Edge maps from a first quadrant spectral sequence

The filtration is finite and preserved by d. Its graded terms are ℤ at (1,0),(0,0), with $d^0=0$; the next differential sends x to 2y, so its kernel is zero and cokernel ℤ/2 by [F3]. Therefore $E^2$ has only (0,0)=ℤ/2 and remains constant. Direct homology gives $H_1=0,H_0=\mathrm{Z}\mathrm{/}2$. The image of $H_0(F_{0}C)=\mathrm{Z}$ onto $H_0(C)$ is all ℤ/2. Thus $F_{-1}{H}_0=0,F_0{H}_0=H_0$, agreeing with [F2].

For n=0 the vertical edge of [F1] is $\mathrm{Z}\mathrm{/}2\to E_{0,0}^{\mathrm{\infty }}\to F_0{H}_0\to H_0$, taking [a] to [a] at every stage. The horizontal edge is $H_0\to H_0\mathrm{/}{F}_{-1}{H}_0\to E_{0,0}^{\mathrm{\infty }}\to E_{0,0}^2$, again [a]↦[a]. For n>0 all these $H_n$ and axis terms vanish by step 1.1, so both edges are the unique zero maps. The normalization $F_n{H}_n=H_n$ holds for every n≥0.