AHSS collapse generally determines only the associated graded object

By [F1] the collapsed page determines exactly the filtration quotients $F^p\mathrm{/}{F}^{p+1}$ (cohomologically) or $F_p\mathrm{/}{F}_{p-1}$ (homologically); nothing in the identification uses or supplies the extension classes, and the collapse hypothesis only asserts the vanishing of the differentials, so it adds no data beyond the pages.

Additively, let $A=\mathbb{Z}\mathrm{/}4$ with the filtration $F^0=A$, $F^1=2\mathbb{Z}\mathrm{/}4\cong \mathbb{Z}\mathrm{/}2$, $F^2=0$, and let $B=\mathbb{Z}\mathrm{/}2\oplus \mathbb{Z}\mathrm{/}2$ with the filtration $F^0=B$, $F^1=0\oplus \mathbb{Z}\mathrm{/}2$, $F^2=0$. Both graded families have $\mathbb{Z}\mathrm{/}2$ in degrees zero and one and zero elsewhere. (They can also be packaged as a finite direct sum, but that packaging is not the definition.) Yet $A$ has an element of order four and every element of $B$ is killed by two, so they are not isomorphic; hence the additive extension data are not determined by the graded pieces.

Multiplicatively, an ambiguity can remain even when the additive groups are known. Set $A={\mathbb{F}}_2[x]\mathrm{/}(x^4)$ and $R={\mathbb{F}}_2[u,v]\mathrm{/}(u^2,v^2)$. Give $A$ the filtration $F^0=A$, $F^1=(x^2)$, $F^2=0$ and $R$ the filtration $F^0=R$, $F^1=(v)$, $F^2=0$, extending by the whole ring below zero and zero above one. Both are multiplicative since the displayed ideals square to zero. In each case the degree-zero quotient is ${\mathbb{F}}_2[a]\mathrm{/}(a^2)$ and the degree-one piece is its free rank-one module generated by $z$; products of two degree-one pieces vanish. Thus their graded rings are both ${\mathbb{F}}_2[a,z]\mathrm{/}(a^2,z^2)$ with filtration degrees $\mathrm{\mid }a\mathrm{\mid }=0$, $\mathrm{\mid }z\mathrm{\mid }=1$. Their additive groups are both ${\mathbb{F}}_2^4$, as witnessed by bases $1,x,x^2,x^3$ and $1,u,v,uv$. But $x\in A$ is nilpotent with $x^2\mathrm{\ne }0$, whereas every nilpotent of $R$ has constant coefficient zero and squares to zero. The rings cannot be isomorphic. These are explicitly computed filtered rings, not claimed AHSS realizations.

If the sole nonzero quotient of a finite decreasing filtration is $F^j\mathrm{/}{F}^{j+1}$, all earlier quotients zero force equality of the preceding stages with the whole group, and all later quotients zero force equality of the following stages with zero. Hence $F^j=H$, $F^{j+1}=0$ and that quotient is $H$ itself. If every quotient vanishes, finite exhaustiveness and separation similarly force $H=0$. Reversing indices gives the increasing case. Multiple quotients do not force ambiguity in every example; they merely permit an extension problem.

Steps 1.1 to 1.3 show that the collapsed pages determine the associated graded object and that distinct filtered objects, additively and multiplicatively, share that associated graded; so the pages alone do not contain general extension data. Step 1.4 records the zero and one-jump exceptions. This establishes the stated limitation without claiming that every individual collapse is ambiguous.