Section4.12Session 15: Objectification of Properties, Part 3
I made some more progess on Session 15 this week, but there’s still much more work to be done. In some ways, my progress through the text seems to parallel the algorithm for finding a presentation of \(X^{\circlearrowright \alpha}\text{.}\) Each section of the text is kind of like a ’generator’ element. The only way for me to find out how many parts Session 15 will take is for me to keep working on it until it reaches a stable point.
Example4.12.1.Exercise 4:.
...\(\alpha\) is itself a map of dynamical systems...
Solution.
We’ve defined the category of \(\mathcal{S}^{\circlearrowright}\) based on the quality of structure preservation such that for an arbitrary map \(X^{\circlearrowright \alpha} \xrightarrow{f} Y^{\circlearrowright \beta}\) we enforce the condition that \(f \alpha = \beta f\text{.}\) For this exercise, we’re essentially considering the special case where \(f = \alpha = \beta\) and \(Y = X\text{.}\)
We know that \(\alpha\) is an endomap on \(X\text{,}\) so we know \(\alpha: X \rightarrow X\) is a valid map in \(\mathcal{S}\text{.}\) To prove that we have a well defined map \(X^{\circlearrowright \alpha} \xrightarrow{\alpha} X^{\circlearrowright \alpha}\) we’ll need to show our structure preservation property still holds. If we substitute \(f = \alpha = \beta\) into \(f \alpha = \beta f\text{,}\) we get \(\alpha \alpha = \alpha \alpha\) which is obviously true.
It really feels like this is stating the obvious, but perhaps we never explicitly proved that \(\mathcal{S}^{\circlearrowright}\)-maps must satisfy this "reflexive property".
Example4.12.2.Exercise 5:.
..if \(\mathbb{N}^{\circlearrowright \sigma} \xrightarrow{f} Y^{\circlearrowright \sigma}\) corresponds to \(y\text{...}\)
Solution.
I’m assuming we’re using the same set-up as Exercise 3 here. Given \(\mathbb{N}^{\circlearrowright \sigma} \xrightarrow{f} Y^{\circlearrowright \sigma}\text{,}\) we know \(f \sigma = \beta f\text{.}\) In particular, this must be true for \(0\text{,}\) establishing that \(f \sigma 0 = f 1 = \beta f 0 = \beta y\text{.}\) We then used induction to establish that \(f(n) = \beta^n y\) for all \(n \in \mathbb{N}\text{.}\)
In Exercise 4, we proved that for any \(X^{\circlearrowright \alpha}\text{,}\) the map \(\alpha\) is itself a well defined map \(X^{\circlearrowright \alpha} \xrightarrow{\alpha} X^{\circlearrowright \alpha}\text{.}\) It follows that \(\mathbb{N}^{\circlearrowright \sigma}\) defines a map \(\mathbb{N}^{\circlearrowright \sigma} \xrightarrow{\sigma} \mathbb{N}^{\circlearrowright \sigma}\text{.}\) Since compositions of \(\mathcal{S}^{\circlearrowright}\)-maps produce \(\mathcal{S}^{\circlearrowright}\)-maps, our composition \(f \circ \sigma\) should be well defined map \(\mathbb{N}^{\circlearrowright \sigma} \xrightarrow{f \circ \sigma} Y^{\circlearrowright \beta}\) such that \((f \circ \sigma) \circ \sigma = \beta \circ (f \circ \sigma)\text{.}\)
When they say "\(\mathbb{N}^{\circlearrowright \sigma} \xrightarrow{f} Y^{\circlearrowright \sigma}\) corresponds to \(y\)", we mean that we can evaluate \(f\) at 0 to get \(y\) or iterate from \(y\) to get \(f\) using \(f(n) = \beta^n y\text{.}\) For \(f \circ \sigma\text{,}\) we can easily evaluate \((f \circ \sigma)(0)\) to establish \(f \sigma 0 = \beta f 0 = \beta y\text{.}\) However, we still need to show that \((f \circ \sigma)(n) = \beta^n (\beta y)\text{.}\)
As in Exercise 3, I think we can do this through induction. We’ve already established that \((f \circ \sigma)(0) = \beta^0(\beta y)\text{,}\) so we know that \((f \circ \sigma)(n) = \beta^n(\beta y)\) for \(n = 0\text{.}\) Consider the case of \(\sigma n = n + 1\text{.}\) Given \((f \circ \sigma)(n) = \beta^n(\beta y)\text{,}\) we can rewrite \((f \circ \sigma)(n+1)\) as \((f \circ \sigma)(\sigma n)\text{.}\) Subsitute \(f \circ \sigma = \beta \circ f\) to get \((\beta \circ f)(\sigma n)\) which can be rewritten as \(\beta( (f \circ \sigma)(n))\) by the associative property. This means that \((f \circ \sigma)(n) = \beta^n(\beta y)\) implies \((f \circ \sigma)(n+1) = \beta( (f \circ \sigma)(n)) = \beta(\beta^n(\beta y)) = \beta^{n+1}(\beta y)\text{.}\)
Having established that both \((f \circ \sigma)(0) = \beta^0(\beta y)\) and \((f \circ \sigma)(n) = \beta^n (\beta y) \implies (f \circ \sigma)(n+1) = \beta^{n+1}(\beta y)\text{,}\) we can use induction to establish that \((f \circ \sigma)(n) = \beta^n (\beta y)\) for the entire domain of \(\mathbb{N}\text{.}\)
We’re provided the following diagram for \(B\text{:}\)
Figure4.12.4.Definition of \(B\)
Here the set \(X\) is the set of people, the operator \(m\) is a map that returns the mother of specified person, and \(B\) is the gender binary map specified above. In order to show \(g\) is an \(\mathcal{S}^{\circlearrowright}\text{,}\) we need to establish that it is both a valid map in \(\mathcal{S}\) and upholds the structure of the endomap \(m\text{.}\)
First of all, let’s address the reality that this is toy model of gender. In this exercise we’re implicitly assuming\(g\) is a valid \(\mathcal{S}\)-map. We’re excluding the existence of non-binary people in this way. We’re also assuming that we have a mother-map, \(m x\text{,}\) that’s defined for all \(x\text{.}\) This model breaks if we "clone" a male, effectively excluding Stormtroopers from our set of people.
There’s also some ambiguity because we haven’t yet named the endomap on \(B\) yet. I think the reason is because the endomap on \(B\) is induced by the maps \(m\) and \(g\text{.}\) For clarity’s sake, let’s define \(\bar{m}: B \rightarrow B\) to be the corresponding endomap in the "gender-binary space". This map \(\bar{m}\) is based on an assumption that the answer to "what gender is your mother?" will always "female".
Under those contraints on \(X\text{,}\) showing \(g\) is an valid \(\mathcal{S}^{\circlearrowright}\)-map basically comes down to establishing the relation \(g m = \bar{m} g\text{.}\) Perhaps this is most easily shown in a table.
Table4.12.5.Table of values for gender map
\(x\)
\(g x\)
\(m x\)
\(g m x\)
\(\bar{m} g x\)
"son"
male
"mother"
female
female
"daughter"
female
"mother"
female
female
Essentially, both \(g m\) and \(\bar{m} g\) are constant maps. We know \(g m = \bar{m} g\) because we’ve defined them both so that they both always return "female". In order for us to have an "objectification in the subjective of gender", all we have to do is to start excluding the people that don’t fit our model!
I guess the lesson here is that if you ask a "yes/no" question you have to get a "yes/no" response. In this case, our standard idempontent map on \(B\) enables us to ask something to the effect of "is this a yes/no question?" which is forced to converge to the value "yes".
