Grothendieck Spectral Sequences and Computations

Let $u:X\rightarrowtail Y$ and $f:X\to I={⨁}_{j=1}^m{I}_j$. By the finite product property, $f$ is determined by the components $f_j={\pi }_{j}f$. For each $j$, injectivity gives $g_j:Y\to I_j$ with $g_{j}u=f_j$. The finite conjunction of these existence assertions follows by induction on $m$ in ordinary first-order logic; it uses no infinite choice.

The product property supplies $g:Y\to I$ with ${\pi }_{j}g=g_j$, and hence ${\pi }_{j}gu={\pi }_{j}f$ for every $j$. Uniqueness in the product property gives $gu=f$, proving injectivity of $I$. For $m=0$, $I=0$ and both maps to it are unique, so the same extension property holds; for $m=1$ the construction is precisely the extension property of $I_1$. Zero summands and zero $X$ or $Y$ obey the same equations.

The possible summands of $T^n$ have $b\le p\le n$, hence form a finite biproduct of injectives. For $n<b$ this is zero. The augmentation takes $K^p$ into $I^{p,0}$; $vϵ=0$ and $hϵ=ϵd_K$ give $De=e{d}_K$.

Adjoin $K^p$ in vertical degree $-1$. Set $A^{p,-1}=K^p$ and $A^{p,q}=I^{p,q}$ for $q\ge 0$, with vertical augmentation $ϵ$. All columns of $A$ are exact. The total object $U^n=T^n\oplus K^{n+1}$ with the signed differential is isomorphic to $\mathrm{Cone}(e{)}^n$, whose differential is $(t,k)\mapsto (Dt+ek,-d_{K}k)$. Explicitly send the $K^{n+1}$ summand of $U^n$ to $(-1{)}^{n+1}k$ in the cone; the $T$ summand is unchanged. This verifies both signs, including negative $b$.

Let $F^{m}U$ be the subcomplex consisting of columns $p\ge m$. The quotient $U\mathrm{/}{F}^{m}U$ has only the columns $b,\dots ,m-1$. Its finite descending column filtration has shifted exact columns as successive quotients, hence it is acyclic by repeated application of the long exact sequence. Fix $n$ and take $m>n+2$. Then $F^{m}U$ is zero in degrees $n-1,n,n+1$, since its least total degree is $m-1$. Therefore $H^n(U)=H^n(U\mathrm{/}{F}^{m}U)=0$. This is a finite argument for each degree and requires neither exact filtered colimits nor any infinite limit.

The degreewise split sequence $0\to T\to \mathrm{Cone}(e)\to K[1]\to 0$ has connecting map $H^n(K)\to H^n(T)$ induced by $e$: lift a cycle to the $K$ summand and its cone differential is its image under $e$. The zero cone cohomology in step 3.1 and the long exact sequence thus make every $H^n(e)$ invertible. Finally apply the bounded-below injective theorem with exactly its DC/supplied-extension hypothesis to obtain K-injectivity. The zero complex and one-column case obey the same construction.

Put $B^p=\mathrm{im}{d}_K^{p-1}$, $Z^p=\ker d_K^p$ and $H^p=Z^p\mathrm{/}{B}^p$. The differential factors as $K^p\twoheadrightarrow B^{p+1}\rightarrowtail Z^{p+1}\rightarrowtail K^{p+1}$. Thus the two exact sequences to resolve are $0\to B^p\to Z^p\to H^p\to 0$ and $0\to Z^p\to K^p\to B^{p+1}\to 0$. At $p=b$, $B^b=0$.

Assemble resolutions $U^{p,\bullet }$ of $B^p$ and $V^{p,\bullet }$ of $H^p$ from successive injective embeddings of their cokernels, setting $U^{b,\bullet }=0$; F2 verifies the assembled complexes. For $0\to B^p\to Z^p\to H^p\to 0$, successive applications of the dual one-degree step F3 assemble $W^{p,\bullet }$ together with a degreewise split exact sequence $0\to U^{p,\bullet }\to W^{p,\bullet }\to V^{p,\bullet }\to 0$. Apply the same step to $0\to Z^p\to K^p\to B^{p+1}\to 0$ to assemble $I^{p,\bullet }$ in $0\to W^{p,\bullet }\to I^{p,\bullet }\to U^{p+1,\bullet }\to 0$. In the supplied-data branch, all embeddings and compatible next-degree lifts just named are part of the supplied successive choices. In the DC branch, they are selected from the nonempty sets supplied by enough injectives and F3, with the serial accounting given below.

Define $h:I^{p,\bullet }\to I^{p+1,\bullet }$ as projection onto $U^{p+1,\bullet }$ followed by inclusion into $W^{p+1,\bullet }$ and then $I^{p+1,\bullet }$. The next projection kills this image, so $h^2=0$. These arrows are cochain maps in the resolution direction, so $hv=vh$. Their kernels, images and cohomology objects in every vertical degree are respectively $W^{p,q}$, $U^{p+1,q}$ and $V^{p,q}$. Their augmentations are exactly the factorizations of $d_K$ in step 1.1. Every required term is injective and the two degreewise sequences split by the horseshoe construction.

Here is the countable-choice accounting for steps 2.1–3.1. Use three successive DC applications, each to a set of finite compatible states with an entire extension relation. In the first pass, enumerate the pairs $(p,q)$ with $p\ge b$, $q\ge 0$ and construct the side resolutions $U^{p,\bullet }$ and $V^{p,\bullet }$ one embedding at a time. At each task only the preceding vertical cokernel of that same side resolution is needed, so an order by increasing $p-b+q$, with the degree-$q-1$ task first, is serial by enough injectives and F2. The zero resolution $U^{b,\bullet }=0$ needs no selections. In the second pass all side resolutions are now available: enumerate $(p,q)$ again and use F3 successively in $q$ to construct $W^{p,\bullet }$ for $0\to B^p\to Z^p\to H^p\to 0$. In the third pass all $W^{p,\bullet }$ and $U^{p+1,\bullet }$ are available: enumerate $(p,q)$ and use F3 successively in $q$ to construct $I^{p,\bullet }$ for $0\to Z^p\to K^p\to B^{p+1}\to 0$. In each horseshoe pass, the degree-$q$ side terms are already fixed and the degree-$q-1$ middle cokernel is already constructed, so every finite state has a next extension. A fixed diagonal enumeration of each countable task set reaches every task; the unions of the three DC chains supply exactly the compatible data used in steps 2.1–3.1. Supplied successive choices give the same three passes in ZF. This does not select resolutions simultaneously for a proper class of complexes.

Set everything to zero for $p<b$ and $q<0$. Step 3.1 now verifies every clause of the Cartan–Eilenberg definition. For the zero complex one may take all data zero, and for a complex in one degree one may take its ordinary injective resolution in that column. Translating $p$ to $p-b$ yields nonnegative indices without an upper bound on $K$.

Under DC apply the K-injective theorem to $T$ and $T^{\mathrm{\prime }}$. In the derived category the isomorphism $Q(e^{\mathrm{\prime }})Q(e{)}^{-1}:T\to T^{\mathrm{\prime }}$ is represented by a unique homotopy class of maps $a:T\to T^{\mathrm{\prime }}$. Its inverse is represented by $a^{\mathrm{\prime }}:T^{\mathrm{\prime }}\to T$. The bijection for maps into $T$ and $T^{\mathrm{\prime }}$ gives $a^{\mathrm{\prime }}a\simeq 1_T$, $a{a}^{\mathrm{\prime }}\simeq 1_{T^{\mathrm{\prime }}}$ and $ae\simeq e^{\mathrm{\prime }}$. In the supplied-data branch these are exactly the comparison maps and homotopies required in the statement.

Additivity of $F$ sends $a^{\mathrm{\prime }}a-1=dH+Hd$ to $F(a^{\mathrm{\prime }})F(a)-1=F(d)F(H)+F(H)F(d)$, and likewise for the other composite. Homotopic maps induce the same map on cohomology because their difference factors through a differential on cycles. Thus $H^n(F(a))$ and $H^n(F(a^{\mathrm{\prime }}))$ are inverse. Any other comparison over $K$ has the same class by the bijection in step 1.1 and hence gives the same cohomology map.

For $f:K\to L$ with total models $T_K,T_L$, represent $Q(e_L)Q(f)Q(e_K{)}^{-1}$ by a cochain map into $T_L$. The representative for a composite and the composite of representatives have identical images in the derived category; the no-roof bijection makes them homotopic. The same holds for identities and for changes of total models. Apply step 2.1 to obtain functorial maps and the natural comparison isomorphism. Zero maps and zero complexes obey these identities, and no uniform choice of representatives for all maps is needed to define their unique homotopy classes.

Call a short exact sequence $0\to X\to Y\to Z\to 0$ of horizontal complexes admissible if the induced sequences on boundaries, cycles and cohomology are also short exact. For such a monomorphism $X\to Y$, the maps $X^p\to Y^p$ and $X^p\mathrm{/}{B}^{p}X\to Y^p\mathrm{/}{B}^{p}Y$ are monic. For the latter assertion, apply the snake lemma to the exact boundary sequences inside the term sequences: an element of $X^p$ mapping into $B^{p}Y$ comes from $B^{p}X$, as a subobject identity. This argument uses kernels and images and holds in an arbitrary abelian category.

Regard $0\to K\to I^{\bullet ,0}\to I^{\bullet ,1}\to \cdots$ as a resolution in horizontal complexes. Its successive image complexes $C_I^q$ fit into admissible sequences $0\to C_I^q\to I^{\bullet ,q}\to C_I^{q+1}\to 0$, with $C_I^0=K$. To check this assertion, use vertical exactness separately on terms, boundaries, cycles and cohomology in F1, or the explicit exactness assumption for the more general source. No source injectivity or splitting is used in this step. In the diagrams for $0\to B\to Z\to H\to 0$ and $0\to Z\to I\to B[1]\to 0$, the snake lemma identifies the induced cokernels with the next cycle, boundary and cohomology objects. This proves the same assertions for every successive image by induction. The analogous statement holds for $J$.

Fix a horizontal row $J^{\bullet ,q}$. Its split sequences decompose it as the locally finite sum of stalk complexes $H^{p,q}$ in degree $p$ and two-term disk complexes $B^{p+1,q}\stackrel{1}{\to }{B}^{p+1,q}$ in degrees $p,p+1$. At each degree there are only three summands. A cochain map $X\to S^p(E)$ is the same as a map $X^p\mathrm{/}{B}^{p}X\to E$. A cochain map $X\to D^p(E)$, with $D^p(E)$ in degrees $p,p+1$, is the same as a map $X^{p+1}\to E$: the degree-$p$ component is that map composed with $d_X^p$. Because each $E$ is injective, these maps extend across the monomorphisms in step 1.1. Assemble the extensions into a map $Y\to J^{\bullet ,q}$. Countably many splittings and extensions are sufficient; take them as supplied or apply DC to finite partial selections in the fixed Hom sets. Thus every row $J^{\bullet ,q}$ has the extension property for admissible monomorphisms.

Extend $K\stackrel{f}{\to }L\to J^{\bullet ,0}$ across $K\to I^{\bullet ,0}$ by step 2.1. After this extension, $v_J{a}^0$ kills $K$, so it descends to $C_I^1$ and extends across $C_I^1\to I^{\bullet ,1}$ into $J^{\bullet ,1}$. Continue: at stage $q$, $v_J{a}^{q-1}$ kills the preceding image because $v_J^2=0$, hence descends and extends. Each extension is a horizontal cochain map. These choices yield $a^q$, with $v_J{a}^{q-1}=a^q{v}_I$, and therefore a bicomplex map over $f$. DC applies to finite partial maps in the set of these Hom groups; supplied extensions give the same recursion without choice.

For two lifts, put $s^0=0$. At vertical degree zero $a^0-a^{\mathrm{\prime }0}$ kills $K$, so it factors through $C_I^1$ and extends to $s^1:I^{\bullet ,1}\to J^{\bullet ,0}$. Inductively, $a^q-a^{\mathrm{\prime }q}-v_J{s}^q$ kills $\mathrm{im}{v}_I^{q-1}$: substitution of the preceding homotopy equation and $v_J^2=0$ verifies this equality. It therefore descends to $C_I^{q+1}$ and extends to $s^{q+1}$ by step 2.1. Thus $a-a^{\mathrm{\prime }}=v_{J}s+s{v}_I$, with every $s^q$ a horizontal cochain map. The same countable-choice accounting applies.

After an additive functor the equation of step 4.1 remains a vertical homotopy, so the induced maps on vertical cohomology coincide. These are the vertical-first $E_1$ maps. On horizontal cohomology, the same equation gives a homotopy for the induced vertical differential, so its cohomology maps coincide on horizontal-first $E_2$. All subsequent maps agree by natural page transitions. The signed total homotopy is $(-1{)}^{p}s$ on bidegree $(p,q)$: the horizontal mixed terms cancel since $s$ commutes with $h$, and the vertical terms are $vs+sv$. It preserves the horizontal-degree filtration and lowers the resolution-degree filtration by one, consistently with the respective starting pages. Applying the result to lifts of identities and composites proves the stated canonicity. Zero rows, zero maps and bottom resolution degree $q=0$ are included by $s^0=0$.

Replace $v$ on column $a$ by $(-1{)}^{a}v$. The mixed composites sum to $(-1{)}^{a}hv+(-1{)}^{a+1}vh=0$, and the new vertical arrow still squares to zero. Then regard the resulting cochain arrows as arrows in the opposite category. Thus the arrow from $C^{a-1,b}$ to $C^{a,b}$ becomes a homological arrow from bidegree $(a,b)$ to $(a-1,b)$ in the opposite category, with analogous vertical arrows. The square-zero and anticommuting identities are unchanged on reversing composition. The total complex there is the opposite of $T$. The increasing column cutoff through $p$ is the quotient $T\mathrm{/}{F}^{p+1}T$ viewed as a subobject in the opposite category; the same holds for the row cutoff.

Apply both homological theorems in that category. Opposite-category homology is original-category cohomology, since kernel and cokernel exchange. Reversing a page arrow of bidegree $(-r,r-1)$ gives bidegree $(r,1-r)$. For the first page, the within-column differential is $(-1{)}^{p}v$. This constant sign does not change its kernel, image, or their canonical quotient, so vertical cohomology has its canonical quotient identification and the $d_1$ on that quotient is induced by $h$. For the row filtration the within-row differential is $h$ and the next differential on horizontal degree $q$ is $(-1{)}^{q}v$.

Translate the abutment precisely. An image subobject of $H_n(T^{\mathrm{o}\mathrm{p}})$ in the opposite category corresponds to the quotient of $H^n(T)$ by $\ker (H^n(T)\to H^n(T\mathrm{/}{F}^{p+1}T))$. The long exact sequence for $0\to F^{p+1}T\to T\to T\mathrm{/}{F}^{p+1}T\to 0$ identifies this kernel with $F^{p+1}{H}^n(T)$. Thus the opposite of the successive quotient between cutoffs $p-1$ and $p$ is exactly $F^p{H}^n\mathrm{/}{F}^{p+1}{H}^n$, proving the asserted abutment rather than an unrelated filtration.

In total degree $n$ all summands have indices between zero and $n$, giving the stated endpoints; negative total degrees vanish. Finite filtrations are exhaustive and separated, and their quotient towers are eventually constant with value $H^n(T)$, so the completion map is an isomorphism. The first-quadrant bounds make both incident differentials eventually zero at each bidegree. These prove strong convergence, including zero and one-summand diagonals. If $a\ge c$, set $a^{\mathrm{\prime }}=a-c$ and replace $v$ by $(-1{)}^{c}v$; then $h+(-1{)}^{a^{\mathrm{\prime }}}((-1{)}^{c}v)=h+(-1{)}^{a}v$. The normalized total degree is $n-c$, and translation back preserves the original target degree. All constructions use finite biproducts and prescribed signs; no choice is used.

In $C^{p,q}=F(I^{p,q})$ filter the signed total complex by the original complex degree $p$. Its graded column has differential $(-1{)}^{p}F(v)$. The kernel and image of this signed differential equal those of $F(v)$, so the canonical cohomology quotient is $H^q(F(I^{p,\bullet }))=R^{q}F(K^p)$ by the supplied injective resolution of $K^p$. The differential to the next column is induced by $F(h)$ and hence is $R^{q}F(d_K^p)$.

Apply finite-diagonal convergence. Its target is $H^{p+q}(\mathrm{Tot}(FI))={\mathbb{R}}_I^{p+q}F(K)$, and the filtration is the image of the cohomology of the subcomplex with columns of degree at least $p$. In total degree $n$ the filtration is all the target at $p=b$ and zero at $p=n+1$; below $n=b$ the target vanishes. Translation uses normalized total degree $n-b$.

For a map of complexes take the comparison of F3, which preserves columns. Its vertical homotopy makes the $E_1$ map independent of the lift, and its total homotopy makes the target map independent as well. Identity and composite comparisons prove naturality. This use requires exactly DC or the supplied comparisons stated above; constructing the sequence for the fixed bicomplex uses no choice. Zero complexes, zero columns and the sole bottom bidegree satisfy the same formulas.

Fix resolution degree $p$. The horizontal sequences $0\to B^{q,p}\to Z^{q,p}\to H^{q,p}\to 0$ and $0\to Z^{q,p}\to I^{q,p}\to B^{q+1,p}\to 0$ split. An additive functor preserves a split sequence, since it preserves the identities of an inclusion and retraction. It follows that the horizontal kernel, image and quotient after $F$ are $F(Z^{q,p})$, $F(B^{q,p})$ and $F(H^{q,p})$. Thus horizontal cohomology of $F(I^{\bullet ,p})$ is canonically $F(H^{q,p})$; the canonical quotient map gives this identification independently of any chosen splitting.

The next differential on $F(H^{q,p})$ is $(-1{)}^{q}F(v)$. The column $H^{q,\bullet }$ resolves $H^{q}K$, so its degree-$p$ cohomology after $F$ is $R^{p}F(H^{q}K)$; the constant sign $(-1{)}^q$ leaves its kernels and images unchanged. This gives the asserted $E_2$, rather than an $E_1$ identification.

The total target is $H^n(\mathrm{Tot}(FI))$ and the filtration is induced by the subcomplex of resolution degrees at least $p$. In total degree $n\ge b$, its endpoints are $F^0{H}^n=H^n$ and $F^{n-b+1}{H}^n=0$. F2 therefore gives finite strong convergence. F3 supplies comparison maps preserving this filtration; their vertical homotopies give identical $E_2$ and target maps. Identity and composite comparisons prove naturality. For $K=0$ with zero data all terms vanish, and the case $n=b$ has only one possible graded quotient.

Let $h^p:I^{p,\bullet }\to I^{p+1,\bullet }$ be the horizontal Cartan–Eilenberg map. Compatibility with the augmentations says that $h^p$ lifts $d_K^p$, so its map on vertical cohomology is the relative derived map $R^{q}F(d_K^p)$. Since $d_1$ is induced by this horizontal map, $d_1=R^{q}F(d_K)$. For $q=0$, left exactness identifies $F(K^p)$ with $\ker (F(I^{p,0})\to F(I^{p,1}))$, and this identification carries $d_1$ to $F(d_K)$. The augmentations therefore give a cochain map $FK\to \mathrm{Tot}(FI)$. On a degree-$n$ cycle its image lies in the extremal column $n$ and has no positive-resolution component. In the column filtration this is exactly the representative of the bottom-row edge from $E_2^{n,0}=H^n(FK)$, with the finite transition quotients killing precisely the later boundaries. Thus the induced cohomology map is that edge.

In the second sequence, $B^{b,p}=0$ and therefore $H^{b,p}=Z^{b,p}\subset I^{b,p}$. Inclusion of these horizontal cycles gives a cochain map from their vertical resolution after $F$, placed starting in total degree $b$, into the total complex. Its cohomology map is $R^{n-b}F(H^{b}K)\to {\mathbb{R}}^{n}F(K)$, using the constant sign $(-1{)}^b$ on the vertical differential. These are the bottom-row representatives for the resolution-degree filtration, so this is its lower edge.

Projection of the total complex onto column $b$ gives a cochain map to that column with its signed differential and original degree placement. The total cocycle equation in column $b+1$ says the horizontal image of the projected vertical class is zero. By the $d_1$ calculation in step 1.1, the cohomology map therefore lands in $\ker (R^{n-b}F(K^b)\to R^{n-b}F(K^{b+1}))$, which is the left-axis $E_2$ term because there is no preceding column. Projection is the quotient by the first positive translated filtration piece, so F3 identifies it with the other first-sequence edge.

Projection onto resolution degree zero sends a total cocycle to a horizontal cohomology class in $F(H^{n,0})$. Its induced vertical differential is zero by the next component of the total cocycle equation. Left exactness identifies this kernel with $F(H^{n}K)$, since $H^{n,\bullet }$ resolves $H^{n}K$. Total boundaries give zero under this map. It is the filtration quotient at resolution degree zero and hence the upper edge. The same equations are morphism equalities on cycle and boundary subobjects, so they do not require selected representatives in an abelian category. For $n=b$ both filtrations have one piece and the arrows agree with the bottom augmentation identification; zero targets cause no exception.

Each $I^q$ is injective, so the hypothesis gives $R^{p}G(F(I^q))=0$ for $p>0$. Additivity makes $F(I)$ a cochain complex, zero in negative degrees. Its cohomology is precisely $R^{q}F(A)$ by definition. In degree zero left exactness identifies its kernel with $F(A)$; positive exactness would additionally require every positive $R^{q}F(A)$ to vanish.

For a witness take $F={\mathrm{Hom}}_{\mathbb{Z}}(\mathbb{Z}\mathrm{/}2,-)$, $G$ the identity of abelian groups and $A=\mathbb{Z}$, with a supplied injective resolution of $A$. Identity is exact, so all its positive derived objects vanish. The projective resolution $0\to \mathbb{Z}\stackrel{2}{\to }\mathbb{Z}\to \mathbb{Z}\mathrm{/}2\to 0$ has rank-one free, hence projective, terms: a map from $\mathbb{Z}$ lifts across an epimorphism by lifting the image of $1$. Applying $\mathrm{Hom}(-,\mathbb{Z})$ gives $\mathbb{Z}\stackrel{2}{\to }\mathbb{Z}$ in degrees zero and one. Its degree-one cohomology is $\mathbb{Z}\mathrm{/}2\mathrm{\ne }0$. F4 identifies this with $H^1(F(I))$. Thus $F(I)$ is not a resolution of $F(A)=0$, even though every one of its terms is $G$-acyclic. This witness is relative to supplied data and uses no choice of an infinite family of lifts.

Apply F2 to $G$ and the complex $F(I^{\bullet })$. Its first page is $E_1^{q,p}=R^{p}G(F(I^q))$. F1 makes this zero for $p>0$; for $p=0$ left exactness identifies it with $GF(I^q)$ and the $d_1$ is its cochain differential. Thus the second page has only the row $p=0$, equal to $H^q(GF(I^{\bullet }))$.

Every differential from page two onward has zero source or target off that row, so the page is stationary. In target degree $n$ its finite filtration has only one potentially nonzero quotient, at $(n,0)$; all earlier successive quotients vanish and the last filtration term is zero. Therefore its bottom edge $H^n(GF(I^{\bullet }))\to H^n(\mathrm{Tot}(GJ))$ is an isomorphism. F3 identifies this exact edge with the augmentation map, proving that augmentation is a quasi-isomorphism.

F4 identifies the source cohomology with $R_{\mathsf{I}}^n(GF)(A)$, and nowhere is $F(I^{\bullet })$ treated as a resolution of $F(A)$. Under DC, or when the relevant change-of-resolution comparisons and homotopies are supplied, this relative identification is independent of $\mathsf{I}$ and natural, so the subscript may then be suppressed. At $n=0$ it agrees with the left-exact augmentation kernel identification, and zero terms or zero complexes satisfy the same one-row argument.

The cohomology of $F(I)$ is $R^{q}F(A)$, with $H^0(F(I))=F(A)$. Substitute this into F2 to obtain the displayed $E_2$. F1 identifies the total target with $R^{p+q}(GF)(A)$ through the actual augmentation quasi-isomorphism. Its proof computes the other filtration as one row, so these are two filtrations of the same total complex, not a claimed equality of their second pages.

Set the lower bound to zero in F3. The bottom horizontal-cycle inclusion gives $R^{n}G(H^0(F(I)))=R^{n}G(F(A))\to H^n(\mathrm{Tot}(GJ))$; projection gives $H^n(\mathrm{Tot}(GJ))\to G(H^n(F(I)))=G(R^{n}F(A))$. Transport both along the augmentation isomorphism of step 1.1. This defines the canonical derived-composite edges and agrees with the finite filtration definition. In degree zero both reduce to $GF(A)$; vanishing edge terms are permitted. The naturality/choice qualifications are exactly those of F1–F3.