An isomorphism on e infinity automatically gives an isomorphism of unfiltered targets

The induced graded map at $p=-m$ is the identity on $k=\mathbb{Z}\mathrm{/}2$ via the coordinate $m$ identification $T_m\mathrm{/}{T}_{m+1}\cong k$. All positive graded pieces are zero. Since both chain differentials vanish, every page differential is zero and these identifications persist on every page, including $E^{\mathrm{\infty }}$ at positions $(-m,m)$. The homology targets are $S$ and $P$ themselves, with the same tails, and the induced target map is the inclusion. Its graded maps are precisely the page maps, so compatibility is satisfied.

The constant-one sequence in $P$ is not in $S$, so this compatible target map is not surjective. Both filtrations are exhaustive ($F_0$ is full) and separated, but the source completion map is this same proper inclusion $S\to P$, hence is not an isomorphism. Thus the finite or complete-target lifting premise of [F3] fails on the source, while the limiting-page isomorphism holds. The index $m=0$ and zero positive levels were included in step 1.1, and the counterexample is choice-free.