If \(S, s, t\) is a given bipointed object...
Solution.
There’s a lot to unpack here. Let’s start with the "bipointed object" in a category \(\mathcal{C}\text{:}\)
Our goal is to demonstrate that "the graph of ’\(X\) fields’ on \(S\) is actually a reflexive graph". Before I get there, I think I want to try to break down the "tempature field" situation to get my bearings.
My conventual notion of a "tempature field" would be like a map \(f: S \longrightarrow T\) from some 3D-space such as \(S = \mathbb{R}^3\) describing the position to a 1D-line such as \(T = \mathbb{R}_{\geq 0}\) describing the temperature at each point in Kelvins. Or at least that’s the basic idea that I’m trying to generalize. Maybe my space is the 2D image of a camera and my temperature line is an interval of temperatures it can detect.
I think this notion of a "heat map" produced by a digital camera is my best lead to go on. Years of programming have taught me to think of this as "an array of bytes". My input space is confined by the height and width of my image and my output space is limited by the precision of the image. Let’s consider a super simple camera that only samples a 4x4 region and has only a 4 temperature resolution.
The key feature that strikes me about this image is that despite this map being non-invertable we still must have a 1-1 correspondance between points in \(S\) with the "arrows" of the map. If we were able to "step out" of the map and look at it from a higher dimensional space we could think about the map as a collection of points \(\mathbf{1} \longrightarrow \mathbb{Z}_4 \times \mathbb{Z}_4 \times \mathbb{Z}_4\) with 3 component properties defined by some maps \(s_x, s_y, t\) which describe the position \(s_x,s_y\) and temperature \(t\text{.}\)
It’s incredible unlikely that our temperature \(t\) alone is going to have an inverse map because neighboring points are likely to have similar temperatures. However, if we have two objects \(x_1,x_2 \in X\) which match for the whole map-triple \((s_x,x_y,t)\) — such that \(s_x x_1 = s_x x_2\text{,}\) \(s_y x_1 = s_y x_2\text{,}\) and \(t x_1 = t x_2\) — that would necessarily imply that they’re the same object \(x_1 = x_2\text{.}\)
By looking at my "camera" in the higher dimensional space \(S \times T\) I can think of it as an "endomap". There’s only one detail which needs to be worked out, and that’s the fact that my original map had 16 arrows to go with each of the \(4 \times 4 = 16\) "dots", we’ve now added in the 4 additional dots from our temperature space \(T\text{.}\) For us to have a well-defined endmap, these dots would need to have arrows also. It seems natural that I’d want these new arrows to "loop".
Maybe it’s not so much a question of if these arrows "should" loop because the properties of our category imply they "must" loop. The fact that any \(f: X \longrightarrow Y\) must preserve source and target for our bipointed object can be expressed as the equations \(s_Y \circ f = f \circ s_X\) and \(t_Y \circ f = f \circ t_X\) as per the the following commutative diagram.
Our "bipointed object" contains 3 arrows, which I’ll call \(a_0,a_1,a_2\text{,}\) and 3 dots, which I’ll call \(d_0,d_1,d_2\text{.}\) Let’s further specify that sources of our arrows are given by \(s a_0 = s a_1 = d_0\) and \(s a_2 = d_1\text{,}\) while the destinations are \(t a_0 = d_1\) and \(t a_1 = t a_2 = d_2\text{.}\)
What strikes me as special about our bipointed object here is that it provides a one-to-one correspondence between "dots" and "arrows". We know we can’t have a retraction from \(\mathbf{1}\) to \(S\) because there’s no retraction from a 2-point space to a 1-point space, but we can start building a common section. Let’s call this map \(i:S \longrightarrow \mathbf{1}\) and define it such that \(i d_1 = i d_2 = d_0\text{.}\) The existence of this common section establishes that we have a reflexive graph. In Exercise 16, we proved that in a reflexive graph \(f_D\) is both determined by \(f_A\) and implies that there’s an arrow forming a "self-loop" at each point.
Conceptually this makes sense for finite sets. If I know I have at least one arrow at each point, then I should be able to follow them. If I have some set of arrows \(A\) and some set of dots \(D\) describing a map \(f\text{,}\) then there can’t be more dots in the image of \(f\) than there were arrows in the source space.
There’s precisely one possible endomap on the space \(\mathbf{1}\) and that’s the identity map. In order for some \(f: \mathbf{1} \longrightarrow S\) to preserve the structure of this endomap, we’d necessarily need to have \(t' f_A = s' f_A\text{.}\) As a result of our category’s structure preservation, it follows that \(f_D \circ t = f_D \circ s\) which our result in Exercise 25 showed to imply that \(f_A\) maps every arrow in \(X\) to a loop in \(Y\text{.}\)
At this point I feel like I’m starting to talk myself in circles, so I’ll take that a clue to move on to the next Session and hope that it helps clear up some of the details. I feel like I can at least see how the the preceeding Exercises were leading up to this problem, which gives me a bit of encouragement moving forward.
