About an argument in Olsson's book

Question

The following picture is from Algebraic spaces and stacks (p.54) by Martin Olsson.

I don't understand how to conclude that $\alpha$ is induced by a nonzero class in the end.

It seems that there might be an exact sequence consists of edge maps like $E_2^{n0}\to H^n\to E_2^{0n}$ if there is a convergent first quadrant spectral sequence $E_2^{rs}\Longrightarrow H^{r+s}$, but this doesn't fall into the usual assumption for such an exact sequence to exist. Any ideas? Another way to explain Olsson's argument is also fine.

Edit. To summarize (by Minseon Shin's answer), the conclusion is: If in addition $E_{2}^{n-i,i} = 0$ for $0<i<n$, then we have an exact sequence \begin{align*} E_{2}^{n,0} \to H^n \to E_{2}^{0,n}.\end{align*} If moreover $E_{2}^{n-1-i,i} = 0$ for $0<i<n$, then we have an exact sequence \begin{align*} 0\to E_{2}^{n,0} \to H^n \to E_{2}^{0,n}.\end{align*} We might have a similar result for exactness on the right.

(This follows from the exact sequences $0\to F^nH^n=E_{\infty}^{n,0} \to H^n \to H^n/F^nH^n=H^n/F^1H^n=E_{\infty}^{0,n}\to 0$ (note $F^1H^n=F^nH^n$) and $E_{2}^{n,0}\to E_{\infty}^{n,0}\to 0, 0\to E_{\infty}^{0,n}\to E_{2}^{0,n}$. If moreover $E_{2}^{n-1-i,i} = 0$ for $0<i<n$, then $E_{2}^{n,0}\xrightarrow{\cong} E_{\infty}^{n,0}$.)

Answer 1

For completeness, the spectral sequence (2.3.14.1) mentioned is \begin{align*} E_{2}^{s,t} = \check{H}^{s}(\mathscr{X},\underline{\mathscr{H}}^{t}(F)) \implies H^{s+t}(C/X,F) \end{align*}

By minimality of $i_{0}$, we have $\underline{\mathscr{H}}^{i}(F) = 0$ for $0 < i < i_0$ so $E_{2}^{i_{0}-i,i} = 0$ for such $i$, thus we have an exact sequence \begin{align*} 0 \to E_{2}^{i_{0},0} \to H^{i_{0}}(C/X,F) \to E_{2}^{0,i_{0}} \;. \end{align*}