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

I think I’d like to start by recapping what I’ve established so far.

We’ve seen that any pair of parallel maps \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\) has an interpretation as a graph which defines an pair of endomaps \(\beta_1,\beta_2\) on \(Y\) depending on which map we consider the source and target. Since swapping the source and target “inverts” the map, we know \(\beta_1 \beta_2 = \beta_2 \beta_1 = 1_Y\text{.}\)

By indexing the objects \(C_i\) of our category \(\mathcal{C}\) in such a way that \(C_0\) is initial and \(C_1\) is terminal, for any object \(C_i\) in the category we can define a unique pair of maps \(C_0 \xrightarrow{f_i} C_i\) and \(C_i \xrightarrow{g_i} C_1\text{.}\)

Given this convention, I can think of my \(X = C_i\) and \(Y = C_j\) for two (possibly equal) indices \(i,j\text{.}\) This would give a unique composition of maps \(C_0 \xrightarrow{f_i} C_i \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} C_j \xrightarrow{g_j} C_1 \text{.}\) Since categories are closed under composition, the compositions \(g_j f\) and \(g_j g\) would both need to be the unique map \(g_i: C_i \rightarrow C_1\text{.}\)

So what happens when we throw \(Y \xrightarrow{h} Z\) into this mix?

Suppose this object \(Z = C_k\) for some index \(k\text{.}\) This gives lets us rewrite \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y \xrightarrow{h} Z\) as \(C_0 \xrightarrow{f_i} C_i \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} C_j \xrightarrow{h} C_k \xrightarrow{g_k} C_1 \text{.}\) Our composition \(g_k h\) would need to be the unique map \(C_j \xrightarrow{g_j} C_1\text{.}\)

If \(f = g\text{,}\) then clearly any map \(Y \xrightarrow{h} Z\) would satisfy \(h f = h g\text{.}\) We only need a specific choice of \(h\) if there’s some \(C_1 \xrightarrow{x} X\) for which \(f x \neq g x\text{.}\) This implicitly defines a pair of points \(C_1 \mathrel{\substack{y_1 \\ \longrightarrow \\ \longrightarrow \\ y_2}} Y\) such that \(y_1 = f x\text{,}\) \(y_2 = g x\text{,}\) and \(y_1 \neq y_2\text{,}\) but \(h y_1 = h y_2\text{.}\)

Maybe this is where it would be helpful to define maps \(X \xrightarrow{\langle f,g \rangle} Y \times Y\) and \(X \xrightarrow{\langle g,f \rangle} Y \times Y\text{.}\) We could also potentially define an operator \(Y \times Y \xrightarrow{\alpha} Y \times Y\) that swaps the order of any \(\langle y_1,y_2 \rangle\) to become \(\langle y_2,y_1 \rangle\text{,}\) and thus establishing \(\langle f,g \rangle = \alpha \langle g,f \rangle\) and \(\alpha^2 = 1_{Y \times Y}\text{.}\)

Well, we’ve already defined two endomaps \(Y \mathrel{\substack{\beta_1 \\ \longrightarrow \\ \longrightarrow \\ \beta_2}} Y \text{.}\) What if we look at how those behave when we look at them as a products \(Y \xrightarrow{\langle 1_Y,\beta_1 \rangle } Y \times Y\) and \(Y \xrightarrow{\langle 1_Y,\beta_2 \rangle } Y \times Y\text{?}\) Or maybe even \(Y \xrightarrow{\langle \beta_1,\beta_2 \rangle } Y \times Y\) and \(Y \xrightarrow{\langle \beta_2,\beta_1 \rangle } Y \times Y\text{?}\)

Out of sheer curiousity, let’s draw out a graph product of what \(Y^{\circlearrowleft \beta_1} \times Y^{\circlearrowleft \beta_2}\) might look like:

I don’t really know what I was looking for here, but it seems interesting that this particular endomap \(Y \times Y \xrightarrow{\beta} Y \times Y\) seems to satisfy \(\beta^3 = \beta\text{.}\)

Let’s see if this property generalizes to a different map. Consider the map \(X^{\circlearrowleft \alpha}\) defined in Session 15. What happens if we take the product \(X^{\circlearrowleft \alpha} \times X^{\circlearrowleft \alpha}\text{?}\)

Okay, that’s pretty ugly. But is it helpful?

This product contains 2 cycles of two (red), 2 cycles of six (purple), and 3 cycles of three (blue). No matter where you start in the graph, you end up in one of those cycles after two steps. After six more steps, you’ll loop through all of points in that cycle (perhaps more than once). We’re guaranteed this \(X^{\circlearrowleft \alpha}\) will satisfy \(\alpha^8 = \alpha^2\text{.}\) Likewise, the respective map \(X \times X \xrightarrow{\bar{\alpha}} X \times X\text{,}\) should also satisfy \(\bar{\alpha}^8 = \bar{\alpha}^2\text{.}\)

I also noticed that the points given by \(\langle x_i,x_i \rangle\) effectively contain a copy of the endomap \(\alpha\) along the diagonal of our product. This was not the case when we took the product of two different maps \(\beta_1,\beta_2\text{.}\) The only points where the diagonal has a well-defined endomap are the points \(\{y_1,y_4,y_7,y_8,y_9\}\text{.}\)

That’s a lot of interesting information, but I feel like I’m straying off topic. Let’s attempt to revise our definition of coequalizer first:

I’m not sure if that’s correct or not, but think there are some improvements over my last one at least. Firstly, I’m not trying to iterate over unique maps. Secondly, in the same way that the equalizer iterates over solutions, the coequalizer iterates over cosolutions.

My guess is that this \(X_h\) is probably something like \(Y+Y\text{.}\) This would give me the two injection maps \(X_h \mathrel{\substack{j_1 \\ \longrightarrow \\ \longrightarrow \\ j_2}} Y\) that could match the maps described in the exercise. I’m wondering if these would need to be “structure preserving” equivalents of the endomaps \(\beta_1,\beta_2\text{?}\)

Since we further have a unique map \(Y \times Y \rightarrow Y+Y\text{,}\) that would essentially give us “jointly monomorphic” property when we compose with \(X \xrightarrow{\langle f, g \rangle} Y \times Y\) because \(q\) is an monomorphism.

That leaves the transitive property. I’m thinking that the idea here would be to define two cosolutions \(Y \xrightarrow{h_1} Z_1\) and \(Y \xrightarrow{h_2} Z_2\text{,}\) then use the universal \(q\) to construct an isomorphism between them. There should be a unique map \(Z_1 \xrightarrow{t_1} T\) a unique map \(Z_2 \xrightarrow{t_2} T\text{,}\) and unique map \(Y \xrightarrow{t} T\text{.}\) Since our definition implies the existence of maps \(Z_1 \xrightarrow{z_1} X_h\) and \(Z_2 \xrightarrow{z_2} X_h\text{,}\) the uniqueness of terminal objects implies \(q = z_1 h_1 = z_2 h_2\text{.}\)