Prove that if \(A\) and \(B\) are objects and \(A \times B = \mathbf{1}\text{,}\) then \(A = B = \mathbf{1}\text{...}\)
Solution.
They asked me to ‘translate’, so that’s what I’ll try to do!
With the statement \(A \times B = \mathbf{1}\text{,}\) we’re assuming we have we have a pair of projection maps \(A \xleftarrow{p_1} \mathbf{1} \xrightarrow{p_2} B\) such that for “every” object \(X\) with maps \(A \xrightarrow{f_1} X \xleftarrow{f_2} B\) there is exactly one map \(X \xrightarrow{f} \mathbf{1}\) with \(f_1 = p_1 f\) and \(f_2 = p_2 f\text{.}\) Our category must allow compositions so we know we have some map \(X \xrightarrow{f} \mathbf{1} \xrightarrow{f_1} A\) in \(\mathcal{C}\text{.}\)
Suppose \(g : X \rightarrow A\) is a another map in \(\mathcal{C}\) such that \(g \neq p_1 f\text{.}\)
For \(g \neq p_1 f\) to be true, there where would need to exist some \(\mathbf{1} \xrightarrow{x} X\) such that \(g x \neq p_1 f x\text{.}\) Or more explicitly:
\begin{equation*}
\mathbf{1} \xrightarrow{x} X \xrightarrow{g} A
\neq
\mathbf{1} \xrightarrow{x} X \xrightarrow{f} \mathbf{1} \xrightarrow{p_1} A
\end{equation*}
However, the composition \(f x\) is a map \(\mathbf{1} \rightarrow \mathbf{1}\) and there’s only one such map: namely the identity \(1_\mathbf{1}\text{.}\) The right side of this can be simplified using the associative property \(p_1 f x = p_1 1_\mathbf{1} = p_1\text{.}\) This implies \(f\) and \(g\) are the same map.
Thus, there is exactly one map \(X \rightarrow A\) and one map \(X \rightarrow B\text{.}\) This implies \(A\) and \(B\) are terminal objects.
And that completes the proof.
