Show that... has the appropriate universal mapping property...
Solution.
So let’s sketch out what we’re looking at.
In order to do so, we need to establish it has the "universal mapping property" with all:
The "universal mapping property" essentially says we have 3 unique isomorphisms for the terminal object \(\mathbf{1}\text{:}\)
\begin{equation*}
\mathbf{1} \rightarrow X \xrightarrow{f_a} C_a \rightarrow \mathbf{1}
\end{equation*}
\begin{equation*}
\mathbf{1} \rightarrow X \xrightarrow{f_b} C_b \rightarrow \mathbf{1}
\end{equation*}
\begin{equation*}
\mathbf{1} \rightarrow X \xrightarrow{f_a} C_c \rightarrow \mathbf{1}
\end{equation*}
For convenience, let’s steal the multi-arrow notation from the previous exercise and denote it simply as \(X \mathrel{\substack{\longrightarrow \\ \longrightarrow
\\ \longrightarrow}} C_a \times C_b \times C_c\text{.}\) By the definition of our products, we can also name two other unique isomorphisms in this manner:
\begin{equation*}
\mathbf{1} \rightarrow P
\mathrel{\substack{p_a \\ \longrightarrow \\ \longrightarrow \\ p_b}}
C_a \times C_b \rightarrow \mathbf{1}
\end{equation*}
\begin{equation*}
\mathbf{1} \rightarrow Q
\mathrel{\substack{q \\ \longrightarrow \\ \longrightarrow \\ q_c}}
P \times C_c \rightarrow \mathbf{1}
\end{equation*}
I think this is similar to what we did with Exercise 15. Our universal mapping property says that we have a unique map \(f = \langle f_a, f_b, f_c \rangle\) satsifying \(X
\mathrel{\substack{\longrightarrow \\ \longrightarrow \\ \longrightarrow}}
C_a \times C_b \times C_c\text{.}\) Our products define unique maps \(X \rightarrow P\) and \(X \rightarrow Q\text{,}\) which allow us to form two unique compositions:
\begin{equation*}
\mathbf{1}
\rightarrow X \rightarrow P \mathrel{\substack{\longrightarrow
\\ \longrightarrow }} C_a \times C_b \rightarrow \mathbf{1}
\end{equation*}
\begin{equation*}
\mathbf{1}
\rightarrow X \rightarrow Q \mathrel{\substack{\longrightarrow
\\ \longrightarrow }} P \times C_c \rightarrow \mathbf{1}
\end{equation*}
We can chain these maps together to get a unique map \(Q \rightarrow P\text{:}\)
\begin{equation*}
Q \mathrel{\substack{\longrightarrow
\\ \longrightarrow }} P \times C_c \rightarrow \mathbf{1}
\rightarrow P
\end{equation*}
Or a unqiue map \(P \rightarrow Q\text{:}\)
\begin{equation*}
P \mathrel{\substack{\longrightarrow
\\ \longrightarrow }} C_a \times C_b \rightarrow \mathbf{1}
\rightarrow Q
\end{equation*}
This is what we’d expect based on our "Uniqueness of Products" theore. It states that if the maps \(P \rightarrow C_i\) and \(Q \rightarrow C_i\) both make products of the same family, then there is exactly one map \(P \xrightarrow{f} Q\) for which \(q_i f = p_i \forall i \in I\) and that this map \(f\) is an isomorphism. Since our \(I = \{a,b,c\}\text{,}\) this gives us three equations \(q_a f = p_a\text{,}\) \(q_b f = p_b\text{,}\) \(q_c f = p_c\text{.}\)
Well, we already have a uniquely defined map \(Q \rightarrow P\) given by \(q\) so we could only possibly have \(f = q^{-1}\text{.}\) It follows that \(q f = 1_P\) and \(f q = 1_Q\text{.}\) Apply \(q\) on the right of the three equations above to get:
\begin{equation*}
q_a f q = q_a 1_Q = q_a = p_a q
\end{equation*}
\begin{equation*}
q_b f q = q_b 1_Q = q_b = p_b q
\end{equation*}
\begin{equation*}
q_c f q = q_c 1_Q = q_c = p_c q = q_c f q = q_c
\end{equation*}
This map \(p_c\) seems to come from nowhere, but is simultaneously well defined by applying the composition \(q_c q^{-1}\text{.}\)
Putting this all together now, any triple-product \(\langle f_a, f_b, f_c \rangle: X \mathrel{\substack{\longrightarrow \\ \longrightarrow
\\ \longrightarrow}} C_a \times C_b \times C_c\) uniquely determines a triple-product \(\langle p_a q, p_b q, q_c \rangle:
Q \mathrel{\substack{\longrightarrow \\ \longrightarrow
\\ \longrightarrow}} C_a \times C_b \times C_c\) via the isomorphisms given by:
\begin{equation*}
\mathbf{1} \rightarrow X \mathrel{\substack{\longrightarrow \\ \longrightarrow
\\ \longrightarrow}} C_a \times C_b \times C_c
\mathrel{\substack{\longrightarrow \\ \longrightarrow}}
(C_a \times C_b) \times C_c
\mathrel{\substack{\longrightarrow \\ \longrightarrow}}
P \times C_c \rightarrow \mathbf{1}
\end{equation*}
and
\begin{equation*}
\mathbf{1} \rightarrow Q \mathrel{\substack{\longrightarrow \\ \longrightarrow
\\ \longrightarrow}} C_a \times C_b \times C_c \rightarrow \mathbf{1}
\end{equation*}
As one last step, let’s see if we can draw a more complete diagram of the situation.
The "Uniqueness of Products" essentially states that we have 3 unique maps defined by the map equivalencies \(f_a = p_a f_P = p_a q f_Q\text{,}\) \(f_b = p_b f_P = p_b q f_Q\text{,}\) and \(f_c = q_c f_Q\text{.}\)
This still feels a little more "hand-wavy" than I’d like, but it makes sense that when the maps are uniquely defined by the domain and codomain that there would be precisely one way to compose them such that \(\mathbf{1} \rightarrow X \mathrel{\substack{\longrightarrow \\ \longrightarrow
\\ \longrightarrow}} C_a \times C_b \times C_c
\mathrel{\substack{\longrightarrow
\\ \longrightarrow}} P \times C_c \rightarrow \mathbf{1}
\text{.}\)
