Draft lesson.

Derived morphisms through quasi-isomorphism roofs

Original text public domain (CC0).

Inverting quasi-isomorphisms creates maps that need not have a chain-map representative between the original complexes. A roof supplies a representative after changing the source. We construct its refinement relation, addition, composition and universal property. We then prove exactly when a source or target makes that change unnecessary.

Let \(\mathcal A\) be an abelian category. Fix an ambient Grothendieck universe in which its objects and arrows form sets, and in which \(\mathbb Z\) is small. Every category and quotient constructed below is in that ambient universe. This is a size assumption, or a permitted enlargement of universe; local smallness alone would not prove that the collection of roofs is a set.

Use Chain homotopies, mapping cones and triangles, with \(C_f^n=Y^n\oplus X^{n+1}\), differential \((d_Yy+fx,-d_Xx)\), and last map the positive projection to \(X[1]\). Its Sections 4 and 6 prove both exact Hom sequences and cone cohomology exactness. Exact complexes, signs and lifting covers supplies the abelian image and quotient constructions. The variance of roofs agrees with the incoming calculus in One-sided fractions and saturation, Section 5. Our roof construction and its checks are given in full here.

Write \(K=K(\mathcal A)\). Equalities in \(K\) mean equality up to chain homotopy. Let \(S\) consist of its quasi-isomorphisms. A chain representative belongs to \(S\) exactly when its cone is acyclic, by the preceding lesson. Identities belong to \(S\), and compositions and the two-out-of-three property follow by applying every cohomology functor. Shifts preserve \(S\).

1. Both square and cancellation axioms

Lemma 1.1 (incoming squares). Given an arrow \(f: X\to Y\) and \(t: V\to Y\) with \(t\in S\), there are \(a: P\to X\) in \(S\) and \(b: P\to V\) such that \(fa=tb\) in \(K\).

Proof. Choose chain representatives. Take the homotopy pullback complex

Squaring the last coordinate gives \(d_Yfx-fd_Xx-d_Ytv+td_Vv=0\); the other coordinates square to zero. The projections \(a(x,v,z)=x\) and \(b(x,v,z)=v\) are chain maps. The degree-minus-one family \(H(x,v,z)=-z\) has \(d_YH+HD=fa-tb\), so gives the asserted equality.

It remains to prove the denominator condition. In the cone of \(a\), coordinates in degree \(n\) are \((x,x',v,z)\) in \(X^n,X^{n+1},V^{n+1},Y^n\), and its differential is

The chain maps to and from \(C_t\) are

For \(r\), the first differential coordinate is \(d_Y(z-fx)-tv\), and the second is \(d_Vv\), exactly the differential of \((z-fx,-v)\) in \(C_t\). Substitution checks \(i\) as well, and \(ri=1\). The family \(B(x,x',v,z)=(0,x,0,0)\), of degree minus one, satisfies

Thus \(C_a\) is homotopy equivalent to the acyclic complex \(C_t\). It is acyclic, so \(a\in S\). \(\square\)

Lemma 1.2 (outgoing squares). Given \(s: X\to U\) in \(S\) and \(f: X\to Y\), there are \(g: U\to P\) and \(t: Y\to P\), with \(t\in S\), such that \(gs=tf\) in \(K\).

Proof. Take the complex

It is the cone of \((s,-f)\), so its differential squares to zero. Let \(g\) and \(t\) be the first and second inclusions. The family \(x\mapsto(0,0,x)\) proves \(gs-tf\) null-homotopic. The cone of \(t\) has coordinates \((u,y,x,y')\) with differential

Its projection \(r(u,y,x,y')=(u,x)\) to \(C_s\) has chain section \(i(u,x)=(u,0,x,fx)\). Their composite \(ri\) is identity, and the family \(B(u,y,x,y')=(0,0,0,y)\) gives

Therefore \(C_t\) is homotopy equivalent to \(C_s\), hence acyclic, proving \(t\in S\). \(\square\)

Lemma 1.3 (both cancellations). If \(t: Y\to Z\) lies in \(S\) and \(tf=tg\), some \(a: X'\to X\) in \(S\) satisfies \(fa=ga\). If \(s: W\to X\) lies in \(S\) and \(fs=gs\), some \(b: Y\to Y'\) in \(S\) satisfies \(bf=bg\).

Proof. In the first case exactness of \(\operatorname{Hom}_K(X,-)\) for the rotated cone triangle of \(t\) factors \(f-g\) through the acyclic complex \(N=C_t[-1]\). Write this factorization \(X\xrightarrow hN\xrightarrow kY\). Rotate the cone triangle of \(h\) backwards to obtain \(X'\xrightarrow aX\xrightarrow hN\to X'[1]\). Its exact cohomology sequence makes \(a\) a quasi-isomorphism, and its consecutive composite \(ha\) is zero in \(K\). Hence \((f-g)a=0\).

In the second case the exact contravariant Hom sequence for \(s\) factors \(f-g\) as \(X\to C_s\xrightarrow kY\). In the cone triangle \(C_s\xrightarrow kY\xrightarrow bC_k\to C_s[1]\), acyclicity of \(C_s\) makes \(b\) a quasi-isomorphism, and \(bk=0\). Thus \(b(f-g)=0\). All factorizations are actual Hom-sequence lifts in the preceding lesson; no projective or injective replacement is assumed. \(\square\)

2. Incoming denominators form a filtered index

For \(X\), let \(I_X\) have objects \(s: U\to X\) in \(S\), and arrows \(a: (s: U\to X)\to(t: V\to X)\) with \(ta=s\) in \(K\). Every such \(a\) belongs to \(S\) by two-out-of-three. Put \(J_X=I_X^{\mathrm{op}}\), so its arrows point toward refinements.

Lemma 2.1. The category \(J_X\) is filtered.

Proof. The identity denominator is an object. Two denominators \(s: U\to X\), \(t: V\to X\) admit, by Lemma 1.1, maps \(a: P\to U\) in \(S\) and \(b: P\to V\) with \(sa=tb\). The composite is in \(S\), and two-out-of-three makes \(b\in S\). This is a common refinement, hence a common target in \(J_X\).

For parallel arrows in \(I_X\), the equation \(ta=tb\), with \(t\in S\), is equalized by precomposing with a denominator from Lemma 1.3. Its composite with the source denominator is still a denominator. Reversing gives the equalizing arrow required in \(J_X\). \(\square\)

Define

For a refinement \(a: V\to U\) with \(sa=t\), the transition sends \(f\) to \(fa\). This is a diagram of abelian groups. Concretely a class is a roof

Two roofs are equal exactly when there is a common refinement \(a: W\to U\), \(b: W\to V\) such that both denominators agree and both numerators agree in \(K\). To see this without assuming a colimit equality test, construct the quotient of the disjoint union of the groups by that relation. Reflexivity and symmetry are immediate. For transitivity, refine the two intermediate refinements to a common one in \(I_X\); if their arrows to the intermediate denominator differ, use its cofiltered equalizing condition before comparing numerators. The denominator and numerator equalities then compose. A compatible family of maps out of the groups factors uniquely through this quotient, which proves that it is the colimit in sets. Addition is computed after moving two representatives to a common refinement. Different choices are equalized by the same filtered conditions; associativity, inverses and zero follow by moving finitely many representatives to one index and using the group laws there. This also proves the abelian-group colimit property in (2.1).

In particular, for ordinary arrows of \(K\),

Here \(Qf\) denotes the roof \((1_X,f)\). A common refinement of its two identity denominators has the same comparison arrow on both sides, giving (2.3).

3. Composition and its complete checks

To compose \((s: U\to X,f: U\to Y)\) and \((t: V\to Y,g: V\to Z)\), choose Lemma 1.1's square

and define the composite to be

Theorem 3.1. This rule is well defined and associative, with identities \((1_X,1_X)\). It is bilinear for the groups (2.1). Thus \(D\) is an additive category and \(Q: K\to D\) is additive.

Proof. First change the square, keeping both roofs fixed. Two choices \((W,a,b)\), \((W',a',b')\) have a common refinement \(d: E\to W\), \(e: E\to W'\) over the denominator target \(U\), so \(ad=a'e\). Then \(tbd=fa d=fa'e=tb'e\). Incoming cancellation for \(t\) gives a further denominator \(k: E'\to E\) with \(bdk=b'ek\). The product roofs have equal denominators and numerators after that refinement.

Next refine the first input by \(u: U'\to U\). Use an incoming square for \(a: W\to U\) and the denominator \(u\): there are \(k: W'\to W\) in \(S\), \(l: W'\to U'\) with \(ak=ul\). Two-out-of-three makes \(l\in S\). The square \(fu l=tbk\) computes the refined product as \((su l,gbk)=(sak,gbk)\), a refinement of the original product. If instead the second input is refined by \(v: V'\to V\), complete \(b: W\to V\) along \(v\): \(bk=vl\), with \(k\in S\). The product for the refined second roof is \((sak,gv l)=(sak,gbk)\). Every equality of roofs is given by common refinements, so these checks prove independence of all representatives.

For associativity take three roofs \((s,f),(t,g),(r,h)\). Suppose first \(fa=tb\) and then \(gbc=rd\), where \(a,c\in S\). One association gives \((sac,hd)\). For the other association choose \(ge=rl\), with \(e\in S\). Refine \(bc\) through \(e\): \(bcu=ev\), with \(u\in S\). Then \(facu=tev\), so a permissible product for the other association is \((sacu,hlv)\). Moreover \(rdu=gbcu=gev=rlv\). Cancellation for \(r\) gives \(k\in S\) with \(duk=lvk\). The first product, refined by \(uk\), and the second, refined by \(k\), have the same denominator and numerator. This proves associativity. Composing a roof \((s,f)\) with the identity of its target uses \(a=1_U,b=f\) and returns \((s,f)\). Composing the identity of its source with that roof uses \(a=s,b=1_U\) and also returns \((s,f)\).

For bilinearity, put two roofs into a common denominator first. If their numerators are \(f_1,f_2: U\to Y\), and the next denominator is \(t: V\to Y\), choose the two incoming squares, then refine their denominators over \(U\) to one arrow \(a: W\to U\) in \(S\). They obey \(f_i a=tb_i\). The square for \(f_1+f_2\) can use \(a,b_1+b_2\); its product numerator is \(g(b_1+b_2)=gb_1+gb_2\). This proves additivity in the first roof. For two second numerators with one denominator, use the same square for the first numerator and that denominator; addition of the second numerators commutes with precomposition by its comparison map. This proves additivity in the second roof.

The image of zero is a zero object. Maps to it have zero numerator. A roof from zero has a denominator \(s: U\to0\). Since \(s1_U=s0_U\), cancellation supplies \(a: V\to U\) in \(S\) with \(a=0\) in \(K\). Its refined numerator is zero, so every map from zero is zero. Finally \(Q\) preserves the biproduct identities \(p_i j_j=\delta_{ij}\) and \(\sum j_i p_i=1\), because it is additive. These identities prove the actual product and coproduct universal properties of the image biproduct. Thus finite biproducts exist in \(D\). \(\square\)

The inverse of \(Qs: U\to X\), for \(s\in S\), is the roof \((s,1_U)\). The product on \(U\) is its identity; the product on \(X\) is \((s,s)\), a refinement of its identity roof. Every roof is consequently \(Qf(Qs)^{-1}\).

4. The universal interface, including chain homotopies

Theorem 4.1. For any category \(\mathcal B\), restriction along \(Q\) is an equivalence onto the full category of functors \(K\to\mathcal B\) that invert \(S\). This includes all natural transformations. The same statement holds for the composite \(C(\mathcal A)\to K\to D\) and functors that invert all chain quasi-isomorphisms.

Proof. For a functor \(F\) inverting \(S\), define

A refinement \(a\) lies in \(S\), so replacing \((s,f)\) by \((sa,fa)\) leaves (4.1) unchanged. Common refinements prove well-definedness. In (3.1), \(F(f)F(a)=F(t)F(b)\). Inverting \(F(a),F(t)\) proves that (4.1) preserves composition; the identity formula is immediate. Every descended functor must use (4.1) because every roof is \(Qf(Qs)^{-1}\), so the extension is unique.

For a natural transformation \(\alpha: F\to G\), retain its object components. Naturality for \(s\), after inverting its images, gives \(\alpha_UF(s)^{-1}=G(s)^{-1}\alpha_X\). Combining this with naturality for \(f\) proves naturality on (4.1). Conversely restriction determines all object components. Thus restriction is fully faithful and has exactly the stated image. No additivity of \(F\) or \(\mathcal B\) was used.

It remains to show that a functor on chain complexes inverting quasi-isomorphisms must identify homotopic maps. Suppose \(f-g=dh+hd\). The contractible cone \(C_{1_X}\) receives its chain inclusion \(j: X\to C_{1_X}\). The map

is a chain map, by the homotopy equation, and \(cj=f-g\). On \(E=X\oplus C_{1_X}\), use the chain inclusions \(i_0(x)=(x,0)\), \(i_1(x)=(x,jx)\), and projection \(p: E\to X\). It is a quasi-isomorphism because its second summand is contractible. Both \(pi_i=1_X\), so any such functor sends \(i_0,i_1\) to the same inverse of its image of \(p\). The chain map \(b(x,z)=gx+cz\) has \(bi_0=g\), \(bi_1=f\), making the two images equal. Hence the functor descends uniquely to \(K\), where the first part applies. Natural transformations descend because the object collection is unchanged, and naturality has already been checked for every chain map and inverse denominator. \(\square\)

We call this localized category \(D(\mathcal A)\), the derived category. This construction concerns its objects and morphisms and their universal property; a localization theorem for its distinguished triangles is a further statement.

Every cohomology functor factors through \(D\), because it inverts quasi-isomorphisms. Therefore

For the forward directions apply these descended cohomology functors. For the first converse, \(0\to X\) is a quasi-isomorphism; the second converse is the already constructed denominator inverse. These prove both zero detection and saturation without confusing quasi-isomorphisms with homotopy equivalences.

There are also outgoing representatives. For an incoming roof \((s,f)\), Lemma 1.2 gives \(gs=tf\) with \(t\in S\), so

Moreover \(Qf=Qg\) for ordinary parallel arrows if and only if some outgoing denominator kills their difference. Indeed (2.3) followed by outgoing cancellation gives that denominator; the converse follows by inverting its image. Thus both directions can be used for computations.

5. Complexes that absorb denominator changes

A complex \(I\) is K-injective if \(\operatorname{Hom}_K(N,I)=0\) for every acyclic complex \(N\). A complex \(P\) is K-projective if \(\operatorname{Hom}_K(P,N)=0\) for every such \(N\). Since shifts of acyclic complexes are acyclic, these conditions include all the shifted Hom groups needed below.

Theorem 5.1. The following are equivalent for \(I\): K-injectivity; every quasi-isomorphism \(s: U\to X\) induces a bijection \(\operatorname{Hom}_K(X,I)\to\operatorname{Hom}_K(U,I)\); and the natural map \(\operatorname{Hom}_K(X,I)\to\operatorname{Hom}_D(X,I)\) is bijective for every \(X\). The dual three conditions hold for K-projective \(P\), with Hom out of \(P\).

Proof. The contravariant exact Hom sequence for the cone of \(s\) has \(\operatorname{Hom}_K(C_s,I)\) and \(\operatorname{Hom}_K(C_s[-1],I)\) on the two sides of the map in question. K-injectivity makes both zero, proving its bijectivity. Every roof \(X\xleftarrow sU\xrightarrow aI\) then has a unique class \(b: X\to I\) with \(bs=a\). It represents the roof. If two such ordinary classes become equal, (2.3) and the same injectivity make them equal in \(K\). This proves the third condition. Conversely, for acyclic \(N\) the group \(\operatorname{Hom}_D(N,I)\) is zero by (4.3), so the third condition gives K-injectivity. The second condition also gives it directly by applying it to \(0\to N\).

For \(P\), the covariant cone Hom sequence makes \(\operatorname{Hom}_K(P,Y)\to\operatorname{Hom}_K(P,V)\) bijective for every quasi-isomorphism \(Y\to V\). Represent every map in \(D\) by (4.4) and use that bijection to remove its outgoing denominator. The outgoing equality test after (4.4) proves injectivity. For the converse apply zero detection to every acyclic target. \(\square\)

Theorem 5.2 (bounded tests). A bounded-below complex of injective objects is K-injective. A bounded-above complex of projective objects is K-projective. No supply of either kind of object in the rest of the category is assumed.

Proof. Let \(I^n=0\) for \(n<m\), and let \(f: N\to I\) with \(N\) acyclic. We construct \(h^n: N^n\to I^{n-1}\) with \(f^n=d_Ih^n+h^{n+1}d_N\). Start with \(h^n=0\) for \(n\le m\). At stage \(n\ge m\), suppose the preceding equations hold. The residual \(r=f^n-d_Ih^n\) kills the incoming differential: its composite with \(d_N^{n-1}\) is \(d_I(f^{n-1}-h^nd_N^{n-1})\), hence zero by the preceding equation and \(d_I^2=0\). At the first stage the same conclusion follows from \(I^{m-1}=0\). Acyclicity makes this incoming image \(Z^nN\). Thus \(r\) factors through the image \(Z^{n+1}N\) of \(d_N^n\). Extend that factor to \(N^{n+1}\to I^n\), using injectivity along \(Z^{n+1}N\hookrightarrow N^{n+1}\), and call it \(h^{n+1}\). This proves the next equation. Induction gives a homotopy on every degree; below \(m\) its equation is zero. Therefore every \(f\) is null-homotopic.

For projectives take \(P^n=0\) for \(n>M\) and a map \(f: P\to N\) with \(N\) acyclic. Start with zero homotopies above \(M\). Descending, the residual \(r=f^n-h^{n+1}d_P\) is a cycle: its outgoing differential is \(f^{n+1}d_P-d_Nh^{n+1}d_P=0\), by the previously established equation and \(d_P^2=0\). The epimorphism \(N^{n-1}\twoheadrightarrow Z^nN\) exists by acyclicity. Projectivity of \(P^n\) lifts \(r\) through it to \(h^n\). This gives \(f^n=d_Nh^n+h^{n+1}d_P\). Descending induction covers all degrees, so \(f\) is null-homotopic. \(\square\)

These are proofs about complexes already supplied. They do not construct an unbounded K-injective resolution from the separate fact that individual objects have injective embeddings.

6. Two worked examples

Example 6.1 (a new map after localization). Put \(M=\mathbb Z/2\) in degree zero. Let \(P\) have \(P^{-1}=P^0=\mathbb Z\) and differential multiplication by two. Reduction in degree zero and zero in degree minus one define a quasi-isomorphism \(q: P\to M\): its only cohomology is \(H^0P=\mathbb Z/2\). Define \(a: P\to\mathbb Z[1]\) to be identity in degree minus one and zero elsewhere. The roof

is nonzero in \(D\). Indeed \(\mathbb Z\) is projective in abelian groups: lift the image of its generator through a surjection and extend by integer multiples. Theorem 5.2 makes \(P\) K-projective, so \(\operatorname{Hom}_D(P,\mathbb Z[1])=\operatorname{Hom}_K(P,\mathbb Z[1])\). Chain maps in this group are integers in degree minus one. A homotopy \(h^0: P^0\to\mathbb Z[1]^{-1}\) changes that integer by \(2h^0\), so the group is \(\mathbb Z/2\) and \(a\) is its nonzero class. Since \(Qq\) is invertible, the roof is nonzero too. But every chain map \(M\to\mathbb Z[1]\) is zero, because the nonzero terms are in different degrees. A derived morphism therefore need not come from \(K\) between the original objects.

Example 6.2 (all vector-space complexes). Over a field, choose a complement \(H^n\) of \(B^nX\) in \(Z^nX\), and a complement \(L^n\) of \(Z^nX\) in \(X^n\). The differential restricts to an isomorphism \(L^n\to B^{n+1}X\): its kernel is zero and its image is all the next boundaries. Identifying these spaces gives

The first and third summands form a contractible complex, with degree-minus-one homotopy sending the first boundary summand to the preceding third summand. Its differential expression is identity on those two summands. Thus \(X\) is homotopy equivalent to its graded cohomology with zero differential, even without boundedness. Each degree uses only a finite direct sum.

For zero-differential complexes the chain maps are all degreewise linear maps and a homotopy has zero differential expression. Conjugating through these equivalences proves that the natural cohomology map identifies

Although the chosen splittings are not canonical, the displayed map is the canonical cohomology map: the equivalences induce the specified identity maps on cohomology. A quasi-isomorphism is consequently already an isomorphism in \(K\). Localizing adds no new maps, so \(K\to D\) is an equivalence and (6.3) also computes the derived Hom. The group example differs because its relevant cycles and boundaries do not all admit such splittings.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: a complex disappears). Let \(A\) be the exact complex \(\mathbb Z\xrightarrow2\mathbb Z\to\mathbb Z/2\) in degrees zero, one and two. Give an inverse roof to \(Q(0\to A)\), and prove directly that \(Q1_A=Q0_A\), although \(1_A\) is not null-homotopic.

Solution. Acyclicity follows from the kernel of reduction being \(2\mathbb Z\), injectivity of multiplication by two, and surjectivity of reduction. Write \(s: 0\to A\). The inverse roof is \(A\xleftarrow s0\xrightarrow{1_0}0\). The product on \(A\) is \((s,s)\), which refines its identity roof. Both the identity and zero roof from \(A\) to itself have refinement through \(s\), and their numerators there are the same map from zero. Hence (2.3) gives \(Q1_A=Q0_A\). A homotopy of the identity would give in degree two a section \(\mathbb Z/2\to\mathbb Z\) of reduction. Such a map is zero, since its image would be killed by two in \(\mathbb Z\). Thus this equality was created by localization and does not hold in \(K\).

Exercise 2 (intermediate: compute a homotopy pullback). In Lemma 1.1 take \(X=\mathbb Z[0]\), \(Y=\mathbb Z/2[0]\), \(V=(\mathbb Z\xrightarrow2\mathbb Z)\) in degrees minus one and zero, and let both \(f\) and \(t^0\) be reduction modulo two. Compute every term and differential of \(P\), and verify its denominator and comparison homotopy.

Solution. The three nonzero terms are \(P^{-1}=\mathbb Z\), \(P^0=\mathbb Z\oplus\mathbb Z\), and \(P^1=\mathbb Z/2\). Formula (1.1) gives

Their composite is zero. The first differential is injective. The degree-zero cycles are pairs with the same parity; modulo pairs \((0,2r)\), projection to \(x\) is an isomorphism with \(\mathbb Z\). It is surjective via \((x,x)\), and two cycle pairs with the same \(x\) differ by an even second coordinate, exactly a boundary. The degree-one map is surjective, since \((0,1)\) maps to one, so its cokernel is zero. Thus projection \(a: P\to X\) is a quasi-isomorphism.

The map \(b: P\to V\) is identity in degree minus one and second projection in degree zero. Its composite with \(t\) is \([v]\), while \(fa=[x]\). The homotopy has only \(H^1: P^1\to Y^0\) nonzero, equal to minus identity. Its expression in degree zero is \(-[v-x]=[x-v]\), and all other components are zero. This is exactly \(fa-tb\). The square commutes in \(K\), with a verified quasi-isomorphism on its required leg.

Exercise 3 (advanced: an injective target reads cohomology). For an injective object \(I\) in degree zero and an arbitrary complex \(X\), prove a natural isomorphism \(\operatorname{Hom}_D(X,I[0])\simeq\operatorname{Hom}_{\mathcal A}(H^0X,I)\). No boundedness of \(X\) is assumed.

Solution. Theorem 5.2 makes the target K-injective, so it suffices to compute Hom in \(K\). A chain map has a component \(a: X^0\to I\) killing \(B^0X\). Restrict it to \(Z^0X\) to get a map from \(H^0X\). Conversely, such a map defines a map \(Z^0X\to I\), which extends along \(Z^0X\hookrightarrow X^0\) by injectivity. The extension kills \(B^0X\), so is a chain map.

If the restriction is zero, \(a\) factors through the quotient \(X^0/Z^0X\), canonically the image \(B^1X\) of \(d_X^0\). Extend this factor along \(B^1X\hookrightarrow X^1\), again by injectivity, to a map \(h^1: X^1\to I\). Then \(a=h^1d_X^0\), the expression of a null-homotopy into \(I[0]\). Conversely every such homotopy vanishes on \(Z^0X\). This proves bijectivity and independence of extensions. Restriction to cycles commutes with precomposition of chain maps and postcomposition of maps of injective targets. The construction therefore gives the stated natural isomorphism, also for derived source morphisms by Theorem 5.1 and the descended cohomology functor.

Exercise 4 (synthesis: two torsion modules). For positive integers \(m,n\), compute \(\operatorname{Hom}_D((\mathbb Z/m)[0],(\mathbb Z/n)[1])\) using roofs, including the equality relation on numerators. Give the classes explicitly for \(m=6,n=10\).

Solution. Replace \((\mathbb Z/m)[0]\) by \(P_m=(\mathbb Z\xrightarrow m\mathbb Z)\) in degrees minus one and zero. Reduction defines a quasi-isomorphism, and its terms are projective, so Theorems 5.1–5.2 compute the answer by \(\operatorname{Hom}_K(P_m,(\mathbb Z/n)[1])\). A chain map is a choice \(a\in\mathbb Z/n\) in degree minus one. A homotopy is a choice \(b\in\mathbb Z/n\) in degree zero, changing the numerator to \(a+mb\). Hence

For the last identification, the quotient is generated by the class of one with the relations \(n=0\) and \(m=0\). Euclidean division reduces these two relations to their greatest common divisor; conversely the integer identity expressing that divisor as a combination of \(m,n\) shows that its relation follows from them. Thus the kernel of \(\mathbb Z\) onto the quotient is exactly their generated ideal. For \(6,10\), the subgroup generated by six is \(\{0,6,2,8,4\}\). There are two numerator classes, the even and odd residues. The zero roof is the even class and a roof with numerator one is the nonzero class. The denominator is the fixed reduction \(P_6\to\mathbb Z/6\), and the equality relation is precisely the computed homotopy relation, by K-projectivity.

References

• The Stacks Project Authors, Localization in categories, for the two fraction variances; Derived categories; and K-injective complexes, for further reading. The constructions and bounded tests used here are proved above.
• Pierre Schapira, An Introduction to Categories and Homological Algebra, Sections 3.1 and 7.1–7.2, for localization and derived categories.