x = g x\text{,}\) it must be true that \(f x - g x = 0\) and \(g x - f x = 0\text{.}\) Even if we have \(f x \neq g x\text{,}\) we’d still be guaranteed to have the sum \((fx - g x) + (g x - f x) = 0\) and \((g x - f x) + (f x - g x) = 0\text{.}\) Perhaps the trick to this will be expressing \(f-g\) and \(g-f\) in terms of the initial and terminal objects.
Let’s come back to the original question, which had to deal with an arbitrary equalizer \(E \xrightarrow{p} X\) to an arbitrary pair of parallel maps \(X \mathrel{\substack{
f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\text{.}\) By definition, \(f p = g p\) and for every \(T \xrightarrow{x} X\) for which \(f x = g x\text{,}\) there is exactly one \(T \xrightarrow{e} E\) for which \(x = p e\text{.}\)
Perhaps we can break this down in four cases:
- \(E\) is neither initial nor terminal
- \(E\) is initial but not terminal
- \(E\) is terminal but not initial
- \(E\) is both initial and terminal
For an initial object \(I\text{,}\) we’d be able to express a “zero map” \(I \xrightarrow{\mathbf{0}_{IT}} T\) through the following composition:
\begin{equation*}
I \xrightarrow{\mathbf{0}_{IT}} T \xrightarrow{x} X \mathrel{\substack{
f \\ \longrightarrow \\ \longrightarrow \\ g}} Y
\xrightarrow{\bar{y}} T
\end{equation*}
Since that map is unique, \(\mathbf{0}_{IT} =
\bar{y} \circ f \circ x \circ \mathbf{0}_{IT} =
\bar{y} \circ g \circ x \circ \mathbf{0}_{IT}\) for all \(T \xrightarrow{x} X\text{.}\)
If \(E\) is an initial object then it would be isomorphic to \(I\text{.}\) This gives us maps \(E \xrightarrow{\mathbf{0}_{IE}} I\) and \(I \xrightarrow{\mathbf{0}_{EI}} E\) satisfying \(\mathbf{0}_{EI} \circ \mathbf{0}_{IE} = 1_E\) and \(\mathbf{0}_{IE} \circ \mathbf{0}_{EI} = 1_I\text{.}\) Effectively, the unique map \(E \xrightarrow{p} X\) would meet the definition of an equalizer because \(f p\) and \(g p\) would both need to be unique map \(E \rightarrow Y\text{.}\)
So what if \(E\) is not an initial object? The definition of \(I\) implies that the map \(I \rightarrow E\) is unique, but \(E\) not being intial implies this map is non-invertible: there is no retraction \(E \rightarrow I\text{.}\) That in turn gives us a pair of maps \(e_1 \neq e_2\) satisfying \(I \rightarrow T \mathrel{\substack{
e_1 \\ \longrightarrow \\ \longrightarrow \\ e_2}} E\) and a unique map \(E \xrightarrow{\bar{e}} T\) such that \(\bar{e} e_1 = \bar{e} e_2 = 1_T\text{.}\) Maybe this is where I need to invoke the product?
I think I’ll need to spend some more time on this.
