I have always wanted to learn spectral sequences, and I finally found some time to do so.
However, I have some problems to understand the very basics of the construction (the answer is probably obvious, but I don't see it). I found the following paper which I will follow, since I find it very detailed:
http://homepages.math.uic.edu/~mholmb2/serre.pdf
My question concerns "The Limit Page"-section on page 2. I understood how he constructed the $E^2$-page: $$E^2=Z^1/B^1,$$ but for some reason not the $E^3$-page. This is taken from the paper, which is the part I can't understand:
Write $\overline{Z}:=\ker d^2$, it as a subgroup of $E^2$, whence, by the correspondence theorem, it can be written as $Z^2/B^1$, where $Z^2$ is a subgroup of $Z^1$. Similarly, write $\overline{B^2}=\operatorname{im} d^2$, which is isomorphic to $B^2/B^1$, where $B^2$ is a subgroup of $Z^2$...
So, let me try to make sense of the above quote, and then someone maybe can correct me if I say something wrong.
- Since $d^2$ is a homomorphism, $\ker d^2$ is a subgroup of $E^2$.
- I actually can't figure out why correspondence theorem implies that $\overline{Z^2}=Z^2/B^1$, for some subgroup $Z^2$ of $Z^1$.
The correspondence theorem says, according to Wikipedia; If $N$ is a normal subgroup of $G$, then there exists a bijection from the set of all subgroups $A$ of $G$ containing $N$, onto the set of all subgroups of the quotient group $G/N$. That is, they define a map $$\phi(A)=A/N.$$
So, if I let $N=\ker d^2$ in the above result, then I would have something like $$\phi(A)=A/\ker d^2,$$ which doesn't really look correct to me. Maybe I want to use some property of the differential to conclude at what he does in what I quoted? I am quite confused right now, so I would be really happy if someone could help me to understand why $\overline{Z^2}=Z^2/B^1$ and why $\overline{B^2}=B^2/B^1$.
Best wishes,
Joel
$\text{ }$
$\text{ }$
Correspondence Theorem: https://en.wikipedia.org/wiki/Correspondence_theorem_(group_theory)
The correspondence theorem is applied to the subgroup $B^1$ of $Z^1$, so that subgroups of $E^2=Z^1/B^1$ correspond to subgroups of $Z^1$ containing $B^1$.