Double Complexes Exact Couples and Convergence — Examples

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.

Each nonzero column is $k\stackrel{1}{\to }k$ in degrees one and zero. Its degree-one kernel and degree-zero cokernel are zero, and all other columns and homology degrees are zero. Thus every column is acyclic, including its degree-zero homology. The first-quadrant support satisfies [F1], so there is no surviving edge complex and assembly gives an acyclic total complex.

For a direct check order total degree one as $C_{1,0}\oplus C_{0,1}$. The total complex is $k\stackrel{x\mapsto (x,0)}{\to }{k}^2\stackrel{(u,v)\mapsto v}{\to }k$ in degrees $2,1,0$. The first map is injective, the second is surjective, and the kernel of the second is $k\oplus 0$, exactly the first image. Hence $H_2=H_1=H_0=0$, with all other degrees already zero. This checks the two endpoints and the middle image-kernel equality independently of the assembly invocation, using explicit maps and no AC.

Homology of each piece equals its degree-zero group and vanishes in all other degrees. The successive quotient at $p=0$ is $\{0,2\}$, isomorphic to $\mathbb{Z}\mathrm{/}2$ by $[a{]}_2\mapsto [2a{]}_4$. At $p=1$ it has cosets $\{0,2\},\{1,3\}$, identified with $\mathbb{Z}\mathrm{/}2$ by parity. All other graded quotients are zero. This proves every displayed $D^1$ and $E^1$ term, including the infinite constant $D^1$ tail.

The $i$ arrow from $D_{0,0}^1$ to $D_{1,-1}^1$ is the inclusion of $\{0,2\}$; at every $p\ge 1$ it is identity into the next $\mathbb{Z}\mathrm{/}4$. At $p<0$ it is the map from zero. The $j$ arrows are the stated identity and parity quotient at $p=0,1$, and zero to zero targets elsewhere. Every $k$ lowers total degree to minus one, where the $D^1$ terms vanish, so $k=0$. These maps have the exact degrees in [F3].

Check exactness at $D^1$ before $j$: at $p=0$, the incoming $i$ image and $\ker j$ are zero; at $p=1$, both are $\{0,2\}$; at $p\ge 2$, both are all of $\mathbb{Z}\mathrm{/}4$; at $p<0$ both are zero. At every $E^1$ term $j$ is onto, so $\mathrm{im}j=\ker k=E^1$. At $D^1$ before $i$, each $i$ is injective, so $\ker i=0=\mathrm{im}k$. All off-diagonal terms give zero equalities. Thus all vertices are explicitly exact. The $d^1=jk$ differentials are zero, and the nonzero $(1,-1)$ term prevents a first-quadrant interpretation. No AC or representative section is used.

Since $k^1=0$, the differential $j^1{k}^1$ is zero and $E^2=E^1$ by the identity cycle quotient. For $D_{p,-p}^2$ take the image of $D_{p-1,1-p}^1\to D_{p,-p}^1$. It is zero for $p\le 0$, the subgroup $\{0,2\}$ for $p=1$, and the whole $\mathbb{Z}\mathrm{/}4$ for $p\ge 2$. These are the displayed terms.

The restricted $i^2$ at $p=1$ is the inclusion $\{0,2\}\hookrightarrow \mathbb{Z}\mathrm{/}4$ and at $p\ge 2$ is identity; at smaller indices it has zero source. For $a\in D_{1,-1}^2=\{0,2\}$ its preimage under the old inclusion is the same element of $D_{0,0}^1$, so $j^2(a)=a\in E_{0,0}^2$. At $p=2$ the old $i^1$ is identity on $\mathbb{Z}\mathrm{/}4$, so $j^2(a)$ is its parity class in $E_{1,-1}^2$. Every other $j^2$ has zero target, and $k^2=0$ by its defining formula. Thus the $j$ index changes by $(-1,1)$ rather than remaining degree zero.

At $D^2$ before $j^2$, when $p=1$ both the incoming $i^2$ image and the kernel are zero; when $p=2$ both are $\{0,2\}$; when $p\ge 3$ both are the whole group; and when $p\le 0$ both are zero. Every $j^2$ onto a nonzero $E^2$ term is surjective, so its image equals $\ker k^2$. All $i^2$ maps are injective, so their kernels are zero, exactly the incoming $k^2$ images. Off-diagonal terms are zero. This verifies all exactness claims of [F3] directly, with the transition indices now one and two. The derivation used only literal subgroup inclusions and quotient maps; no chosen section or AC occurs.

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.

Only total degree zero has nonzero components. Consequently ${\mathrm{Tot}}_0^{\oplus }C=S$ and ${\mathrm{Tot}}_0^{\mathrm{\Pi }}C=P$ by [F1, F2]. All other total degrees and every total differential vanish. In either complex every degree-zero element is a cycle and the boundary subgroup is zero. Thus the homology groups in degree zero are $S$ and $P$ respectively, and all other homology groups are zero.

The canonical comparison has degree-zero component the finite-support inclusion. The tuple $(1,1,\dots )$ has infinite support, so it is not in the image; its homology class is unchanged because there are no boundaries. Even an abstract homology isomorphism is impossible: $S$ has a bijection with $\mathbb{N}$, while every purported enumeration $e:\mathbb{N}\to P$ misses the sequence $y_j=1-e(j{)}_j$. Therefore a bijection $S\to P$ would contradict [F2].

The support includes $(j,-j)$ for every positive $j$, so is not first quadrant and is infinite on its sole nonzero diagonal. Hence neither first-quadrant finite-diagonal comparison nor finite-filtration convergence is contradicted. The index $j=0$ contributes one copy of $k$, the omitted total degrees are genuinely zero, and all constructions and diagonalization are explicit and choice-free.

In the first example every graded quotient is $k\mathrm{/}k=0$, so every page is zero. But $H_0(C)=k\mathrm{\ne }0$ and $F_p{H}_0(C)=k$ for every $p$, by the identity inclusion of the whole subcomplex. Its filtration is exhaustive, nonseparated, and has zero associated graded. Thus its zero limiting page can identify with its zero graded homology without implying zero homology. The filtration is not finite, since no piece is zero.

In the second example the inclusion preserves every tail, so the pieces are subcomplexes. Both chain filtrations are exhaustive ($F_0$ is full) and separated, but the degree-one completion map is the nonsurjective $S\to P$. At $p=-m$ each of the degree-one and degree-zero graded groups is $T_m\mathrm{/}{T}_{m+1}=k$ via coordinate $m$, and the graded differential is the identity. At $p>0$ both graded groups are zero. Therefore $E^1(K)=0$ and every later page vanishes.

The inclusion is injective, so $H_1(K)=0$, while $H_0(K)=P\mathrm{/}S$. The constant-one sequence represents a nonzero class. For any $x\in P$ and any $m$, delete its first $m$ coordinates to obtain $t\in T_{m}P$. The difference $x-t$ has finite support, so $[x]=[t]$ in $P\mathrm{/}S$. Thus the homology image of every tail subcomplex is all of $P\mathrm{/}S$, and $F_p{H}_0(K)=P\mathrm{/}S$ at every index. The chain filtration was separated, but its homology filtration is not. This example fails chain completeness and finite bounds; the first fails separation already on chains. Neither is a counterexample to a theorem that requires these missing hypotheses. Empty deleted prefixes at $m=0$, zero other degrees and every positive filtration index obey the same calculations, without AC.

The differential squares to zero since the group below degree zero vanishes. The sole intermediate piece $kc[0]$ is a subcomplex, so the filtration on $C$ is by subcomplexes. The projection is a chain map: $f(da)=f(b)=0=d(f(a))$, and it preserves the specified pieces, including the zero lower tail and full upper tail. Both filtrations are finite in every degree, with the common bounds minus one and one.

On ${\mathrm{gr}}_0$ the map is the identity $kc[0]\to kc[0]$, hence an isomorphism on its sole homology group. On ${\mathrm{gr}}_1$, the source is $ka\stackrel{a\mapsto b}{\to }kb$ and the target is zero, because $F_{1}D=F_{0}D$. The source has zero kernel in degree one and zero cokernel in degree zero, so the graded map is again a quasi-isomorphism. Every other graded complex is zero on both sides. Therefore all hypotheses of the finite branch of [F1] hold.

Apply [F1] to conclude that $f$ is a quasi-isomorphism. Directly, $d:C_1\to C_0$ is injective with image $kb$, giving $H_1(C)=0$ and $H_0(C)=(kb\oplus kc)\mathrm{/}kb\cong kc$ via $[xb+yc]\mapsto yc$. The induced map $H_0(f)$ is this same isomorphism, while all other homology maps are isomorphisms between zero groups. This checks the specific map rather than only the isomorphism type of the target. All coefficients, zero terms, filtration endpoints and one-dimensional graded pieces have been computed explicitly; no AC or splitting choice is required.