Collapse with projective associated graded pieces splits the finite filtration noncanonically

The quotient map $q_p:F_{p}H\to G_p$ is surjective. Apply [F1] with the identity of $G_p$ to obtain a linear section $s_p:G_p\to F_{p}H$, with $q_p{s}_p=1$. Then ${\varphi }_p:F_{p-1}H\oplus G_p\to F_{p}H$, $(x,y)\mapsto x+s_p(y)$, is linear. If its value is zero, applying $q_p$ gives $y=0$ and then $x=0$. For any $z\in F_{p}H$, take $y=q_{p}z$; the remainder $z-s_p(y)$ lies in $\ker q_p=F_{p-1}H$, so $z$ is in its image. Thus ${\varphi }_p$ is an isomorphism restricting to the given inclusion on the first summand.

Starting from $F_{a}H=0$, apply step 1.1 successively at the finitely many indices $a+1,\dots ,b$. This gives $H\cong {⨁}_{a<p\le b}{G}_p$ and sends each partial sum through $p$ onto $F_{p}H$. The inverse is therefore filtered too. Only finitely many sections are selected, by finite induction, so no arbitrary-index choice or AC is needed. Zero pieces require only the zero section; the zero module and a single nonzero stage are included. Bounds below $a$ and above $b$ add zero graded pieces and do not change the conclusion.

Under the spectral-sequence hypothesis, the target filtration is finite by assumption, and the supplied abutment isomorphisms in [F3] identify its projective limiting terms with the modules $G_p$. Step 2.1 then applies degree by degree. No claim that collapse alone forces finiteness, projectivity or a determination of the extension was used.

Noncanonicity occurs already for $H=\mathbb{Z}\oplus \mathbb{Z}$ with filtration $0\subset \mathbb{Z}\oplus 0\subset H$. Both graded pieces are projective: given a surjection of abelian groups and a map from $\mathbb{Z}$, lift the image of $1$ to one element and extend by integer multiples. The quotient onto the second coordinate has distinct sections $s_0(y)=(0,y)$ and $s_1(y)=(y,y)$. The automorphism $T(x,y)=(x+y,y)$ preserves the filtration and induces the identity on both graded pieces, but takes $s_0$ to $s_1$. More strongly, every section has $s(1)=(t,1)$ for an integer $t$, and $T(s(1))=(t+1,1)\mathrm{\ne }s(1)$. Thus no section can be invariant under all automorphisms of the given filtered data; a canonical splitting does not follow. This uses only the elementary integer-module operations, not AC.