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..... Guys, I would like to make a toast... Everyone at home get your drinks. *Ahem*.... I would like to dedicate this toast... to California. Our beautiful Golden State... A TOAST FOR CALIFORNIA. *Cling* ALL HAIL CALI. WOOOOOOOOOOOOOOOOOOO. And YOU KNOW WHAT? .....You guys. Yes, all you Cali viewers sitting at home, you guys have done a fantastic job through all this. So I'd also like to dedicate this toast to YOU, because this wouldn't have been possible without your compliance. If all you wonderful Golden citizens didn't obediently bend over and follow all the mandates till now, they might've found it too impractical to enact something like this. BUT YOU BENT OVER, AND NOW HERE WE ARE. *CLING* ALL HAIL YOU. THANK YOU SAN FRANCISCO. WOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOOO..



Hi I'm done now. sorry for bein a gay. I try to NOT follow ~the nyooz~, but ~the nyooz~ is getting a lil too close to home （ｉДｉ）. anyway, here's prop 6.8 for today's exercise （ﾉ´д｀）

A

SO, UM... I'm not going to be OVERTECHNICAL with this exercise. I'm going to come up with the examples, and I think they're intuitively obvious enough....... I'M LAZY. E.g. Here's what I came up with for (a) (dim X ≥ 2):

π : A2 -{(0, 0)}	→ P1
(x,y)	(x,y)

Actually as I was typing this I realize that this is actually the map from which we construct the "Cone over P¹" from 2.14 (a), (b), (c). I think the very statement of that exercise makes it pretty clear that we can't extend the morphism (cause we have to explicitly union in the (0, 0) to make the cone). It's PRETTY INTUITIVE.

Okay, you know what, here's a taste of the technical details. Here's how it goes. *AHEM* If you had an extension ϕ of π, then (0, 0) has to get mapped to some point in P¹. And this point can't satisfy ALL homogenous polynomials in P¹. There is some closed set Z(f) ⊂ P¹ such that (0, 0) Z(f). Now, for ϕ to be continuous, the inverse image of this set has to be closed. ϕ^-1(Z(f)) consists of all the affine versions of the homogenous points that satisfy f. But here's the CATCH: Since f is homogenous, (0, 0) has to satisfy f. But (0, 0) ϕ^-1(Z(f)). So we have the situation Z(f)≠ϕ^-1(Z(f)). (the former Z being taken in affine space). But what closed set would satisfy all the constraints of f but exclude (0, 0) and only (0, 0)? I think it's pretty "obvious" that there isn't any, even though it's kinda hard to prove LOL.

B

Okay, here's what I came up with (Y is affine). Actually, it's just the classic isomorphism of projective space onto one of its open affines:

ϕ : P1 -{(1, 0)}	→ A1
(x,y)	x∕y

Yeah, I think it's PRETTY obvious that this can't be extended. LE CAN'T DIVIDE BY 0. Technical details are similar to part A.

The "technical details" of both of these have to do with proving a set is NOT closed.... IDK if there's a convenient way to do this but it just seems really really annoying to me. Whatevs. Not my prob for an exercise where I just have to come up with examples. NOT MY PROB!!!