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

The lack of a question means we’re proving the given statement, so let’s peice together what we’ve been given.

Our point is defined as a \(\mathcal{C}\)-map \(\mathbf{1} \rightarrow X\text{,}\) where \(\mathbf{1}\) is a terminal object in \(\mathcal{C}\text{.}\) By the definition of terminal object, there exists exactly one \(\mathcal{C}\)-map \(X \rightarrow \mathbf{1}\) for each object of \(X\text{.}\) To help keep these ideas separate, let’s call the point \(\mathbf{1} \xrightarrow{x} X\) and the "terminal map" \(X \xrightarrow{\bar{x}} \mathbf{1}\text{.}\)

If we have some arbitrary map \(X \xrightarrow{f} Y\text{,}\) then we should be able to compose this map with our "point map". Namely, if we have a map \(\mathbf{1} \xrightarrow{x} X\) in our category \(\mathcal{C}\) then our composition \(f \circ x\) should also be in the category. This effectively gives us a map \(\mathbf{1} \xrightarrow{f x} Y\) since we have \(\mathbf{1} \xrightarrow{x} X \xrightarrow{f} Y\text{.}\)

So how do we prove \(f x\) is a point? I think we basically already have done so by establishing the existance of a map \(\mathbf{1} \xrightarrow{f x} Y\) through composition. We might have to also prove that there’s a corresponding map \(Y \xrightarrow{\bar{f x}} \mathbf{1}\) too, which I think we can effectively do by following the arrows backwards since each of these single element sets are isomorphic to every other one by the "Uniqueness of Terminal Objects" theorem.

I think we’re effectively splitting the domain up into points. If we have a map \(X \xrightarrow{f} Y\text{,}\) this terminal object lets us break apart \(X\) into a bunch of individual points \(\mathbf{1} \xrightarrow{x_i} X\text{,}\) and consequently allows us to break up \(f\) into smaller maps where the domain and codomain are both single element sets. For any map \(f: X \rightarrow Y\text{,}\) each point \(x_i\) in \(X\) could be used to construct a map \(f_i: \{x_i\} \rightarrow \{f x_i\}\text{.}\) Each such "mini-map" must be invertable, so there’s precisely one map from \(\{f x_i\}\) to \(\mathbf{1}\) and it’s given by the composition \(\bar{x_i} f_i^{-1}\text{.}\)

I’m thinking that this is actually a special case of Exercise 1 where \(Y = X\) and \(f = 1_X\text{.}\) If we have a point \(x \in X\text{,}\) that implies the existance of a map \(1 \xrightarrow{x} X\text{.}\) If we compose this map with the identity map, the result should also be a point in \(X\) since the category is closed under the composition \(\mathbf{1} \xrightarrow{x} X \xrightarrow{1_X} X\text{.}\)

Since \(x\) is a point of \(X\text{,}\) there must be a unique \(\mathcal{C}\)-map \(X \xrightarrow{\bar{x}} \mathbf{1}\text{.}\) If we take this map and compose it with \(1_X\text{,}\) then the result should also be a \(\mathcal{C}\)-map.

In other words, the requirement of an identity map in our category ensures that these maps are unique. Since we have a sequence of maps \(\mathbf{1} \xrightarrow{x} X \xrightarrow{\bar{x}} \mathbf{1}\text{,}\) it also should follow that the composite \(\mathbf{1} \xrightarrow{x} X \xrightarrow{1_X} X \xrightarrow{\bar{x}} \mathbf{1}\) is also in our category. Note that \(1_X \circ x = x\) and \(\bar{x} \circ 1_X = \bar{x}\) since we’re using the identity map.

I suppose the part we haven’t addressed yet is the uniqueness property. If we had a point \(x\) of \(X\) that points to two different elements of \(x_1,x_2\text{,}\) then we don’t have a well-defined "map" because each point of a map can only point to one value. On the other hand, if we had two points \(x_1, x_2 \in X\) that both point to some point \(x \in X\) then \(x\) is not really a point because the map \(X \xrightarrow{\bar{x}} \mathbf{1}\) needs to be unique by the definition of \(\mathbf{1}\) as a terminal object.

So now we don’t just have a set \(X\) alone, but one that’s also equipped with an endomap \(\alpha\text{.}\) If this set has a terminal object \(\mathbf{1}\text{,}\) that implies the existence of exactly one \(\mathcal{S}^{\circlearrowright}\)-map \(\boxed{X^{\circlearrowright\alpha}} \xrightarrow{f_x} \boxed{\mathbf{1}^{\circlearrowright\beta}}\) for each element \(x\) in \(X\text{.}\) However, since there’s only one element in \(\mathbf{1}\) the only possible endomap on \(\mathbf{1}\) is the identity map \(\beta = 1_\mathbf{1}\text{.}\) It follows from the structure preserving property of \(\mathcal{S}^{\circlearrowright}\) that \(f_x \alpha = \beta f_x\text{,}\) and substituting \(\beta = 1_\mathbf{1}\) gives us \(f_x \alpha = f_x\text{.}\)

Effectively, this map \(f_x\) must be invariate with respect to the operation \(\alpha\text{.}\) If \(x\) is a fixed point under \(\alpha\text{,}\) then it’s obvious that \(f_x \alpha x = f_x x\) will hold true because \(\alpha x = x\text{.}\) If \(x\) is not a fixed point, then \(\alpha x \neq x\) implies that \(\alpha x\) is a different point in \(X\) and must have it’s own unique map \(f_{\alpha x}\) such that \(f_{\alpha x} \alpha = f_{\alpha x}\text{.}\)

In this category, an \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\)-map requires that both "source and target" must be preserved. Namely, \(s' f_D = f_A s\) and \(t' f_D = f_A t\text{.}\) If the codomain of any such a map \(\boxed{X^{\downarrow_\bullet^\bullet \downarrow}} \xrightarrow{f} \boxed{Y^{\downarrow_\bullet^\bullet \downarrow}}\) is a one-element space \(Y=\mathbf{1}\text{,}\) then the only possible choice for of maps for \(s'\) and \(t'\) is \(s' = t' = 1_\mathbf{1}\text{.}\)

It follows that \(f_A s = s' f_D = 1_\mathbf{1} f_D = f_D\) and \(f_A t = t' f_D = 1_\mathbf{1} f_D = f_D\text{.}\) This, in turn, implies that \(f_A s = f_A t = f_D\) for every "point" \(\mathbf{1} \xrightarrow{x} X\text{.}\) In other words, each "point" in \(X\) refers to a "dot" which is both the source and target of an "arrow". Such points are effectively our "self-loops" in \(X\text{.}\)

I’m a little uncertain about this actually. I was expecting to see a loop of multiple points rather than something that looks like a fixed point.

Suppose we have a pair of maps \(X \mathrel{\substack{f \\ \longrightarrow \\ \longrightarrow \\ g}} Y\text{,}\) and each point \(\mathbf{1} \xrightarrow{x} X\) has the property that \(f x = g x\text{.}\)

It seems like the fact that \(f = g\) is self-evident to a degree. Two maps are the same if the domains and codomains are the same, and they agree on the value of each possible input. We already established that \(x\) being a point in \(X\) means that \(f x\) and \(g x\) are necessarily points in \(Y\text{.}\)

So why would this fail in \(\mathcal{S}^{\circlearrowright}\) or \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\text{?}\) I’d suspect it has something to do with the way the our dots can degenerate into arrows. Consider the following endomap:

If we take each of the "points" and replace them with an "self-loop", we’d get something like this:

I think the problem here arises when we try to distinguish between the self-loop that "is" \(x_2 \circlearrowright\) and the self-loop that connects \(x_2 \circlearrowright\) to itself. We’ve defined terminal objects in such a way that there would be exactly one map from \(X \rightarrow \mathbf{1}\text{,}\) but here we essentially have two different arrows which both have a source of \(x_2\) and a target of \(x_2\text{.}\)

This has got me second guessing my answers to Session 17 Exercise 1. I’m starting to wonder if I was supposed to classify (b) is terminal but (a) as not, because (a) admits a loop that (b) doesn’t.