A spectral sequence collapses when one differential is zero

In [F4] take $C_2=\mathrm{Z}x$, $C_1=\mathrm{Z}y$, dx=y, and all other groups and differentials zero. Give x filtration degree 2 and y filtration degree 0: a generator is present precisely at or above its degree. This is finite and d preserves filtration because 0≤2. From [F2], $E^0$ has ℤ at (2,0) and (0,1), zero elsewhere. The differential lands two filtration levels lower, so $d^0=0$.

At the x position, $A_{2,2}^1=A_{2,2}^2=\mathbb{Z}x$ and both $B^1,B^2$ are zero. At the y position $A_{0,1}^1=A_{0,1}^2=\mathbb{Z}y$; the potential boundary sources $F_0{C}_2$ and $F_1{C}_2$ are zero, and $F_{-1}{C}_1=0$. Thus both $E^1$ and $E^2$ have the same two groups. The $d^1$ target from x is (1,0), which is zero, so $d^1=0$; [F3] instead gives $d^2[x]=[y]$, the nonzero identity between the two copies of ℤ.

At r=3, the x numerator is zero since dx is not in $F_{-1}{C}_1$; the y denominator is all ℤy since it now includes $d(F_2{C}_2)$. Every page term is therefore zero. Thus $d^1=0$ but the sequence does not collapse at $E^1$ by [F1], since $d^2\mathrm{\ne }0$.