The category \(\mathbf{1}/\mathcal{S}\) of pointed sets...
Solution.
We defined this category back in Exercise 9, to have as objects maps \(\mathbf{1} \xrightarrow{x} X\) in \(\mathcal{S}\text{,}\) and as maps tbe maps \(X \xrightarrow{f} Y\) in \(\mathcal{S}\) from \(\mathbf{1} \xrightarrow{x_0} X\) to \(\mathbf{1} \xrightarrow{y_0} Y\) in \(\mathcal{S}\) satisfying \(f x_0 = y_0\text{.}\) We showed that under these condiditions “any terminal object is also inital” and that \(X\) itself is not inital.
Our hint was to “determine the nature of sums within the category”, so lets do that next. The best analogy I have for sum here is the disjoint union of sets. Let’s consider the breakfast ordering scenario from way back in Article 1. Suppose John, Mary, and Sam are at a dinner that only serves a single choice of either coffee or eggs.
If \(X = \{\text{John}, \text{Mary}, \text{Sam}\}\) and \(Y = \{\text{eggs}, \text{coffee}\}\text{,}\) then every map \(X \xrightarrow{f} Y\) corresponds with an endomap \(X+Y \xrightarrow{\bar{f}} X+Y\) that we can make by taking \(X+Y\) as the union of \(X \cup Y\) and mapping each point in \(Y\) to itself.
Now suppose our situation is a little bit different. The server comes asks "What does everyone want?". Sam says "coffee" and Mary says "eggs". John, however, is a little distracted and reveals his true feelings by blurting out "Mary". The server assumes that John wants the same order as the only woman at the table and puts in a second order of eggs for John. Effectively, our server acts as a composition of maps between what they ordered and what they get.
My hunch is that the treatment of \(\text{Mary}\) as one object or two is going to present problems in \(\mathbf{1}/\mathcal{S}\text{.}\) Consider what happens when we try to embed this composition as an endomap.
We should have exactly one map from \(\{\text{John}\} \rightarrow \{\text{Eggs}\}\) but effectively have a second one that passes through \(\{\text{Mary}\}\text{.}\) The only way we can resolve this in \(\mathbf{1}/\mathcal{S}\) is to separate out two unique Mary maps \(\mathbf{1} \rightarrow \{\text{Mary}_\text{person}\}\) and \(\mathbf{1} \rightarrow \{\text{Mary}_\text{order}\}\text{.}\)
I think the key fact here is that every \(\mathbf{1}/\mathcal{S}\)-map is inherently invertable. By the uniqueness of terminal objects, each \(\mathbf{1} \xrightarrow{x_0} X, \mathbf{1} \xrightarrow{y_0} Y\) must correspond with a unique inverse map \(X \xrightarrow{\bar{x_0}} \mathbf{1}, Y \xrightarrow{\bar{y_0}} \mathbf{1}\text{.}\) Given that \(f x_0 = y_0\text{,}\) precomposing with \(\bar{x_0}\) gives us \(f = y_0 \bar{x_0}\text{.}\) If we define \(g = x_0 \bar{y_0}\text{,}\) then it follows that \(g \circ f = 1_X\) and \(f \circ g = 1_Y\text{.}\) It’s easier to see this with a diagram:
The uniqueness of our maps \(f,g\) basically depends on whether or not \(X = Y\text{.}\) If \(X = Y\text{,}\) then we must have \(f = g = 1_X = 1_Y\text{.}\) If \(X \neq Y\text{,}\) then then we know \(g = f^{-1}\text{.}\)
I’m thinking about \(f\) as uniquely defining a dual map \(X \xrightarrow{\bar{f}} \mathbf{2}\) by the property that \(\bar{f} x_i = 1\) if \(f x_i = x_i\) for every \(x_i \in X\) and \(\bar{f} x_i = 2\) if there exists some \(x_i \in X\) for which \(f x_i \neq x_i\text{.}\) This allows me to define an something like an injection map \(X \xrightarrow{f_p} Y \times \mathbf{2}\) that records for each \(\mathbf{1} \xrightarrow{y_i} Y\) whether the point we came from is the same or different.
Perhaps this is setting up something similar to Russell’s Paradox: "Does the set of all sets not containing itself contain itself?". A set which contains itself, shouldn’t, and a set which doesn’t contain itself, should. Maybe we can use the notion that the map \(\mathbf{1} \rightarrow \mathbf{2}\) admits a retraction but \(\mathbf{2} \rightarrow \mathbf{1}\) does not to further show that \(f\) either fixes all points or no points.
To get a better idea of what’s going on, lets draw an external diagram showing both the sum and product:
I’m thinking this diagram is uniquely defines a map from \(A \times X \rightarrow Y+B\) satisfying all commutative triangles. It seems reasonable to conclude that \(A = \mathbf{1}\) and \(B = \mathbf{0}\) would satisfy this property, but our result from Exercise 9 said that \(\mathbf{1} \in X\) implies \(\mathbf{1} = \mathbf{0}\) and \(X \neq \mathbf{0}\text{.}\) Maybe we can use that to contradict the distributive property somehow.
Let’s try breaking down each side of the distributive property with some diagrams.
Breaking this down like this doesn’t quite seem right. The way we’ve defined a sum, we “come from” those sets rather than “go to” them. My sum of \((A \times B) + (A \times C)\) should probably look more like this:
Likewise, our \(A\times(B+C)\) should be updated as well:
Alright, let’s try to combine this idea with our universal mapping property. Let’s use the convention \(A=C_0,B=C_1,C=C_2\) to index our objects \(C_i\) for \(i \in I = \{0,1,2\}\text{.}\) Our definition of product family \(P = A \times B \times C\) with maps \(X \xrightarrow{p_0} A, X \xrightarrow{p_1} B
X \xrightarrow{p_2} C\) implies for any maps \(X\xrightarrow{f_0} A,
X\xrightarrow{f_1} A, X\xrightarrow{f_2} C, \)there’s a unique map \(X \rightarrow P\) such that all triangles below commute.
Likewise, our definition of a sum family says that for \(S = A + B + C\) with maps \(A \xrightarrow{j_0} Y, B \xrightarrow{j_1} Y
C \xrightarrow{j_2} Y\) implies for any maps \(A\xrightarrow{g_0} Y,
A \xrightarrow{g_1} Y, C \xrightarrow{g_2} Y, \)there’s a unique map \(S \rightarrow Y\) such that all triangles below commute.
Combining those two diagrams together:
There for any set of \(f_i,g_i\text{,}\) there should be exactly one pair \(f,g\) such that everything commutes.
I’m not exactly sure where to take it from here, so maybe I should just let it simmer for a while.
