🍔 is the set of integers modulo 2 (more literally, if two integers differ by an even integer you consider them the same). I can write out more in a bit.

Edit: this previous post has some good comments, and you can find some of the notation and the answer on the wikipedia page for cohomology ring (they use F~2~ for integers mod 2 and RP^n^ instead of P^n^(R)). I don't know enough algebraic topology to actually know why that's the answer but I can at least answer these:

The first definition of 🍕 with the contravariant thing also doesn’t parse for me, what does that “-” mean in the function arguments?

I assume it's shorthand for saying that if you define f(x) = 🍕(x, B) then f : C --> Set is contravariant.

In the definition of 🌭, what is the n (or the P)? ChatGPT started yapping about real projective space, but I’m not sure if that’s correct.

It's not the notation I'm used to (I'd also think of power sets first), but I think it's n-dimensional real projective space.

🍕(--, B) : C -> Set denotes the contravariant hom functor, normally written Hom(--, B). In this case, C is a category, and B is a fixed object in that category. The -- can be replaced by either an object or morphism of C, and that defines a map from C to Set.

For any given object X in C, the hom-set Hom(X, C) is the set of morphisms X -> B in C. For a morphism f : X -> Y in C, the Set morphism Hom(f, B) : Hom(Y, B) -> Hom(X, B) is defined by sending each g : Y -> B to gf : X -> B. This is the mapping C -> Set defined by Hom(--, C), and it's a (contravariant) functor because it respects composition: if h : X -> Y and f : Y -> Z then fh : X -> Z and Hom(fh, C) = Hom(h, C)Hom(f, C) sends g : Z -> B to gfh : X -> B.

--

P^(n)(R) AKA RP^n is the n-dimensional real projective space.

--

The caveat "phi is a morphism" is probably just to clarify that we're talking about "all morphisms X -> Y [in a given category]" and not simply all functions or something.

--

For more context, the derived functor of Hom(--, B) is called the Ext functor, and the exactness of that sequence (if the typo were fixed) is the statement of the universal coefficient theorem (for cohomology): https://en.wikipedia.org/wiki/Universal_coefficient_theorem The solution to this problem is the "Example: mod 2 cohomology of the real projective space" on that page. It's (Z/2Z)[x] / <x^(n+1)> or 🍔[x]/<x^(n+1)>, i.e. the ring of polynomials of degree n or less with coefficients in 🍔 = Z/2Z, meaning coefficients of 0 or 1.

All I can say is that P(ℝ) refers to a power set of ℝ (all rational numbers). Although I don't know what n stands for in Pⁿ(ℝ)

Basically P(A), where A = {1,2,3}, equal {Φ,1,2,3,(1,2),(2,3),(1,3),(1,2,3)}

Maths also uses my life to solve too