Collapse solves all extension problems

Put the filtered group $A=\mathbb{Z}\mathrm{/}4$ of [F1] in chain degree zero, using [F2]. Since the chain differential is zero, [F3] gives $E_{p,-p}^r=F_{p}A\mathrm{/}{F}_{p-1}A$ and zero elsewhere for every $r$. Thus its only nonzero page objects are the two $\mathbb{Z}\mathrm{/}2$ pieces at $p=0,1$. By [F4] every page differential is induced by the zero map, so the sequence collapses from page zero by [F5]. Its filtration is finite, and [F6] identifies its filtered abutment with $H_0(C)=A=\mathbb{Z}\mathrm{/}4$.

The projection $\mathbb{Z}\mathrm{/}4\to (\mathbb{Z}\mathrm{/}4)\mathrm{/}\{0,2\}\cong \mathbb{Z}\mathrm{/}2$ has fiber $\{1,3\}$ over $1$. Each element of this fiber doubles to $2\mathrm{\ne }0$, so neither can be the image of $1$ under an additive section from $\mathbb{Z}\mathrm{/}2$. Hence the quotient extension has no section in the sense of [F7]. The extension problem remains unsolved by the stable pieces even though the sequence has already collapsed.