I'm currently reading Timothy Chow's article defining spectral sequences, and I'm stuck on one step. Here is a summary:
We have a finite-dimensional chain complex $... \overset{\partial}{\longrightarrow} C_{d+1} \overset{\partial}{\longrightarrow} C_d \overset{\partial}{\longrightarrow} ...$ which has a two-layered filtration, ie, for any $d$, we have:
$0 = C_{d, 0} \subset C_{d,1} \subset C_{d,2} = C_d$. A fortiori, there also is such a filtration for the set of boundaries $B_d$, and the set of cycles $Z_d \subset C_d$.
We go on to define an associated complex by $E_{d,p}^0 := C_{d,p}/C_{d, p-1}$ with induced map $\partial^0$. And its homology $E_{d,p}^{1} := H_d(E_{d,p}^0) = \dfrac{\ker\{\partial^0 : E_{d,2}^0 \longrightarrow E_{d-1,2}^0\}}{\text{im}\{\partial^0 : E_{d+1,2}^0 \longrightarrow E_{d,2}^0\}}$.
So far, so good. Now, the paper (page 3) tries to analyse the difference between that homology, and that of the original complex (whose $d$-th homology group would simply be $Z_d/B_d$).
Now here is the part that I struggle to understand (I attached a big screenshot for context but really the part I struggle with is the first line):
I don't see why the space of boundaries of $E_{d,2}^1$ would be $B_{d,2} + C_{d,1}$. Here was my train of thought:
The said space of boundaries is $\text{im}\{\partial^0 : E_{d+1,2}^0 \longrightarrow E_{d,2}^0\} = \text{im}\{\partial^0 : C_{d+1,2}/C_{d+1,1} \longrightarrow C_{d,2}/C_{d,1}\}$. Now, $\partial^0$ is constant on $C_{d+1,1}$, so the latter image is the same as $\text{im}\{\partial^0 : C_{d+1,2} \longrightarrow C_{d,2}/C_{d,1}\}$.
And since $\partial(C_{d+1,2}) = B_{d,2} \subset C_{d,2}$, then shouldn't we have:
$\text{im}\{\partial^0 : C_{d+1,2} \longrightarrow C_{d,2}/C_{d,1}\} = \dfrac{B_{d,2}}{B_{d,2}\cap C_{d,1}} = \dfrac{B_{d,2}}{B_{d,1}}$
Surely that is wrong, but I don't understand how to get to the result mentioned in the paper.
Moreover, the space of boundaries should be a subspace of $E_{d,2}^0 = C_{d,2}/C_{d,1}$. I don't see how we can view $B_{d,2} + C_{d,1}$ as a subspace of the latter.
Thanks for reading, I hope this was clear! :)
I believe what is meant is not that $B_{d,2}+C_{d,1}$ is a subspace of $E^0_{d,2}$. Let me try to rephrase the image you linked. We are looking at the following diagram (sorry, I don't know how to draw inclusion arrows with AMScd).
$$\require{AMScd} \begin{CD} C_{d+1,2} @>>> C_{d,2} @>f>> C_{d-1,2}\\ @AAA @AAA @AAA\\ C_{d+1,1} @>>> C_{d,1} @>>> C_{d-1,1} \end{CD}$$
What the author says is that $E^1_{d,2}$ is isomorphic to $\frac{f^{-1}(C_{d-1,1})}{B_{d,2}+C_{d,1}}$. To see that, complete the diagram as follows.
$$\require{AMScd} \begin{CD} 0 @. 0 @. 0\\ @AAA @AAA @AAA\\ E^0_{d+1,2} @>v>> E^0_{d,2} @>u>> E^0_{d-1,2}\\ @AAA @AAgA @AAA\\ C_{d+1,2} @>>> C_{d,2} @>f>> C_{d-1,2}\\ @AAA @AAA @AAA\\ C_{d+1,1} @>>> C_{d,1} @>>> C_{d-1,1} \end{CD}$$
Then $$\begin{align*} \DeclareMathOperator{\im}{im}\ker(u)/\im(v) &= g^{-1}(\ker(u))/g^{-1}(\im(v))\\ &= f^{-1}(C_{d-1,1})/(B_{d,2}+C_{d,1})\text{.} \end{align*}$$