Draft lesson.

Injective replacements and right derived functors

Original text public domain (CC0).

A test for complexes of injectives does not construct an injective replacement. We build one by replacing successive terms and pushing the differential forward. Each degree stabilizes after finitely many steps. This gives a resolution functor in the homotopy category and a right derived functor with its full transformation property. Examples compute both a finite injective resolution and higher derived groups that never stop.

Let \(\mathcal A\) be an abelian category with enough injectives: each object admits a monomorphism into an injective object. Work in an ambient Grothendieck universe containing its objects, arrows and \(\mathbb Z\) as sets. Choices of replacements are made in that universe. Write \(K^+(\mathcal A)\) for the homotopy category of complexes whose terms vanish below some integer, and \(\mathcal I\) for the full additive subcategory of injective objects. No upper bound is assumed.

Derived morphisms through quasi-isomorphism roofs, Section 5, proves that every bounded-below complex of injectives is K-injective, and that Hom into it is the same in \(K\) and \(D\). Triangles after quasi-isomorphism localization proves the triangulated localization, exact-sequence triangles and all their comparison properties. Chain homotopies, mapping cones and triangles supplies the actual Hom-complex and cone formulas. Exact complexes, signs and lifting covers, Section 3, proves that a pushout of a monomorphism is monic and supplies categorical covering lifts. These are the precise earlier results used below.

1. Replace one term without changing cohomology

Let \(E\) be a complex and \(u: E^n\hookrightarrow J\) a monomorphism. Take the pushout of \(u\) and \(d_E^n\), denoting its maps by \(a,b\):

Lemma 1.1. There is a complex \(E'\) and a termwise monic quasi-isomorphism \(v: E\to E'\), with \(E'^n=J\), \(E'^{n+1}=P\), and all other terms unchanged.

Proof. Define its affected differentials by

The last map exists uniquely by the pushout property, since \(d_E^{n+1}d_E^n=0\). The composite at degree \(n-1\) is \(au d_E^{n-1}=b d_E^n d_E^{n-1}=0\); the next composite is zero by definition. The composite at degree \(n+1\) vanishes after both \(a\) and \(b\), hence is zero by the pushout property. All other complex equations were unchanged.

The components of \(v\) are \(u,b\) in the affected degrees and identities elsewhere. Equations (1.1)–(1.2) make it a chain map. The map \(b\) is monic because it is a pushout of \(u\), so \(v\) is termwise monic. Its quotient is zero except in degrees \(n,n+1\). Both quotient terms identify with \(C=J/u(E^n)\): the second identification is the cokernel comparison for a pushout. Its induced differential is identity, since it is induced by \(a\). Thus the quotient is the contractible complex \(C\xrightarrow1C\). The short exact sequence of complexes and its cohomology sequence make \(v\) a quasi-isomorphism. This proves the assertion with all differentials and quotient maps specified. \(\square\)

2. Construct and compare complete replacements

Theorem 2.1. If \(X^n=0\) for \(n<m\), there is a termwise monic quasi-isomorphism \(q_X: X\to I_X\), where \(I_X^n=0\) for \(n<m\) and every \(I_X^n\) is injective.

Proof. Starting at \(m\), apply Lemma 1.1 in successive degrees \(m,m+1,m+2,\ldots\), embedding the term currently in that degree into an injective object. A step changes only its own degree and the following one. Consequently degree \(k\), and its outgoing differential, have reached their final values after the step in degree \(k+1\). Take these stabilized terms and differentials to define \(I_X\). Every complex equation and every chain-map equation for the induced map from \(X\) occurs in finitely many degrees, so is an equation at some finite stage. The induced map in each degree is a finite composite of monomorphisms, hence monic. The terms below \(m\) stay zero, and each other final term is an injective object chosen at its own step.

For each fixed \(k\), the terms in degrees \(k-1,k,k+1\) have stabilized after finitely many steps. The induced \(H^k(q_X)\) is therefore the cohomology map of that finite composite of quasi-isomorphisms, hence invertible. This argument uses no exactness of an infinite limit or colimit. It proves the claimed quasi-isomorphism in every degree. \(\square\)

Choose one such map for each bounded-below complex, taking \(q_I=1_I\) when the complex already consists of injective objects. This normalization is possible by taking identity embeddings and unchanged pushouts. For \(f: X\to Y\) in \(K^+(\mathcal A)\), there is a unique homotopy class \(j(f): I_X\to I_Y\) with

Indeed the target \(I_Y\) is K-injective, so precomposition with \(q_X\) is bijective on Hom groups. Its surjectivity constructs the class and its injectivity proves uniqueness. Precomposing with \(q_X\) checks \(j(1)=1\), \(j(gf)=j(g)j(f)\), and \(j(f+g)=j(f)+j(g)\). Thus \(j\) is an additive resolution functor into \(K^+(\mathcal I)\), and \(q\) is natural in the homotopy category. This makes no assertion that arbitrary choices of chain representatives commute strictly.

Let \(D^+(\mathcal A)\) be the full subcategory of \(D(\mathcal A)\) consisting of objects whose cohomology vanishes below some integer. It is closed under shifts and distinguished triangles by the long exact cohomology sequence, so is triangulated with the inherited triangles.

Every such object has a bounded-below representative. If \(H^kX=0\) for \(k<m\), form \(T\) with zero terms below \(m\), term \(T^m=\operatorname{coker}d_X^{m-1}\), and unchanged terms above \(m\). The outgoing differential factors through that cokernel. The map \(X\to T\), quotient in degree \(m\) and identity above it, is a chain map. In degrees above \(m\) it preserves cohomology; in degree \(m\) the kernel of the induced differential is exactly \(Z^mX/B^mX\), and below it both cohomologies vanish. Thus it is a quasi-isomorphism. Composing it with Theorem 2.1 gives a quasi-isomorphism \(q_X: X\to I_X\) also for every object of \(D^+\). On bounded-below complexes retain the previous choices.

Theorem 2.2. The exact functor

is an equivalence. Its inverse can be chosen as \(J(X)=I_X\), with comparisons determined by \(q_X\).

Proof. For two bounded-below injective complexes, the K-injective Hom test identifies their Hom in \(K\) with their Hom in \(D\); hence (2.2) is fully faithful. The replacements just constructed give essential surjectivity. Finite sums of injectives are injective, since maps into each summand can be extended separately. Shifts and cones of bounded-below injective complexes consequently stay in \(K^+(\mathcal I)\), and images of their cone triangles are distinguished. Thus (2.2) is exact.

Define \(J\) on an arbitrary derived arrow \(\alpha: X\to Y\) as the unique homotopy class satisfying

Full faithfulness gives existence and uniqueness, and composition, addition and identity follow directly. These equations make \(Qq_X\) a natural isomorphism from identity to \(QJ\); the normalization gives \(JQ=1\) on injective complexes. On the image of a map of bounded-below complexes, (2.3) agrees with (2.1).

The inverse is exact as well. Choose a chain representative between the injective replacements of the first two vertices of a distinguished triangle. Its cone is a bounded-below injective complex. The triangle morphism axiom compares its image cone triangle to the original triangle using the isomorphisms on the first two vertices. The two-vertex criterion makes the third comparison invertible, so every such triangle is isomorphic to an image injective cone triangle. This proves preservation by \(J\), with shift comparisons induced by full faithfulness. Their naturality and coherence follow by applying the faithful functor \(Q\), where they are the already fixed shift comparisons. \(\square\)

3. The total functor and its transformation property

Let \(G: K^+(\mathcal A)\to\mathcal T\) be an exact functor into a triangulated category. Restrict it to \(K^+(\mathcal I)\) and define

for bounded-below complexes \(X\). The functor \(RG\) is exact by Theorem 2.2. Equation (2.1) proves naturality of \(\eta\).

Theorem 3.1. For any functor \(H: D^+(\mathcal A)\to\mathcal T\), composition with \(\eta\) is a bijection

No additivity or exactness of \(H\) is needed for this assertion.

Proof. Given \(\gamma: G\to HQ\), define its extension on every object of \(D^+\) by

For a derived arrow \(\alpha: X\to Y\), put \(t=J\alpha\). Naturality of \(\gamma\) on this arrow of \(K^+(\mathcal I)\) gives \(\gamma_{I_Y}G(t)=H(Qt)\gamma_{I_X}\). Substituting (2.3) proves \(\beta_Y RG(\alpha)=H(\alpha)\beta_X\). Thus the formula is natural for all derived arrows, including roofs.

For a bounded-below \(X\), naturality of \(\gamma\) for \(q_X\) gives \(\gamma_{I_X}G(q_X)=H(Qq_X)\gamma_X\), so \(\beta_{QX}\eta_X=\gamma_X\). Conversely any extension must agree with \(\gamma\) at an injective complex, because there \(q_I=1\) and \(\eta_I=1\). Naturality for the isomorphism \(Qq_X\) then forces (3.3). This proves both uniqueness and the bijection. \(\square\)

This is the precise universal transformation property meant here by the right derived functor. If replacements are changed, apply (3.2) to the other pair \((RG',\eta')\) and conversely. The two resulting comparisons are inverse, since their composites extend the same transformation as identity. With three choices, the comparison through the middle choice equals the direct comparison by the same uniqueness. Thus independence of resolutions includes its compatibility with transformations and its cocycle law.

For an additive functor \(F: \mathcal A\to\mathcal B\) between abelian categories, apply this to \(G=Q_{\mathcal B}K(F)\), where \(F\) is applied term by term. Additivity makes it preserve differential signs, chain homotopies, shifts and finite biproducts. Its cone comparison sends the two summands to the two summands of the image cone, so is an isomorphism of complexes respecting the structural maps. Hence \(G\) is exact, and

No enough-injectives hypothesis on \(\mathcal B\) is needed for this construction. Applying \(F\) to zero terms keeps this image complex bounded below.

4. Cohomology, long exact sequences and derived Hom

Define \(R^nF(A)=H^n(RF(A[0]))\). If \(F\) is left exact, the construction gives

To verify these assertions, the replacement of \(A[0]\) can start in degree zero, so its negative terms vanish. Its degree-zero kernel is \(A\); left exactness identifies the kernel of its image differential with \(F(A)\), which is its degree-zero cohomology. The identification is induced by \(F(q_A)\), so is natural. For an injective object the normalized replacement is the object itself in degree zero, proving positive vanishing.

A termwise short exact sequence of bounded-below complexes gives a natural distinguished triangle in \(D^+(\mathcal A)\). Apply the exact \(RF\), then the cohomology sequence in \(D^+(\mathcal B)\). This gives every position and connecting map of its long exact derived-cohomology sequence. In particular a short exact sequence of objects gives

The zero at the start follows from negative vanishing, and naturality of the connecting maps follows from naturality of the exact-sequence triangles and the functor \(RF\).

For modules over a ring \(R\) and a fixed left module \(M\), take the left exact additive functor \(F=\operatorname{Hom}_R(M,-)\) to abelian groups. Left exactness follows by factoring a map through a kernel whenever its next composite is zero. For any \(X\in D^+(R)\), the Hom complex from \(M[0]\) to \(I_X\) has in degree \(n\) just \(\operatorname{Hom}_R(M,I_X^n)\), with differential induced by \(d_{I_X}\). The Hom-complex identity in the prerequisite and the K-injective test give naturally

For the last identification use the inverse of \(Qq_X[n]\), and the equality of Hom in \(K\) and \(D\) into the shifted K-injective complex. For \(X=A[0]\) these groups are \(\operatorname{Ext}^n_R(M,A)\) by definition. Thus the injective-resolution and derived-morphism definitions coincide, including their natural maps and all shift signs.

5. Change a functor and compare compositions

A natural transformation \(\alpha: F\to F'\) of additive functors acts on replacements by the termwise chain map \(\alpha_{I_X}\). Naturality on a chain representative of \(Jt\) proves that these define \(R\alpha: RF\to RF'\). They commute with the replacement transformations since \(\alpha_{I_X}F(q_X)=F'(q_X)\alpha_X\). Identities and composition are termwise identities and composition; comparisons between different replacement choices are compatible by Theorem 3.1. This proves functoriality in the original functor as well as in its argument.

One useful composition case has a direct proof. Suppose \(\mathcal B\) also has enough injectives and \(F\) carries injective objects of \(\mathcal A\) to injective objects of \(\mathcal B\). For an additive \(L: \mathcal B\to\mathcal C\), the complex \(F(I_X)\) is already a bounded-below injective complex. Its normalized replacement in \(\mathcal B\) is itself. Consequently

On arrows both sides apply \(LF\) to \(Jt\), so this comparison is natural. It is compatible with the replacement transformations, and a change of resolutions changes it by the canonical comparisons just proved.

For example, if \(F\) is an additive right adjoint of an exact functor \(A: \mathcal B\to\mathcal A\), then \(F\) preserves injectives. A map \(M\to F(I)\) corresponds to \(A(M)\to I\). A monomorphism \(M\hookrightarrow N\) is sent by exact \(A\) to a monomorphism, through which the latter map extends by injectivity. Adjunction turns the extension into \(N\to F(I)\), proving the assertion. Formula (5.1) concerns this injective-preserving hypothesis. The more general hypothesis that \(F(I)\) is merely acyclic for \(L\) requires a further acyclic-complex theorem; it is not inferred from this argument.

6. Two worked examples

Example 6.1 (integers resolved by divisible groups). A divisible abelian group is injective. Here is the extension proof. For a map on a subgroup \(A\subseteq B\), order its extensions to intermediate subgroups by inclusion. The union over a chain is an extension, so Zorn gives a maximal one. If \(x\) lies outside its subgroup, the integers \(n\) with \(nx\) in that subgroup form an ideal \(r\mathbb Z\), \(r\ge0\). If \(r>0\), choose \(e\) in the divisible group with \(re\) equal to the image of \(rx\); if \(r=0\), choose \(e=0\). Extend by sending \(a+nx\) to the old image of \(a\) plus \(ne\). Two such expressions differ by a multiple of the defining relation \(rx\), so have the same image. This contradicts maximality unless the subgroup was all of \(B\). Thus every map extends, proving injectivity.

Both \(\mathbb Q\) and \(\mathbb Q/\mathbb Z\) are divisible: divide a rational representative by a positive integer.

This also proves that abelian groups have enough injectives. A nonzero element of finite order maps nontrivially from its cyclic subgroup to \(\mathbb Q/\mathbb Z\) by sending a generator of order \(n\) to \(1/n\); an element of infinite order can be sent to \(1/2\). Injectivity extends that map to the whole group. Evaluation into the product of \(\mathbb Q/\mathbb Z\) over all such homomorphisms is therefore monic. That product is divisible, by coordinatewise division and choice, hence injective by the preceding proof.

Hence

is an injective resolution in degrees zero and one. For \(M=\mathbb Z/m\), \(m>1\), its Hom into \(\mathbb Q\) is zero, since a rational number killed by \(m\) is zero. Its Hom into \(\mathbb Q/\mathbb Z\) is the subgroup killed by \(m\), consisting of the classes \(a/m\) and naturally isomorphic to \(\mathbb Z/m\). Applying Section 4 gives \(\operatorname{Ext}^0_{\mathbb Z}(M,\mathbb Z)=0\), \(\operatorname{Ext}^1_{\mathbb Z}(M,\mathbb Z)=\mathbb Z/m\), and zero in every higher degree. By (4.3) this agrees with the roof computation of the extension morphism.

Example 6.2 (derived groups in every degree). Let \(R=k[\varepsilon]/(\varepsilon^2)\) over a field and let \(M=k=R/(\varepsilon)\). The module \(J=\operatorname{Hom}_k(R,k)\), with \((a\phi)(r)=\phi(ra)\), is injective as an \(R\)-module. Indeed the correspondence between \(R\)-linear maps \(N\to J\) and linear functionals on \(N\) sends a map to evaluation at one; its inverse sends \(\ell\) to the map \(n\mapsto(r\mapsto\ell(rn))\). These formulas verify inverse bijections and naturality. A functional on a vector subspace extends by completing a basis, so this correspondence proves the required module extension property.

The same correspondence works with any vector space in place of \(k\). Therefore \(\operatorname{Hom}_k(R,N)\) is injective for every module \(N\), and \(n\mapsto(r\mapsto rn)\) embeds \(N\) into it: evaluation at one is a left inverse as a linear map. This verifies enough injectives in this example, rather than just injectivity of one chosen module.

The pairing \((r,a)\mapsto\lambda(ra)\), where \(\lambda\) extracts the coefficient of \(\varepsilon\), identifies \(R\) with \(J\). In the basis \(1,\varepsilon\) its matrix is \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), invertible over any field, and the defining action makes the identification \(R\)-linear. Thus \(R\) is injective. There is an injective resolution

The first map is injective, and the image and kernel of every subsequent multiplication map are both \(k\varepsilon\). Every term is the injective module just proved. Applying \(\operatorname{Hom}_R(k,-)\) replaces each term by \(k\varepsilon\simeq k\), and every differential becomes zero. Therefore \(\operatorname{Ext}^n_R(k,k)=k\) for every \(n\ge0\). The input is in degree zero; its higher derived groups have no upper bound. Enough injectives supplies a bounded-below resolution, not a resolution of finite length.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: one pushout by hand). For the complex \(\mathbb Z\xrightarrow2\mathbb Z\) in degrees zero and one, replace the degree-zero term by its embedding into \(\mathbb Q\). Compute the pushout, both new differentials and the induced cohomology map.

Solution. The pushout is \(P=(\mathbb Q\oplus\mathbb Z)/\langle(z,-2z): z\in\mathbb Z\rangle\). The map

It kills every relation. Its inverse sends \((r,[n])\) to \([(r-n/2,n)]\); changing \(n\) by \(2z\) changes the pair by \((-z,2z)\), a relation, proving well-definedness. The two composites are identities. In these coordinates the map from \(\mathbb Q\) is \(q\mapsto(q,0)\), and the map from the old degree-one term is \(n\mapsto(n/2,[n])\). The new degree-zero differential is the former map, and its next differential is zero. The square commutes since the latter sends \(2z\) to \((z,0)\). Both old and new degree-zero cohomology are zero. The new degree-one cohomology is \(\mathbb Z/2\), and the old cokernel maps to it by reduction, an isomorphism. This computes the entire one-term replacement and its quasi-isomorphism.

Exercise 2 (intermediate: uniqueness is up to homotopy). Over a field take \(X=k[0]\), and \(I^0=k\oplus k\), \(I^1=k\), with differential \((x,u)\mapsto u\) and zero elsewhere. Embed \(X\) by \(x\mapsto(x,0)\). Give two different chain endomorphisms extending identity on \(X\), and compute their comparison homotopy.

Solution. All vector spaces are injective by extending a linear map from a subspace after choosing a complement. The displayed embedding is a quasi-isomorphism: its image is the degree-zero kernel and the degree-one differential is surjective. One endomorphism is identity. The other is projection onto the first summand in degree zero and zero in degree one. Both are chain maps and agree on \(X\), but differ on the contractible second part. Set \(h^1(v)=(0,v)\) and all other homotopy components zero. Then \(dh+hd\) is \((x,u)\mapsto(0,u)\) in degree zero and identity in degree one, exactly the difference. Thus they give the same unique class of (2.1), despite differing as chain maps. This is why that equation defines a functor in \(K\).

Exercise 3 (advanced: an exact functor needs no correction). If \(F: \mathcal A\to\mathcal B\) is exact and additive, prove that it descends termwise to an exact functor on \(D^+\), naturally isomorphic to \(RF\). Determine all \(R^nF\) on objects.

Solution. Exactness preserves kernels, cokernels and images. Thus it preserves the cycle, boundary and quotient descriptions of cohomology, giving naturally \(H^n(FX)\simeq F(H^nX)\). It therefore sends every quasi-isomorphism to a quasi-isomorphism. On a roof define its descended map by the image numerator followed by the inverse image denominator; the proved roof universal property verifies refinements and composition. It preserves cones and shifts by additivity, so is exact. Each \(F(q_X)\) is a quasi-isomorphism, hence the transformation in (3.1) is an isomorphism in \(D^+(\mathcal B)\). This identifies the descended functor with \(RF\), naturally and with the specified triangle and shift comparisons. On an object in degree zero, its only cohomology is \(F(A)\) in degree zero, so \(R^0F=F\) and \(R^nF=0\) for every \(n\ne0\).

Exercise 4 (synthesis: no finite injective replacement). For the ring and module of Example 6.2, show that no bounded-above and bounded-below complex of injective modules can be quasi-isomorphic to \(k[0]\). Also verify the dimension-shifting recurrence using the short exact sequence \(0\to k\to R\to k\to0\).

Solution. If such a complex \(I\) existed with zero terms above \(N\), the K-injective test would identify every \(\operatorname{Ext}^n_R(k,k)\) with \(H^n\operatorname{Hom}_R(k,I)\). The latter is zero for \(n>N\), contradicting the nonzero groups computed in every degree in Example 6.2. A quasi-isomorphism in either direction, or merely an isomorphism in the derived category, gives the same contradiction by the natural identification (4.3).

In the sequence \(0\to k\to R\to k\to0\), the first map is \(1\mapsto\varepsilon\) and the second is reduction. Applying \(\operatorname{Hom}_R(k,-)\), the first map identifies \(k\) with \(k\varepsilon\); the second map is zero, since it kills that submodule. Positive derived groups at \(R\) vanish by injectivity. The long exact sequence therefore gives an isomorphism from \(\operatorname{Hom}_R(k,k)=k\) to \(\operatorname{Ext}^1_R(k,k)\), and for every \(n\ge1\) an isomorphism

These are the connecting maps of the proved derived triangle, so their recurrence is natural. They recover the nonvanishing in all degrees and the obstruction to a finite injective replacement.

References

• The Stacks Project Authors, Injective resolutions, Lemma 13.18.3, and Right derived functors and injective resolutions, Section 13.20, for further reading. The pushout construction, comparison and universal transformation property used here are proved above.
• Pierre Schapira, An Introduction to Categories and Homological Algebra, Sections 7.2–7.3, for the resolution and derived-functor framework.