Formulate and prove in two ways...
Solution.
Alright, I think I need to start here by assuming that the category \(\mathcal{C}\) has an initial object \(I\) and terminal object \(T\text{.}\) To make the definitions of these simplier, I’m going to introduce the notation \(|I \xrightarrow{s_T} T| = 1\) to denote there exists precisely one map \(s_T\) with domain \(I\) and codomain \(T\text{.}\)
\begin{equation*}
\exists I \in \mathcal{C}: \forall X \in \mathcal{C}:
|I \xrightarrow{s_X} X| = 1
\end{equation*}
\begin{equation*}
\exists T \in \mathcal{C}: \forall X \in \mathcal{C}:
|X \xrightarrow{t_X} T| = 1
\end{equation*}
We can rewrite these definitions using the reflexive property to demonstrate that \(|I \rightarrow I| = 1\) and \(|T \rightarrow T| = 1\text{.}\) Since every object in the category has an identity map, these two endomaps must be the respective identities \(1_I\) and \(1_T\text{.}\)
As a category, \(\mathcal{C}\) must also have an associative property such that for any maps \(A \xrightarrow{f} B \xrightarrow{g} C \xrightarrow{h} D\) we must have \(h \circ (g \circ f) = (h \circ g) \circ f\text{.}\) In particular, let’s consider the case where \(A = I\text{,}\) \(D = T\) and \(g = 1_X\text{.}\) Since there is only one map \(I \xrightarrow{s_T} T\text{,}\) both sides of the associative property must equal to this map. Namely, \(t_X \circ (1_X \circ s_X) = (t_X \circ 1_X) \circ s_X = s_T = t_X \circ s_X\text{.}\)
Now suppose we have an arbitrary pair of maps \(X \xrightarrow{f_1} B_1\) and \(X \xrightarrow{f_2} B_2\text{.}\) Use the definition of terminal object, we can define two unique maps: \(B_1 \xrightarrow{t_1} T\) and \(B_1 \xrightarrow{t_2} T\text{.}\) Likewise, the definition of initial object gives us two more: \(I \xrightarrow{s_1} B_1\) and \(I \xrightarrow{s_2} B_2\text{.}\) Let’s draw these out in a commutative diagram.
The uniqueness of our maps \(s_T,s_X,t_X\) means that we can use them to describe some equivalences between our other maps.
\begin{equation*}
s_1 = f_1 \circ s_X
\end{equation*}
\begin{equation*}
s_2 = f_2 \circ s_X
\end{equation*}
\begin{equation*}
s_T = t_1 \circ s_1 = t_2 \circ s_2
\end{equation*}
\begin{equation*}
t_X = t_1 \circ f_1 = t_2 \circ f_2
\end{equation*}
Since we already know \(s_T = t_X \circ s_X\text{,}\) we can combine the results above to say \(s_T = t_1 \circ f_1 \circ s_X = t_2 \circ f_2 \circ s_X\text{.}\)
Now let’s consider a pair of products \(P\) and \(Q\) in this category. We’ll need three maps for each: \(X \xrightarrow{p} P,
P \xrightarrow{p_1} B_1,
P \xrightarrow{p_2} B_2\) and \(X \xrightarrow{q} Q,
Q \xrightarrow{q_1} B_1,
Q \xrightarrow{q_2} B_2\) respectively. The uniqueness of initial and terminal objects implies we have a set of unique maps \(I \xrightarrow{s_P} P\text{,}\) \(I \xrightarrow{s_Q} Q\text{,}\) \(P \xrightarrow{t_P} T\text{,}\) and \(Q \xrightarrow{s_Q} T\) These maps can be arranged together in the following diagram:
Suppose for a moment that the unique map \(I \xrightarrow{s_T} T\) is an isomorphism. The existence of a map \(T \xrightarrow{s_T^{-1}} I\) would immediately define an isomorphism \(P \leftrightarrows Q\) through the following unique compositions:
\begin{equation*}
P \xrightarrow{t_P} T \xrightarrow{s_T^{-1}} I
\xrightarrow{s_Q} Q
\end{equation*}
\begin{equation*}
Q \xrightarrow{t_Q} T \xrightarrow{s_T^{-1}} I
\xrightarrow{s_P} P
\end{equation*}
Having such maps would also define an isomorphism between \(B_1 \leftrightarrows B_2\text{,}\) uniquely determined by the following compositions:
\begin{equation*}
B_1 \xrightarrow{t_1} T \xrightarrow{s_T^{-1}} I
\xrightarrow{s_2} B_2
\end{equation*}
\begin{equation*}
B_2 \xrightarrow{t_2} T \xrightarrow{s_T^{-1}} I
\xrightarrow{s_1} B_1
\end{equation*}
Since that would essentially complete our proof, I think we can proceed with the assumption that there is no inverse for \(s_T\text{.}\) The non-existence of an inverse implies that we’re either missing a section \(T \xrightarrow{s} I \xrightarrow{s_T} T = 1_T\) or retraction \(I \xrightarrow{s_T} T \xrightarrow{r} I = 1_I\text{.}\)
All of our maps come from \(\mathcal{S}\text{,}\) where the initial object is the empty set \(\emptyset\text{.}\) In \(\mathcal{S}\text{,}\) there is a map from \(\emptyset \rightarrow \mathbf{1}\) but not from \(\mathbf{1} \rightarrow \emptyset\text{.}\) If the nature of initial and terminal objects is preserved, then we must have uniquely defined isomorpisms \(\emptyset \leftrightarrows I\) and \(\mathbf{1} \leftrightarrows T\text{.}\) The fact that \(s_T\) is not invertable means there must be at least one point that gets lost in the process.
I think this allows use to define a unique composition sequence:
\begin{equation*}
I \rightarrow T \rightarrow \mathbf{1} \rightarrow X \rightarrow T
\end{equation*}
If that’s is a unique composition, so is the following:
\begin{equation*}
X \rightarrow T \rightarrow \mathbf{1} \rightarrow X
\end{equation*}
I’ve been wondering if this map sequence \(I \rightarrow T \rightarrow \mathbf{1}\) is the reason why the authors introduced the “heavy arrow” notation in Session 22. There is at most one such map the composition is equal to the identity map \(1_X\text{.}\) This map is an epimorphism because it can be precomposed to any object without changing it.
This means we can still use the same techique to form our bijections, it just has one additional stop along the way:
\begin{equation*}
P \xrightarrow{t_P} T \rightarrow \mathbf{1} \rightarrow X
\xrightarrow{q} Q
\end{equation*}
\begin{equation*}
P \xrightarrow{t_P} T \rightarrow \mathbf{1} \rightarrow X
\xrightarrow{q} Q
\end{equation*}
\begin{equation*}
B_1 \xrightarrow{t_1} T \rightarrow \mathbf{1} \rightarrow X
\xrightarrow{f_2} B_2
\end{equation*}
\begin{equation*}
B_2 \xrightarrow{t_2} T \rightarrow \mathbf{1} \rightarrow X
\xrightarrow{f_1} B_1
\end{equation*}
Assuming that’s correct, that would complete our direct proof. However, stepping out through the singleton \(\mathbf{1}\) in \(\mathcal{S}\) feels like a “hack”. I can’t help but wonder if there’s a better approach to this. Maybe I can somehow use \(\mathbb{N}\) as an initial object to enumerate the points of \(X\text{?}\) That would certainly make things a lot easier.
