If you are having trouble reading that, it says

"Yo, this notation is getting kinda complicated." I know, right. I'm barely hanging
on as well. Here's what all that nonsense means:
"Ah, the ring of of germs of regular functions. I see... It all makes sense now..."
That's the spirit, reader. Fake it till you make it. And don't worry, I did
not use (3.1.1) to verify that this is an equivalence relation! You what
they say: "To use it, you have to make an ass out of u and me"... What?
PART (a):
The idea for this whole exercise is not to be insightful and thoughtful about the
meaning of what's happening, but to just plunge into the symbols and meme your
way through it. In mathematics, we call these exercises "computational." I prefer
to call them symbo bimbo bashes (whose adjectivalization is "symbo bimbo
bashful". Make sure to memorize this: I will test you on this in a future post), as
I, by virtue of experiencing qualia and all that, am above the mere manipulation
of formal symbols and can find certain combinations of words more amusing and
vivid than others, although I have been referred to in the past as "not human".
Now, remember, we already know that ϕ : X → Y induces a morphism that looks
like this:
ϕ* :
(Y )  →
(X)  

(f : U → k) 
(f ∘ ϕ : ϕ^{1}(U) → k)   
which coaxes us into defining
ϕ_{P}* as
ϕ_{P}* :
_{ϕ(P),Y }  →
_{P,X}  

< U,f >  →< ϕ^{1}(U),ϕ * (f) >  

 =< ϕ^{1}(U),f ∘ ϕ >   
Our job is show that this is well defined. So suppose
< U,f >=
< V,g > in
_{ϕ(P),Y }. Then
ϕ_{P} * (< U,f >)  =< ϕ^{1}(U),f ∘ ϕ >    

 =< ϕ^{1}(U) ∩ ϕ^{1}(V ),f ∘ ϕ >    

 =< ϕ^{1}(U ∩ V ),f ∘ ϕ >    

 =< ϕ^{1}(U ∩ V ),g ∘ ϕ >  (since f = g on U ∩ V )    

 =< ϕ^{1}(U) ∩ ϕ^{1}(V ),g ∘ ϕ >    

 =< ϕ^{1}(V ),g ∘ ϕ >    

 = ϕ_{P} * (< V,g >)     
so yea, it's welldefined. And the ring homomorphism properties are trivial. Done.
PART (b):
Starting with the "only if". Suppose
ϕ an isomorphism. Then
ϕ is automatically a
homeomorphism. And
ϕ_{P}^{1}* is clearly and inverse for
ϕ_{P}*. So that direction is
easy...
For the other direction, note that all we need to do is show that
ϕ^{1} satisfies the
regularity property. I.e. given
U, open in
X and
f :
U → k a regular function, is
f ∘ ϕ^{1} :
ϕ(
U)
→ k regular?
Time for some symbo bimbo bashing. Since
ϕ_{P}* is an isomorphism, we have access
to the inverse map
ϕ_{P}^{1}* :
_{P,X} →
_{ϕ(P),Y } for any
P ∈ X. Let
P ∈ U. Then
since
< U,f >∈
_{P,X}, we have that
ϕ_{P}^{1} * (
< U,f >) =
< ϕ(
U)
,f ∘ϕ^{1} > is in
_{ϕ(P),Y }. I.e.
f ∘ ϕ^{1} is regular. DONE.
PART (c):
A bit tougher than the first two parts. So let's say
ϕ_{P}* (
< U,f >) =
ϕ_{P}* (
< V,g >).
I want to show that
< U,f >=
< V,g >, i.e. that
f =
g on
U ∩ V .
First thing, let's unwrap the given equality:
ϕ_{P} * (< U,f >)  = ϕ_{P} * (< V,g >)    

< ϕ^{1}(U),f ∘ ϕ >  =< ϕ^{1}(V ),g ∘ ϕ >    

f ∘ ϕ >  = g ∘ ϕ >  (on ϕ^{1}(U ∩ V ))    

f  = g  (on U ∩ V ∩ ϕ(X))    

    
The key point is that we are basically given that
f =
g on a significant subset
S =
U ∩ V ∩ ϕ(
X) of
U ∩ V . In fact, since
ϕ(
X) is dense in
Y , this subset should
be
dense on
U ∩ V shouldn't it? It's been a while since we've invoked our
fave
closure property on this blog. Let's use it!
cl_{U∩V }(S)  = U ∩ V ∩ cl_{Y }ϕ(X)  

 = U ∩ V ∩ Y  

 = U ∩ V  

  
Yep!
So f = g on "almost all" of U ∩ V . What about the points that are not in S?
Like, let's say R was a point in U ∩ V but not in S. We want f = g on R as well.
Well, since f and g are regular let's say that W is a neighborhood of R such that
f = f_{1}∕f_{2} and g = g_{1}∕g_{2} are rational. Let's set h to be the crossmultiplication
of those polynomials,
so, h ∘ ϕ = 0 at least on S, a dense subset of U ∩ V . Around this point, I was
reminded of something that popped up earlier in the text:
Ooohh. It's.. (3.1.1), haha. And I will use it here. Not directly, but using
the logic as inspiration. You see, asking "where is h = 0 in U ∩ V is the
same as computing Z(h) ∩ (U ∩ V ). Now, from our work we know that
So what happens if we take the closure in U ∩ V on both sides? We get
Z(h) ∩ (U ∩ V )  ⊃ U ∩ V  

Z(h) ∩ (U ∩ V )  = U ∩ V  

  
(why did the left side stay the same? Cause by the definition of the subspace
topology, it's a closed subset of U ∩V , so taking the closure does nothing). Hence
h = 0 on all of U ∩ V , including that pesky point R. So f(R) = g(R), as needed.
FIN.
And just in case you're wondering, I am too lazy to figure out how to make the
sheaf symbols
turn out pink. You know what? They might as well
stay black for the entire rest of the blog. We'll call it an inside joke, just
like the long gaps and delays between exercises are an "inside joke".