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$\newcommand{\Inj}{\operatorname{Inj}}$$\newcommand{\Prop}{\operatorname{Prop}}$Consider the "category of propositions" or "of properties" $\Prop$, where objects are logical propositions (statements? properties? predicates?) $P, Q$ and there is an arrow $P \to Q$ if and only if $P \implies Q$ ($P$ implies $Q$). (We could also consider the opposite category - it doesn't really matter.)

Given an arbitrary set $S$, we can consider the "power set category", "the slice category $\Inj /S$, where $\Inj$ is the wide subcategory of $\operatorname{Set}$ with morphisms restricted to injections" (quoted from nLab).

For "any" proposition $P$ (again, see note below), i.e. object of $\Prop$, the axiom schema of specification implies that there is a corresponding object $\{S | P \} \hookrightarrow S$ of $\Inj/S$, where $\{S|P\} \overset{def}{=} \{s \in S: P(s) \text{ is true}\}$. i.e. that "set builder notation" makes sense/is valid.

(Cf. notation here, where $P$ is described as "a formula with one free variable", or here where $P$ is described as a "property". $P(s)$ is meant to denote that $s$ has property $P$ or that formula $P$ is true with $s$ as the value of the free variable. E.g. $P$ is "$\cdot$ has an inverse" and $P(s)$ is "$s$ has an inverse".)

Questions:

  1. What is the name of the principle/axiom (schema) that allows us to extend this to an actual functor $\Prop \to \Inj/S$, i.e. defined on arrows/morphisms, not just on objects?

More concretely, we know that/it's "obvious" that, if $P \implies Q$, then $\{S | Q\} \subseteq \{S | P\}$, and therefore we have a commuting triangle of injections, i.e. a morphism in $\Inj/S$, and it's straightforward to check that this relationship is functorial.

Intuitively, the more restrictive a property is, the fewer elements satisfy it.

  1. Does the axiom schema of specification only imply the definition of the functor on objects, or does it actually also imply the definition of the functor also on arrows/morphisms?

  2. What is the name of this functor?

Notes:

  • I know that it is important to be specific about what logical language the objects of $\Prop$ can be written in. Cf. the nLab article. However, the functorial relationship doesn't seem to depend on the precise definition of $\Prop$ (although please do explain if it does), and I have no training in formal logic and so wouldn't understand a precise definition of $\Prop$ anyway.

  • With $\Prop$ defined as above, i.e. $P \to Q$ exactly when $P \implies Q$, this functor is contravariant, but again we could just as easily define $\Prop$ as the opposite of this category and then the functor would be covariant.

  • It seems that this functor can be interpreted as "the reason why" the same rules of Boolean algebra apply both to logical propositions and to power sets.

  • "Morally" there should probably be some terminology for this in terms of ETCS and/or topoi, because those are apparently related not only to foundations of set theory but also categorical logic, but I won't pretend to know/understand enough about any of those to be able to find the answer by myself currently.

  • Assume a fixed model of set theory is chosen (e.g. axiom of choice or continuum hypothesis is or is not valid, it shoudn't matter). Maybe some notion of "syntactic category", or "categorical notion of theory". At least according to nLab, "One may interpret mathematical logic as being a formal language for talking about the collection of monomorphisms into a given object of a given category: the poset of subobjects of that object. More generally, one may interpret type theory and notably dependent type theory as being a formal language for talking about slice categories, consisting of all morphisms into a given object." So maybe "the functor" I have in mind is just any functor from a category describing a given theory to $\Inj/S$?

Related questions:

hasManyStupidQuestions
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    For one thing, I think this functor needs to be defined in the context of a fixed model of set theory (and the set $S$ should be an element of that model), because a statement can be true in some models and false in others. E.g. consider $S=\Bbb N$ and $P(x)=$"the continuum hypothesis holds". – Karl Feb 16 '22 at 16:49
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    @Karl Please feel free to assume that your favorite fixed model of set theory is chosen. Presumably any "reasonable" model of set theory includes "set builder notation", thus the axiom schema of specification. Anyway I edited the post to address the concern. – hasManyStupidQuestions Feb 17 '22 at 00:52
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    Assuming classical logic, the poset of propositions has two elements true, false, and can be embedded into any power set by mapping false to the empty set and true to the maximal subset. I think that's the functor you're looking for. writing ${x | P(x) }$ doesn't really make sense since propositions don't take arguments. – Jonas Frey Feb 17 '22 at 01:04
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    Presumably by "propositions" OP means unary predicates and has an arrow $P\to Q$ when $\forall x (P(x)\implies Q(x))$ is true (i.e. satisfied by the selected model) (or maybe provable?). My logic background is patchy, but I think the ambiguity there is part of the problem in attempting to describe this functor. – Karl Feb 17 '22 at 01:07
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    "The idea" is that $P$/$Q$ is supposed to be a "property" (e.g. "is an open subset", "is differentiable", etc.) such that for each element of $S$ it is either true or false (in classical logic) that the property is satisfied. In Wikipedia it talks about elements $x$ "obeying a formula $\phi$ with one free variable" as $\phi(x)$, in nLab it talks about a "property P" and elements $a$ of $X$ for which it holds, written $P(a)$. I'll try to clarify the post. – hasManyStupidQuestions Feb 17 '22 at 01:11
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    Maybe "class" is what you want, so that the functor acts on classes by taking their intersection with $S$. – Karl Feb 17 '22 at 01:32
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    Yes, I think that is one way it can be defined, although I feel that might be "kicking the can down the road a little bit"? E.g. if we define ${S|P}$ to be the intersection of $S$ with the class of all sets satisfying property $P$, and $P \implies Q$, then we get ${S | Q} \subseteq {S|P}$ because the class of all sets satisfying property $Q$ is a subclass of all sets satisfying property $P$. That might be easier to prove or relate to a given axiom - idk. Really what I'm interested in is the name for $(P \implies Q) \iff {S|Q} \subseteq {S|P}$, and if it can be formalized as a functor – hasManyStupidQuestions Feb 17 '22 at 01:37
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    Yeah, I think this "kicking the can down the road" problem is kinda the whole point. If you define predicates semantically (as classes) then the functor is sort of trivial, but if you define predicates syntactically (as formulae with one free variable) then how do you know what the arrows are? Do they represent provable entailment in some deductive system? Just trying to state the question clearly gets us into logic and model theory. – Karl Feb 17 '22 at 01:45
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    The propositions in Prop will have to be written in a formal language. If you are not precise about what a proposition is and what not, then you are asking for inconsistencies. Now once you have a formal language, you can usually form a syntactic category, and models of your language are functors from this syntactic category into other categories which provide semantic for your language. So my best guess for what you are looking for is Lawveres functorial models/semantic. A good book to look at is Categorical Logic and Type Theory by Jacobs – Nico Feb 17 '22 at 19:51
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    I will write a detailed answer once I understand enough about the topic! You will have to wait for 2 or 3 months though. – Nico Feb 17 '22 at 19:52
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    @Karl I think this is a very good argument. Purely speculatively, maybe it might even effectively be an axiom/property of a formal theory that such a functor exists (e.g. how Zorn's lemma isn't really a good axiom, but equivalent to axiom of choice in ZF theory). And then one could study logics/etc. where the functor doesn't exist and their properties. The problems you mention also sound similar to what the Wikipedia article on ZF theory describes as the reason for using the "modern" schema of replacement instead of the older Zermelo schema of specification – hasManyStupidQuestions Feb 18 '22 at 01:02
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    @Merle Haha, no worries, I should start reading the book myself too! The reference is good. I think you are right. The functor seems to follow directly from specification and transitivity of subsets, the former apparently requiring us (according to the Wikipedia articles on ZF as a whole and the axiom schema itself) to be specific about a given 1st-order logic language, the second (according to proofwiki https://proofwiki.org/wiki/Subset_Relation_is_Transitive) requiring us to be precise about 0th order logic seemingly. Trying to define them as categories doesn't escape that need for rigor. – hasManyStupidQuestions Feb 18 '22 at 01:12

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