Draft lesson.

Signed bifunctors and sum and product total complexes

Original text public domain (CC0).

A separately additive functor combines two differentials. Its two translation comparisons must anticommute, so the mixed curvature terms cancel. For commuting double complexes the same cancellation appears as a sign on the second differential. We construct both diagonal totals, their homotopies and whole cone comparisons, and specify the normalization needed to identify a Hom total with the usual Hom complex.

Use Universes and small categories for size conventions. All categories below are additive. An infinite coproduct or product is used only where its existence is assumed. Translations, coherent powers and universal triangle comparisons, Sections 1–3 supplies coherent inverse powers and the generator odd equation. Differential objects, curvature and cone homotopies, Sections 1–5 supplies the actual homotopy quotient and all triangle axioms, including nonzero curvature. The ordinary complex and Hom conventions are those of Graded complexes, contravariant signs and stupid truncations.

1. Odd interchange for every integer power

Let the translations on \(\mathcal A,\mathcal B,\mathcal D\) be \(T,S,U\). A functor \(F:\mathcal A\times\mathcal B\to\mathcal D\) is separately additive if fixing either object gives an additive functor. Thus each fixed-variable morphism map is a group homomorphism, and \(F(f,g)=F(1,g)F(f,1)=F(f,1)F(1,g)\) with the appropriate intermediate objects. This does not mean additivity on the product category: \(F(f+f',g+g')\) generally has four terms.

Specify natural isomorphisms

with the typed odd equation

Both routes have source \(F(TX,SY)\) and target \(U^2F(X,Y)\). All minus signs act in additive Hom groups.

We calculate with strictly invertible translations. This loses no generality: the strict models and additive evaluation equivalences in the differential-object prerequisite, Section 1, transport the functor in both variables and lift each comparison through full faithfulness of the target equivalence. Naturality, (1.2), curvature equations and homotopy equations are reflected by that full faithfulness. In the original categories every power reduction uses the coherent comparisons constructed in the translation prerequisite, Section 2.

Iterate \(a\) to comparisons \(a_r:F(T^rX,Y)\to U^rF(X,Y)\), with \(a_0=1\) and

For positive \(r\) this defines the next comparison; for negative \(r\) solve it backwards using inverses. In particular \(a_{-1,TX,Y}=U^{-1}(a_{X,Y}^{-1})\). Define \(b_s\) the same way. These iterations satisfy

for all integers. To check the first equation, its two sides agree for \(t=0\); (1.3) advances equality from \(t\) to \(t+1\), and cancellation of invertible factors reverses that step. This proves it upwards and downwards; the second proof is identical.

Lemma 1.1. For all \(r,s\in\mathbb Z\),

Proof. First fix \(s=1\). Starting at \(r=0\), expand the next \(a\) by (1.3), use (1.2) for the last generator crossing, and use the preceding equality for the remaining crossings. Naturality moves the intermediate comparison past the other variable. This advances the scalar from \((-1)^r\) to \(-(-1)^r\). Because every comparison is invertible, the same expanded equality can be canceled backwards; it also advances from \(r+1\) to \(r\), including the step from zero to minus one. Thus a block of \(r\) first-variable shifts crosses one second-variable shift with scalar \((-1)^r\) for every integer \(r\).

For this fixed \(r\), expand a block of \(s+1\) second-variable shifts by the second equation of (1.4). Moving its last generator past the first-variable block multiplies the previous scalar by \((-1)^r\). Start at \(s=0\), and cancel the same invertible factors to run the recurrence downwards as well as upwards. The scalar is therefore \((-1)^{rs}\) for all integers \(s\). All reductions have the same endpoints by (1.4); in nonstrict models their coherence is the power cancellation theorem cited above. This proves (1.5) with its actual comparison maps. \(\square\)

For \(u:X\to T^rX'\), write \(\widehat F(u,1)=a_{r,X',Y}F(u,1)\); for \(v:Y\to S^sY'\), use \(\widehat F(1,v)=b_{s,X,Y'}F(1,v)\). Naturality and (1.5) give the useful morphism form

Indeed the ordinary functor square for \(u,v\) commutes; naturality attaches the four power comparisons to that square, leaving exactly (1.5) at its far corner. Both sides now go from \(F(X,Y)\) to \(U^{r+s}F(X',Y')\). This includes negative-degree homotopies.

2. Curvature, homotopies and the entire cone in each variable

For differential objects \(d_X:X\to TX\), \(d_Y:Y\to SY\), put

Using (1.6) at \((r,s)=(1,1)\), the two mixed terms in \(U(D)D\) cancel. The unmixed terms use (1.4) and functoriality. Therefore

The target of each term is \(U^2F(X,Y)\). Square-zero inputs give a square-zero output. For differential maps \(f:X\to X'\), \(g:Y\to Y'\), substitute their two differential equations and use naturality of \(a,b\) to obtain \(D_{X',Y'}F(f,g)=U(F(f,g))D_{X,Y}\). Identities and composition are inherited from \(F\), and fixed-variable additivity remains true. Thus (2.1) defines a bifunctor on differential objects and on complexes.

Suppose a differential map \(f:X\to X'\) is a boundary

Use \(H=a_{-1,X',Y}F(h,1):F(X,Y)\to U^{-1}F(X',Y)\). In its output boundary \(U(H)D+U^{-1}(D_{X',Y})H\), the two first-variable terms equal \(F(f,1)\) by (1.4) and separate additivity. The other two terms cancel by (1.6) with \((r,s)=(-1,1)\): explicitly \(U(H)D_2=-U^{-1}(D_{2,X',Y})H\). For a boundary in the second variable use \(b_{-1}F(1,h')\), with the same argument and exchanged variables. No square-zero hypothesis was used. Since the resulting boundary is the differential map \(F(f,1)\), it is automatically an admissible curvature homotopy by the prerequisite's Lemma 2.1. The two null-homotopy ideals are therefore preserved. Differences of representatives, and then composition of representatives, prove well-defined separately additive bifunctors on \(K_d\) and \(K_c\).

The comparison \(a:F(TX,Y)\to UF(X,Y)\) is also a differential isomorphism when \(TX\) has differential \(-T(d_X)\) and the output shift has differential \(-U(D)\). Naturality accounts for the first-variable term, and (1.2) makes the second-variable term acquire the required minus sign. For \(b:F(X,SY)\to UF(X,Y)\), naturality accounts for the second-variable term and (1.2) for the first. Their inverse comparisons give the inverse shifts.

Fix \(Y\) and take a differential map \(f:X\to X'\). Recall the ordered cone \(C_f=TX\oplus X'\), with differential \(\bigl(\begin{smallmatrix}-T(d_X)&0\\T(f)&d_{X'}\end{smallmatrix}\bigr)\). Separate additivity supplies the actual biproduct comparison, and the cone map is

In these coordinates its differential is

For the first diagonal entry the input differential contributes the negative translated \(D_1\), and odd interchange contributes the negative translated \(D_2\). The second diagonal is (2.1); the off-diagonal is the naturality square for \(a\) at \(f\). Thus (2.4) is a differential isomorphism. It takes the inclusion to the second biproduct inclusion, and its first projection is \(aF(p_f,1)\), exactly the required last arrow. In the second variable use \(F(X,C_g)\) and \(\operatorname{diag}(b,1)\): now the shifted \(D_2\) is negative by the input cone and the shifted \(D_1\) is negative by (1.2), giving the same matrix and the last arrow \(bF(1,p_g)\). Naturality follows on each biproduct leg. Since distinguished triangles are isomorphic to these whole cone triangles, these comparisons prove separate triangulatedness on both \(K_d\) and \(K_c\).

3. Commuting double complexes and their two shifts

A commuting double complex consists of objects \(B^{i,j}\) and maps

Every composite in the last equation has target \(B^{i+1,j+1}\). A morphism is a family of maps commuting with both differentials. Identities, composition, addition and finite biproducts are defined componentwise and preserve these equations, giving an additive category \(C^2(\mathcal C)\).

Regarding \(j\) as the inner complex degree, the objects \((B^{i,\bullet},d'')\) are complexes and \(d'^{i,\bullet}\) are chain maps, by the commuting square. Their consecutive composite is zero. This gives a complex in \(C(\mathcal C)\). Conversely unpack such a complex: its terms give \(d''\), its outer maps give \(d'\), and precisely (3.1) results. These assignments are inverse on objects and morphisms after the harmless family identifications. Regarding \(i\) as the inner degree instead gives the column identification. Thus both are equivalences with \(C(C(\mathcal C))\), without imposing anticommuting squares.

The two outer shifts consequently have formulas

On the right each differential is evaluated at the reindexed source. Substitution verifies (3.1). Reindexing backwards with the same negative differential gives inverse functors. In particular these shifts negate different directions.

4. The two totals use different universal properties

If countable coproducts exist, define

Each input summand has only two outgoing components, so the coproduct universal property defines \(D\). Its square on that summand has components \((d')^2\), \((d'')^2\), and

These vanish by (3.1); equality on every injection proves \(D^2=0\).

If countable products exist, define

Here the projections on the right are from the degree-\(p-1\) product. Each output uses exactly two incoming projections. This defines a map into the product, and requires no infinite sum of input coordinates. On any output projection of \(D^2\) the same-direction terms vanish; the two mixed coefficients are \((-1)^{i-1}\) and \((-1)^i\), and the commuting square makes them cancel. The product universal property therefore proves square zero. We use coordinate language only to abbreviate these projection equations.

A double-complex map induces either total map by its component maps and the corresponding universal property. The two differential squares commute componentwise. Components also prove identity, composition and additivity, so both totals are additive functors wherever defined.

If for every \(p\) only finitely many \(B^{i,p-i}\) are nonzero, finite biproducts suffice for both constructions. A coproduct over a family with only finitely many nonzero objects is its finite biproduct: maps from the zero entries are uniquely zero. The product assertion follows by its projections. The canonical map from the sum to the product has component identity on a matching index and zero otherwise. On the finite nonzero indices these are exactly inverse biproduct coordinates, so it is an isomorphism. The two definitions of \(D\) have the same entries under this map. The map also commutes with every double-complex morphism; zero entries can be included in the union of the two finite supports. Thus the finite-diagonal total is a natural single construction, needing neither countable products nor countable coproducts.

Symbols such as \(x_i\otimes y_j\) in a general additive category mean a chosen component or biproduct leg; there need not be elements or a tensor product. Every calculation above is an equation of actual morphisms. Actual tensor elements will be used only in the module example below.

5. Transposition, cones and double homotopies

Define the transpose by

at the reindexed positions. Exchanging both differential directions is necessary for their types. The squares and zero composites are inherited from (3.1), and \(v^2=1\). Reindex each total and multiply its \((i,j)\) component by \((-1)^{ij}\), obtaining

The original \(d'\) edge becomes a second-direction edge with coefficient \((-1)^j\) in the transposed total. Its comparison equation is \((-1)^{(i+1)j}=(-1)^{ij+j}\). The original \(d''\) edge becomes a first-direction edge; its equation is \((-1)^i(-1)^{i(j+1)}=(-1)^{ij}\). Thus \(r_BD=D r_B\). The same component signs define the inverse, since \(v^2=1\) and \((-1)^{ij+ji}=1\). Every component of a double-complex map commutes with these scalars, proving naturality. These arguments use injections for sums and projections for products, and apply to the finite total too.

For a chain map \(f:X\to Y\), take \(B^{-1,j}=X^j\), \(B^{0,j}=Y^j\), all other columns zero, \(d'=f\), and \(d''=d_X,d_Y\). The commuting square is the chain-map equation. Its total in degree \(p\) is \(X^{p+1}\oplus Y^p\), with differential

This is exactly the ordered cone. The inclusion from the zero column is \(y\mapsto(0,y)\), and the projection to the minus-one column is the positive map \((x,y)\mapsto x\) into \(X[1]\). Thus all three cone arrows, not just its object, agree.

A double null homotopy for \(f:B\to E\) consists of maps

with \(d''_Et_1=t_1d''_B\), \(d'_Et_2=t_2d'_B\), and

on each bidegree, with the shifted indices specified by (5.4). Its total homotopy sends a component \((i,j)\) to the two components \((i-1,j),(i,j-1)\) by

In \(D_Eh+hD_B\), the first-direction terms from \(t_1\) have coefficient one. Its mixed terms have coefficients \((-1)^{i-1},(-1)^i\), so cancel by its cross-commutation condition. The second-direction terms from \(t_2\) have coefficient \((-1)^{2i}=1\). Its mixed terms have coefficients \((-1)^i,(-1)^{i+1}\), so cancel by the other cross-commutation condition. What remains is (5.5), proving the homotopy equation. For a sum, every input has two outgoing homotopy entries; for a product, every output has two incoming entries. Hence both maps and calculations are legitimate. The finite case uses the same biproduct entries. Differences of maps use (5.5) for their difference.

6. Total bifunctors and every stated bounded case

Let \(F:\mathcal C\times\mathcal C'\to\mathcal C''\) be separately additive, now without translation data on these base categories. For complexes \(X,Y\), put

An additive functor sends a zero map to zero, so the same-direction composites are zero. The two mixed composites agree by the ordinary functor square. Thus (6.1) is a commuting double complex, natural on pairs of chain maps. Section 4 gives its sum or product total whenever the corresponding countable objects exist. The total action on a pair \((f,g)\) is componentwise \(F(f^i,g^j)\). Holding either variable fixed, separate additivity of these components and the universal property of the total prove separate additivity of the functor.

Here are the finite-diagonal cases, requiring only finite biproducts in \(\mathcal C''\). If \(X^i=0\) for \(i<a\) and \(Y^j=0\) for \(j<b\), a possibly nonzero degree-\(p\) entry has \(a\le i\le p-b\), a finite interval; the output is zero for \(p<a+b\). If instead \(X^i=0\) for \(i>a\), \(Y^j=0\) for \(j>b\), it has \(p-b\le i\le a\), and the output is zero for \(p>a+b\). If \(X\) is bounded in \([a,b]\), the possible \(i\)'s lie in that interval whatever \(Y\) is; bounded \(Y\) works by interchanging the indices. Each zero input entry gives a zero output entry because the fixed-variable functor is additive. Hence these arguments also apply when some further entries happen to vanish. Section 4 proves the natural sum/product identification in every case.

To retain the signs and cone data, first extend \(F\) to graded objects by

The first translation comparison is reindexing: \(F(X[1]^i,Y^j)=F(X^{i+1},Y^j)\) goes to that same entry in \(F_\star(X,Y)[1]\) with coefficient one. The second sends \(F(X^i,Y^{j+1})\) to its corresponding entry with coefficient \((-1)^i\). Each is an isomorphism, with the inverse reindexing and the same scalar. This is proved on injections or projections, and gives natural comparisons. On an entry \(F(X^{i+1},Y^{j+1})\), the two routes of the generator square have coefficients \((-1)^{i+1}\) and \((-1)^i\), so are negatives. Thus (6.2) is a signed bifunctor of graded categories with translation, in the precise sense of (1.2).

Its differential (2.1) is exactly \(F(d_X,1)+(-1)^iF(1,d_Y)\), the commuting total differential of Section 4. Theorem of Section 2 therefore descends it through homotopies in both variables, with its whole cone comparison and the actual last arrow. For completeness a first-variable input homotopy \(s:X\to X'[-1]\) gives total homotopy entries \(F(s^i,1)\); the second-variable homotopy \(t\) gives \((-1)^iF(1,t^j)\). These are (5.6), and also agree with the inverse translation comparisons of (6.2).

Finite biproducts can be regrouped with countable sums or products in the cone comparison. For products the isomorphism \(\prod_q(A_q\oplus B_q)\simeq(\prod_qA_q)\oplus(\prod_qB_q)\) is defined by its two families of projections; its inverse combines the two coordinate projections with each finite biproduct pairing. Checking every projection proves the two inverse equations. For sums, define the comparison and its inverse on each of the two types of injections, and check those injections. Consequently (2.4)–(2.5) apply literally to (6.2); no exactness of an infinite total functor is assumed.

In the finite cases, bounds survive signed shifts and finite biproducts. For a cone, its first-variable support is the union of the support of \(X[1]\) and \(X'\); it is still bounded below, bounded above or bounded as required. The corresponding statement holds in the second variable. Thus every homotopy and cone calculation stays in the indicated domains. We have proved separately triangulated bifunctors

using the finite total, and on all of \(K(\mathcal C)\times K(\mathcal C')\) using a sum or product total when those countable objects exist. This uses the full triangulated structures proved in the prerequisites, not just completion of individual arrows.

Define the reversed functor \(G:\mathcal C'\times\mathcal C\to\mathcal C''\) by \(G(Y,X)=F(X,Y)\), including the same reversal on pairs of maps. Its double complex on \((Y,X)\) is precisely the transpose of (6.1). The isomorphism (5.2) gives the natural chain isomorphism

Its inverse, chain equation and naturality were all proved in Section 5. This compares two specified functors; it does not assume that \(F\) itself has a symmetry.

7. The literal Hom double complex and its normalization

For an additive \(\mathcal C\), use the target-first base bifunctor \(F(Y,Z)=\operatorname{Hom}_{\mathcal C}(Z,Y)\) on \(\mathcal C\times\mathcal C^{\mathrm{op}}\), valued in abelian groups. Degree reversal of \(X\) into \(\mathcal C^{\mathrm{op}}\), as in the ordinary-complex prerequisite, Section 4, has term \(X^{-j}\) and differential represented in \(\mathcal C\) by \((-1)^jd_X^{-j-1}\). Its Hom double complex is

The second component has target \(\operatorname{Hom}(X^{-j-1},Y^i)\). Squaring \(d'\) gives \(d_Y^2\), while squaring \(d''\) gives \((-1)^j(-1)^{j+1}f d_X^{-j-1}d_X^{-j-2}=0\). Postcomposition and precomposition commute, with the unchanged scalar \((-1)^j\), so (3.1) holds. These checks also follow from (6.1), but display all indices.

In the product total write \(j=-k\), \(i=k+p\), so its degree-\(p\) group is the full product of \(\operatorname{Hom}(X^k,Y^{k+p})\). Formula (4.3) gives the literal total differential

In fact the incoming precomposition has \(j=-k-1\) and \(i=k+p+1\), so the product of its two signs is \((-1)^{i+j}=(-1)^p\). This is a plus where the standard Hom differential uses a minus. Define the actual comparison

where \(H\) is the standard Hom complex with \((\partial f)^k=d_Y^{k+p}f^k-(-1)^p f^{k+1}d_X^k\). Its chain equation is

Every component square of \(J\) is identity, so it is an isomorphism. For chain maps \(a:X'\to X\), \(b:Y\to Y'\), both constructions act by \(f^k\mapsto b^{k+p}f^ka^k\). This commutes with \((-1)^k\), proving full two-variable naturality. Equivalently replace the vertical differential in (7.1) by \((-1)^{j+1}\) times precomposition; its ordinary product coordinates give \(H\) directly. Multiplication by \((-1)^j\) is an isomorphism from the literal bicomplex to this normalized one: it commutes with \(d'\), and its vertical coefficient changes by a minus. Its total is exactly (7.3).

We now verify the full shift and triangle interfaces of this comparison. Write \(H(X,Y)\) source first in formulas, and \(\mathcal H(Y,X)=H(X,Y)\) for its target-first bifunctor on \(K(\mathcal C)\times K(\mathcal C)^{\mathrm{op}}\). The opposite translation is \(X\mapsto X[-1]\), with the complete triangle definition in Connecting signs and opposite triangle structures, Section 3. The natural standard comparisons are

For the first, both differentials are \(-d_Y f-(-1)^pf d_X\) after reindexing. For the second, the source differential is \(d_Yf+(-1)^pf d_X\), while the target shifted differential is \(-d_Yg-(-1)^pgd_X\). The degree-dependent scalar changes sign between \(p\) and \(p+1\), so these two chain equations agree. The inverse-source-shift comparison \(H(X[1],Y)\to H(X,Y)[-1]\) is \(f\mapsto((-1)^pf^{k-1})_k\); applying the second comparison after shifting back gives the same signs twice, hence identity. The first comparison has the obvious inverse reindexing. On \(H(X[-1],Y[1])\), the two routes to \(H(X,Y)[2]\) have coefficients \((-1)^{p+1}\) and \((-1)^{p+2}\). Thus they satisfy the generator odd equation, with the precise opposite translation.

For a map \(a:Y\to Z\), the ordered cone has \(C_a^l=Y^{l+1}\oplus Z^l\). Decompose a map into that biproduct into its two components. This defines a natural isomorphism

Its first differential is \(-d_Yu-(-1)^pu d_X=-\partial_H^{p+1}u\); its second is \(a u+\partial_H^p v\). These are precisely the cone entries. Inclusion agrees with the second coordinate inclusion, and projection agrees with the first line of (7.5) after \(H(X,p_a)\). Thus every arrow of the target-variable cone triangle is preserved.

For the other variance take \(a:X\to X'\), and set \(A=H(X',Y)\), \(B=H(X,Y)\), \(g=H(a,Y):A\to B\). Decomposing maps from \(C_a^l=X^{l+1}\oplus X'^l\) gives the natural chain isomorphism

Indeed the two source boundary components are \(d_Yu+(-1)^pu d_X-(-1)^pv a\) and \(\partial_Av\). Reindex the first by \(k=l+1\). Multiplication by its degree-\(p+1\) output sign \((-1)^p\) makes its \(v a\) entry \(-g(v)\), and the remaining entries become \(-\partial_B\) of the signed \(u\). The second remains \(\partial_Av\). These are the differential entries of \(C_g[-1]\). Reversing the permutation and multiplying by the same scalar gives the inverse.

To check the exact opposite cone triangle, inverse-rotate the original cone triangle. In \(K(\mathcal C)\) it is \(C_a[-1]\xrightarrow{-p_a[-1]}X\xrightarrow aX'\xrightarrow{j_a}C_a\). The definition of the opposite structure makes

a distinguished opposite triangle. For a degree-\(p\) family on \(C_a[-1]^l=X^l\oplus X'^{l-1}\), write its components as \((u^l,v^l)\). Combining the second line of (7.5) with \(\Psi[1]\) gives the comparison

Precomposition by \(-p_a[-1]\) sends a family \(b\) on \(X\) to \((-b,0)\), and (7.9) takes it to \((0,b)\), the positive cone inclusion. Precomposition by \(j_a[-1]\), followed by the second comparison in (7.5), takes \((u,v)\) to \(((-1)^{p+1}v^{k+1})_k\), the positive first projection in (7.9). The first arrow is \(g\) by definition. Hence all three squares to the whole cone triangle commute, proving separate triangulatedness in the opposite variable as well. The actual map homotopies in the ordinary-complex prerequisite, Section 3, ensure descent through \(K\); their formulas also follow by conjugating the total homotopies of Section 6.

Finally give the literal total \(R\) these precise interfaces by transport along \(J\). For either shifted input \(Z\), its shift comparison is \(J_{X,Y}[1]^{-1}\lambda_H J_Z\), where \(\lambda_H\) is the relevant line of (7.5). For any of the triangles just checked, its cone comparison is the composite of \(J\) at the third object, (7.6) or (7.9), and the inverse cone isomorphism induced by \(J\) at the first two objects. A commuting square of chain maps induces the cone map on the two diagonal blocks, as proved in the cone prerequisite; naturality of \(J\) makes that square commute. These typed composites are natural chain isomorphisms. Conjugating each of the three triangle squares just proved shows that the literal total preserves the same entire triangles, with its transported last-arrow comparison. The odd shift equation is preserved by the same conjugation. Thus identifying the two Hom conventions requires the specified \(J\), and its shift and cone interfaces are now proved, not left implicit.

8. Balanced tensor over an algebra

Let \(k\) be a commutative ring, \(R\) a unital \(k\)-algebra, \(M\) a right \(R\)-module and \(N\) a left \(R\)-module. Construct \(M\otimes_RN\) from the free \(k\)-module on pairs \((m,n)\), quotienting by additivity in both variables, \(k\)-linearity and the relations \((mr,n)=(m,rn)\). A \(k\)-bilinear balanced map kills precisely these imposed relations, so extends uniquely through the quotient. This proves the balanced universal property. Simple tensors generate it. A pair of module maps therefore induces the tensor map, and checking generators proves its identity, composition and separate-additivity laws. Finite direct sums distribute by this same universal property, or by the resulting biproduct maps.

The category of \(k\)-modules has countable coproducts, given by finite-support coordinate families. Section 6 applies without boundedness: the tensor total is

Each component map is balanced since the input differentials are module maps. The same-direction terms in its square are zero and the mixed coefficients \((-1)^{i+1},(-1)^i\) cancel. Checking simple tensors and then every injection proves square zero. Its homotopies and both complete cone comparisons are those proved in Section 6. In a noncommutative algebra there need not be a map interchanging \(m,n\) into the same tensor construction: their module sides differ. The always valid comparison (6.4) is with the functor whose argument order has been reversed. Any additional symmetry requires specified module and algebra data.

9. Four graded exercises with complete solutions

Exercise 1 (foundation: a signed square). Put \(\mathbb Z\) at bidegrees \((0,0),(1,0),(0,1),(1,1)\), with each horizontal map multiplication by two and each vertical map multiplication by three, and put zero elsewhere. Calculate the total differentials and the transpose comparison.

Solution. Both paths across the square multiply by six, and each same-direction square goes through zero. In increasing first-index order the total has \(\mathbb Z\), \(\mathbb Z_{(0,1)}\oplus\mathbb Z_{(1,0)}\), \(\mathbb Z\) in degrees zero, one, two. Its matrices are \(D^0=(3,2)^{\mathsf t}\), \(D^1=(2,-3)\), whose product is zero. The transpose exchanges the middle entries; its corner signs are positive except at \((1,1)\), where the sign is negative. Its degree-two comparison is therefore minus identity. The two chain equations are checked by permuting \((3,2)\) to \((2,3)\) in the first square and by negating \((2,-3)\) to \((-2,3)\) in the second. An unsigned transpose fails the second equation.

Exercise 2 (intermediate: sums and products can differ). Put \(B^{i,-i}=\mathbb Z\) for every integer \(i\), all other entries zero, and both differentials zero. Are the two totals isomorphic? Explain the finiteness assumption that would make their comparison an isomorphism.

Solution. Both are concentrated in degree zero, with groups \(\bigoplus_{i\in\mathbb Z}\mathbb Z\) and \(\prod_{i\in\mathbb Z}\mathbb Z\). The canonical map is inclusion of finite-support families. The constant-one family is outside its image. In fact the groups are not even abstractly isomorphic: the sum is countable, being the union over finite supports of countable sets of integer tuples. The product contains all binary sequences indexed by the nonnegative integers; Cantor's diagonal argument proves that set uncountable. Finite support of the diagonal of objects, as in Section 4, would reduce both universal objects to one finite biproduct. The finite support of each element of a sum does not impose that condition on a double complex.

Exercise 3 (advanced: contraction survives either total). Suppose a commuting double complex has maps \(s^{i,j}:B^{i,j}\to B^{i,j-1}\) satisfying \(d''s+sd''=1\) and \(d's=sd'\). Prove that every existing sum or product total is contractible. Why is using \(s\) without a bidegree sign generally wrong?

Solution. Use \(t_1=0,t_2=s\) in (5.4)–(5.5), so the double boundary is the identity. The total homotopy is \(h|_{(i,j)}=(-1)^is\). Its vertical boundary terms are \(d''s+sd''=1\), and its horizontal terms are \((-1)^id's+(-1)^{i+1}sd'=0\). Thus \(Dh+hD=1\). For products each output of \(h\) has a single incoming projection, so the same proof does not need finite-support input families. If instead \(h=s\), its vertical boundary is \((-1)^i1\), and its horizontal terms add as \(d's+sd'\). Even with zero horizontal differential the identity fails at an odd first index over \(\mathbb Z\): take the two-term vertical complex \(\mathbb Z\xrightarrow1\mathbb Z\) in column \(i=1\). Its total differential is minus identity, so the unsigned contraction gives minus identity. The signed contraction gives identity.

Exercise 4 (synthesis: detect the Hom normalization by evaluation). Let \(X^0=X^1=\mathbb Z\), \(d_X^0=1\), with other terms zero, and let \(Y^0=\mathbb Z\) alone. Describe the literal product total \(R(X,Y)\), the standard \(H(X,Y)\), and \(J\). In the usual Hom-tensor evaluation, show why using the literal coordinates without \(J\) gives a wrong chain equation.

Solution. In degree minus one both groups consist of the map \(X^1\to Y^0\); in degree zero both consist of the map \(X^0\to Y^0\). All other groups vanish. Formula (7.2) makes the differential from minus one to zero multiplication by minus one. The standard formula makes it multiplication by plus one. The comparison \(J\) is minus identity in degree minus one, since that component has source index one, and identity in degree zero. Thus \(J^0D_R=\partial J^{-1}\).

Take \(f:X^1\to Y^0\) the identity in degree minus one and \(x=1\in X^0\). Their tensor lies in degree minus one. Unnormalized evaluation of it is zero, because \(f\) has no component on \(X^0\). But the tensor differential has evaluations \(-1\) from \(D_Rf\otimes x\) and \(-1\) from \((-1)^{-1}f\otimes d_Xx\), summing to \(-2\); the chain equation fails. The normalized evaluation is \(f\otimes x\mapsto(-1)^kf^k(x)\) for \(x\in X^k\), namely the ordinary evaluation after \(J\otimes1\). Its two boundary evaluations here are \(-1\) and \(+1\), so cancel. Generally (7.4) and the full evaluation calculation in the ordinary-complex prerequisite, Section 5, prove its chain equation. This example also shows why the two differential formulas cannot be identified by unsigned coordinates.

References

• The Stacks Project Authors, Double complexes and associated total complexes, for the commuting-square convention and diagonal sum definition. The product and curvature arguments above are given explicitly.
• The Stacks Project Authors, Hom complexes, for the standard Hom differential on module complexes. Section 7 specifies the comparison from the literal signed degree-reversal convention and proves it for arbitrary additive categories.
• Differential objects, curvature and cone homotopies, for the full triangulated structures used by the cone comparisons; Connecting signs and opposite triangle structures, for the opposite triangle definition used in (7.8).