Oh, so there really is a function between, e.g., $K'$ and $M$? It will be a homomorphism, right?
@loch I don't think they necessarily the kernel and cokernel. That was the first part of the theorem, and they were denoted by $K$ and $C$ (no apostrophe)$.
yes - if it says it's an exact sequence or more generally a chain complex there are maps despite he didn't give them names!
oh - then i guess you're supposed to prove that they do correspond to the kernel and cokernel using exactness
of course i guess when i said inclusion maybe one should be more careful and just say that the image of $K'\rightarrow M$ is equal to the kernel of $f$, and hence probably it's more correct to say $K'$ is isomorphic to the kernel
lol I emailed my analysis I professor to see if she could help me self-learn analysis over the summer. at first I asked my analysis ii professor but he told me he's spending the summer in France
turns out my analysis one prof is married to my analysis ii prof and theyre both in france together rip
@TedShifrin Well, there's a battery of universities which have a good undergrad program in mathematics called the "IITs" but their entrance procedure is really hard (it's a Physics+Chemistry+Mathematics test with a limited amount of time and hard problems), so I didn't take it. There's only a handful of others which have their independent admission exams out of those
Use $f(z)=(az+b)/(cz+d)$, solve the equations $f(z_i)=w_i$ for $i=1,2,3$. You can divide through and get rid of one of the $a,b,c,d$. For example, you can assume $d=0$ or $d=1$. What must happen if $d=0$?
Uhm I have a book which computes $\dfrac{w-w_1}{w-w_3}=\dfrac{cz_1+d}{cz_3+d}\dfrac{z_2-z_3}{z_2-z_1} $ and $\dfrac{w_2-w_3}{w_2-w_1}=\dfrac{cz_1+d}{cz_3+d}\dfrac{z_2-z_3}{z_2-z_1}$
Now it is time to learn what you said earlier, Ted. Let see what is the cross-ration and how the Möbius transformations come from in terms of linear algebra.
you need algebraic number theory to understand the statements
for the proofs, there are different approaches
you can use representation theory and complex analysis (for the global case), group cohomology (works for both local and global, I think), harmonic analysis on locally compact groups, formal groups (for local) ...
@TedShifrin if I want to find a linear transformation which maps $|z-i| \leq 2$ into $\Im(z) \geq 0$, is not uniquer right? If unique for three particular points.
so Ted you would write $\begin{bmatrix} A & B \\ C & D \end{bmatrix} \begin{bmatrix} z \\ 1 \end{bmatrix} = \begin{bmatrix} w \\ 1 \end{bmatrix}$ for $z, w \neq \infty$ and solve the equations that way?
Oh, well, typically our courses were 150 minutes of lecture per week, so in 8 weeks that's 20 hours of lecture. Yours is 21 hours. Pretty much the same.
@TedShifrin im reasonably good at hard work i have gotten 5 A+ with all 3rd and 4th year math classes and my friend always takes 6 classes a semester about the same grades as me but he actually a genius.
I got offered a summer course in algebra at anther university should i take it? its a week of 8hrs of abstract algebra all expenses including travel and lodging paid? @TedShifrin @MatheinBoulomenos
Anonymous
@MatheinBoulomenos Thanks. I'm trying to convey what I understand from that answer. If I draw the Hasse diagram, I need to show that every pair of subgroups must have an unique greatest lower bound and unique least upper bound. Now, according to the answerer some of groups will form a chain structure (not sure why?). Each of those chains will contain the intersection of groups which form the chain and also the union of those groups.
If $A$ is a closed set then $X \setminus A$ is open, and no point $p \in X\setminus A$ can be a limit point of $A$, because to be a limit point of $A$ every neighbourhood of $p$ needs to contain at least one point of $A \setminus \lbrace p \rbrace = A$, and since $X \setminus A$ is a neighbourhood of $p$ (being that it is open), this is an example of a neighbourhood of $p$ without the required property
no credit its covering stuff i dont already know its in august so my doctor should let me, but i am a bit slow at getting used to how people teach i don't wanna waste the position someone else who could learn more could take
@user48929 I haven't read the answer in detail, but it's not that difficult. The main point is the easy fact that any intersection of subgroups will again be a subgroup. This intersection gives you the greatest lower bound since any subgroup that is contained in some family of subgroups is also contained in their intersection
for upper bounds if you have some subgroups $(U_i)_{i \in I}$, then you can take the intersection of all subgroups that contain the union $\bigcup_{i \in I} U_i$ (this is equal to the subgroup generated by the union). You get that this a least upper bound basically by definition
Anonymous
10:46 PM
@MatheinBoulomenos Yeah, I was thinking along that line too. But I am not being able to convince myself that always such chain structures will exist in the Hasse diagram
but for every subset, there's a smallest subgroup that contains it (the subgroup generated by that subset) if you take the union of some subgroups as the subset, taking the subgroup generated by it will give you the least upper bounds
my impression of measure theory so far has been that you prove some big theorems that are used a lot and then you can forget about most constructions basically
Anonymous
@MatheinBoulomenos I'm not sure what you mean by subgroup generated by a subset. Let's consider $\Bbb Z_2\times \Bbb Z_2$. The members of the group are $\{(0,0),(0,1),(1,0),(1,1)\}$. Say I take an arbitrary subset $\{(0,0),(0,1),(1,1)\}$. What would be the subgroup generated by it?
