A collapse with a noncanonical extension choice

Put $H=\mathbb{Z}\mathrm{/}4$ in cochain degree one, with zero differential, $F^{0}H=H$, $F^{1}H=2\mathbb{Z}\mathrm{/}4$ and $F^{2}H=0$. The associated graded is $\mathbb{Z}\mathrm{/}2$ at $(0,1)$ and $(1,0)$, zero elsewhere. Every differential is zero, so all pages equal this graded object, and actual cohomology is $H$ with exactly the displayed finite filtration. A section of $H\to H\mathrm{/}(2\mathbb{Z}\mathrm{/}4)$ would send the element of order two to an element killed by two lifting the odd coset. Its only lifts are $1$ and $3$, both of order four. Thus no section exists; equivalently $H$ has order-four elements while $(\mathbb{Z}\mathrm{/}2{)}^2$ does not. This is F1's extension obstruction with no choice assumption.

Similarly place $k^2$ in degree one, $F^1=k{e}_1$, with zero differential. Its stationary entries are $k$ at the same two positions, and its finite image filtration gives the actual abutment. Each line $L_a=k(a{e}_1+e_2)$ is a complement, since its intersection with $k{e}_1$ is zero and every vector is the sum of elements in those two lines. Conversely every complement has this form by normalizing the second coordinate of a nonzero vector in it. The shear $e_1\mapsto e_1$, $e_2\mapsto e_2+e_1$ sends $L_a$ to $L_{a+1}$ and fixes no complement, while inducing identity on both graded pieces. Hence existence and even an explicit choice $L_0$ do not give naturality.

In arbitrary dimension F2 uses AC for bases and complements, followed by only finitely many filtration splittings. In the displayed two-dimensional example the formula already supplies complements in ZF. In both cases the full target has zero/full filtration endpoints $F^2=0,F^0=H$, so convergence has no hidden issue; the unresolved question from page data alone is the extension or its choice of section. With just one nonzero quotient this particular extension obstruction disappears.