If both \(E, p\) and \(F, q\) are equalizers...
Solution.
The text has defined an equalizer of two maps \(f,g: X \rightarrow Y\) as a map \(E \xrightarrow{p} X\) if \(fp = gp\) and for each \(T \xrightarrow{x} X\) for which \(fx=gx\text{,}\) there is exactly one \(T \xrightarrow{e_E} E\) for which \(x = p e_E\text{.}\)
Since \(F, q\) is also an equalizer of the same pair \(f,g\text{,}\) then \(F \xrightarrow{q} X\) must satisfy \(fq = gq\) and for each \(T \xrightarrow{x} X\) for which \(fx=gx\text{,}\) there is exactly one \(T \xrightarrow{e_F} F\) for which \(x = q e_F\text{.}\)
We’re asked to prove that there is a unique map \(F \xrightarrow{e} E\) for which \(p e = q\) is an isomorphism.
By the definition of \(T\) as a terminal object, there is a unique map \(E \xrightarrow{\bar{e_E}} T\) and a unique map \(F \xrightarrow{\bar{e_F}} T\text{.}\) I claim that \(e_E \circ \bar{e_F}\) is actually the unique map \(F \xrightarrow{e} E\) for which \(p e = q\) and \(e_F \circ \bar{e_E}\) is its unique inverse \(E \xrightarrow{e^{-1}} F\text{.}\) To establish this, I need to show \(pe=q\text{,}\) \(e^{-1} e = 1_F\text{,}\) and \(e e^{-1} = 1_E\text{.}\)
Suppose \(F \xrightarrow{pe} X \neq F \xrightarrow{q} X\text{.}\) There would need to exist some some \(T \xrightarrow{f'} F\) such that \(p e f' \neq q f'\text{.}\) Expanding \(e\) with my above defintion gives us \(p e_E \bar{e_F} f' \neq q f'\text{.}\) Since \(\bar{e_F} f'\) is a map \(T \rightarrow T\) it could only be the unique map \(1_T\text{.}\) It follows that \(p e_E \bar{e_F} f' = p e_E 1_T = p e_E \neq q f'\text{.}\) Since \(q f'\) is a map \(T \xrightarrow{x} X\text{,}\) the definition of \(E, p\) as an equalizer says that for there is exactly one map \(p e_E = x\) for all \(x\) satisfying \(fx = gx\text{,}\) contradicting our assumption that \(pe \neq q\) and implying that in fact \(pe = q\text{.}\)
Expanding \(e^{-1} e\) using those definitions above gives us \((e_F \circ \bar{e_E}) \circ (e_E \circ \bar{e_F})\) . We can regroup these terms using the associative property to get \(e_F \circ (\bar{e_E} \circ e_E) \circ \bar{e_F}\text{.}\) Since \((\bar{e_E} \circ e_E)\) is a map \(T \rightarrow T\text{,}\) it could only be the unique map \(1_T\text{.}\) Our expression for \(e^{-1} e\) becomes \(e_F \circ 1_T \circ \bar{e_F}\) or simply \(e_F \circ \bar{e_F}\text{.}\)
Suppose \(e_F \circ \bar{e_F} \neq 1_F\text{.}\) There would need to be some \(T \xrightarrow{f'} F\) such that \(e_F \bar{e_F} f' \neq f'\text{.}\) However, \(\bar{e_F} f': T \rightarrow T\) could only possibly be the map \(1_T\text{.}\) It follows that \(e_F \neq f'\text{,}\) but this contradicts our definition of \(e_F\) as an equalizer if we post-compose by \(q\) to get \(q e_F \neq q f'\text{.}\) The composition \(q f'\) is a map \(T \rightarrow X\) \(q e_F = x\) so any \(q f' \neq q e_F\) would contradict the definition of \(e_F\) as an equalizer.
Likewise, expanding \(e e^{-1}\) gives us \((e_E \circ \bar{e_F}) \circ (e_F \circ \bar{e_E})\text{.}\) Once again, the terms in the middle can be regrouped to get \(e_E \circ (\bar{e_F} \circ e_F) \circ \bar{e_E}\) and we can use the fact that \(\bar{e_F} \circ e_F = 1_T\) to simplify down to \(e_E \bar{e_E}\text{.}\) If \(e_E \bar{e_E}\) were anything other than \(1_E\text{,}\) it would contradict our definition of \(e_E\) as an equalizer.
