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Stream: learning: questions

Topic: Generalizing the sum of product of powersets

Bassel El Mabsout (Aug 01 2020 at 16:13):

I'm trying to get a better grasp of CT, so I'm trying to take an equivalence relation that I'm aware of, that looks like it is generalizable to categories with certain properties.

$$\sum_{p \in P(S)} \prod_{x \in p} x = \prod_{x \in S} (x + 1)$$

here the set S would be a finite set of integers for example (see: https://math.stackexchange.com/a/2646380) and P(S) is the power set. This seems to me like it would have a very nice category theoretic interpretation. If we have a closed cartesian category (as the category used for the "lower" part currently named S) with finite limits and there's a way to talk about a generalized powerset of that (which seems to be power objects or maybe specifically a subobject classifer). Then I think the equivalence can become an isomorphism somewhere that may be true in things other than Set. Am I on the right track? Am I way off?

Javier Prieto (Aug 02 2020 at 13:57):

Here's a non-answer: If you interpret $S$ as a (small, finite?) category my first guess for the "generalized powerset" would be the functor category $\Omega^S$ with $\Omega$ being the subobject classifier in $\mathrm{Cat}$, but apparently this doesn't exist.

I think it'd be better to interpret $S$ as an object in a sufficiently nice category (maybe a topos?) and let the $x$'s have type $x\colon 1 \to S$; but again, I'm just guessing.

Spencer Breiner (Aug 02 2020 at 14:27):

Hi Bassel. You can think of this as a bijection between two sets of morphisms in Set. The trick is to represent your set of sets $S$ as a bundle $\pi:X\to S$. For each element $x\in S$, we think of $\pi^{-1}(x)\subseteq X$ as the elements of $x$ (called the fiber of the bundle).

For the LHS fix a subset/monic $P\rightarrowtail S$. Then an element of the product of $\prod_{x\in P} x$ gives an arrow $s:P\to X$ with $\pi(s(x))=x$ (called a partial section).

On the RHS we are working with a different bundle: $\pi':X+S \to S$; note that this adds one element to each fiber of the bundle. An element of the right-hand product gives a map $s':S\to X+S$, again satisfying $\pi'(s'(x))=x$ (a total section).

We can easily go back and forth between these. Given $s$ we get
$s'(x)= \textsf{if }x\in P\textsf{ then } s(x)\textsf{ else } x$
On the other hand, given $s'$ we can define $P=\{x\in S\ |\ s'(x)\in X\}$.

I'm not sure exactly what is needed to make it go through in general. Probably just (extensive?) coproducts and complemented subobjects (in order to use if-then-else reasoning).

Bassel El Mabsout (Aug 03 2020 at 04:25):

Thank you for your response! I'm gona have to reread this a few times to grok exactly what you've presented but I already learned a few things from the first few passes!