Tricoloring continued...
Solution.
Part (a) asks us to show that “an induced tricoloring is a tricoloring”. Suppose there’s a map \(X_D \xrightarrow{c} \mathbf{3}\) that tricolors the dots of graph \(X\text{.}\) By definition of tricoloring, for any arrow \(A \xrightarrow{a_X} X_A\) in \(X\) we must have \(c s a_X \neq c t a_X\text{.}\)
Given that our map \(Y \xrightarrow{f} X\) is a \(\mathcal{S}^{\downarrow_\bullet^\bullet \downarrow}\)-map, it must preserve structure. Namely, if \(s': Y_A \xrightarrow Y_D\) and \(t': Y_A \xrightarrow Y_D\) then \(s f a_Y = f s' a_Y\) and \(t f a_Y = f t' a_Y\) for every \(A \xrightarrow{a_Y} Y_A\text{.}\)
It should follow from these that \(c' = c \circ f\) is a tricoloring on \(Y\text{.}\)
Suppose \(c'\) is NOT a “tricoloring”. There must exist some arrow \(a_Y \in Y_A\) with \(c' s' a_Y = c' t' a_Y\) such that the source and target have the same color. Expand using our definition of \(c'\) to get \(c f s' a_Y = c f t' a_Y\text{.}\) By our structure preservation property, this is equivalent to \(c s f a_Y = c t f a_Y\text{.}\) However, \(a_X = f a_Y\) is an arrow in \(X_A\text{.}\) The definition of tricoloring states that every \(c s a_X \neq c f a_X\) so the statement \(c s (f a_Y) = c t (f a_Y)\) generates a contradiction. Our assumption must be false, establishing that the tricoloring induced by composition \(c \circ f\) is in fact a tricoloring.
I kind of wish my Python work was reliable enough to answer part (b), but for now I think I’d like to do some research on references to “Fatima” that might yield clues.
- The book is literally dedicated to Fatima.
- The name shows up in Session 1, but as a member of a set.
- In Session 8, Fatima comments that maps having the same domain and codomain is a prerequisite condition for being the same map.
- In Session 9, Fatima is used as an example for someone’s congressional representative. The idea being that the representative is a fixed point of the endomap.
- In Session 11, Fatima suggests using the associative property with the structure preservation properties. This is in the context of a map from a 3-cycle to a given graph.
- In Session 20, Fatima suggests that the terminal object \(T\) translates into \(1\) in arithmetic. Fatimia also points out that this element corresponds with map to a “loop” .
-
In Session 22, Fatima asks about the heavy arrow notation which is defined such that “any problem of factoring a map through such a map has exactly one solution”. There is exactly one map \(\bar{x}\) such that \(\bar{x} p = x\) in the following diagram:
- Also in Session 22, Fatima points out that the two arrows in the graph of the sum are incident at a dot.
Based on this information, I think I can expect my solution to have 3 dots to represent the 3 colors. If the number \(1\) corresponds to a map to the terminal object, perhaps my other dots might be defined by loops of other sizes?
Consider the following graph:
An fixed point shape can only get mapped to \(a\text{.}\) Any 2-cycle shape can only get mapped to \(a,b\text{.}\) Any product-like shape would need to be mapped to \(b\text{.}\) Any sum-like shape would need to be mapped to \(a\text{.}\) Any cycle-like shape would have points like \(c\text{.}\)
Out of curiousity, let’s depict this graph with a matrix also:
\begin{equation*}
\begin{bmatrix} 1 & 1 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0 \end{bmatrix}
\end{equation*}
Hey! Look! It’s a glider from Conway’s Game of Life! I wonder if that’s a coincidence?
1
en.wikipedia.org/wiki/Glider_(Conway%27s_Game_of_Life)If I’m correct, then my “tricoloring” map applied to \(X^{\circlearrowleft \alpha}\) from Session 15 should look something like this:
The part of this that I’m unsure of is how to treat the cycle. It seems like there’s two ways of mapping the 2-cycle to the 2-cycle, which would imply that my coloring is not necessarily unique. Maybe there’s someway I can “step out” and refer to the initial object to sort the two out? Or maybe this is the intended meaning of “induced by exactly one map”? Once you decide how to map the loop, then you have exactly one map?
I think this is as close to an answer as I’m going to get for now.
