I'm a beginner in spectral sequences, and I have some questions which I'm confused while reading Bott&Tu - Differential forms in algebraic topology, chapter 14, pp.156-160. https://www.maths.ed.ac.uk/~v1ranick/papers/botttu.pdf
Q1. In p.157, in the middle of the page there is a sentence "In the graded case we get from this short exact sequence a long exach sequence of cohomology groups $$\cdots \to H^k(A) \to H^k(A) \to H^k(B) \to H^{k+1}(A)\to \cdots,$$ which we may write as ~~~."
1-1. Is the graded case means: $K=\bigoplus_{k\in \Bbb Z} C^k$ with $D:C^k\to C^{k+1}$ as described in p.156?
1-2. If this is right, does $H^k(A)$ mean the cohomology groups of the complex $\cdots \to \bigoplus_p K_p\cap C_k \xrightarrow{D} \bigoplus_p K_p \cap C^{k+1}\to \cdots$?
1-3. Does $H(A)$ means $\bigoplus_k H^k(A)$ in the last display of p.157?
Q2. In p.159, there is a sentence "~~ (14.5) $H(K)=F_0\supset F_1\supset \cdots$ making $H(K)$ into a filtered complex." The definition of a filtered complex is given in p.156, which is a differential complex with a sequence of subcomplexes? But why $H(K)$ is a differential complex? What is the differential operator of $H(K)$?
Q3. In the second paragraph of p.160, the book is considering the case that $K$ has a grading $K=\bigoplus_{n\in \Bbb Z} K^n$. Then it says that $\{K^n\cap K_p\}$ is a filtration of $K^n$. But I cannot understand this, because the differential operator $D$ maps $K^n$ into $K^{n+1}$, so $K^n\cap K_p$ is not a subcomplex of $K^n$ isn't it? Even $K^n$ cannot be a differential complex, I think.
I am too confused with this section. Any helps will be greatly appreiciated.
Q1. Note that here it is not assumed that the differential complex has a grading. From the example on page 158 where they write out $A_1=H(A)$ we can see that \begin{align}H(A)=\oplus_{p\in \mathbb{Z}}H(K_p) \end{align} Q2. The differential $D:C \to C$ induces a differential in cohomology, which is the zero map as any cohomology class is represented by an element in the kernel of $D$.
Q.3 Indeed $K^n$ is in general not a subcomplex. I would guess that what they wanted to say there is that the grading induces a grading $K_p^{\bullet}$ for each $p\in \mathbb{Z}$.
If one assumes that there is a grading as in 3. then 1. would turn into \begin{align}H^k(A)=\oplus_{p\in \mathbb{Z}}H^k(K_p) \end{align}