Given two parallel maps \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\text{...}\)
Solution.
Let’s start by using the same parallel maps \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\) that I used in Exercise 8, shown here:
My equalizer (of questionable accuracy) for these maps looked like this:
In particular, I’m not really sure how I’m supposed to handle the pair \(x_3,x_4\text{.}\) Something tells me my equalizer is supposed to either group these objects together to satisfy that “exactly one” property or just exclude them.
Setting those concerns aside for now, let’s start with visualizing the graphs \(\Gamma_f = \langle ?,f \rangle\) and \(\Gamma_g = \langle ?,g \rangle\text{.}\) Let’s rotate the points of \(Y\) and lay out our product in an array:
Our graphs \(\Gamma_x,\Gamma_y\) can be added to this diagram as maps \(X \rightarrow X \times Y\) as so:
Note that there’s precisely three points where the arrows of \(f,g\) match up with each other, and those correspond with the three arrows in the depiction of my equalizer:
This interpretation leads to believe that my equalizer from Exercise 8 is, in fact, the correct one. The product also gives me a way of defining the equalizer more explicitly:
\begin{equation*}
e(x) =
\begin{cases}
\langle x_1,y_1 \rangle & \text{if}\, x = x_1 \\
\langle x_3,y_4 \rangle & \text{if}\, x = x_3 \\
\langle x_4,y_4 \rangle & \text{if}\, x = x_4
\end{cases}
\end{equation*}
