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"Glue"ing? ...Seriously? Anyway, this is somewhat close to being "not worth doing because it's just about the existence of a convenient object," but I think it's worth doing for the sake of construction. So let's try2 NATURALLY construct this badboy and not worry too much abt verifying its validity. Ad then, we'll be done with section 1. WOOT SEASON FINALE SUCKAS. So here's how I went about it:

Attempt #1: Sexting with sections

Okay, so this was my first idea. I defined

F(U) = PU Fi(Ui U)

Got U? Just iterate over all the Fi elements where they intersect U and stick 'em together. Seems naturalish, but here's the PROBLEM: What if they don't overlap in a consistent way? What if I pick s Fi(Ui U) and t Fj(Uj U) where s and t don't restrict to the same thing (thru ϕij) in a neighborhood of Ui Uj U? That's not good. The reason it's not good is if I go down to U = Ui, i'm getting extra elements that aren't in Fi(Ui) (so we won't satisfy the Fi F|Ui condition)

"Alrite, how about adding an extra stipulation that the components DO pairwise overlap consistently, using the ϕij's?" Okay that seems fair. Maybe that actually works, in fact. HOWEVER, I'm gonna ghost this idea. This "extra stipulation" in the name of "consistency" reminds of another stud, especially since after getting pounded the other night, my brain is STALKED.

Attempt #2: U can talk the talk, but can you stalk the stalk?

Remember the definition of sheafification?


Okay. I'm going to basically copy this definition. I'm gonna copy it and it's GONNA WORK!!!

We construct the sheaf F as follows. For any open set U, let F(U) ⊂{s : U PU FP} where " FP" can be taken as any stalk ( Fi)P where P Ui. The choice of stalk doesn't matter cause they're isomorphic thx to the ϕij's. And, the s must satisfy

  1. For each P U,s(P) FP. Okay, THIS is where we use the triple interesection property given in the exercise, although it's easy to miss. The selection of s(P) doesn't matter on the which Fi stalk we use, because it represents the "same" element in each stalk.
  2. Given P U, assume we've picked i as the representative index. Then we must have a neighborhood V U of P and an element t Fi(V ) such that Q V : tQ = s(Q). The validity of this property again doesn't depend on the choice of stalk because the sheaves are locally isomorphic (so such a t exists so long as we sufficiently refine)


Yep. There's a STALKIFIED condition. I think the fact that this sheaf satisfies the properties is "obvious". Other than uniqueness, which...


Err, woops. remember that brilliant two-birds-with-one-horn trick I came up with? Yeah... I probably shouldve done that here..... You know, using the conditions to construct a unique object.... TOO LATE.


So that's that for section 1 of Hartshorned II. Kind of a blistering short "finale," but those ain't so bad are they? Next section, we finally define the famous scheme! WE ARE ENTERING THE HEART OF MODERN ALGEBRAIC GEOMETRY. VARITIES? ZERO SETS OF POLYNOMIALS? PSSHHHHHHHHH. I CAME HERE FOR THE 900IQ ABSTRACT STUFF NOT BORING CONCRETE STUFF. MUAHAHAHAHAHA. SEEYA IN SCHEMELAND.

Oh, and since this is a season finale, I made a closing picture on my reflections on section 1.

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