II.1.21cde
1/14/2021
AWWW, YES. We're finally relating what we've learned here a bit back
to chapter 1. EXCITING! Get ready for some EBIC NOSTALGIABAIT! But first,
>Verify that this presheaf is a sheaf
BOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOORING. Skipping this
one. It's NOT that I can't do it. It's just NOT WORTH IT. Bleh.
Got stuck lol [Might add more detail later].
Okay, FINALLY here's one I did. We wanna show the exactness of
 (1) 
which, as usual, we'll do at the stalk level. But first let's make these maps...
Errr... Actually by definition
_{
} is a subsheaf of
_{X}, so ϕ can just be the
inclusion. For ψ on the other hand, we need to, given an open set U, define
ψ(U) : _{X}(U)  → (U) 
Okay. Let's go ahead and unwrap the RHS:
(U)  = i_{*} _{P}(U) ⊕ i_{*} _{Q}(U)  
= 
Okay lol, so before we move on, what does i_{*}
_{P} look like? I mean, first of all, the
regular functions on a singleton are necessarily just gonna be constant functions.
So if P ∈ U, then it just becomes k. Otherwise U ∩{P} = ∅, yielding a 0...
Wait a second....
i_{*} _{P}(U) ⊕ i_{*} _{Q}(U)  = (k or 0) × (k or 0) 
(× is the same as ⊕ for finite products of groups) AND THUS, the natural map
for ψ(U) is just plain ol' evaluation
f  f(P) × f(Q)  (taking f(L) = 0 if f not defined at L)  
Okay, now let's drop down to the stalk level and get exactness there:
 (2) 
Now, erm... I'm just gonna assume the direct product commutes over stalks...
EXERCISE LEFT 2 READER:
_{R}  = (i_{*} _{P}(U) ⊕ i_{*} _{Q}(U))_{R}  
= (i_{*} _{P}(U))_{R} ⊕ (i_{*} _{Q}(U))_{R} 
AWW YEA, MULTISTALK DRIFTING. And thanks to our work in 1.17, I
know that these stalks are just k or 0, depend on R's inclusion in the closures of
P and Q.... HOWEVER.... Recall this is X = P^{1}, in which singletons are
already closed!! So really, this depends on whether R is P or Q. And since P
and Q are distinct...
_{R}  = 
Either way, ψ_{P} is clearly surjective (trivial for
_{R} = 0 and the constant
functions cover
_{R} = k). So now all we have to verify for exactness of (2) is...
imϕ_{R} ker ψ_{R} 
Which I'll split into cases.
Then
_{R} = 0 and ker ψ_{R} = (
_{X})_{R}. Oh, probably a good time to
mention now that by virtue of ϕ_{P} being an injection, we can identify
imϕ_{P} = (
_{Y })_{R}. OKAY. Now since S = {P,Q} is closed, we can obtain
a neighborhood V or R that does not intersect S. Then CLEARLY,
_{Y }(V )  = _{X}(V ) 
(since the restrict to 0 condition is vacuously true), and the same for all smaller
neighborhoods of R, hence yielding
( _{Y })_{R}  = ( _{X})_{R} 
i.e. imϕ_{R} = ker ψ_{R}. Done
Assume wlg R = P. Then
_{P} = k, and
ker ψ_{P} = {< V,f >∈ _{X}f(P) = 0} 
Which if ya think about it, is basically just a description of (
)_{P}. So we're done,
haha.
So here's an example of the famous OOPSIE that we've kept run into about
surjectivity. We gotta show that the map of global sections
ψ(X) : _{X}(X)  → (X) 
is NOT surjective (despite ψ being surjective as a whole). Actually, this is ez.
Since both P and Q are in X, I know that
(X) = k ×k. And a theorem I.3.4
all the way back from Chapter I tells us that
_{X}(X) = k. So our map is
ψ(X) : k  → k × k 
where, following the definition of ψ, the image of ψ is the diagonal, a proper
subset of k × k. DONE.
Okay, first thing's first, let's go ahead and make the injection
This becomes obvious when you work out
. Going by the definition of constant
sheaf, we can write
(U)  = {f : U → Kf continuous}  (3) 
Alright, so I fucked this up a couple of times confused as to what was going on...
but then I realized something. Remember, the constant sheaf gives K a discrete
topology, right? I also I know that if f ∈
(U), it has to be continuous. Now
let's say s ∈ K is in the image of f (there is at least one such element). Now
take another element t ∈ K. By continuity, f^{1}(s) and f^{1}(t) have to both be
open. By assumption, f^{1}(s) is nonempty, but what about f^{1}(t). If it's NOT
empty, then it intersects f^{1}(s), because open sets here are dense. But if s≠t, we
get a contradiction (f has points that map to two different things), so we must
have s = t. So basically, all f does is pick out a point of K... so it can just be
taken as an element of K. So (for nonempty U) I can just rewrte (3) as
(U)  = K 
LMFAO. What a simplification. Also note, given open U

(was unsure about that last equality at first, but it's thanks to good ol'birational
equivalence)
Wee, so that's our injection.
And now we have to deal with this business about... what's that? Skyscraper
sheaves?
∕  ∑ _{P∈X}i_{P}(K∕ _{P})  (4) 
Okay, this looks scary, Let's plug in an open U ⊂ X and see what happens.
(U)∕ (U)  ∑ _{P∈X}i_{P}(K∕ _{P})(U)  
i.e.K(U)∕ (U)  ∑ _{P∈U}K∕ _{P}  
OHHHHHHHH BOY. This is some WILD stuff, lol. So sending an element from
LHS → RHS is evidently extremely trivial, but the reverse direction? We have to
match up tons of copies of K to a SINGLE element of K, and our only hope is
taking advantage of our quotient.
WHAT. THE. FUCK. I need to "glue" a bunch of K∕
_{P} elements into a
SINGLE K(U)∕
(U) element? This honestly doesn't even seem remotely
possible because there's no restriction on the selections of each K∕
_{P}
component. Yes, we are actually trying to prove that you can literally just pick a
ton of random elements from K, and somehow when you quotient all of them by
_{P}s, it glues together into something.
Alright: For the first time in Hartshorned II so far, I am going to look up a
solution. Maybe I should make a counter for how many times I do this (update: it's now on the list page)
UPDATE: I'M RETARDED.
I SOMEHOW forgot to try checking
this at the stalk level. The mantra of sheaves has been "stalks are nice" and I
FORGOT?!!?!:!?!? HOW?????? I know why: Because I was so used to going down
to stalks when it came to the question of exactness; I forgot it would also be
useful for the question of isomorphism. Let's start from (4) and try this again.
∕  ∑ _{P∈X}i_{P}(K∕ _{P})  
i.e. ( ∕ )_{Q}  (∑ _{P∈X}i_{P}(K∕ _{P}))_{Q}  
i.e. _{Q}∕ _{Q}  ∑ _{P∈X}(i_{P}(K∕ _{P}))_{Q}  
i.e. K∕ _{Q}  (i_{P}(K∕ _{Q}))_{Q}  
i.e. K∕ _{Q}  K∕ _{Q}  (from our work on skyscrapers) 
AND THUS THAT QUESTIoN MARK IS NO MORE. DONE. HOW
CONCRETELY THEY ACTUALLY MANAGE TO GLUE TOGETHER IDK,
BUT THE STALKS TELL ME IT'S TRUE, THEREFORE IT'S
TRUE. God bless stalks.
And yes, I know we're on part D and I could make some joke about Initial "D",
but what do we do when the punny butts are before us? Do we bite? When the
dillydallying Dstring butt's beckoning you to sacrifice your morals and senses at
its cheeky altar: Remember the wise magazine that once said "NO TACCHI!"
and the wise man who said that "it pays to be conservative at times."
Maintain your cool outside–protect the warm cave excavated in your chest.
Hold steady. Stay the course; Stay the mandate. Keep your eyes on the
road. Take 10 towels and 10 hands and wipe your face. You'll be fine,
trust me. And when it does confront you, you can recite this: Modem
talking, modern walking in the streets (New desire). Take me higher,
lift me higher with your sneed (I need child). Get your sweet & tight, if
you wanna see me, stalking on the net (I know the way you like it). Get
your NFTs, cause I need no money. All I wanna groom IS. YOU. BABY.
CUNNY IN THE
90S
IS A NEW WAY I LIKE /tv/
CUTE. AND. FUNNY IN THE 90S
COME ON CHILD RUN TO ME
WE. LOVE. CUNNY IN THE 90S
IS A NUDE WAY TO SET ME FREE
IT'S. JUST. CUNNY IN THE 90S
YES I WANNA UOHHHHH
YES I WANNA SEE
Personally, I have another "first Cousin problem."
Note: holeinmyheart is now doublecanceled. It's not clear what he was thinking
here, but for some reason he decided to double down in order to perform the art of
multicancel drifting, thus ending his internet career. He is finally free.
WHEN THE SUN GOES
DOWN,
WILL YOU HEAR THE CRY,
OR JUST WALK ON BY,
AND LEAVE WITHOUT A SOUND.

And uhhh.... WE'RE DONE, LOL. Thanks to part D,
(U) is a subgroup of
(X), so this is just the first isomorphism theorem... In fact, does the sections
even have to be global? I can replace X with open U in X and it would still
work, wouldn't it? Guess that's it.
Next exercise is the last one of this section. Hartshorned II is almost at its first season
finale. Yayyyyy!!! Also shoutout to cyuucat
for the tip on how2glowtext. (Although I'm sure they disavow my use of it)