Draft lesson.

Derived Hom and bounded resolution pairs

Original text public domain (CC0).

A resolution of one input can suffice for a derived bifunctor, but this needs a uniform exactness condition in the other input. We prove the two-input criterion first. We then construct derived Hom with its contravariant source, identify its cohomology with derived-category morphisms, and construct derived tensor over an algebra whose multiplication need not commute. Explicit cyclic-group computations retain the comparison maps and their signs.

Use the universe conventions. All abelian categories below are locally \(\mathcal U\)-small. A larger ambient universe houses denominator categories and formal objects when necessary. Complexes are cohomological: \(d:X^i\to X^{i+1}\), and \(X[1]^i=X^{i+1}\) has differential \(-d_X\). Write \(Q\) for localization at quasi-isomorphisms. The bounded denominator comparison and its equality tests identify \(D^+,D^-,D^b\) with their full subcategories of \(D\).

Our total-complex conventions and complete cone comparisons are those of Signed bifunctors and sum and product total complexes, Sections 5–8. We use the entire adapted-pair localization theorem, including its units, its incoming counterpart and its negative shift exchange. These are separately additive functors; additivity on a product category is not the assertion being made.

1. The normalized product Hom complex

For any two complexes in an additive category define

The factors are abelian groups. Their countable product exists in \(\mathsf{Ab}_{\mathcal U}\). Applying the differential twice gives two zero square terms and two opposite multiples of \(d_Yf^{i+1}d_X\); thus \(\partial^2=0\). For chain maps \(a:X'\to X\), \(b:Y\to Y'\), the action is \(f^i\mapsto b^{i+n}f^ia^i\). The chain-map equations prove that this commutes with (1.1). Checking components proves identities, composition and separate additivity.

Degree-zero cycles are exactly chain maps: their equation is \(d_Yf=fd_X\). A degree-minus-one family \(h\) has boundary \(d_Yh+hd_X\), exactly a chain homotopy difference. Consequently there is a natural group isomorphism

Naturality follows from the explicit precomposition and postcomposition action, which takes a homotopy to the composed homotopy.

For precision, the literal target-first Hom double complex in the total-complex prerequisite has terms \(\operatorname{Hom}(X^{-q},Y^p)\), first differential postcomposition by \(d_Y\), and second differential \((-1)^q\) times precomposition by \(d_X\). Its literal product total \(R\) has differential \(d_Yf+(-1)^n f d_X\). The natural chain isomorphism

has inverse itself: \(\partial J=JD_R\), since the source-index multiplier changes sign on the precomposition term. Here the superscript on \(R^n,H^n\) denotes a complex term, not its cohomology. Section 7 of that prerequisite proves the full cone and shift interfaces after this normalization, including the last arrow in the opposite-source triangle. We keep those interfaces.

In standard coordinates the natural unit-shift comparisons are

In the last two formulas \(n\) is the degree in the source Hom complex. The target comparison is reindexing and has the same differential on both sides. For a source shift, its differential reverses the sign of precomposition; the degree-dependent multiplier also reverses sign from \(n\) to \(n+1\), giving the chain equation. The last two maps are inverse after shifting back. Their two orders with a target shift differ by a minus sign, as proved for the entire separately triangulated bifunctor in the prerequisite. Iterating the first comparison gives every target integer shift.

2. An absolute Hom value represents derived morphisms

Suppose the product Hom bifunctor, with target \(D(\mathsf{Ab}_{\mathcal U})\), is right localizable at a pair \((Y,X)\) in the following precise sense. Its formal value is represented by an object \(Z\) of that target category. Denote this representing object, with its specified formal identification, by \(R\operatorname{Hom}_{\mathcal C}(X,Y)\).

Let \(O_Y\) be the outgoing quasi-isomorphisms \(t:Y\to Y_t\). Let \(I_X\) be the incoming quasi-isomorphisms \(s:X_s\to X\), with arrows oriented toward refinements, so precomposition is covariant on \(I_X\). The two-index Hom formula proves that both indices are filtered, including parallel-arrow equalizers. The formal product localization gives the stipulated representation

This is a formal colimit of represented objects. Its representation is stronger than the mere existence of an ordinary colimit. The absolute representation theorem says, pointwise, that every functor out of the target sends (2.1) to the corresponding represented formal value. Its proof applies the ind-extension of that functor, which preserves the presenting formal diagram and constant objects. No preservation of ordinary colimits by that functor is required.

Apply this assertion to cohomology \(H^0\), followed if necessary by inclusion into abelian groups in the ambient universe of the indices. The formal value is now represented by \(H^0Z\). Realize this formal diagram by the ordinary filtered colimit in those ambient abelian groups. Realization carries a constant object to itself, and (1.2) identifies each diagram entry. Hence

The second isomorphism sends the class of a numerator \(f\) to \((Qt)^{-1}Qf(Qs)^{-1}\). Its full proof establishes surjectivity by roofs and injectivity by common denominators and a further refinement killing the numerator difference. It also proves addition and naturality for original arrows and their inverse denominators. Thus (2.2) is the canonical comparison, with both variances retained.

The same argument works on the pair of bounded domains \(K^-(\mathcal C)^{\mathrm{op}}\times K^+(\mathcal C)\). Incoming sources may be bounded above and outgoing targets bounded below: the bounded denominator comparison preserves every numerator and every finite equality diagram in these indices. Therefore their two-index group colimit is still the Hom group in the full \(D(\mathcal C)\), even when the two representatives do not belong to one common bounded half.

The target-shift comparison (1.4) transports (2.1) to a represented formal value at \((X,Y[r])\), for every integer \(r\). Translation is an invertible functor, so it also transports the stipulated representation. Taking (2.2) at that pair gives

For heart objects define \(\operatorname{Ext}_{\mathcal C}^r(A,B)\) by the Hom group on the right. Thus (2.3) identifies it with the corresponding cohomology of derived Hom whenever the stipulated formal value exists. Negative integers are included. For heart objects their negative groups vanish by the disjoint-bound Hom theorem in the bounded denominator prerequisite. This conditional argument does not assert existence of all unbounded Hom values in an arbitrary abelian category.

3. Two classes that compute a bounded bifunctor

Let \(F:\mathcal A\times\mathcal B\to\mathcal E\) be separately additive, with all three categories abelian. For bounded-below complexes set

on the \((p,q)\) component. If \(X\) vanishes below \(a\) and \(Y\) below \(b\), then \(a\le p\le n-b\); each diagonal is finite and the total vanishes below \(a+b\). Only finite biproducts in \(\mathcal E\) are required. The functor square makes the two unsigned differentials commute, and the two total mixed coefficients are opposite. This proves square zero. First-variable homotopies total with coefficient one and second-variable homotopies with coefficient \((-1)^p\). Sections 5–6 of the total-complex prerequisite prove their homotopy equations, both entire cone triangles and the negative exchange of the two shift comparisons.

Let \(\mathcal J\subset\mathcal A\), \(\mathcal J'\subset\mathcal B\) be full additive classes, including zero objects and finite biproducts. Call the pair F-injective if:

1. Every bounded-below complex in \(\mathcal A\) has an outgoing quasi-isomorphism into a bounded-below complex of \(\mathcal J\)-objects; the corresponding statement holds for \(\mathcal B,\mathcal J'\).
2. For \(I\in K^+(\mathcal J)\), \(J\in K^+(\mathcal J')\), the total \(T_F(I,J)\) is acyclic whenever either \(I\) or \(J\) is acyclic.

These requirements are exactly the adapted-pair hypotheses for \(K^+(\mathcal J),K^+(\mathcal J')\). Both are full triangulated subcategories: shifts and cones use only class terms and finite biproducts, and all homotopies between them have components in the full classes.

Theorem 3.1. An F-injective pair gives an absolutely right derived, separately triangulated bifunctor

If \(s:X\to I\), \(t:Y\to J\) are any two class replacements, there is a natural computation \(R^+F(QX,QY)\simeq QT_F(I,J)\). It retains both shift comparisons and their negative exchange. Its unit \(\eta:QT_F\to R^+F(Q\times Q)\) satisfies

Proof. Put \(\mathcal M=K^+(\mathcal J)\cap N^+(\mathcal A)\), and define \(\mathcal M'\) similarly. Replacement density gives the exact equivalences \(K^+(\mathcal J)/\mathcal M\simeq D^+(\mathcal A)\) and \(K^+(\mathcal J')/\mathcal M'\simeq D^+(\mathcal B)\), with the full denominator equality tests. By condition 2, an intrinsic quasi-isomorphism in either class subcategory has an acyclic image cone under the fixed-variable total functor. That functor therefore sends it to an isomorphism in \(D^+(\mathcal E)\).

Descend the restricted total functor through these two class localizations and compose with their exact quasi-inverses. The adapted-pair theorem proves that this descent is exact separately, not just an ordinary functor on objects. Naturality extends to inverse denominators by multiplying the ordinary naturality equations by their invertible images. Faithfulness of restriction then preserves both shift comparisons and their negative exchange. Its product replacement proof gives absolute universality after every further target functor, so (3.2) represents its formal localization.

The theorem also supplies the isomorphic unit on class pairs. Naturality along \((s,t)\) is exactly (3.3), which specifies the intended computation. For two choices of replacements, put their outgoing denominators over a common target, equalize any homotopy-class comparisons, and replace the common target in each class. The resulting intrinsic denominators are inverted by the descended total. Thus the two computations agree through that common class pair. Their comparison is independent of the refinement by the same equality test. Triple comparisons agree after a common refinement, so satisfy the cocycle equation; compositions of arbitrary localized arrows agree there as well. Equivalently, they are the unique comparisons preserving the unit in the absolute universal property. This proves independence, coherence and naturality for every fraction in both variables. \(\square\)

Write \(R^rF=H^rR^+F\). This definition refers to the bounded-below domains in (3.2).

4. Fixed-variable tests and a single replacement

For a separately additive fixed-variable functor \(G\), a class is G-injective when it supplies all bounded-below replacements and \(G(E)\) is acyclic for each acyclic bounded-below class complex. This is the definition and full computing theorem in Universal derived functors and compatible composition, Section 4.

Proposition 4.1. Suppose \(\mathcal J'\) is \(F(A,-)\)-injective for every \(A\in\mathcal J\), and \(\mathcal J\) is \(F(-,B)\)-injective for every \(B\in\mathcal J'\). Then the pair is F-injective.

Proof. Specialize the first assumption to \(A=0\in\mathcal J\). Even though \(F(0,-)=0\), the definition of an injective class still includes all outgoing replacements in \(\mathcal B\). This gives the second density requirement. Specializing the second assumption to \(B=0\) gives the first density requirement.

If \(I\in K^+(\mathcal J)\) is acyclic, each first-direction complex \(F(I,J^q)\) is acyclic by the second assumption. The double complex in (3.1) has finite diagonals. Apply the complete finite-diagonal comparison theorem to its map from the zero double complex: its first iterated cohomology is zero, so its total is acyclic. This theorem holds in every abelian target; its proof uses kernels, images, truncations and exact cohomology sequences.

If \(J\) is acyclic instead, the first assumption makes every second-direction complex \(F(I^p,J)\) acyclic. Use the transposed comparison theorem, whose total multiplier is \((-1)^{pq}\). Its total is again acyclic. The two density requirements and these two acyclicity assertions are precisely the pair definition. \(\square\)

Corollary 4.2. Suppose \(\mathcal J\) is cogenerating in \(\mathcal A\), \(F(A,-)\) is exact for every \(A\in\mathcal J\), and \(\mathcal J\) is \(F(-,B)\)-injective for every object \(B\in\mathcal B\). Then \(R^+F\) exists and any class replacement \(s:X\to I\) computes

For \(A\in\mathcal A\), \(B\in\mathcal B\) in degree zero, this is the same unit-normalized value as \(R^+(F(-,B))(A)\).

Proof. Use \(\mathcal J'=\mathcal B\). The identity map supplies every second replacement. An exact additive functor takes any acyclic complex to an acyclic complex: it preserves the kernel, image and cycle/boundary quotient, by Exact bifunctors, Lemma 1.1. Therefore all of \(\mathcal B\) is \(F(A,-)\)-injective. Proposition 4.1 applies, and (3.3) with \(t=1_Y\) gives (4.1). Cogeneration also gives class replacements with the same lower bound by Relative replacements, Theorem 2.1.

When \(Y=B[0]\), the only second index is zero. Thus (3.1) is literally the complex \(F(I,B)\), with its usual differential. Both the fixed-variable derivation and (4.1) use this class complex and the same map \(F(A,B)\to F(I,B)\). Their absolute universal units give the unique identification preserving that map. For naturality in \(B\), use the same replacement of \(A\): the comparisons are induced by \(F(I,-)\) on maps. For a denominator in the second complex variable, its cone is acyclic and the exact fixed-class terms make its total cone acyclic; hence its image is invertible. This proves descent and naturality in that variable too, as in the full single-family corollary of the adapted-pair prerequisite. \(\square\)

5. The opposite bounded domains of derived Hom

Apply the pair theorem to the target-first bifunctor \(F(Y,Z)=\operatorname{Hom}_{\mathcal C}(Z,Y)\) on \(\mathcal C\times\mathcal C^{\mathrm{op}}\). Degree reversal takes bounded-above source complexes to bounded-below complexes in the opposite category. The actual reversal, its signs and its opposite triangulation are those used in (1.3) and in Section 7 of the total-complex prerequisite. In particular their normalized Hom total is (1.1), with its complete separately triangulated structure.

Suppose full additive classes \(\mathcal J,\mathcal P\subset\mathcal C\) make \((\mathcal J,\mathcal P^{\mathrm{op}})\) an F-injective pair. Theorem 3.1 and normalization give

Written target first, its domain is \(D^+(\mathcal C)\times D^-(\mathcal C)^{\mathrm{op}}\). The source reverses arrows and shifts; it is not a second covariant bounded-below input in \(\mathcal C\).

For any \(s:P\to X\) in \(K^-(\mathcal C)\), \(t:Y\to I\) in \(K^+(\mathcal C)\), with class terms and both maps quasi-isomorphisms, the value is \(Q_{\mathsf{Ab}}H(P,I)\). The comparison is specified by the pair unit and the natural coordinate normalization, so includes maps, shifts and the negative exchange. If \(X^i=0\) above \(b\) and \(Y^j=0\) below \(a\), degree \(n\) of the unreplaced Hom complex has possible indices

The same statement, with the actual replacement bounds, applies to \(H(P,I)\). Hence each diagonal is finite, the total is zero below the difference of its bounds, and its sum and product constructions are canonically the same finite biproduct. The identifications commute with the differential, with chain maps and with (1.3)–(1.4), since all are coordinate maps.

In fact one may retain the original bounds. Replacement density for \(A[0]\) gives a map \(A\to I^0\) whose factor to \(H^0I\) is an isomorphism; that map is monic. Thus \(\mathcal J\) is cogenerating. Apply the same argument in \(\mathcal C^{\mathrm{op}}\) to see that \(\mathcal P\) is generating. The same-bound replacement theorem and its opposite then provide \(Y\to I\) zero below \(a\) and \(P\to X\) zero above \(b\). Therefore (5.1)'s values lie in \(D^{\ge a-b}\).

Absolute universality lets us apply Section 2 on these bounded domains. Its bounded denominator equality tests prove

for every integer. For bounded \(X,Y\), the left group is \(\mathcal U\)-small: terms of its representing Hom complex are finite products of \(\mathcal U\)-small groups, and kernels, images and quotients stay \(\mathcal U\)-small. Equation (5.3) for \(n=0\), together with full faithfulness of \(D^b\to D\), proves that all Hom sets of \(D^b(\mathcal C)\) are \(\mathcal U\)-small under the pair hypothesis. This is local smallness; no small set of all objects or isomorphism classes has been claimed.

6. Enough injectives or enough projectives

Proposition 6.1. If \(\mathcal C\) has enough injectives, (5.1) exists. For bounded-above \(X\), bounded-below \(Y\), and any bounded-below injective replacement \(t:Y\to I\), it is computed by \(H(X,I)\). If \(\mathcal C\) has enough projectives instead, any bounded-above projective replacement \(s:P\to X\) computes it by \(H(P,Y)\).

Proof. For an injective object \(J\), \(\operatorname{Hom}_{\mathcal C}(-,J)\), regarded covariantly on \(\mathcal C^{\mathrm{op}}\), is exact: extension across a monomorphism gives its final surjectivity, and the kernel equations give the other two positions. Finite biproducts of injectives are injective by extending the finitely many components. Enough injectives and the replacement theorem supply every bounded-below injective replacement.

For each fixed object \(A\), the injective class is \(\operatorname{Hom}(A,-)\)-injective. Indeed an acyclic bounded-below injective complex \(E\) is contractible by Relative replacements, Lemma 4.1: its complete upward homotopy construction applies to the identity of \(E\). Applying the additive functor \(\operatorname{Hom}(A,-)\) to that contraction proves its image contractible and hence acyclic. Corollary 4.2 with target class the injectives and the entire opposite category as second class now applies. Its identity second replacement gives precisely \(H(X,I)\) after (1.3). This computation includes (5.3) and all pair-unit naturality, not just the object value.

For the projective assertion apply this argument to \(\mathcal C^{\mathrm{op}}\) and interchange the two inputs with their specified signed comparison. Projective objects are injective in the opposite category; enough projectives gives bounded-above replacements in \(\mathcal C\). Equivalently, \(\operatorname{Hom}(P,-)\) is exact for a projective object \(P\), and an acyclic bounded-above projective complex is contractible by the dual downward construction. The class of projectives is therefore adapted for \(\operatorname{Hom}(-,A)\) in the opposite input, for every fixed \(A\). The identity replacement in the target gives \(H(P,Y)\), with the source comparison and opposite cone signs of (1.4). Formula (5.2) proves finite diagonals in both computations.

If both types of replacement exist, there are actual chain maps

For the first, every fixed projective term gives an exact target Hom functor, so the comparison of target-direction complexes is a quasi-isomorphism. For the second, every fixed injective term gives an exact opposite-source Hom functor, so the comparison in the source direction is a quasi-isomorphism. Finite diagonals and the corresponding orders of the comparison theorem make both total maps quasi-isomorphisms. Their squares commute by precomposition/postcomposition. The unit normalization identifies them with the derived comparison maps, proving the claimed canonical agreement. \(\square\)

7. Projectives and derived tensor on both module sides

Let \(k\) be a commutative unital ring, \(R\) a unital central \(k\)-algebra, \(N\) a right \(R\)-module and \(M\) a left \(R\)-module. Write

Here \(\mathsf{Mod}(R)\) means left modules; right modules are \(\mathsf{Mod}(R^{\mathrm{op}})\). This notation names the four derived categories with their cohomological shifts. The existence of the unbounded category itself does not assert existence of every unbounded derived bifunctor by the bounded criteria of this lesson.

The balanced tensor construction gives \(N\otimes_RM\) in \(\mathsf{Mod}(k)\). Its relations are additive and \(k\)-linear in both inputs and impose \(nr\otimes m=n\otimes rm\). The balanced universal property constructs maps and proves right exactness in either variable by its preservation of cokernels. Keep these module sides throughout; no flip of right and left factors in a fixed noncommutative tensor functor is assumed.

Every left module is a quotient of the free left module with basis its underlying set; send \(e_m\) to \(m\). A map out of that free module lifts across any epimorphism by choosing one lift per basis element, proving projectivity. A projective module is a direct summand of such a free module: lift its identity across the free cover to obtain a section. The identical argument with right bases gives enough right projectives. These constructions remain in \(\mathcal U\).

For a free left module \(R^{(S)}\), its tensor with \(N\) is naturally \(N^{(S)}\), by the balanced maps on the individual basis entries. Direct sums of short exact module rows are exact: kernels and images are checked coordinatewise, and a finite-support element of the final sum lifts using finitely many coordinate lifts. Thus tensor with a free left module is exact in the right variable. A direct summand is exact as well: each tensor row splits into its two summand rows, and kernel, image and quotient calculations respect that decomposition. Hence every projective left module is flat. The right-module proof is the same, retaining the reversed module sides.

Every bounded-above module complex has an incoming replacement by a bounded-above free complex with the same upper bound. The full downward pullback construction is Universal derived functors, Lemma 8.2; apply it to \(R\) or \(R^{\mathrm{op}}\), respectively. At degree \(n\) it takes a free cover of the pairs \((x,z)\in X^n\oplus Z^{n+1}P\) satisfying \(d_Xx=fz\). Its projections define the new differential and map. Lifts of \((x,0)\) make the cycle map surjective; lifts of \((x,z)\) when \(fz=d_Xx\) kill precisely the kernel in the next cohomology degree. Downward induction therefore gives the required quasi-isomorphism without any finite resolution-length assumption.

The full projective classes form an incoming adapted pair for tensor. They supply the two replacements just proved. For an acyclic upper-bounded complex in either class and an upper-bounded complex in the other, each fixed opposite projective term tensors the acyclic complex exactly. The double complex has finite diagonals: if the upper bounds are \(u,v\), then

for its possible total entries. Acyclic rows or columns and the two orders of the comparison theorem make the total acyclic. This checks both null-image conditions of the incoming pair theorem. It consequently supplies the absolutely left derived, separately triangulated bifunctor

If \(q:Q\to N\), \(p:P\to M\) are upper-bounded right and left projective replacements, respectively, it is computed by \(\operatorname{Tot}(Q\otimes_RP)\), with differential \(d_Q\otimes1+(-1)^i1\otimes d_P\) on \(Q^i\otimes_RP^j\).

Both single replacements also compute it, through the actual maps

For the left map, each left projective term \(P^j\) gives an exact functor \(-\otimes_RP^j\); it takes \(q\) to a quasi-isomorphism by preservation of the actual cohomology quotient. Common upper bounds give finite diagonals for both double complexes, so the first-direction comparison theorem proves that total map a quasi-isomorphism. For the right map, each \(Q^i\) tensors exactly, and the transposed comparison theorem proves the same assertion. This proof does not require projectivity or flatness of the unreplaced inputs.

These maps are the normalized derived computations. Write \(G(A,B)=Q_k\operatorname{Tot}(A\otimes_RB)\), \(L(A,B)=A\otimes_R^{\mathbf L}B\), and let \(\sigma:L(Q\times Q)\to G\) be the pair counit. It is invertible at \((Q,P)\). Naturality gives

The right-hand comparison maps were proved invertible above, and both \(L\)-maps invert the denominators. Thus \(\sigma_{N,P}\) and \(\sigma_{Q,M}\) are invertible. The standard double-replacement identification is \(G(Q,P)\to L(N,M)\) given by \(L(q,p)\sigma_{Q,P}^{-1}\). Compose it with \(\sigma_{N,P}L(1_N,p)^{-1}\). Functoriality and the first equation of (7.6) show that this composite is exactly \(G(q,1_P)\), the left arrow in (7.4). Composing instead with \(\sigma_{Q,M}L(q,1_M)^{-1}\) gives exactly \(G(1_Q,p)\), its right arrow. This proves both single-replacement identifications with their actual maps. Common incoming refinements and the pair theorem's universal counit prove independence of resolutions, naturality in all derived arrows and compatibility with composition. Therefore the three complexes represent canonically the same value, rather than merely having isomorphic cohomology groups.

For modules concentrated in degree zero define

The same-bound free replacements are zero above degree zero. Hence the derived tensor has no positive cohomology; \(\operatorname{Tor}_r\) is zero for \(r<0\). Its degree-zero cohomology is canonically \(N\otimes_RM\): a single left projective resolution has cokernel \(M\) at degree zero, and right exactness makes its tensor cokernel precisely that ordinary tensor. These are objects of \(\mathsf{Mod}(k)\); no ring commutativity or finite generation of the inputs has been used.

8. Four graded exercises with complete solutions

Exercise 1 (introductory: the two Hom shift orders). Over a field of characteristic different from two, compute both comparison routes from \(H(X[1],Y[1])\) to \(H(X,Y)\), using (1.4), when \(X=Y=k[0]\). Explain the sign using the degree in the source of each comparison. Also verify the source-shift chain equation before specializing to zero differentials.

Solution. Put \(A=H(X[1],Y)\), \(B=H(X,Y)[-1]\), and let \(\lambda^n(f)^i=(-1)^nf^{i-1}\). In \(A\) the source differential is \(d_Yf+(-1)^n f d_X\), because \(d_{X[1]}=-d_X\). The target differential in \(B\) is \(-d_Yg+(-1)^{n-1}g d_X\). Substituting \(g=(-1)^nf\) makes this \((-1)^{n+1}d_Yf-f d_X\); applying \(\lambda^{n+1}\) to the source differential gives the same expression, with the indicated reindexing. Thus \(\lambda\) is a chain map.

For \(X=Y=k[0]\), the input \(H(X[1],Y[1])\) is \(k\) in degree zero; its generator is the map between the two degree-minus-one copies of \(k\). If the source shift is removed first, \(\lambda\) is applied in degree zero, with scalar one. The target comparison applied next has scalar one, so the route through \(H(X,Y[1])[-1]\) gives positive identity on \(H(X,Y)\). If the target shift is removed first, its comparison gives an element of degree one in \(H(X[1],Y)\), shifted by \([1]\). Now the source comparison has scalar \((-1)^1=-1\); the route through \(H(X[1],Y)[1]\) therefore gives negative identity. Both iterated integer shifts have the original zero differential and the same degree-zero term. Their two comparison maps are negatives and are distinct in this characteristic. This checks the exchange on nonzero cohomology.

Exercise 2 (intermediate: cyclic derived Hom). For positive integers \(m,n\), compute every \(\operatorname{Ext}_{\mathbb Z}^r(\mathbb Z/m,\mathbb Z/n)\) using a source projective replacement. Keep the actual Hom differential.

Solution. Let \(P^{-1}=P^0=\mathbb Z\), \(d_P^{-1}=m\), with the quotient augmentation to \(\mathbb Z/m[0]\). Multiplication by \(m\) is injective and its cokernel is \(\mathbb Z/m\), so this is a projective replacement. For \(B=\mathbb Z/n[0]\), (1.1) gives \(H(P,B)^0=H(P,B)^1=\mathbb Z/n\) and zero terms elsewhere. Its differential is minus multiplication by \(m\): at source degree zero, \(d_Bf=0\) and the precomposition term is \(-f d_P\).

Put \(d=\gcd(m,n)\), \(m=dm_0\), \(n=dn_0\), with \(m_0,n_0\) coprime. A class \(x\bmod n\) lies in the kernel precisely when \(n\mid mx\), equivalently \(n_0\mid x\). Thus its kernel is the subgroup generated by \(n/d\), canonically identified with \(\mathbb Z/d\) by \(t\mapsto t(n/d)\bmod n\). Its cokernel is \((\mathbb Z/n)/m(\mathbb Z/n)\). Bezout's identity makes the subgroup generated by \(m\bmod n\) equal to the subgroup of multiples of \(d\), so reduction modulo \(d\) identifies this quotient with \(\mathbb Z/d\). The sign of the image does not change that subgroup.

By (5.3), \(\operatorname{Ext}^0\) and \(\operatorname{Ext}^1\) are these two groups, and all other integer \(\operatorname{Ext}^r\) vanish. The degree-zero identification is ordinary Hom: a homomorphism is determined by the image of \(1\), which must be killed by \(m\). If either \(m\) or \(n\) equals one, then \(d=1\) and both groups are zero, as required for a zero input.

Exercise 3 (advanced: both cyclic tensor comparison maps). Compute \(\mathbb Z/n\otimes_{\mathbb Z}^{\mathbf L}\mathbb Z/m\) by replacing both inputs. Give the full total differential and the induced cohomology maps to each single replacement.

Solution. Take right \(Q^{-1}\xrightarrow{n}Q^0\) and left \(P^{-1}\xrightarrow{m}P^0\), each with terms \(\mathbb Z\) and its quotient augmentation. Order the total degree-minus-one coordinates as \(a\in Q^0\otimes P^{-1}\), \(b\in Q^{-1}\otimes P^0\). The total is

The minus sign in the first column comes from the odd degree of \(Q^{-1}\); the second composite is \(mn-nm=0\). Let \(d=\gcd(m,n)\). The first map is injective. Coprimality of \(m/d,n/d\) shows that the kernel of the row consists exactly of multiples of \(v=(n/d,-m/d)\). Its incoming image consists of multiples of \(dv\). Thus \(H^{-1}=\mathbb Z/d\), with class of \(v\) as generator. The final image is \(d\mathbb Z\) by Bezout, giving \(H^0=\mathbb Z/d\); every other cohomology group is zero.

The left map in (7.4) kills total degree \(-2\), sends \((a,b)\) to \(a\bmod n\) in degree \(-1\), and sends \(c\) to \(c\bmod n\) in degree zero. Its target is \(\mathbb Z/n\xrightarrow{m}\mathbb Z/n\). The first column maps to \(n\bmod n=0\), and the row maps to \(ma+nb\bmod n=ma\); hence both chain equations hold. On \(H^{-1}\) it sends \([v]\) to \(n/d\bmod n\), a generator of the kernel of multiplication by \(m\), of order \(d\). On \(H^0\), reduction modulo \(d\) identifies the induced map with identity of \(\mathbb Z/d\).

The right map kills degree \(-2\), sends \((a,b)\) to \(b\bmod m\), and sends \(c\) to \(c\bmod m\). Its target is \(\mathbb Z/m\xrightarrow{n}\mathbb Z/m\). The first column maps to \(-m\bmod m=0\), and the row maps to \(ma+nb\bmod m=nb\), proving both chain equations. On \(H^{-1}\), it sends \([v]\) to \(-m/d\bmod m\), a generator of the kernel of multiplication by \(n\), again of order \(d\). Its degree-zero map is identity after reduction modulo \(d\). Both maps are therefore quasi-isomorphisms with their actual signs. It follows that \(\operatorname{Tor}_0\) and \(\operatorname{Tor}_1\) are \(\mathbb Z/d\), and all other integer Tor groups vanish, including the cases where one input is zero.

Exercise 4 (advanced: why the unbounded Hom uses a product). Let \(X^i=Y^i=k\) for every integer \(i\), with zero differentials. Compare degree-zero cohomology of the product Hom complex with the direct-sum version. Locate the identity chain map.

Solution. Each term of the product Hom complex is \(\prod_{i\in\mathbb Z}k\), and every differential is zero. Thus \(H^0H(X,Y)=\prod_i k\). Degree-zero chain maps are arbitrary families of scalars on the degree terms, and all homotopy differences are zero; this also verifies (1.2) directly. The proposed sum version instead has \(H^0=\bigoplus_i k\), consisting of finite-support families. Its map to the product is the coordinate inclusion, which is injective but misses the constant-one family. That family is the identity chain map of \(X=Y\). Therefore a sum construction fails even to represent every homotopy-category endomorphism in this example. Bounds of the opposite types in (5.2) give finite support of each diagonal of objects and make sum and product equal there. Finite support of individual sum elements supplies no such bound in this unbounded example.

References

• The linked prerequisite lessons give the complete formal localization, resolution, finite-diagonal and sign interfaces used in the proofs.
• The Stacks Project, Derived hom, Section 15.75, especially equation 15.75.0.2 and Lemma 15.75.2, gives the module-derived Hom and projective-source formulation.
• The Stacks Project, Derived tensor product, Section 15.60, especially Definition 15.60.1, Lemma 15.60.7 and Definition 15.60.13, gives the flat-complex formulation and its bounded-above case.