Draft lesson.

Hereditary cohomology and integral tensor splittings

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

Level: advanced. Prerequisites: abelian categories, cochain complexes, quasi-isomorphisms, exact Hom sequences and bounded derived tensor.

The cohomology of a complex usually forgets how its degrees are joined together. When every second extension group vanishes, those joins can be split in the derived category. Over the integers this also gives an explicit tensor calculation: ordinary tensor products occupy one degree, and first Tor groups occupy the preceding degree. We construct both splittings, retain their signs, and explain their dependence on choices.

Conventions and prerequisites

Work in a fixed Grothendieck universe containing the integers, with the axiom of choice; use a larger ambient universe for localizations when required. All categories are locally small in the stated universe. Complexes are cohomological, and

An object placed in degree \(r\) is therefore \(A[-r]\). An object of \(D^+(\mathcal A)\) has cohomology zero in sufficiently negative degrees; an object of \(D^-(\mathcal A)\) has cohomology zero in sufficiently positive degrees. Smart truncations give representatives with the corresponding term bounds.

The following three lessons supply the exact interfaces we use. Injective tests and finite homological dimension, Sections 1–4 proves the Ext criteria, their exact-sequence dimension shifts, and the finite-resolution interpretation. Relative replacements and finite resolution bounds, Sections 1–4 constructs bounded-below injective replacements and proves that maps into such complexes are computed in the homotopy category. Derived Hom and bounded resolution pairs, Sections 1–7 constructs bounded-above free replacements and balanced derived tensor, including its actual single-replacement comparison maps. Its normalized Hom complex computes Ext from a projective resolution to a heart object.

Freely accessible comparisons are Stacks, Tag 059Y, which records the free-subgroup fact used in Section 3, Tag 01D6, for divisible and injective abelian groups, and Tag 013K, for injective resolutions. The arguments below give the subgroup and splitting constructions explicitly.

1. Hereditary quotients of injective objects

Call an abelian category hereditary here if

By the single-degree criterion in the injective-tests prerequisite, it is equivalent to require just degree two to vanish for every pair. Enough injectives is a separate assumption.

Lemma 1.1. In a hereditary abelian category, every quotient of an injective object is injective.

Proof. Write the quotient as a short exact row

For every \(T\), its exact Hom sequence has the segment

The outside groups vanish by injectivity and (1.1). Thus \(\operatorname{Ext}^1(T,B)=0\) for every \(T\). The injective criterion says precisely that \(B\) is injective. This uses the quotient with its actual kernel; it does not require the kernel to be injective. \(\square\)

We also need the elementary closure under direct summands. If \(J\) is a retract of an injective \(I\), compose a map into \(J\) with its inclusion into \(I\), extend there, and then compose with the retraction. This extends the original map into \(J\). Finite biproducts of injectives are injective because their finitely many component maps extend separately. No closure under infinite sums is being assumed.

2. Splitting every bounded-below complex into its cohomology

Theorem 2.1. Let \(\mathcal A\) be hereditary and have enough injectives. For every \(X\in D^+(\mathcal A)\), form the complex

It is bounded below. There is a generally noncanonical isomorphism

Moreover \(C(X)\), with the degree inclusions, is the coproduct of the objects \(H^rX[-r]\) in \(D^+(\mathcal A)\). This coproduct assertion needs neither infinite coproducts in \(\mathcal A\) nor a coproduct assertion in the unbounded derived category.

Proof of the splitting. Choose a term-bounded-below representative \(X_0\) and a quasi-isomorphism \(q:X_0\to I\) to a bounded-below injective complex, by the relative-replacement theorem. Put

Each \(B^r\) is a quotient of \(I^{r-1}\), hence is injective by Lemma 1.1. This includes the zero boundaries below the lower bound. Its monomorphism into \(I^r\) therefore has a retraction \(\rho_r:I^r\to B^r\): extend the identity of \(B^r\) across that monomorphism. Let \(J^r=\ker\rho_r\). The split exact row gives

Because \(d_I^r\) kills \(B^r\), its restriction to \(J^r\) has the same image as the whole differential. Its factorization through that image is an epimorphism

Let \(K^r=\ker\delta_r\). Equation (2.4) identifies \(Z^r\) with \(B^r\oplus K^r\). Quotienting by the first summand gives an isomorphism

Use its inverse and the inclusion \(K^r\hookrightarrow J^r\hookrightarrow I^r\) to define \(v^r:H^rI\to I^r\). Its image is in the cycle kernel, so the components form a chain map

By (2.6), \(H^rv\) is the identity of \(H^rI\) for every integer \(r\). Thus \(v\) is a quasi-isomorphism. The isomorphisms \(H^rq:H^rX_0\to H^rI\) give a chain isomorphism \(C(X)\to C(I)\). The two quasi-isomorphisms into \(I\) prove (2.2).

The construction can also be viewed as two-term resolution blocks. Define \(E_r\) by the exact coresolution

placing \(J^r\) in degree \(r\) and \(B^{r+1}\) in degree \(r+1\). At degree \(r\), the only blocks present are \(E_r\) and \(E_{r-1}\). Equations (2.4)–(2.5) identify this degreewise finite assembly with \(I\), including its differential. If the same coresolution is first placed in degrees zero and one and then shifted by \([-r]\), its differential becomes \((-1)^r\delta_r\). The chain isomorphism to \(E_r\) is identity in degree \(r\) and multiplication by \((-1)^r\) in degree \(r+1\). These signs reconcile the resolution notation with the actual complex.

Proof of the coproduct property. There is a common lower bound \(a\) with \(H^rX=0\) for \(r<a\). The complex (2.1) exists simply by specifying its single object in each degree. For \(Y\in D^+(\mathcal A)\), choose one bounded-below injective replacement \(Y\to T\). The K-injective Hom test in the relative-replacement prerequisite gives

A chain map from \(C(X)\) to \(T\) is exactly a family of maps \(H^rX\to T^r\) killed by \(d_T^r\). A homotopy between two such maps is exactly a family \(h^r:H^rX\to T^{r-1}\) with difference \(d_T^{r-1}h^r\). The equations for distinct \(r\) do not interact, because the source differential is zero. Consequently restriction to the degree inclusions gives

For surjectivity, choose representatives of the homotopy classes independently and assemble their components. For injectivity, choose the homotopies independently and assemble them. These are choices in sets in the fixed universe, and there is only one source object in each degree. Applying the same K-injective Hom test to each factor gives the required product of derived Hom sets. The map is restriction along the stated inclusions, so is natural in \(Y\). This proves the universal property.

Thus one may write \(C(X)=\bigoplus_r H^rX[-r]\) in \(D^+\), with precisely that meaning. The retractions in (2.4) are choices; the resulting isomorphism (2.2) is not asserted to be natural. Exercise 2 supplies a concrete obstruction to any natural choice in abelian groups. \(\square\)

3. Why abelian groups have homological dimension one

The tensor calculation will apply to arbitrary abelian groups, including groups whose free presentations have uncountable rank. We give the needed rank-independent argument.

Lemma 3.1. Every subgroup of a free abelian group is free.

Proof. Well-order a basis and write

for a subgroup \(L\subset F\). The coefficient at \(e_\alpha\) maps \(L_{\alpha+1}\) to a subgroup of \(\mathbb Z\), with kernel \(L_\alpha\). Every subgroup of \(\mathbb Z\) is \(d\mathbb Z\) for a unique integer \(d\geq0\): for a nonzero subgroup choose its least positive element and apply division with remainder.

If the coefficient image is zero, \(L_{\alpha+1}=L_\alpha\). Otherwise let its positive generator be \(d_\alpha\), and choose \(l_\alpha\in L_{\alpha+1}\) with that coefficient. For each \(l\in L_{\alpha+1}\), its coefficient is \(m d_\alpha\) for a unique integer \(m\), and \(l-m l_\alpha\) belongs to \(L_\alpha\). A nonzero multiple of \(l_\alpha\) has nonzero \(e_\alpha\)-coefficient. Hence

at every nonzero-image stage. At a limit ordinal \(\lambda\), finite support gives \(F_\lambda=\bigcup_{\alpha<\lambda}F_\alpha\), and therefore \(L_\lambda=\bigcup_{\alpha<\lambda}L_\alpha\). The same observation applies at a limit final index \(\kappa\). Transfinite induction using (3.2) now shows that the chosen \(l_\alpha\)'s span \(L\). They are independent: a nonzero finite linear combination has a largest index with nonzero coefficient, and its coefficient at the corresponding \(e_\alpha\) cannot vanish. This is a basis of \(L\), without a countability assumption. \(\square\)

Every abelian group \(M\) is the quotient of the free group \(F_0\) on its underlying set, by the map taking its basis element \(e_m\) to \(m\). Its kernel \(F_1\) is free by Lemma 3.1. Thus

is a length-one free resolution. A free group is projective: lift a map across an epimorphism by choosing a lift for each basis image. The projective criterion in the injective-tests prerequisite therefore shows that all second and higher Ext groups vanish. Alternatively the normalized Hom complex of (3.3), placed in degrees minus one and zero, has terms only in degrees zero and one when the target is a heart object; the derived-Hom comparison identifies its cohomology with Ext. For \(M=N=\mathbb Z/2\), that complex is \(\mathbb Z/2\xrightarrow{0}\mathbb Z/2\). Its first cohomology is nonzero. The homological dimension is consequently exactly one.

To apply Theorem 2.1 in this category, we also verify enough injectives. An abelian group \(J\) is divisible if multiplication by every positive integer is surjective. If \(J\) is injective, extend the map \(m\mathbb Z\to J\) sending \(m\) to a prescribed \(x\) across \(m\mathbb Z\subset\mathbb Z\). Its value on one is a solution of \(m y=x\), proving divisibility.

Conversely take a map \(g:A\to J\) with \(A\subset B\) and \(J\) divisible. Order its partial extensions by extension. Unions of chains are compatible extensions, so Zorn's lemma supplies a maximal one, still denoted \(g:A'\to J\). If \(b\in B\setminus A'\), the integers \(q\) with \(q b\in A'\) form an ideal \(d\mathbb Z\). If \(d=0\), set \(y=0\); otherwise choose \(y\) with \(d y=g(d b)\). The formula

is well-defined. Indeed, equality of two presentations gives \((q-q')b\in A'\); if \(d>0\), write \(q-q'=t d\) and use \(g(t d b)=t d y\), while if \(d=0\), then \(q=q'\). Formula (3.4) strictly extends the maximal partial map, a contradiction. Thus divisible groups are injective.

The group \(\mathbb Q/\mathbb Z\) is divisible, since a representative can be divided by any positive integer. Its characters detect nonzero elements of any group \(G\). On a cyclic subgroup generated by an element of finite order \(m>1\), send the generator to \(1/m+\mathbb Z\). On an infinite cyclic subgroup, send it to \(1/2+\mathbb Z\). These characters extend to \(G\) by injectivity. Evaluation is therefore a monomorphism

The product is injective, since each coordinate of a map extends across a monomorphism and the coordinate extensions assemble into a product map. Its index set and product belong to the fixed universe. This proves enough injectives, and completes all hypotheses for Theorem 2.1 in abelian groups.

4. Free resolution blocks for a bounded-above complex

For tensor we use bounded-above free complexes. Choose a free replacement \(p:P\to X\) for \(X\in D^-(\mathsf{Ab})\), with \(P^i=0\) for \(i>u\). The construction and incoming comparison maps are in Section 7 of the derived-Hom prerequisite. Define

Both are subgroups of free groups, so are free by Lemma 3.1. In the exact row

choose a section \(s_r\) of the epimorphism, using projectivity of \(B^{r+1}\). The isomorphism \((z,b)\mapsto z+s_r(b)\) identifies \(P^r\) with \(Z^r\oplus B^{r+1}\). Under these identifications the differential is

Indeed \(d_P^r z=0\), and \(d_P^rs_r(b)\) is the actual inclusion of its image \(B^{r+1}\) into \(P^{r+1}\). That image lies in \(Z^{r+1}\), by square zero, so has no second coordinate in the next decomposition.

For every \(r\), let \(E_r\) be the two-term complex

The quotient \(Z^r\to H^rP\) is an augmentation to \(H^rP[-r]\), and is a quasi-isomorphism. At degree \(k\), only \(E_k\) and \(E_{k+1}\) contribute, with terms \(Z^k\) and \(B^{k+1}\). Equation (4.3) proves a chain isomorphism between their degreewise finite assembly and \(P\). Blocks above \(u\) are zero, since \(P^{u+1}=0\) also forces \(B^{u+1}=0\).

For signs, let \(A_r\) denote the ordinary free resolution \(B^r\to Z^r\) in degrees minus one and zero. The complex \(A_r[-r]\) has differential \((-1)^r\iota_r\). Its chain isomorphism to \(E_r\) multiplies degree \(r-1\) by \((-1)^r\) and degree \(r\) by one. It commutes with the augmentation to \(H^rP[-r]\). This convention will be used for the tensor shifts.

Every assertion so far is about the actual bounded-above complex and two summands per degree. An unrestricted infinite sum of resolutions is not required to make (4.3)–(4.4) meaningful.

5. A tensor pair and its shift signs

For two abelian groups \(M,N\), choose free resolutions \(A\to M\), \(B\to N\) in degrees minus one and zero as in (3.3). The total differential on a term \(A^i\otimes B^j\) is

The two mixed terms have opposite coefficients, while the two square terms vanish, so \(D^2=0\). By the single-replacement comparison of the derived-Hom prerequisite, the augmentation induces a quasi-isomorphism

The comparison is the actual map \(1_A\otimes(B\to N)\), with the total signs in (5.1). Its quasi-isomorphism follows also directly from that prerequisite's finite-diagonal proof: each free term of \(A\) tensors the quasi-isomorphism \(B\to N\) exactly.

The target of (5.2) has two terms. Consequently

The first equality is right exactness of tensor, applied to the augmentation cokernel. The second is the definition of Tor as negative derived-tensor cohomology, evaluated using this free resolution. This also proves that higher Tor groups vanish for arbitrary abelian groups.

We retain the shift comparison explicitly. A component of \(A[-r]\otimes B[-s]\) coming from original degrees \(i,j\) lies in total degree \(i+j+r+s\). Multiplication on that component by \((-1)^{s i}\) defines a chain isomorphism

For the first differential, its coefficient in the source is \((-1)^r\); the multiplier at the next first degree is \((-1)^{s(i+1)}\). Their product equals \((-1)^{r+s}(-1)^{s i}\), the target shifted coefficient. For the second differential, the source coefficient is \((-1)^{i+r+s}\); the multiplier stays \((-1)^{s i}\) because only \(j\) changes. The target coefficient is also \((-1)^{r+s}(-1)^i\). This proves the chain equation for both summands, for all integers \(r,s,i,j\). The inverse uses the same multipliers.

Combining (5.3)–(5.4), the tensor of the shifted groups has ordinary tensor in cohomological degree \(r+s\) and first Tor in degree \(r+s-1\). This degree distinction is the reason that Tor contributes to a different diagonal in the next theorem.

6. The integral tensor splitting with finite diagonals

Theorem 6.1. For \(X,Y\in D^-(\mathsf{Ab})\) and every integer \(n\), there is a generally noncanonical group isomorphism

Each displayed sum is finite after zero summands are omitted. No finite-generation hypothesis on the groups and no lower bound on the two complexes is required.

Proof. Choose bounded-above free replacements \(P\to X\), \(Q\to Y\), zero above \(u,v\), respectively. The derived-tensor prerequisite identifies the value with \(\operatorname{Tot}(P\otimes Q)\), including its natural counit computation. Its degree-\(n\) terms have indices

Thus the total diagonal is finite. Decompose \(P\) into the free blocks \(E_r\) of Section 4, and \(Q\) into blocks \(F_s\). Their actual chain isomorphisms induce a chain isomorphism of the tensor total to the assembly of \(\operatorname{Tot}(E_r\otimes F_s)\). Tensor distributes over the finite biproducts in each original entry, and (6.2) has only finitely many entries. To verify that the new block assembly is also finite in each degree, note that a pair block can have terms only in degrees

For it to occur in degree \(n\), one must have \(r+s\in\{n,n+1,n+2\}\), \(r\leq u\), \(s\leq v\). Such pairs satisfy \(n-v\leq r\leq u\) and \(n-u\leq s\leq v\), so are a finite set. This proves the claimed degreewise finiteness without interchanging infinite sums or infinite cohomology operations.

The differential preserves each pair label \((r,s)\), by (4.3) in both variables. Hence kernels and images in each degree are the finite direct sums of the block kernels and images; quotienting gives the corresponding finite direct sum of their cohomologies. More explicitly, the finite sets of blocks meeting degrees \(n-1,n,n+1\) have finite union. Compute these three degrees using that union and insert zero terms for the other blocks. Ordinary finite biproduct exactness then proves the stated cohomology decomposition.

Each block \(E_r\) is isomorphic to \(A_r[-r]\), with the sign specified in Section 4, and similarly for \(F_s\). Their augmentations resolve \(H^rP[-r]\) and \(H^sQ[-s]\). Equations (5.3)–(5.4) therefore give

In particular the possible term in degree \(r+s-2\) contributes zero cohomology. The quasi-isomorphisms identify \(H^rP,H^sQ\) with \(H^rX,H^sY\). Summing (6.4) over the finite sets proves (6.1). When \(n>u+v\), both displayed sets are empty and the total also has no term, so the theorem includes these degrees as well.

The derived object and the identifications arising from resolution comparison are independent of the resolutions, by the incoming pair counit theorem in the prerequisite. The block decomposition, however, used the sections in (4.2). Those sections need not be compatible with maps of complexes. Formula (6.1) asserts existence of the displayed group isomorphism; it does not assign a natural splitting to every pair. \(\square\)

7. Four graded exercises with complete solutions

Exercise 1 — shifted cyclic groups (18 points)

Let \(m,l\geq1\), let \(r,s\) be arbitrary integers, and put \(g=\gcd(m,l)\). Compute all cohomology groups of

Give a generator for the kernel defining Tor, and include the coprime and zero-group cases.

Solution. Resolve \(\mathbb Z/m\) by \(\mathbb Z\xrightarrow{m}\mathbb Z\) in degrees minus one and zero. Tensoring this single resolution with \(\mathbb Z/l\) gives

Its cokernel is \(\mathbb Z/g\). Indeed the quotient of \(\mathbb Z\) by the subgroup generated by \(m,l\) is \(\mathbb Z/g\), by the Bezout identity. Its kernel consists of residue classes \(a\) with \(l\mid m a\). Write \(m=g m'\), \(l=g l'\), where \(m',l'\) are coprime. The condition is \(l'\mid a\); the kernel is generated by \(l'=l/g\) modulo \(l\), and this element has order \(g\). Thus both groups in (5.3) are \(\mathbb Z/g\).

Use the chain comparison (5.4) to shift the tensor by \([-(r+s)]\). Therefore the only possibly nonzero groups are

The differential signs in the shifted resolutions are \((-1)^r m\) and \((-1)^s l\), and (5.4) accounts for them; changing the sign of a multiplication map changes neither its kernel nor its cokernel. If \(g=1\), both groups vanish, so the derived tensor is the zero object. If \(m=1\) or \(l=1\), one input is already zero and \(g=1\), giving the same conclusion. The argument covers every integer shift, including negative ones.

Exercise 2 — why the cohomology isomorphism is not natural (16 points)

Put \(A=\mathbb Z/2\). Show that there is a nonzero arrow \(f:A[-1]\to A\) in the derived category that induces zero in every cohomology degree. Deduce that the identity functor on \(D^+(\mathsf{Ab})\) cannot be naturally isomorphic to the functor \(X\mapsto C(X)\), acting by cohomology on arrows. Also construct a nonidentity automorphism of \(A\oplus A[-1]\) inducing the identity on every cohomology group.

Solution. The resolution \(\mathbb Z\xrightarrow{2}\mathbb Z\to A\) gives the normalized Hom complex

in degrees zero and one. The differential is zero since multiplication by two is zero on \(A\); the possible Hom sign has no effect on that zero map. Its first cohomology is \(A\), so \(\operatorname{Ext}^1(A,A)=A\). By shifting the derived Hom group, choose its nonzero class as

The source has its only nonzero cohomology in degree one, and the target in degree zero. Thus each degreewise cohomology map of \(f\) is zero. The assignment \(C\) is a functor: it takes the maps on cohomology as its zero-differential chain maps, so respects identity and composition. In particular \(C(f)=0\).

If there were a natural isomorphism \(\eta:\operatorname{Id}\to C\), naturality would give \(\eta_A f=C(f)\eta_{A[-1]}=0\). Invertibility of \(\eta_A\) would force \(f=0\), a contradiction. An isomorphism in the reverse direction has an inverse in this direction, so is equally impossible.

On \(V=A\oplus A[-1]\), let \(N\) have the sole off-diagonal component \(f\) from the second summand to the first. Then \(N^2=0\), since the projection of the first summand to the second is zero. Hence \(u=1_V+N\) has inverse \(1_V-N\). It is not identity: projecting \(N\) onto the first summand after including the second recovers \(f\ne0\). Nevertheless \(H^jN=0\) for every \(j\), so \(H^ju=1\). Cohomology alone does not select an isomorphism in Theorem 2.1.

Exercise 3 — infinitely many cohomology groups, finite answers (18 points)

Let \(X=Y\) be the zero-differential complex having \(\mathbb Z/2\) in every degree \(r\leq0\) and zero in degrees \(r>0\). Compute \(H^n(X\otimes^{\mathbf L}_{\mathbb Z}Y)\) for every integer \(n\). Explain why the computation is finite in each degree.

Solution. These are bounded-above complexes with upper bound zero. For \(n\leq0\), the pairs \(r,s\leq0\) with \(r+s=n\) are indexed by \(r=n,n+1,\ldots,0\), so there are \(1-n\) pairs. Each contributes \(\mathbb Z/2\) to the tensor sum. The pairs with \(r+s=n+1\) have \(-n\) members when \(n\leq-1\), and none when \(n=0\). Each contributes \(\operatorname{Tor}_1(\mathbb Z/2,\mathbb Z/2)=\mathbb Z/2\), as computed in Exercise 1. Hence

For example the degrees zero, minus one and minus two have one, three and five copies, respectively. An actual free replacement is assembled from one block \(\mathbb Z\xrightarrow{2}\mathbb Z\) in degrees \(r-1,r\) for each \(r\leq0\), with its augmentation to the degree-\(r\) group. There are two free summands in each negative degree and one in degree zero. The tensor total has the finite diagonal \(n\leq i\leq0\) in degree \(n\leq0\). Equivalently its block labels satisfy (6.3) with \(u=v=0\). Thus neither the infinitely many nonzero cohomology groups nor the infinitely many blocks causes an infinite sum in a fixed degree. The original complex has no lower bound, so this example also tests the full bounded-above scope of Theorem 6.1.

Exercise 4 — read the Tor contribution inside a concrete complex (20 points)

Let \(P\) be the free complex

with other terms zero. Determine its cohomology and its two free blocks. Compute \(H^n(P\otimes^{\mathbf L}_{\mathbb Z}\mathbb Z/6)\) both directly and from (6.1). Check the shift sign in the degree-one block.

Solution. The composite of the displayed differentials is zero. Multiplication by two is injective, so \(H^{-1}P=0\). The kernel of \(d^0\) consists of the first coordinate, and its quotient by the image of \(d^{-1}\) is \(\mathbb Z/2\). The cokernel of \(d^0\) is \(\mathbb Z/3\). Thus

The first coordinate in degree zero, together with degree minus one, is the block \(E_0:\mathbb Z\xrightarrow{2}\mathbb Z\) in degrees minus one and zero. The second coordinate in degree zero, together with degree one, is \(E_1:\mathbb Z\xrightarrow{3}\mathbb Z\) in degrees zero and one. Their direct sum is literally (7.7). If the ordinary resolution \(A_1:\mathbb Z\xrightarrow{3}\mathbb Z\) is placed in degrees minus one and zero and shifted by \([-1]\), its differential is \(-3\). Multiplication by minus one in degree zero and by one in degree one gives a chain isomorphism \(A_1[-1]\to E_1\): both sides of its chain equation are multiplication by \(-3\).

Since \(P\) is bounded above and free, the single-replacement comparison computes its derived tensor by the actual complex

Its degree-minus-one kernel is generated by \(3\) modulo six, and is \(\mathbb Z/2\). Its degree-zero cycles have arbitrary first coordinate and second coordinate in \(\{0,2,4\}\). Its boundaries are \(\{0,2,4\}\) in the first coordinate and zero in the second. The quotient is \(\mathbb Z/2\oplus\mathbb Z/3\). Finally its degree-one cokernel is \((\mathbb Z/6)/\{0,3\}=\mathbb Z/3\). All other degrees vanish.

Formula (6.1) gives exactly these groups. Degree minus one receives \(\operatorname{Tor}_1(\mathbb Z/2,\mathbb Z/6)=\mathbb Z/2\). Degree zero receives \((\mathbb Z/2)\otimes\mathbb Z/6=\mathbb Z/2\) and \(\operatorname{Tor}_1(\mathbb Z/3,\mathbb Z/6)=\mathbb Z/3\), because the latter pair has cohomology indices summing to one. Degree one receives \((\mathbb Z/3)\otimes\mathbb Z/6=\mathbb Z/3\), with no Tor term. This checks the adjacent diagonal, both actual free blocks, and the sign required by the shifted resolution.