L&S Cat Theory: Attempted solutions to "Conceptual Mathematics" by Lawvere and Schanuel

I’ve been thinking a lot about the "family tree" problem from Section 4.7 and am wondering if combining two different "binaries" might give me what I’m looking for here. Consider the set \(\mathbf{G} \times \mathbf{C}\) we defined in Session 12 Exercise 3:

What if we choose some map \(\mathbf{P} \rightarrow \{\text{he-bear},\text{she-bear}\}\) as our sum \(X \rightarrow \mathbf{1}+\mathbf{1}\text{?}\)

We have a unique pair of injection maps \(\{\text{he-bear}\} \xrightarrow{j_1} \{\text{he-bear},\text{she-bear}\}\) and \(\{\text{she-bear}\} \xrightarrow{j_2} \{\text{he-bear},\text{she-bear}\}\) defined by \(j_1 ( \text{he-bear} )= \text{he-bear}\) and \(j_2 ( \text{she-bear} ) = \text{she-bear}\text{.}\) For any object \(Y\) and maps \(\{\text{he-bear}\} \xrightarrow{g_1} Y, \{\text{she-bear}\} \xrightarrow{g_2} Y \text{,}\) our definition of a sum says there should be exactly one map \(\{\text{he-bear},\text{she-bear}\} \xrightarrow{g} Y\) for which \(g_1 = g j_1\) and \(g_2 = g j_2\text{.}\)

Let’s take \(Y\) to be our product \(\mathbf{G} \times \mathbf{C}\text{.}\) As a product, we should have a unique set of three maps:

By defining \(X\) by a sum, we also force another set of three unique maps:

For any person in \(P\text{,}\) consider the compositions \(P \xrightarrow{f} P \xrightarrow{m} P \xrightarrow{f} P\) and \(P \xrightarrow{m} P \xrightarrow{f} P \xrightarrow{m} P\text{.}\)

First, consider a person \(p_1 \rightarrow \text{he-bear}\text{.}\) In the first composition, we’d have \(f p_1 \rightarrow \text{he-wolf}\text{,}\) \(m f p_1 = \text{she-wolf}\) and \(f m f p_1 \rightarrow \text{he-bear}\text{.}\) In the second composition, we’d have \(m p_1 \rightarrow \text{she-wolf}\text{,}\) \(f m p_2 \rightarrow \text{he-bear}\text{,}\) and \(m f m p_2 \rightarrow \text{{she-bear}}\text{.}\)

Next, consider another person \(p_2 \rightarrow \text{she-bear}\text{.}\) In the first composition, we’d have \(f p_1 = \text{he-wolf}\text{,}\) \(m f p_1 = \text{she-wolf}\) and \(f m f p_1 = \text{he-bear}\text{.}\) In the second composition, we’d have \(m p_1 = \text{she-wolf}\text{,}\) \(f m p_2 = \text{he-bear}\text{,}\) and \(m f m p_2 = \text{{she-bear}}\text{.}\)

This gives us an isomorphism between the set of labels \(\{\text{he-bear},\text{she-bear}\}\) and the compositions which fix those respective points \(\{f m f,m f m\}\text{.}\) Since our definition of product requires the maps \(\{\text{he-bear},\text{she-bear}\} \rightarrow \{\text{he-bear}\}\) and \(\{\text{he-bear},\text{she-bear}\} \rightarrow \{\text{she-bear}\}\) to be unique, we’d be unable to use these factor to \(X\) completely because we’d be excluding all of the people in the wolf clan that get "stepped over" through the composition.

This doesn’t feel like a very rigorous solution, but I think as close I’m going to get for now. Let’s keep moving forward.

Okay, I see four different sums we’re going to need: \(A+B\text{,}\) \(B+C\text{,}\) \((A+B)+C\) and \(A+(B+C)\text{.}\) We’ll start with the easy sums first.

For \(A+B\) with maps \(A\rightarrow A+B, B\rightarrow A+B\) to be a sum, each \(Y\) and pair of maps \(A \rightarrow Y\) and \(B \rightarrow Y\) should have exactly one map \(A+B \rightarrow Y\) such that \(A \rightarrow A+B \rightarrow Y = A \rightarrow Y\) and \(B \rightarrow A+B \rightarrow Y = B \rightarrow Y\text{.}\)

For \(B+C\) with maps \(B\rightarrow B+C, C\rightarrow B+C\) to be a sum, each \(Y\) and pair of maps \(B \rightarrow Y\) and \(C \rightarrow Y\) should have exactly one map \(B+C \rightarrow Y\) such that \(B \rightarrow B+C \rightarrow Y = B \rightarrow Y\) and \(C \rightarrow B+C \rightarrow Y = C \rightarrow Y\text{.}\)

Now let’s start composing our sums.

For \((A+B)+C\) with maps \(A+B\rightarrow (A+B)+C, C\rightarrow (A+B)+C\) to be a sum, each \(Y\) and pair of maps \(A+B \rightarrow Y\) and \(C \rightarrow Y\) should have exactly one map \((A+B)+C \rightarrow Y\) such that \(A+B \rightarrow (A+B)+C \rightarrow Y = A + B \rightarrow Y\) and \(C \rightarrow (A+B)+C \rightarrow Y = C \rightarrow Y\text{.}\)

For \(A+(B+C)\) with maps \(A \rightarrow A+(B+C), B+C\rightarrow A+(B+C)\) to be a sum, each \(Y\) and pair of maps \(A \rightarrow Y\) and \(B+C \rightarrow Y\) should have exactly one map \(A+(B+C) \rightarrow Y\) such that \(A \rightarrow A+(B+C)\rightarrow Y = A \rightarrow Y\) and \(B+C \rightarrow A+(B+C) \rightarrow Y = B+C \rightarrow Y\text{.}\)

Maybe here we can define a "triple sum" \(A+B+C\) as a "family" by a unique set of 3 maps \(A \xrightarrow{j_1} A+B+C\text{,}\) \(B \xrightarrow{j_2} A+B+C\text{,}\) and \(C \xrightarrow{j_3} A+B+C\text{.}\) These should have the property that for any object \(Y\) and maps \(A \xrightarrow{g_1} Y,B \xrightarrow{g_2} Y, C \xrightarrow{g_3} Y\text{,}\) we have exactly one map \(A+B+C \xrightarrow{g} Y\) with \(g_1 = g j_1\text{,}\) \(g_2 = g j_2\text{,}\) and \(g_3 = g j_3\text{.}\)

If we choose \(Y = (A+B)+C\text{,}\) the set of the 3 unique maps \(A \xrightarrow{k_1} (A+B)+C\text{,}\) \(B \xrightarrow{k_2} (A+B)+C\text{,}\) and \(C \xrightarrow{k_3} (A+B)+C\) should have some corresponding unique map \(A+B+C \xrightarrow{k} (A+B)+C\) with the property that \(k_1 = k g_1\text{,}\) \(k_2 = k g_2\text{,}\) \(i_3 = k g_3\text{.}\)

Likewise, choosing \(Y = A+(B+C)\) gives us unique maps \(A \xrightarrow{l_1} A+(B+C)\text{,}\) \(B \xrightarrow{l_2} A+(B+C)\text{,}\) and \(C \xrightarrow{l_3} A+(B+C)\) corresponding to the unique map \(A+B+C \xrightarrow{k} A+(B+C)\) satisfying \(l_1 = l g_1\text{,}\) \(l_2 = l g_2\text{,}\) \(l_3 = l g_3\text{.}\)

It seems like it would make sense to build an isomorphism by pairing up each \(j_i\) with the respective \(k_i,l_i\) pair for \(i \in I = \{1,2,3\}\text{.}\) That’s at least starting to resemble a dual definition for to the "product family". Perhaps we can define a "sum family" as follows: