In \(\mathcal{S}\text{,}\) there are many...
Solution.
The authors discussed this notion of factoriong in the context of products before. Specifically, they considered an example of a system changing over time:
\begin{equation*}
\frac{T \rightarrow B_1 \times B_2}
{T \rightarrow B_1, T \rightarrow B_2}
\end{equation*}
This notation was meant to convey the existence of a unique isomorphism between the product and pairs of factors.
The "dual" notion of factoring for sums would probably look like
\begin{equation*}
\frac{T \rightarrow B_1 + B_2}
{T \rightarrow B_1, T \rightarrow B_2}
\end{equation*}
but what actually follows from the definition of sum is something to the effect of
\begin{equation*}
\frac{B_1 + B_2 \xrightarrow{j} Y}
{B_1 \xrightarrow{j_1} Y, B_2 \xrightarrow{j_2} Y}
\end{equation*}
So what happens if we have an arbitrary map \(X \xrightarrow{f} \mathbf{1}+\mathbf{1}\text{?}\) Maybe we can substitute \(B_1 = B_2 = \mathbf{1}\) into our definition. Doing so says that any pair of maps \(\mathbf{1} \xrightarrow{j_1} \mathbf{1}+\mathbf{1}\) and \(\mathbf{1} \xrightarrow{j_2} \mathbf{1}+\mathbf{1}\) would imply that for any \(Y\) and pair of maps \(\mathbf{1} \xrightarrow{g_1} Y,\mathbf{1} \xrightarrow{g_2} Y\text{,}\) we will have exactly one map \(\mathbf{1}+\mathbf{1} \xrightarrow{g} Y\) with the property that \(g_1 = g j_1\) and \(g_2 = g j_2\text{.}\)
If that’s true for all \(Y\text{,}\) why not try assigning \(Y = \mathbf{1} + \mathbf{1}\text{?}\) The existence of a unique map defined by \(\mathbf{1} + \mathbf{1} \rightarrow \mathbf{1} + \mathbf{1}\) would need to be the identity map so we can conclude \(g = 1_{\mathbf{1} + \mathbf{1}}\text{.}\) It follows that we’d need to have \(g_1 = j_1\) and \(g_2 = j_2\text{.}\)
So we start with some points \(x\) in \(X\) which are each isomorphic to \(\mathbf{1}\)
\begin{equation*}
\mathbf{1} \xrightarrow{x} X \xrightarrow{\bar{x}} \mathbf{1}
\end{equation*}
By mapping each \(x\) with \(y = f x\text{,}\) the corresponding \(\bar{y}\) should uniquely satisfy
\begin{equation*}
\mathbf{1} \xrightarrow{x} X \xrightarrow{f}
\mathbf{1} + \mathbf{1} \xrightarrow{\bar{y}} \mathbf{1}
\end{equation*}
but we actually have two such maps given by
\begin{equation*}
\mathbf{1} \xrightarrow{x} X \xrightarrow{f}
\mathbf{1} + \mathbf{1} \xrightarrow{\bar{j_1}} \mathbf{1}
\end{equation*}
and
\begin{equation*}
\mathbf{1} \xrightarrow{x} X \xrightarrow{f}
\mathbf{1} + \mathbf{1} \xrightarrow{\bar{j_2}} \mathbf{1}
\end{equation*}
That said, I’m still struggling to come up with a concrete example which meets this criteria. Maybe that statement that \(X \neq \mathbf{0},\mathbf{1}\) is supposed to be a hint. If we have a unique map \(\mathbf{0} \rightarrow \mathbf{1}\) and a unique map \(\mathbf{1} \rightarrow \mathbf{1}+\mathbf{1}\) the resulting composition \(\mathbf{0} \rightarrow \mathbf{1}+\mathbf{1}\) should be unique as well. However, while there should be a unique inverse \(\mathbf{1} + \mathbf{1} \rightarrow \mathbf{1}\text{,}\) there can’t exist a inverse satisfying \(\mathbf{1} + \mathbf{1} \rightarrow \mathbf{0}\) because we need someplace for the map to send the points to.
