Draft lesson.

Finite diagonal totals and signed double truncations

Original text public domain (CC0).

A commuting double complex has two cohomology directions. A comparison on one iterated cohomology becomes a comparison on the total, provided every diagonal has finite support. We prove this in any abelian category, keeping the eight truncation functors, the identity-cone quotient and the total shift signs explicit. The lower and upper bounds in the proof depend on the total degree being examined.

We use Signed bifunctors and sum and product total complexes, Sections 3–5, for commuting squares, shifts and totals; Smart truncations and image boundary homotopies, Sections 1–3, for the four one-variable truncations and their exact sequences; and Cohomology in abelian categories with translation, for cohomology and its natural exact sequences. Size conventions are those of Universes and small categories.

1. The finite diagonal category and its exact total

Let $\mathcal A$ be an abelian category. A commuting double complex has objects $X^{i,j}$, square-zero arrows $d':X^{i,j}\to X^{i+1,j}$ and $d'':X^{i,j}\to X^{i,j+1}$, and

d'd''=d''d'.\tag{1.1}

Both composites in (1.1) go from $X^{i,j}$ to $X^{i+1,j+1}$. A morphism commutes with both arrows. Say that $X$ has finite diagonal support if

S_n(X)=\{i\in\mathbb Z:X^{i,n-i}\ne0\} \quad\hbox{is finite for every }n\in\mathbb Z.\tag{1.2}

No bound on either coordinate over the whole complex is imposed. Denote the full subcategory of such double complexes by $C_f^2(\mathcal A)$.

Proposition 1.1. This is an abelian category with componentwise kernels and cokernels. Its total functor to $C(\mathcal A)$ is exact.

Proof. For a double-complex map $f:X\to Y$, each kernel $K^{i,j}\to X^{i,j}$ carries uniquely induced differentials, because the two differentials take kernels of $f$ into kernels of its next component. Composing either induced differential twice, or comparing the two mixed composites, and then postcomposing with the kernel inclusion gives zero or (1.1) in $X$. Cancellation of the inclusion proves all double-complex equations in $K$. The kernel universal property in every component proves the double-complex kernel universal property. The cokernel construction is the dual argument: factor the outgoing differentials through the cokernel projections, and precompose the resulting equations with their epimorphic projections.

The componentwise coimage-to-image comparison of $f$ is an isomorphism in $\mathcal A$. The naturality equations for kernel and cokernel constructions make these comparisons a double-complex map. Their componentwise inverses commute with the differentials by cancellation, so they form its inverse. Componentwise zero objects and finite biproducts complete the abelian-category axioms. On diagonal $n$, a kernel or cokernel can be nonzero only at a coordinate in $S_n(X)\cup S_n(Y)$. This finite set also bounds finite biproducts. Thus all the constructions remain in $C_f^2(\mathcal A)$.

Define

\operatorname{Tot}(X)^n=\bigoplus_{i+j=n}X^{i,j},\qquad D\iota_{i,j}=\iota_{i+1,j}d'+(-1)^i\iota_{i,j+1}d''.\tag{1.3}

Only finitely many nonzero terms are summed. Zero terms may be omitted, and finite biproduct universal properties make the resulting object independent of the omission. The two mixed terms of $D^2$ on the $(i,j)$ injection are $(-1)^id'd''$ and $(-1)^{i+1}d''d'$; they cancel. The unmixed terms are zero. Hence $D^2=0$. Applying $f^{i,j}$ on each diagonal summand defines $\operatorname{Tot}(f)$, and its chain equation follows on each injection from the two equations for $f$. This also proves the identity, composition and additive laws.

For a short exact sequence $0\to X\to Y\to Z\to0$, choose on each diagonal the finite union of the three supports and insert zero terms elsewhere. The finite biproduct of the component sequences is short exact. To justify that assertion directly, a map into the biproduct lies in the kernel of the sum of the outgoing maps exactly when each of its finitely many projections lies in the corresponding kernel; the componentwise factorizations uniquely assemble. A map out of the biproduct which kills the incoming sum factors uniquely through the outgoing sum by the dual argument on injections. Thus the sum of the kernels is the kernel, and the sum of the cokernels is the cokernel. Applying this in every degree proves exactness of (1.3). It uses no infinite product or coproduct. $\square$

2. Both directions and all eight truncations

Let $F_I$ regard $i$ as the outer degree and $j$ as the inner degree. Its outer term in degree $i$ is the complex $(X^{i,\bullet},d'')$, and its outer differential is the chain map $d'$. Equation (1.1) is exactly that chain-map equation. Thus

F_I:C^2(\mathcal A)\longrightarrow C(C(\mathcal A))\tag{2.1}

is an equivalence: unpacking a complex of complexes reverses the construction on objects and morphisms. Let $F_{II}$ instead use $j$ as the outer degree, the complexes $(X^{\bullet,j},d')$ as its terms, and $d''$ as its outer differential. This is the second equivalence, with a different outer degree.

Write $U_q=\tau_{\le q}$, $V_q=\widetilde\tau_{\le q}$, $W_q=\widetilde\tau_{\ge q}$, and $L_q=\tau_{\ge q}$ for the four one-variable functors. Their first-direction versions are

U^I_q=F_I^{-1}U_qF_I,\quad V^I_q=F_I^{-1}V_qF_I,\quad W^I_q=F_I^{-1}W_qF_I,\quad L^I_q=F_I^{-1}L_qF_I.\tag{2.2}

Define $U^{II}_q,V^{II}_q,W^{II}_q,L^{II}_q$ with $F_{II}$ in the same formula. These are all eight functors; $F_I$ and $F_{II}$ themselves are not truncations.

For concrete first-direction formulas set

Z_I^{q,j}=\ker d'^{q,j},\quad B_I^{q,j}=\operatorname{im}d'^{q-1,j}, \quad Q_I^{q,j}=\operatorname{coker}(B_I^{q,j}\hookrightarrow X^{q,j}).\tag{2.3}

Commutation induces a $d''$ on each of these families. For instance, $d''$ on cycles is the unique factor of $d''$ through the next kernel; on images it is the unique map making the image inclusions commute; on quotients it is the unique map making the quotient projections commute. Square zero follows by cancelling those inclusions or projections. Componentwise kernels and cokernels make $C(\mathcal A)$ abelian, so these are also the kernels, images and quotients computed in that category.

The terms of (2.2) are

\begin{aligned} (U^I_qX)^{i,j}&=\begin{cases}X^{i,j}&i<q,\\Z_I^{q,j}&i=q,\\0&i>q,\end{cases}& (V^I_qX)^{i,j}&=\begin{cases}X^{i,j}&i\le q,\\B_I^{q+1,j}&i=q+1,\\0&i>q+1,\end{cases}\\ (W^I_qX)^{i,j}&=\begin{cases}0&i<q-1,\\B_I^{q,j}&i=q-1,\\X^{i,j}&i\ge q,\end{cases}& (L^I_qX)^{i,j}&=\begin{cases}0&i<q,\\Q_I^{q,j}&i=q,\\X^{i,j}&i>q.\end{cases} \end{aligned}\tag{2.4}

At the changed first-direction boundary, $U^I_q$ uses the factor of $d'$ into the kernel, $V^I_q$ the epimorphism onto its image, $W^I_q$ the image inclusion, and $L^I_q$ the factor of $d'$ out of the quotient. The second-direction differential is the induced $d''$ just described, with no added scalar. The corresponding formulas for direction II interchange $i,d'$ with $j,d''$. The one-variable chain equations and their naturality hold in the abelian category $C(\mathcal A)$, so (2.4) gives actual commuting double complexes and additive functors on their morphisms.

All eight preserve (1.2). A kernel, image subobject or quotient at an unchanged boundary position is zero when the corresponding original term is zero. The apparently reindexed term $(W^I_qX)^{q-1,j}=B_I^{q,j}$ is a quotient of $X^{q-1,j}$, so it too is zero when that original term is zero. The remaining terms are original or zero. This proves a support inclusion on each diagonal for each functor. The direction-II proof has the same argument with the two indices exchanged.

The natural first-direction maps are

U^I_qX\longrightarrow V^I_qX\longrightarrow X \longrightarrow W^I_qX\longrightarrow L^I_qX.\tag{2.5}

They are identities on the common original terms. At the changed terms they are respectively the cycle inclusion, image inclusion, image epimorphism and quotient epimorphism. In particular the map $X\to L^I_qX$ is zero for $i<q$, the quotient map for $i=q$, and the identity for $i>q$. There is also a natural map $V^I_qX\to X$, the identity for $i\le q$, the image inclusion for $i=q+1$, and zero above. These are the maps used for stabilization below.

The complex $H_I^q(X)$ has terms

H_I^q(X)^j=Z_I^{q,j}/B_I^{q,j}\tag{2.6}

and differential induced by $d''$; its degree is $j$, not $q$. Define $H_I(X)$ as the double complex with this family in first degree $q$, zero first differential, and that second differential. Then $H_{II}H_I(X)$ has objects $H^j(H_I^i(X))$ and zero differentials. In the other order, $H_{II}^q(X)$ is a complex in degree $i$ with differential induced by $d'$, and $H_IH_{II}(X)$ has objects $H^i(H_{II}^j(X))$. These are two constructions; they need not be isomorphic.

The one-variable lower truncation calculation, applied through $F_I$, gives natural isomorphisms of complexes in degree $j$:

H_I^p(L^I_aX)=\begin{cases}0&p<a,\\H_I^p(X)&p\ge a.\end{cases}\tag{2.7}

At $p=a$, the kernel of the differential on $X^{a,j}/B_I^{a,j}$ is precisely $Z_I^{a,j}/B_I^{a,j}$; its incoming differential is zero. At $p=a+1$, the differential out of that quotient has the same image as the original $d'^{a,j}$, since its composite with the quotient epimorphism is $d'^{a,j}$. Above this both neighboring differentials are original, and below it every term is zero. These descriptions prove (2.7) and commute with the induced $d''$, so the displayed isomorphisms include the vertical differentials and are natural under double-complex maps.

3. The exact identity-cone quotient and its total signs

For $X\in C_f^2(\mathcal A)$ fix $q$, and put $B^j=B_I^{q+1,j}$, with induced differential $\delta$. The first-direction truncation sequence is

0\longrightarrow U^I_qX\longrightarrow V^I_qX \longrightarrow R_q(X)\longrightarrow0.\tag{3.1}

Here $R_q(X)$ has copies of the complex $B$ at first degrees $q$ and $q+1$, the positive identity as first differential between them, and $\delta$ as second differential on both copies. All other first degrees are zero. The quotient map in first degree $q$ is the image epimorphism $X^{q,j}\to B_I^{q+1,j}$; in degree $q+1$ it is the identity; and elsewhere it is zero.

Proof of (3.1). The inclusion is the cycle inclusion in first degree $q$ and the identity below. Its quotient in first degree $q$ is the coimage of $d'^{q,j}$, naturally identified with its image $B_I^{q+1,j}$. In degree $q+1$ the quotient is that same image; its incoming differential becomes the identity by the image factorization. The $d''$ equations for the image maps identify both induced second differentials with $\delta$. This proves degreewise exactness and both differential equations of the sequence. It is also the exact sequence of complexes of complexes obtained from $U_qF_I(X)\to V_qF_I(X)$. Thus there is no ambiguity about its category or its maps. $\square$

For an ordinary complex $E$, use $E[r]^n=E^{n+r}$ with differential $(-1)^rd_E$. The ordered cone of $f:E\to F$ has terms $E^{n+1}\oplus F^n$ and differential

d_{C_f}=\begin{pmatrix}-d_E&0\\f&d_F\end{pmatrix}.\tag{3.2}

Set $J=B[-q-1]$. Then the actual total quotient has a natural chain isomorphism

\operatorname{Tot}R_q(X)\simeq C_{1_J}.\tag{3.3}

Proof, including the signs. In total degree $n$, order the two first-degree pieces as $q,q+1$. The total term is

B^{n-q}\oplus B^{n-q-1}=J^{n+1}\oplus J^n.

In these coordinates (1.3) gives

D=\begin{pmatrix}(-1)^q\delta&0\\1&(-1)^{q+1}\delta\end{pmatrix}.\tag{3.4}

Since $d_J=(-1)^{q+1}\delta$, (3.4) is exactly (3.2) for the identity of $J$. Thus (3.3) is the identity on these reindexed summands. It involves the total shift of the ordinary complex $B$, not an untyped identification of the two different complex categories.

The degree-minus-one map

h^n(a,b)=(b,0)\tag{3.5}

denotes a biproduct matrix with the identity from the second copy to the preceding first copy. Matrix multiplication gives $Dh+hD=1$: its first coordinate is $a+((-1)^q+(-1)^{q+1})\delta b=a$, and its second is $b$. This is a categorical matrix calculation, valid without interpreting $a,b$ as elements. It contracts (3.3). $\square$

Applying the exact total functor to (3.1) gives a short exact sequence with this contractible quotient. Its natural long exact cohomology sequence therefore proves

\alpha_q(X):\operatorname{Tot}(U^I_qX)\longrightarrow \operatorname{Tot}(V^I_qX)\quad\hbox{is a quasi-isomorphism}.\tag{3.6}

All image, quotient and cone identifications above are natural in $X$; hence so is (3.6).

4. Successive truncations and the shifted cohomology complex

There is a second natural exact sequence:

0\longrightarrow\operatorname{Tot}(V^I_{q-1}X) \longrightarrow\operatorname{Tot}(U^I_qX) \longrightarrow H_I^q(X)[-q]\longrightarrow0.\tag{4.1}

Proof. Before totalization, $V^I_{q-1}X\to U^I_qX$ is the identity in first degrees below $q$, and the inclusion $B_I^{q,j}\to Z_I^{q,j}$ in degree $q$. It is zero above. Its quotient is therefore the double complex supported in first degree $q$, with second-degree terms $H_I^q(X)^j$. Its second differential is induced by $d''$, exactly the differential $\delta_H$ of (2.6); its first differential is zero. The quotient map on $Z_I^{q,j}$ kills the image of the incoming $d'$, so it commutes with the first differential. The quotient universal property and the image naturality prove its second equation. This establishes the short exact sequence as double complexes. Exactness of totalization follows from Proposition 1.1.

The quotient total has term $H_I^q(X)^{n-q}$ in degree $n$, with differential $(-1)^q\delta_H$. The ordinary shifted complex $H_I^q(X)[-q]$ has precisely those terms and that differential, since $(-1)^{-q}=(-1)^q$. The identification in (4.1) is the identity on those coordinates. Its projection is the cycle-to-cohomology quotient in the first-degree-$q$ summand and zero on every other summand. This specifies the exact maps and the required sign. $\square$

5. The comparison theorem with degreewise bounds

Comparison lemma. In a commuting comparison of five-term exact sequences in an abelian category, isomorphisms at the four outside positions imply an isomorphism at the middle position.

Proof. First consider a comparison of short exact sequences, with vertical maps $l:L\to L'$, $m:M\to M'$, $n:N\to N'$, where $l,n$ are invertible. Write their kernel inclusions as $i,i'$ and their quotient maps as $q,q'$. If $mt=0$, then $nqt=q'mt=0$, so $qt=0$ and $t=is$ uniquely. Now $i'ls=mis=0$, whence $s=0$ and $t=0$. Apply this to any test arrow $t$, including a difference of two arrows, to prove that $m$ is monic. Dually, if $tm=0$, then $ti'l=tmi=0$, so $ti'=0$ and $t=sq'$ uniquely. The equation $snq=sq'm=0$, with $q$ epic and $n$ invertible, gives $s=0$ and $t=0$. Applying this to every test arrow proves that $m$ is epic. The abelian coimage-to-image axiom makes a map which is both monic and epic invertible.

For an exact row $A_0\to A_1\to A_2\to A_3\to A_4$, kernel/image factorization gives the natural short exact row

0\longrightarrow\operatorname{coker}(A_0\to A_1) \longrightarrow A_2\longrightarrow\ker(A_3\to A_4)\longrightarrow0.\tag{5.0}

The left term is the coimage, hence image, of $A_1\to A_2$ by exactness at $A_1$; its inclusion is the kernel of the right map by exactness at $A_2$. The right term is the image of $A_2\to A_3$ by exactness at $A_3$, so the arrow onto it is epic. These identifications are natural under a comparison by kernel and cokernel uniqueness. The outside four isomorphisms induce isomorphisms on the cokernel and kernel in (5.0), since applying their inverse squares supplies the inverses. The short exact comparison just proved then makes the middle arrow invertible. $\square$

Theorem 5.1. Let $f:X\to Y$ be a morphism in $C_f^2(\mathcal A)$. Suppose that it induces an isomorphism

H_{II}H_I(f):H_{II}H_I(X)\longrightarrow H_{II}H_I(Y).\tag{5.1}

Then $\operatorname{Tot}(f)$ is a quasi-isomorphism.

Proof. By (2.6), condition (5.1) means exactly that

H_I^p(f):H_I^p(X)\longrightarrow H_I^p(Y) \quad\hbox{is a quasi-isomorphism of complexes in degree }j \quad\hbox{for every }p.\tag{5.2}

It asserts isomorphisms on every $H^j$ of those complexes, rather than an isomorphism of the complexes themselves. Formula (2.7), including its naturality, shows that $g=L^I_a(f)$ satisfies (5.2) for every integer $a$: below $a$ the maps are $0\to0$, and at or above $a$ their cohomology is that of the original maps.

Fix a total degree $k$. The union

\bigcup_{n\in\{k-1,k,k+1\}}\bigl(S_n(X)\cup S_n(Y)\bigr)\tag{5.3}

is finite. Choose $a$ strictly less than every integer in it; if the union is empty, any integer works. The natural maps

\beta_X:\operatorname{Tot}X\longrightarrow\operatorname{Tot}(L^I_aX), \qquad \beta_Y:\operatorname{Tot}Y\longrightarrow\operatorname{Tot}(L^I_aY)\tag{5.4}

are isomorphisms in degrees $k-1,k,k+1$. Indeed those diagonals have no nonzero original summand in first degree at or below $a$. The quotient at first degree $a$ is zero since its original term is zero. Every remaining nonzero summand, and the map on it, is unchanged. The map is a chain map by (2.5), so the two arrows in these three consecutive total degrees commute with the original differentials. A chain map which is an isomorphism in those three degrees induces an isomorphism on $H^k$: its middle isomorphism carries the kernel of the outgoing differential to that kernel, and the preceding isomorphism carries the incoming image to that image. It therefore carries their quotient isomorphically. This proves the cohomology assertion of (5.4).

Put $A=L^I_aX$, $B=L^I_aY$, and $g=L^I_a(f)$. Their first degrees below $a$ vanish. We claim that

\operatorname{Tot}(U^I_qg)\quad\hbox{and}\quad \operatorname{Tot}(V^I_qg)\quad\hbox{are quasi-isomorphisms for every }q.\tag{5.5}

For $q<a$, both double truncations are zero: even the added image in $V^I_{a-1}$ is the image of the zero arrow from first degree $a-1$. At $q=a$, and subsequently for each $q$, apply (4.1) to $A$ and $B$. The resulting diagram of short exact total sequences has right map $H_I^q(g)[-q]$, which is a quasi-isomorphism by (5.2). If its left map $\operatorname{Tot}(V^I_{q-1}g)$ is a quasi-isomorphism, the natural long exact cohomology sequences make the middle map $\operatorname{Tot}(U^I_qg)$ a quasi-isomorphism as well. In each degree apply the comparison lemma to the segment whose outside positions are the preceding right cohomology, the current left cohomology, the current right cohomology and the following left cohomology. All four maps are isomorphisms, so the middle total cohomology map is an isomorphism.

Finally the square with horizontal arrows $\alpha_q(A),\alpha_q(B)$ from (3.6) commutes. On cohomology its horizontal arrows and its left vertical arrow are isomorphisms, so its right vertical arrow $\operatorname{Tot}(V^I_qg)$ is an isomorphism. Starting with the zero left map at $q=a$, this proves the induction for all $q\ge a$, hence (5.5).

The complexes $A,B$ still have finite diagonal support by Section 2. Choose $b\ge a$ strictly above every first coordinate occurring in their total diagonals $k-1,k,k+1$. If those supports are empty, choose any $b\ge a$. The natural maps

\gamma_A:\operatorname{Tot}(V^I_bA)\longrightarrow\operatorname{Tot}A, \qquad \gamma_B:\operatorname{Tot}(V^I_bB)\longrightarrow\operatorname{Tot}B\tag{5.6}

are isomorphisms in the same three degrees. The retained terms below or at first degree $b$ are unchanged; all original terms above $b$ on those diagonals are zero. The extra image in first degree $b+1$ is a subobject of the original term there and is therefore zero too. The maps (5.6) induce isomorphisms on $H^k$ by the same kernel-and-image argument used for (5.4). Naturality gives a commutative square between (5.6) and $g$. Its left map is a quasi-isomorphism by (5.5), so $H^k(\operatorname{Tot}g)$ is an isomorphism. The natural square between (5.4) and $f$ now proves the same for $H^k(\operatorname{Tot}f)$.

This argument applies to every integer $k$. The bounds $a,b$ may change with $k$; the induction takes place only after the chosen lower truncation. There is no claim that a single finite strip represents the whole total, and no quadrant hypothesis has been used. Thus $\operatorname{Tot}f$ is a quasi-isomorphism in all degrees. $\square$

The theorem is also valid with the directions exchanged. To see the sign in the exchange, define $X^{\mathrm{tr},i,j}=X^{j,i}$, with $d'_{\mathrm{tr}}=d''$ and $d''_{\mathrm{tr}}=d'$. The natural total isomorphism

\theta_X:\operatorname{Tot}(X^{\mathrm{tr}})\longrightarrow\operatorname{Tot}(X) \quad\hbox{acts on }X^{p,q}\hbox{ by }(-1)^{pq}.\tag{5.7}

On its transposed coordinate $(q,p)$, the first outgoing arrow is $d''$. After multiplying at the destination, its scalar is $(-1)^{p(q+1)}$, equal to $(-1)^{pq}(-1)^p$, the scalar of the second outgoing total arrow of $X$. The other outgoing arrow has scalar $(-1)^q$ before the destination multiplier $(-1)^{(p+1)q}$; their product is $(-1)^{pq}$, the scalar of the first outgoing arrow of $X$. Thus (5.7) is a chain map, its inverse has the same multipliers, and the formula is natural. Applying Theorem 5.1 to the transposed map and conjugating by $\theta$ proves the exchanged version.

6. Four graded exercises with complete solutions

Exercise 1 (introductory: the isolated-line shift). Let $E$ be the integer complex $\mathbb Z\xrightarrow{2}\mathbb Z$ in degrees $0,1$. Place it in first degree $r$ of a double complex, with zero first differential. Compute its total differential and cohomology for every integer $r$, and give the comparison with $E[-r]$.

Solution. The only total terms are $\mathbb Z$ in degrees $r,r+1$, and (1.3) makes the differential multiplication by $(-1)^r2$. The shifted complex $E[-r]$ has the same terms and differential, so the comparison is the identity in both degrees. Multiplication by $2$ or $-2$ is injective and has image $2\mathbb Z$. Therefore $H^{r+1}=\mathbb Z/2\mathbb Z$, $H^r=0$, and every other cohomology object is zero. Negative values of $r$ use the same parity formula; for instance $r=-1$ gives $-2$, while $r=2$ gives $2$. This verifies the shift sign rather than silently replacing it by a positive differential.

Exercise 2 (intermediate: a signed identity cone over the integers). Let $B$ be $\mathbb Z\xrightarrow{3}\mathbb Z$ in degrees $0,1$. Take copies of $B$ in first degrees $q,q+1$, joined by the identity, with zero terms elsewhere. Write the complete total and an explicit contraction. Relate this to (3.1).

Solution. In degrees $q,q+1,q+2$ the terms are $\mathbb Z,\mathbb Z^2,\mathbb Z$. In the middle term order the first-degree-$q$ piece before the first-degree-$q+1$ piece. With $s=(-1)^q$, the two differentials are

t\longmapsto(3st,t),\qquad (a,b)\longmapsto a-3sb.\tag{6.1}

Their composite is zero. The first is injective, the second is surjective, and its kernel is the set of $(3sb,b)$, exactly the first image. A contraction is $h^{q+1}(a,b)=b$, $h^{q+2}(c)=(c,0)$, and zero elsewhere. In degree $q$, $hD(t)=t$. In degree $q+1$, $Dh(a,b)+hD(a,b)=(3sb,b)+(a-3sb,0)=(a,b).$ In degree $q+2$, $Dh(c)=c$. Thus it is contractible for both parities and all shifts. For this double complex $X$, the first differential from degree $q$ is the identity, so $U^I_qX=0$ and $V^I_qX=X$. Its image complex is $B$, making (3.1) the identity-cone quotient itself. The vertical complex has nonzero cohomology $\mathbb Z/3\mathbb Z$, while its horizontal identity cone has zero total cohomology.

Exercise 3 (advanced: finite diagonals without a quadrant). For each $m\in\mathbb Z$, put a copy of $\mathbb Q$ at $(i,j)=(m^3,m-m^3)$, and put zero elsewhere. Set both differentials to zero. Prove finite diagonal support, show that neither coordinate has a global lower or upper bound, and give explicit stabilization bounds for a fixed total degree $k$.

Solution. The point indexed by $m$ has total degree $m$. Thus each diagonal has exactly one nonzero term, and the total is $\mathbb Q$ in every integer degree with zero differential. The values $m^3$ go to both infinities as $m$ does. The values $m-m^3$ go to minus infinity for positive $m$ and to plus infinity for negative $m$. Hence neither coordinate is bounded in either direction. For the adjacent diagonals $k-1,k,k+1$, choose

a=(k-1)^3-1,\qquad b=(k+1)^3+1.\tag{6.2}

The cube function is increasing on the integers, so $a$ is strictly below and $b$ strictly above all three first coordinates. Since every differential is zero, lower smart truncation discards the points with first coordinate below $a$, and upper image-extended truncation discards the points above $b$; its added images are zero. Both maps preserve the three displayed total terms and their zero differentials, hence $H^k=\mathbb Q$. No fixed finite pair $a,b$ can retain all points, because the cubes are unbounded. This illustrates why the comparison proof chooses bounds after fixing $k$.

Exercise 4 (advanced: the two iterated orders can differ). Over $\mathbb Q$, take one-dimensional terms with basis labels $a,b,c,d$ at $(0,1),(1,1),(1,0),(2,0)$. Let $d'a=b$, $d'c=d$, $d''c=b$, and let every other differential be zero. Check the double-complex equations, compute both iterated cohomologies and the total, and decide whether the converse of Theorem 5.1 follows.

Solution. Both first-direction pairs are identity maps followed by zero, and the only nonzero second arrow is also an identity followed by zero. Thus both squares of differentials vanish. The only mixed square which starts at a nonzero term and could have both routes is the one from $c$; both $d'd''c=d'b$ and $d''d'c=d''d$ are zero. All other routes are zero, so (1.1) holds. First-direction cohomology is zero in both occupied second degrees, because the pairs $a\to b$ and $c\to d$ are isomorphisms. Consequently $H_{II}H_I=0$.

In direction II the pair $c\to b$ is acyclic. The surviving second-direction cohomology consists of $\mathbb Q a$ at $(0,1)$ and $\mathbb Q d$ at $(2,0)$. Their induced first differentials are zero because the intermediate cohomology terms vanish. Thus $H_IH_{II}$ has exactly these two nonzero objects. The total has $\mathbb Q^2$ in degree $1$, with coordinates $(a,c)$, and $\mathbb Q^2$ in degree $2$, with coordinates $(b,d)$. Its only differential and inverse are

D=\begin{pmatrix}1&-1\\0&1\end{pmatrix},\qquad D^{-1}=\begin{pmatrix}1&1\\0&1\end{pmatrix}.\tag{6.3}

The minus sign comes from the first degree $1$ of $c$. Hence the total is contractible, using $D^{-1}$ as its degree-minus-one homotopy. The map $0\to X$ satisfies (5.1) and has acyclic total, as the theorem predicts. For the transposed double complex, the same acyclic total comparison holds, but its $H_{II}H_I$ equals the nonzero $H_IH_{II}(X)$. Therefore the theorem's iterated-cohomology condition is sufficient and is not necessary for a total quasi-isomorphism. This also rules out treating either iterated cohomology as the total cohomology without further hypotheses.

References

The three linked prerequisite lessons give the complete ordinary truncation, cone, cohomology and totalization interfaces used above.
The Stacks Project, Section 12.25, Spectral sequences: double complexes, for the two iterated-cohomology conventions and the finite diagonal hypothesis. The proof of Theorem 5.1 here uses the explicit exact truncation sequences instead of a spectral-sequence argument.