Any map \(p\) which is an equalizer...
Solution.
Given that \(E \xrightarrow{p} X\) is an equalizer of the pair \(X \substack{
f \\ \longrightarrow \\ \longrightarrow \\ g
} Y\text{,}\) we know that for each \(T \xrightarrow{x} X\) for which \(f x = g x\) there is exactly one map \(T \xrightarrow{e} E\) for which \(x=pe\text{.}\)
To demonstrate that \(E \xrightarrow{p} X\) is a monomorphsim, we need to show injectivity: that for any pair of maps \(T \xrightarrow{e_1} E\) and \(T \xrightarrow{e_2} E\text{,}\) \(p e_1 = p e_2\) implies \(e_1 = e_2\text{.}\)
The contrapositive of this definition says that \(e_1 \neq e_2\) implies \(p e_1 \neq p e_2\text{.}\)
So let’s assume \(p\) is not a monomorphism.
In that case, we’d have arbitrary pair of maps \(T \xrightarrow{e_1} E\) and \(T \xrightarrow{e_2} E\) with \(e_1 \neq e_2\text{,}\) but for which \(p e_1 = p e_2\text{.}\)
If \(e_1 \neq e_2\text{,}\) there exists some \(\mathbf{1} \xrightarrow{s} T\) such that \(e_1 s \neq e_2 s\text{.}\) But there is only one map \(\mathbf{1} \xrightarrow{s} T\) by the definition of terminal objects. Since all terminal objects are isomorphic, we have a unique map \(T \xrightarrow{r} \mathbf{1}\) such that \(r s = 1_\mathbf{1}\) and \(s r = 1_T\text{.}\)
Let’s use these maps to define two points \(\mathbf{1} \xrightarrow{x_1} X\) and \(\mathbf{1} \ \xrightarrow{x_2} X\) by the compositions \(x_1 = p e_1 s\) and \(x_2 = p e_2 s\text{.}\) Since there’s a unique inverse map \(X \xrightarrow{\bar{x}} \mathbf{1}\) corresponding to each point \(\mathbf{1} \xrightarrow{x} X\text{,}\) the following compositions must both produce the only map \(\mathbf{1} \rightarrow \mathbf{1}\text{,}\) our identity map \(1_\mathbf{1}\text{:}\)
\begin{equation*}
\mathbf{1} \xrightarrow{s} T \xrightarrow{e_1} E
\xrightarrow{p} X \xrightarrow{\bar{x_1}} \mathbf{1}
\end{equation*}
\begin{equation*}
\mathbf{1} \xrightarrow{s} T \xrightarrow{e_2} E
\xrightarrow{p} X \xrightarrow{\bar{x_2}} \mathbf{1}
\end{equation*}
Since we’ve already assumed \(p e_1 = p e_2\text{,}\) that means that \(\bar{x_1} = \bar{x_2}\text{.}\) This brings us to a contradiction because the inverse maps are supposed to be unique for \(x_1 \neq x_2\text{.}\) There is no retraction for the map \(\{x_1,x_2\} \rightarrow \mathbf{1}\text{.}\) It follows that our assumption was wrong, and \(p\) is, in fact, a monomorphism.
I’m thinking that the big idea here is that having two points \(\mathbf{1} \xrightarrow{x_1} X\) and \(\mathbf{1} \xrightarrow{x_2} X\) and two maps \(X \substack{
f \\ \longrightarrow \\ \longrightarrow \\ g
} Y\) allows use to define four points \(\mathbf{1} \rightarrow Y\text{,}\) but if the “points in” are isomorphic to the “points out”, then those four points need collapse to form pairs of two after we apply \(f,g\text{.}\) Either \(f x_1 = g x_1\) and \(f x_2 = g x_2\text{,}\) or \(f x_1 = g x_2\) and \(f x_2 = g x_1\text{.}\)
