If \(B \xrightarrow{\alpha} A \xrightarrow{\beta} B\text{...}\)
Solution.
We’re given that \(1_B = \beta \alpha\) and \(f = \alpha \beta\) is idempotent. Recall that \(A \xrightarrow{f} A\) is idempotent if and only if \(f f = f\text{.}\) This fact follows from associative and identity properties:
\begin{equation*}
f f = (\alpha \beta) (\alpha \beta) = \alpha (\beta \alpha) \beta
= \alpha 1_B \beta = \alpha \beta = f\text{.}
\end{equation*}
We’re asked to prove that \(\alpha\) is an equalizer for the pair \(f, 1_A\text{.}\) Per the definition, \(B \xrightarrow{\alpha} A\) is an equalizer of \(f, 1_A\) if \(f \alpha = 1_A \alpha\) and for each \(T \xrightarrow{x} A\) for which \(f x = 1_A x\text{,}\) there is exactly one \(T \xrightarrow{e} B\) for which \(x = \alpha e\text{.}\)
Let’s start by demonstrating that \(f \alpha = 1_A \alpha\text{.}\)
Given that \(f = \alpha \beta\text{,}\) we can precompose by \(\alpha\) to get \(f \alpha = (\alpha \beta) \alpha\text{.}\) By the associative property, \((\alpha \beta) \alpha = \alpha (\beta \alpha)\text{,}\) and \(\beta \alpha = 1_B\text{.}\) Subsitute and rewrite the identity maps to get \(f \alpha = \alpha 1_B = \alpha = 1_A \alpha\text{.}\)
Step one: complete. Our next one look a little trickier.
For every \(T \xrightarrow{x} A\) for which \(f x = 1_A x\) (or simply \(f x = x\)), we need to prove there is exactly one map \(T \xrightarrow{e} B\) for which \(x = \alpha e\text{.}\)
First, we need to demonstrate that some such \(e\) must exist, then prove that we can’t have more than one such map.
It seems obvious that for any \(T \xrightarrow{x} A\) satisfying \(f x = x\) the existance of some \(T \xrightarrow{e} B\) would be guaranteed by the composition \(\beta x\text{.}\) Clearly \(T \xrightarrow{x} A \xrightarrow{\beta} B\) has the appropriate domain and codomain. Given \(e = \beta x\text{,}\) it follows that we can postcompose both sides by \(\alpha\) to show \(\alpha e = \alpha \beta x = f x = x\text{.}\) The only remaining part of the question is: why can’t we have more than one such map?
Suppose we have a second map \(T \xrightarrow{e'} B\) satisfying \(\alpha e' = x\) for some \(e' \neq e\text{.}\) For us to have \(e' \neq e\) we’d need to have a point \(\mathbf{1} \xrightarrow{t} T\) such that \(e' t \neq e t\text{.}\) However, there is precisely one map \(\mathbf{1} \rightarrow T\) by the definition of terminal objects and that map needs to be an isomorphism.
Since we know \(\alpha e' = \alpha e = x\text{,}\) we can postcompose by \(\beta\) and apply \(\beta \alpha = 1_B\) to get \(\beta \alpha e' = e'= \beta \alpha e = e = \beta x\) This contradicts our assumption that \(e' \neq e\) and proves \(e\) is unique.
This feels a bit like a “vacuous truth”. If there’s no \(T \xrightarrow{x} A\) such that \(f x = x\text{,}\) then it doesn’t matter what the conclusion of the conditional statement is.
