First quadrant support alone identifies the abutment without a filtration

These data satisfy [F1]: all differential composites vanish and the homology of each page is that same page. The support is first quadrant. Put $H_1=A=\mathbb{Z}\mathrm{/}4$ with $F_{-1}A=0$, $F_{0}A=\{0,2\}$, $F_{1}A=A$, constant beyond these endpoints. The degree-zero graded piece is $k$ by $[a{]}_2\mapsto [2a{]}_4$; the degree-one quotient is $k$ by parity of the representative. Alternatively put ${\tilde{H}}_1=B=k^2$ with $F_{-1}B=0$, $F_{0}B=k\times 0$, $F_{1}B=B$. Its two pieces are $k$ by the first coordinate in the subobject and the second coordinate in the quotient. Take every other target degree zero. Thus the same specified stationary page has finite normalized abutment data to either target.

Every element of $B$ is killed by $2$, whereas $2[1{]}_4=[2{]}_4\mathrm{\ne }0$ in $A$. An additive isomorphism $A\to B$ would send $[2{]}_4$ to zero, contradicting injectivity. Therefore the two possible targets are not even isomorphic as unfiltered objects. Support alone cannot select between them or supply the missing extension data. The all-zero spectral sequence is also first quadrant but contains no target as part of its definition; the nonzero example above proves underdetermination even after fixing graded identifications. Zero other degrees, both finite endpoints and all four residues have been checked, without AC.