The two spectral sequences of a two by two double complex

In the column sequence the vertical differential is zero, so $E^1$ consists of the four copies of $k$. Its $d^1$ is identity from $(1,1)$ to $(0,1)$ and zero from $(1,0)$ to $(0,0)$. Thus $E^2$ is $k$ at $(0,0),(1,0)$ and zero elsewhere. No higher differential has both source and target among those two positions, so these are also the limiting terms.

In the row sequence, the row with vertical index one is the identity complex $k\to k$ and has zero homology. The row with vertical index zero has zero horizontal differential, so its two homology terms are $k$. With the required transposition these lie at spectral positions $(0,0),(0,1)$. The induced vertical $d^1$ is zero, and all later differentials have zero endpoints. Thus this $E^1$ page is already stationary.

Order total degree one as $C_{1,0}\oplus C_{0,1}$. Then the total complex is $k\stackrel{x\mapsto (0,x)}{\to }{k}^2\stackrel{0}{\to }k$ in degrees $2,1,0$. The first map is injective, its image is $0\oplus k$, and the second map has kernel $k^2$ and zero image. Consequently $H_2=0$, $H_1=k^2\mathrm{/}(0\oplus k)\cong k$ by the first coordinate, and $H_0=k$. Other degrees are zero.

The surviving $H_1$ class is represented by $C_{1,0}$. Row cutoff zero already includes it, giving $F_0^{\mathrm{r}\mathrm{o}\mathrm{w}}{H}_1=H_1$ and $F_{-1}^{\mathrm{r}\mathrm{o}\mathrm{w}}{H}_1=0$. Column cutoff zero includes only the degree-one summand $C_{0,1}$, whose class in total homology is zero, so $F_0^{\mathrm{c}\mathrm{o}\mathrm{l}}{H}_1=0$ and $F_1^{\mathrm{c}\mathrm{o}\mathrm{l}}{H}_1=H_1$. This accounts for the limiting positions $(0,1)$ versus $(1,0)$ despite the common total target. The first-quadrant finite bounds, axes and all omitted zero degrees have been checked; no representatives or maps were selected using AC.