A first quadrant five term exact sequence

The only nonzero differential sends the weight-zero vector $b$ to the weight-two vector $c$. Thus it preserves every decreasing piece, and $d^2=0$ because the next degree is zero. All filtration pieces are full for $p\le 0$ and zero for $p\ge 3$, so the finite construction applies. The $E_0$ terms are $k[a]$ at $(1,0)$, $k[b]$ at $(0,1)$ and $k[c]$ at $(2,0)$; all other positions vanish. This support is first quadrant.

The differential raises weight by two on $b$ and vanishes on $a,c$, so its graded $d_0$ is zero. The $d_1$ of $[b]$ is its differential projected to weight one in degree two, which is zero, and all other $d_1$ maps are zero as well. Hence $E_2=E_1=E_0$ as the displayed graded groups. On page two, $db=c\in F^2{C}^2$ gives $d_2[b]=[c]$ by [F2]. Both classes generate their copies of $k$, so this map is identity under those identifications. The class $[a]$ has zero differential and receives none, and every later nonzero term is therefore just $k[a]$ at $(1,0)$.

Directly, the kernel of $d:C^1\to C^2$ is $ka$, there are no incoming boundaries in degree one, and $d(C^1)=kc=C^2$. Thus $H^1=k[a]$ and $H^2=0$. The class $[a]$ is represented in $F^{1}C$, so $F^1{H}^1=H^1$ and $F^2{H}^1=0$. This is the normalized finite abutment of [F1]; all degree-two target pieces are zero.

In [F3], the map $E_2^{1,0}\to H^1$ sends $[a]$ to $[a]$ and is identity. The next map $H^1\to E_2^{0,1}$ factors through $H^1\mathrm{/}{F}^1{H}^1=0$, so is zero. The transgression is the identity $[b]\mapsto [c]$ from step 2.1, and the last map has zero target $H^2$. Thus at the first $k$ the kernel is zero, at the second $k$ the zero-map kernel is all $k$ and equals the preceding image, at the third $k$ the identity-map kernel is zero and equals the preceding image, and at the fourth $k$ the next kernel is all $k$ and equals the transgression image. The terminal target is zero. This checks the printed sequence at every position, including its nonzero $d_2$ and both ends, with no AC.