$E_{p,q}^r$ spectral sequence. Find a l.e.s. $\cdots \to A_{p+1}\to E_{p+1,0}^2\xrightarrow{d_{p+1,0}^2} E_{p-1,1}^2\to A_p\to E_{p,0}^2\to\cdots$ Let $E_{p,q}^r$ be a spectral sequence which converges to $A_n$. Let $E_{p,q}^r=0$ for $q\ge 2$. How to construct a long exact sequence
$$\cdots \to A_{p+1}\to E_{p+1,0}^2\xrightarrow{d_{p+1,0}^2} E_{p-1,1}^2\to A_p\to E_{p,0}^2\to \cdots \text{?}$$
I would say, for all $p\ge 0$ the maps $A_{p}\to E_{p,0}^2$ have to be the canonical projection, because: $E_{p,q}^r$ converges to $A_n$ means that there exists a sequence $$0=F_0A_n\subseteq F_1A_n\subseteq \cdots \subseteq F_{n-1}A_n\subseteq F_nA_n=A_n$$ such that $\bigcup\limits_{l=0,\ldots,n}F_lA_n=A_n$ and $E_{pq}^\infty\cong \frac{F_pA_{p+q}}{F_{p-1}A_{p+q}}$.
In our case is: $E_{p+1,0}^r\cong \frac{F_{p+1}A_{p+1}}{F_p A_{p+1}}=\frac{A_{p+1}}{F_p A_{p+1}}$ for all $r\ge 2$. Therefore we can take $A_{p+1}\to E_{p+1,0}^2$ the canonical projection I would say.
But I'm not sure if its range is the kernel of $d_{p+1,0}^2$.
The next step is then to consider $E_{p-1,1}^2\to A_p$. Again, I have $E_{p-1,1}^2\cong \frac{F_{p-1}A_{p}}{F_{p-2}A_{p}}$ and we have the inclusion $F_{p-1} A_p \to A_p$. But here I'm stuck. How to continue to construct the long exact sequence?
Best.
 A: The convergence $E_{p, q}^r \Rightarrow A_{p + q}$ guarantees a filtration of $A_n$ for each $n$, 
$$
0 = F_{-1} A_n \subseteq F_0 A_n \subseteq \cdots \subseteq F_{n - 1} A_n \subseteq F_n A_n = A_n
$$
such that $E_{p, q}^\infty \cong F_p A_{p + q} / F_{p - 1} A_{p + q}$. The condition $E_{p, q}^r = 0$ for $q \geq 2$ tells us that 
$$
0 = F_{-1} A_n = F_0 A_n = \cdots = F_{n - 2} A_n \subseteq F_{n - 1} A_n \subseteq F_n A_n = A_n,
$$
since $0 = E_{n - 2, 2}^2 = E_{n - 2, 2}^\infty \cong F_{n - 2} A_n / F_{n - 3} A_n$ which implies that $F_{n - 2} A_n = F_{n - 3} A_n$, and then we use induction. In particular, 
$$
E_{p - 1, 1}^\infty \cong F_{p - 1} A_p / F_{p - 2} A_p = F_{p - 1} A_p.
$$
Since $E_{p, q}^2 = 0$ for $q \geq 2$, the spectral sequence has converged by the third page with $E_{p, 0}^\infty = E_{p, 0}^3 = \operatorname{ker} d_{p, 0}^2$ and $E_{p, 1}^\infty = E_{p, 1}^3 = \operatorname{coker} d_{p + 1, 0}^2$.
Associated to $d_{p + 1, 0}^2$, for each $p$, there is the associated exact sequence,
$$
0 \to \operatorname{ker} d_{p + 1, 0}^2 \to E_{p + 1, 0}^2 \xrightarrow{d_{p + 1, 0}^2} E_{p - 1, 1}^2 \to \operatorname{coker} d_{p + 1, 0}^2 \to 0.
$$
Finally, the observation that 
$$
E_{p, 0}^\infty \cong F_p A_p / F_{p - 1} A_p = A_p / F_{p - 1} A_p
$$
gives the short exact sequence 
$$
0 \to F_{p - 1} A_p = E_{p - 1, 1}^\infty \to A_p \to E_{p, 0}^\infty \to 0.
$$
Looking at the isomorphisms given by looking at the third page of the spectral sequence, this short exact sequence corresponds with 
$$
0 \to \operatorname{coker} d_{p, 0}^2 \to A_p \to \operatorname{ker} d_{p, 0}^2 \to 0.
$$
Splicing these sequences together gives the desired long exact sequence.
