Double Complexes Exact Couples and Convergence

Fix $n$ and $p+q=n$. Substitute the defining formula for $d$ twice and distribute composition over addition. This gives $d_{n-1}{d}_n{\iota }_{p,q}^n={\iota }_{p-2,q}^{n-2}{h}_{p-1,q}{h}_{p,q}+{\iota }_{p-1,q-1}^{n-2}(v_{p-1,q}{h}_{p,q}+h_{p,q-1}{v}_{p,q})+{\iota }_{p,q-2}^{n-2}{v}_{p,q-1}{v}_{p,q}$. Each term is an arrow from $C_{p,q}$ to $T_{n-2}$.

The first and last composites vanish by the two square-zero axioms; the middle parenthesis vanishes by anticommutation. Therefore $d_{n-1}{d}_n{\iota }_{p,q}^n=0$ for every $p+q=n$, including when any of the source or target components is zero.

The zero arrow $T_n\to T_{n-2}$ has these same composites with every injection. Coproduct uniqueness therefore gives $d_{n-1}{d}_n=0$. Since $n$ was arbitrary, all chain identities hold. The argument also covers an all-zero or a single-supported diagonal and uses no exactness of infinite coproducts or representative selections.

Fix $n$ and let $I_n=\{(p,q):p+q=n,C_{p,q}\mathrm{\ne }0\}$. A finite biproduct of the objects indexed by $I_n$ exists. Adjoining the unique maps from and to each omitted zero object makes its coproduct and product structures satisfy the universal properties for the whole diagonal: those omitted components impose no conditions on a family of maps. This constructs $T_n$ and $P_n$, including $I_n=\mathrm{\varnothing }$, when both are zero.

Define ${\eta }_n:T_n\to P_n$ by ${\pi }_{a,b}^n{\eta }_n{\iota }_{p,q}^n=1_{C_{p,q}}$ if $(a,b)=(p,q)$ and zero otherwise. Successive coproduct and product universal properties give its existence and uniqueness. Under the identifications in the preceding step it is precisely the finite biproduct comparison, so it is invertible. If $I_n$ has one element, it is the identity on that component.

Test $d_n^P{\eta }_n$ and ${\eta }_{n-1}{d}_n^T$ by precomposing with ${\iota }_{p,q}^n$ and postcomposing with ${\pi }_{a,b}^{n-1}$. Both composites are $h_{p,q}$ for $(a,b)=(p-1,q)$, $v_{p,q}$ for $(a,b)=(p,q-1)$, and zero otherwise, by the two differential formulas. Universal-property uniqueness gives $d_n^P{\eta }_n={\eta }_{n-1}{d}_n^T$.

Multiplying this equation by the inverses gives $d_n^T{\eta }_n^{-1}={\eta }_{n-1}^{-1}{d}_n^P$. Hence the degreewise inverse is also a chain map. All components of the comparison are uniquely specified, and its inverses are unique; no simultaneous choice of lifts or representatives is involved. The conclusion holds for the zero complex and for a complex supported on one row or column as well.

The abelian group identities for coordinate addition on $P$ hold at each index by the identities in $k$. The zero sequence has empty support; negatives preserve support and the support of a sum is contained in the union of the two supports, so $S$ is a subgroup. For maps $f_j:G\to k$, the unique map $G\to P$ is $g\mapsto (f_j(g){)}_j$. For maps $g_j:k\to G$, define $S\to G$ by $x\mapsto {\sum }_{j\in \mathrm{supp}(x)}{g}_j(x_j)$. This sum is finite; extending the summation set by zero terms proves additivity, and the identity $x={\sum }_j{\iota }_j(x_j)$ proves uniqueness. These are precisely the product and coproduct properties.

The inclusion is injective. The constant-one sequence belongs to $P$ but has infinite support, so does not belong to $S$. Each unit sequence belongs to $S$, and different indices give different unit sequences.

Encode $x\in S$ by $b(x)={\sum }_j{2}^j{x}_j\in \mathbb{N}$. If $x\mathrm{\ne }y$, their finite union of supports has a largest differing index $r$. The magnitude of the contribution there is $2^r$, while the sum of the magnitudes at lower indices is at most ${\sum }_{j<r}{2}^j=2^r-1$; the latter identity follows by starting with $0=1-1$ and adding $2^r$ at the next index. Hence $b(x)\mathrm{\ne }b(y)$. Conversely, successive unique divisions of any nonnegative integer by $2$ give its binary digits; the nonzero quotients strictly decrease, so after finitely many divisions the quotient is zero. Substituting the equations $a_j=2a_{j+1}+r_j$ back gives $a_0={\sum }_j{2}^j{r}_j$. Thus $b$ is a bijection $S\to \mathbb{N}$.

The first-$m$-coordinates map $A\to k^m$ is onto by extension by zero, for either $A=S$ or $P$, and has kernel $T_{m}A$. It therefore induces a bijective homomorphism $A\mathrm{/}{T}_{m}A\to k^m$: equality of images means the difference lies in $T_{m}A$, and every tuple is represented by its zero extension. For $m=0$ the quotient and empty tuple group are zero. If $x\in {\bigcap }_m{T}_{m}A$, take $m=j+1$ to conclude $x_j=0$ at each index, so $x=0$.

For any map $e:\mathbb{N}\to P$, the sequence $y_j=1-e(j{)}_j$ lies in $P$ and differs from $e(j)$ at coordinate $j$. Thus $e$ is not onto. If an injection $P\to \mathbb{N}$ existed, inversion on its image and the zero sequence as value off that image would define a surjection $\mathbb{N}\to P$, which has just been excluded. Hence $P$ is uncountable and cannot be bijective with $S$.

Let $L$ be the subgroup of ${\prod }_{m\ge 0}{k}^m$ consisting of tuples $(z^{(m)}{)}_m$ for which truncating $z^{(m+1)}$ gives $z^{(m)}$. For any compatible cone of homomorphisms into $k^m$, the map sending an element to its tuple of cone images is the unique homomorphism into $L$ inducing that cone. Thus $L$ is the inverse limit. The homomorphism $P\to L$ sends a sequence to its initial segments. Its inverse sends a compatible tuple to $x_j=z_j^{(j+1)}$; compatibility proves that all its first-$m$ coordinates equal $z^{(m)}$. These formulas are mutually inverse and select no representatives.

The quotient identifications in step 2.2 commute with truncation, so they identify both $\widehat{S}$ and $\widehat{P}$ with $L\cong P$. The completion map sends $x$ to its initial segments, hence becomes the original inclusion for $S$ and the identity for $P$. Step 2.2 proves separatedness for both, and step 1.2 proves the first inclusion is proper. All constructions use explicit coordinates, finite sums, or uniquely specified digits; AC has not been used.

The row-$p$ quotient of total degree $p+q$ is $C_{q,p}$. The arrow $h$ stays in this row and $v$ enters the preceding row, which is zero in the quotient. Thus $E_{p,q}^0=C_{q,p}$ and $d^0=h$. This remains valid when either index is negative, as the component is then zero.

Taking homology gives $E_{p,q}^1=H_q^h(C_{\ast ,p})$. A horizontal cycle $x$ has total differential $dx=vx$, so the page-one rule gives $d^1[x]=[vx]$. This is a well-defined horizontal homology class: $h(vx)=-v(hx)=0$, and if $x$ changes by $hy$ then $vx$ changes by $vhy=-hvy$, a horizontal boundary. In an abelian category these calculations mean preservation of kernel and image subobjects; they may be checked after epic pullback and descend uniquely. No global representatives are selected.

Since $v^2=0$, the induced page-one arrows square to zero, and their homology is precisely $H_p^v(H_q^h(C))$. The next-page isomorphism therefore gives the displayed $E^2$. Both $E^0$ and all later subquotients vanish off the first quadrant. The differential bidegrees are those of the filtered construction, $(-r,r-1)$, with no extra sign in $d^1$ because $h+v$ was already the total differential.

In degree $n\ge 0$, $F_{-1}^{\mathrm{r}\mathrm{o}\mathrm{w}}{T}_n=0$ and $F_n^{\mathrm{r}\mathrm{o}\mathrm{w}}{T}_n=T_n$; negative degrees are zero. Thus the bounded-filtration theorem applies degree by degree to this spectral sequence and identifies its stationary page with the associated graded of the displayed image filtration. That filtration is zero at $p=-1$ and all of $H_n(T)$ at $p=n$ for $n\ge 0$. For $n=0$ there is only one possible quotient, and for the zero complex all pages and quotients are zero. The argument uses no infinite exactness or AC.

Define $D_{a,b}=C_{b,a}$, with $h_{a,b}^D=v_{b,a}^C$ and $v_{a,b}^D=h_{b,a}^C$. The two square-zero identities for $D$ are those for $v^C,h^C$ respectively, and its mixed sum is the mixed sum for $C$ with the two summands exchanged. Thus $D$ is again a first-quadrant anticommuting double complex. In each total degree, permutation of the finite summands gives an isomorphism $\mathrm{Tot}(D)\to T$. It commutes with total differentials because it changes $h^D+v^D$ into $v^C+h^C=h^C+v^C$. The row cutoff $b\le p$ in $D$ becomes the column cutoff in $C$.

Apply the row theorem to $D$. Its horizontal homology at $(q,p)$ is $H_q^v(C_{p,\ast })$, and its induced vertical differential is the original $h^C$. Its second page is therefore $H_p^h(H_q^v(C))$, with the same filtered bidegrees. These are the pages of the column-filtered $T$: the filtration-preserving isomorphism in step 1.1 identifies their graded objects and differentials, and natural page transitions propagate the identification through every page. No sign twist is needed in the transposition.

The same filtered isomorphism sends $H_n(F_p^{\mathrm{r}\mathrm{o}\mathrm{w}}\mathrm{Tot}(D))\to H_n(\mathrm{Tot}(D))$ to $H_n(F_p^{\mathrm{c}\mathrm{o}\mathrm{l}}T)\to H_n(T)$, hence sends their images to each other. Naturality of finite convergence identifies the stationary page in step 2.1 with the stated associated graded of these images. The bounds are $-1$ and $n$ for $n\ge 0$; for negative $n$ the complex is zero. This includes $n=0$, a zero complex and a complex in one column. All maps are specified by finite permutations and universal properties, so no choice assumption enters.

Both convergence theorems identify the unfiltered target in degree $n$ with $H_n(T)$ for the same total complex $T$ and the same differential $h+v$. They make different specified filtrations on that object; equality of targets alone asserts neither equality of those filtrations nor equality of the pages.

For the early-page witness, take $C_{1,0}=C_{0,0}=k$, $h_{1,0}=1_k$, and every other component and arrow zero. The only row complex is $k\stackrel{1}{\to }k$, whose kernel in degree one and cokernel in degree zero are both zero. Thus the row $E^1$ page is zero. Each nonzero column is a single $k$, so column $E_{1,0}^1=E_{0,0}^1=k$, with $d_{1,0}^1=1_k$. Hence its $E^1$ page is nonzero but its $E^2$ page is zero. The total complex is also $k\stackrel{1}{\to }k$ and has zero homology.

For the filtration witness, instead take just $C_{1,0}=k$ with all arrows zero. Then $H_1(T)=k$. The row filtration has $F_0^{\mathrm{r}\mathrm{o}\mathrm{w}}{H}_1(T)=k$, since vertical index zero is already included. The column filtration has $F_0^{\mathrm{c}\mathrm{o}\mathrm{l}}{H}_1(T)=0$ and $F_1^{\mathrm{c}\mathrm{o}\mathrm{l}}{H}_1(T)=k$. Thus the filtrations on the same nonzero target differ; the jumps occur at different filtration degrees.

These witnesses establish the two possible failures while step 1.1 proves the common-target assertion. All witnesses have finite support and specified zero or identity maps, with no representative choices. The zero complex would give equal zero pages and filtrations, which is consistent with the claim that differences can occur.

Since $C_{p,-1}=0$, $B_p=C_{p,0}\mathrm{/}\mathrm{im}(v:C_{p,1}\to C_{p,0})$. Anticommutation gives $hv=-vh$, so $h$ preserves the indicated boundary images and induces a differential on $B$; its square is induced by $h^2=0$. The prescribed $\rho$ commutes with differentials on $C_{n,0}$ by this definition. On a summand $C_{p,1}$, its only potentially surviving output under $\rho d$ is a vertical boundary and is therefore killed. On summands with $q>1$ both outputs have positive vertical degree and are killed. Hence $\rho$ is a chain map.

Regard $B$ as a double complex $D$ in row zero with horizontal differential that of $B$. The same formulas give a morphism $C\to D$ and a column-filtered total map equal to $\rho$. Its map on vertical $H_0$ is the identity $B_p\to B_p$, and its maps on positive vertical homology are isomorphisms $0\to 0$ by hypothesis. Thus the induced map on $E^1$ is an isomorphism. Natural homology transitions imply successively that its maps on $E^r$ for all $r\ge 1$ are isomorphisms.

The page $E^1$ is supported on $q=0$; $E_{p,0}^2=H_p(B)$. For $r\ge 2$, an outgoing differential from $(p,0)$ lands at positive second coordinate $r-1$ and an incoming source has negative second coordinate $1-r$, so both maps are zero. Thus $E^2=E^{\mathrm{\infty }}$. In degree $n\ge 0$ the finite homology filtration has all quotients zero except possibly the one at $p=n$. A quotient $F_p\mathrm{/}{F}_{p-1}=0$ means $F_p=F_{p-1}$; starting at $F_{-1}=0$ and applying this finitely often gives $F_{n-1}=0$, while $F_n=H_n$. Consequently the sole graded piece canonically equals $H_n$, for both total complexes.

By naturality of the abutment, the isomorphism on that sole graded piece induced in step 2.1 is exactly $H_n(\rho )$ under these canonical identifications, rather than an unspecified isomorphism of the two homology objects. Equivalently it is the horizontal edge of the column sequence: for $D$ the edge is the identity and the edge square for $\rho$ commutes. Hence $H_n(\rho )$ is invertible. Negative-degree homologies are zero and $n=0$ uses $F_{-1}=0$, so $\rho$ is a quasi-isomorphism in all degrees.

Exchange the two coordinates and the arrows $h,v$. Their anticommuting sum and total complex are unchanged under summand permutation, while columns become rows. The same projection and proof give the row assertion. If columns are completely acyclic, then also $B_p=0$ for every $p$, so the first quasi-isomorphism has zero target; the row conclusion follows in the same manner. No surviving nonzero edge is asserted in these completely acyclic cases. All maps are canonical quotients and finite-filtration maps, requiring no AC.

At the component $E_{p-r,q+r-1}$, exactness says that the image of $j_{p-1,q}$ lies in the kernel of $k_{p-r,q+r-1}$. Therefore $k_{p-r,q+r-1}{j}_{p-1,q}=0$ as an arrow from $D_{p-1,q}$ to $D_{p-r-1,q+r-1}$. This is valid even when any component or arrow is zero.

Substituting both differential formulas and using associativity gives $d_{p-r,q+r-1}{d}_{p,q}=j_{p-r-1,q+r-1}(k_{p-r,q+r-1}{j}_{p-1,q})k_{p,q}=0$. Its target is $E_{p-2r,q+2r-2}$, as required for the square of a map of bidegree $(-r,r-1)$. This works at $r=1$ and all larger integers, with no choice or convergence assumption.

If $a$ is locally $ix$, then $ia=i(ix)$ lies in $\mathrm{im}i$. Descent of this membership shows that $i$ restricted to $D^{\mathrm{\prime }}\subseteq D$ factors through the target $D^{\mathrm{\prime }}$. Its factorization is unique because that inclusion is monic. This defines $i^{\mathrm{\prime }}$ without any preimage choice.

The composite $j:D\to E$ lands in $\ker d$ since $dj=jkj=0$. Follow it by $\ker d\to E^{\mathrm{\prime }}$. This map kills $\ker i$: an arrow into $\ker i=\mathrm{im}k$ locally has form $kz$, and its image is $[jkz]=[dz]=0$. Hence it descends through $D\mathrm{/}\ker i\cong \mathrm{im}i$ to $j^{\mathrm{\prime }}$, uniquely. Explicitly, if $ix=iy$ locally, then $x-y=kz$ after a further epic pullback, so $[jx]-[jy]=[dz]=0$. Thus its formula is independent of the preimage.

On $\ker d$, the map $k$ lands in $\ker j=\mathrm{im}i$, because $jke=de=0$. Changing a cycle representative by a boundary $dz$ changes its image by $kdz=kjkz=0$. Thus this restricted map kills the boundary image and descends uniquely to $k^{\mathrm{\prime }}:E^{\mathrm{\prime }}\to D^{\mathrm{\prime }}$. Equality after the epic cycle quotient also proves independence for arbitrary maps into $E^{\mathrm{\prime }}$, not just element representatives.

For $i^{\mathrm{\prime }}$ the degree remains $(1,-1)$. To compute $j^{\mathrm{\prime }}$ on $D_{p,q}^{\mathrm{\prime }}$, its local $i$-preimage is at $D_{p-1,q+1}$ and $j$ sends it to $E_{p-r,q+r}$, giving degree $(-r,r)$. The cycle restriction and quotient for $k^{\mathrm{\prime }}$ preserve the original degree $(-1,0)$. The constructions above still apply when any image or homology object is zero, and for $r=1$ give $\mathrm{deg}{j}^{\mathrm{\prime }}=(-1,1)$. Every lift was a finite local epic pullback used to prove a canonical factorization; no global representative selection or AC was used.