Draft lesson.

First derived cohomology and composition through truncation

Level: advanced. Prerequisites: smart truncations, bounded derived categories, universal right derived functors and compatible computing classes.

A complex with one cohomology object is canonically a shifted object of its abelian category. With two cohomology objects, its truncation triangle retains an additional connecting morphism. These observations compute shifted resolutions and derived compositions. A separate denominator argument identifies the first derived cohomology even when no injective resolution class is available.

Conventions and prerequisites

All functors between abelian categories are additive. Complexes are cohomological: the differential increases degree. Our signed shift has \(X[m]^n=X^{n+m}\), differential \((-1)^m d_X\), and

The notation \(R^+F\) means the universally defined bounded-below right derived functor, with its unit and specified exact shift comparison. For a complex or derived object \(X\), put \(R^nF(X)=H^n(R^+F(X))\). For an abelian object this means its degree-zero complex. Every integer \(n\) is allowed. Formal denominator diagrams use the ambient universe of the universal-localization prerequisites; an enlarged universe is used when necessary, without asserting that the original denominator index is small.

Cohomological bounds and truncation adjunctions, Sections 2–5 proves the full bounded denominator comparisons, the natural singleton-cohomology identifications and the actual truncation triangles. Universal derived functors, Sections 2, 4, 5 and 9 supplies absolute formal representation after every further functor, exactness of \(R^+F\), computation on an F-injective class and the normalized compatible-composition isomorphism. These statements include naturality after inversion of quasi-isomorphisms.

1. A complex with one cohomology object

Lemma 1.1. Suppose \(Y\in D^+(\mathcal A)\) has zero cohomology outside degree \(p\). There is a canonical natural isomorphism

Proof. For an actual representative, write \(U_pY=\tau_{\le p}Y\). Its degree-\(p\) term is \(Z^pY\); all terms above \(p\) vanish. The inclusion and the cycle quotient give chain maps

The second map is zero below \(p\) and is \(Z^pY\twoheadrightarrow Z^pY/B^pY\) at \(p\). It kills the image of the preceding differential, so it is a chain map. Both maps induce the identity on the only possible cohomology object. Their cones are acyclic, hence both are invertible in \(D^+\). Define \(\beta_Y=Q(q_Y)Q(U_pY\to Y)^{-1}\).

Kernel inclusions and cycle quotients commute with every chain map. The truncation prerequisite proves their descent through homotopies. The resulting squares therefore commute for homotopy-category arrows and for inverses of quasi-isomorphisms. Every derived arrow is a fraction of such arrows, so (1.1) is natural on the full subcategory with this concentration condition. It induces the identity on \(H^p\). \(\square\)

Theorem 1.2. Let \(\mathcal J\) be an F-injective full additive class in \(\mathcal A\), and let \(Y\) be a bounded-below complex of its objects. If its cohomology is zero outside \(p\), and \(A=H^pY\), then

Here \(F\) may be any additive functor for which this class is F-injective; in particular the conclusion applies to left exact \(F\).

Proof. The computing-class theorem makes the unit \(\eta_Y: QF(Y)\to R^+F(QY)\) invertible. Apply the exact functor \(R^+F\) to (1.1), then use its specified shift comparison. This gives an isomorphism \(b_Y: R^+F(QY)\to R^+F(A)[-p]\). The actual computation is

Taking degree \(k+p\) and using (0.1) proves (1.3). The argument uses the sign-preserving shift comparison; it does not identify shifted termwise differentials by an unsigned relabelling.

Both \(b_Y\) and the unit are natural. A quasi-isomorphism between class complexes has an acyclic class cone, whose F-image is acyclic. Thus its F-image is a quasi-isomorphism. Formula (1.4) is consequently natural also after class localization, and comparisons between computing representatives agree with the normalized unit comparison. For a heart object the general heart theorem gives \(R^kF(A)=0\) for \(k<0\). In (1.3) this implies vanishing of the F-image cohomology below \(p\), even if the original complex has nonzero terms there. \(\square\)

2. The first cohomology without a resolution-class assumption

Theorem 2.1. Let \(F: \mathcal A\to\mathcal B\) be additive and suppose its universal \(R^+F\) exists. If \(H^jX=0\) for \(j<p\), then \(R^nF(X)=0\) for \(n<p\). If \(F\) is also left exact, there is a natural isomorphism

Neither assertion assumes enough injectives or an F-injective class.

Proof. The natural quasi-isomorphism \(X\to\tau_{\ge p}X\) lets us replace \(X\) by a complex whose terms vanish below \(p\). The lower-bound denominator theorem says that the outgoing quasi-isomorphisms \(s: X\to Y_s\) with \(Y_s^j=0\) for \(j<p\) form a cofinal subcategory of the entire outgoing index. It preserves the common refinements and equality witnesses as well as objects.

The universal derived value represents the formal diagram \(QF(Y_s)\). Absolute representation, proved in the universal-derived prerequisite, says that applying any further functor gives a formal diagram represented by that functor applied to the derived value. Apply this to \(H^n: D^+(\mathcal B)\to\mathcal B\). For \(n<p\), every \(H^n(F(Y_s))\) is zero because the terms of \(F(Y_s)\) vanish there. This is the constant formal zero object, so its representing object \(R^nF(X)\) is zero.

At degree \(p\) there are no incoming boundaries in \(Y_s\), and left exactness gives

These are the actual kernel comparison maps. Since \(s\) is a quasi-isomorphism, \(F(H^p(s))\) identifies every object in (2.2) with \(F(H^pX)\). For an index arrow \(a: Y_s\to Y_t\) with \(as=t\) in the homotopy category, its cohomology satisfies \(H^p(a)H^p(s)=H^p(t)\). Thus all transition maps become identities under those identifications. The filtered nonempty index is connected; at every test object its constant Yoneda diagram has colimit the same Yoneda object. Absolute representation therefore identifies \(R^pF(X)\) with \(F(H^pX)\).

For a representative already zero below \(p\), the identity denominator shows that the resulting map is \(H^p(\eta_X)\) after the kernel identification \(F(H^pX)=H^p(F(X))\). For a general representative use the natural quasi-isomorphism \(X\to\tau_{\ge p}X\) and its invertible derived image. This specifies (2.1) independently of the representative. Unit naturality and naturality of the kernel comparison give its commutative squares for homotopy maps; their squares for inverse quasi-isomorphisms follow by inversion. Fractions now prove naturality for every derived arrow between objects with this lower bound. \(\square\)

This proof uses representation of a formal diagram after \(H^n\). It requires no ordinary colimit in \(\mathcal B\) and no assertion that \(H^n\) preserves ordinary colimits. The left exactness hypothesis enters precisely in (2.2).

3. A derived composition with one cohomology degree

Let \(F: \mathcal A\to\mathcal B\) and \(G: \mathcal B\to\mathcal C\) be additive. Suppose there are full additive classes \(\mathcal J,\mathcal J'\) such that \(\mathcal J\) is F-injective, \(\mathcal J'\) is G-injective and \(F(\mathcal J)\subset\mathcal J'\). The complete compatible-class theorem supplies

This particular comparison is normalized by the units: if \(\theta,\eta,\eta'\) are the units for the composite, F and G, respectively, then \(cQ\theta=(R^+G\eta)(\eta'F)\). It is a natural isomorphism of exact functors with their specified shift comparisons. Neither F nor G is required to be left exact here.

Theorem 3.1. Put \(Z=R^+F(X)\). If \(Z\) has zero cohomology outside degree \(q\), then

In particular, for \(X=A[0]\) and \(q\ge0\) this is \(R^j(GF)(A)\simeq R^{j-q}G(R^qF(A))\).

Proof. Apply \(R^+G\) to the canonical isomorphism \(\beta_Z: Z\to H^qZ[-q]\) of Lemma 1.1. Composing it with \(c_X\) and the shift comparison gives

Every map is the specified natural isomorphism. Degree \(j\) in (3.3) is degree \(j-q\) of the last unshifted object, proving (3.2). All its squares commute for derived maps between objects satisfying this same concentration condition. The formula includes negative \(j\); heart values of an additive universally derived functor have zero negative cohomology. \(\square\)

4. Two cohomology degrees and the actual connecting map

Keep the compatible classes of Section 3. Suppose \(Z=R^+F(X)\) has cohomology only in degrees \(a<b\). Set \(A=H^aZ\), \(B=H^bZ\). Zero values at either of these degrees are permitted.

Theorem 4.1. There is a natural long exact sequence

Its connecting map is induced by the actual truncation connecting morphism

Proof. The natural truncation triangle is \(\tau_{\le a}Z\to Z\to\tau_{\ge a+1}Z\to(\tau_{\le a}Z)[1]\). Its first vertex has only cohomology A in degree a, and its third has only cohomology B in degree b. Lemma 1.1 therefore transports it, naturally, to

The last arrow is the transported positive cone projection with the actual inverse cone-to-quotient quasi-isomorphism of the truncation prerequisite. It is part of the triangle data.

Write \(\sigma_T^m: R^+G(T[m])\to R^+G(T)[m]\) for the specified shift comparison. Apply the exact functor \(R^+G\) to (4.3) and identify its middle vertex with \(R^+(GF)(X)\) by (3.1). The three arrows of the resulting distinguished triangle are exactly

The standard target cohomology sequence of this actual distinguished triangle is (4.1), with \(\delta^k=H^k(\gamma_Z)\). Indeed (0.1) makes the three outer degrees \(k-a\), \(k-b\) and \(k-a+1\). Exactness includes every integer k and every connecting position. The original truncation triangle, Lemma 1.1, c and the shift comparisons are natural. Therefore (4.4), and hence its entire long exact sequence, is natural for derived maps between objects with this fixed pair of possible degrees. \(\square\)

Shifting (4.2) by b identifies it with an element of \(\operatorname{Ext}^{b-a+1}_{\mathcal B}(B,A)\), where Ext is defined by derived Hom. A gap between a and b does not make this group or the connecting map zero. Exercise 4 gives an explicit nonzero example.

5. Module categories and a computable injective object

We record the supplies of objects needed for the following examples. Let R be any unital algebra over a field k, and let M be a left R-module. Give \(T_M=\operatorname{Hom}_k(R,M)\) the left action \((r\phi)(s)=\phi(sr)\). For every left R-module N, evaluation at 1 gives a bijection

Its inverse sends a k-linear map \(\ell\) to \(n\mapsto(s\mapsto\ell(sn))\). These formulas prove R-linearity and both inverse equations. A k-linear map defined on a subspace extends to the entire vector space by extending a basis. Thus the right-hand contravariant Hom functor is exact, and \(T_M\) is injective as an R-module. The R-linear map \(M\to T_M\), \(m\mapsto(s\mapsto sm)\), is monic because evaluation at 1 is m. Hence every such module category has enough injectives, for arbitrary M and arbitrary R.

For \(R=k[\varepsilon]/(\varepsilon^2)\), set \(S=R/(\varepsilon)\). The linear functional \(\lambda\) extracting the coefficient of \(\varepsilon\) gives an R-linear isomorphism

In the basis \(1,\varepsilon\), the pairing matrix is \(\begin{pmatrix}0&1\\1&0\end{pmatrix}\), whose determinant is nonzero in every characteristic. Thus (5.2) is bijective, and R itself is injective by (5.1). The augmentation \(S\to R\), \(1\mapsto\varepsilon\), extends to an injective resolution

Its exactness follows from \(\ker(\varepsilon: R\to R)=\varepsilon R=\operatorname{im}(\varepsilon: R\to R)\). Enough injectives and the computing-class theorem justify applying any additive functor termwise to a bounded-below injective resolution. For \(G=\operatorname{Hom}_R(S,-)\), every term of \(G(5.3)\) is k and every differential is zero: a map from S has image in \(\varepsilon R\), annihilated by multiplication by \(\varepsilon\).

6. Four graded exercises with complete solutions

Exercise 1 — A resolution beginning in any integer degree (introductory)

For the dual-number algebra and functor G of Section 5, fix any \(p\in\mathbb Z\). Let Y have R in every degree at least p, zero terms below p, and differential multiplication by \(\varepsilon\). Identify its only cohomology and compute \(H^{k+p}(G(Y))\) for every integer k. Check the shifted-resolution formula.

Solution. At p there are no incoming boundaries, so \(H^pY=\ker\varepsilon=\varepsilon R\simeq S\), by \(1\mapsto\varepsilon\). At every later degree the kernel and image are both \(\varepsilon R\), so the cohomology is zero; it is also zero below p. Every term is injective by (5.2). The complex G(Y) has k in degrees at least p and zero differential. Consequently \(H^{k+p}(G(Y))\) is k when \(k\ge0\) and zero when \(k<0\). Applying G to (5.3) gives exactly \(R^kG(S)=k\) for \(k\ge0\) and zero for \(k<0\), agreeing with Theorem 1.2 for every p, including negative p. The identification uses (1.4); it is unaffected by the signed shift differential because its natural exact shift comparison is retained.

Exercise 2 — Concentration alone does not compute a derived functor (intermediate)

With the same R, S and G, let \(Y=S[0]\). Compare its termwise G-image with its derived G-image. Explain precisely which hypothesis of Theorem 1.2 would fail if one tried to use the class of all R-modules.

Solution. Y has only \(H^0Y=S\). Its termwise image is \(G(Y)=k[0]\), whose positive cohomology is zero. In contrast, (5.3) and its zero G-differential give \(R^nG(S)=k\) for every \(n\ge0\). In particular the putative shifted formula at \(p=0,k=1\) would wrongly identify k with zero.

The full class of modules supplies identity replacements, but it fails the second F-injective-class requirement. The bounded-below augmented complex \(0\to S\to R\to R\to\cdots\), with S in degree zero and the first R in degree one, is acyclic. Its G-image begins \(k\xrightarrow\sim k\xrightarrow0 k\xrightarrow0\cdots\); at degree two its kernel is k and its incoming image is zero. Thus that image is not acyclic. This checks the failed class condition directly. Theorem 2.1 still applies to Y and identifies its first derived cohomology with \(G(S)=k\); it makes no assertion that later groups vanish.

Exercise 3 — A compatible composition concentrated in degree one (advanced)

Let \(R=k[t]\), \(S=R/(t)\), \(F=\operatorname{Hom}_R(S,-): R\text{-Mod}\to k\text{-Vect}\), and \(G=V\otimes_k-\) for any k-vector space V. Verify compatible classes and compute every \(R^j(GF)(R)\).

Solution. Section 5 gives enough injectives in R-Mod, whose injective class is F-injective. Every short exact sequence of k-vector spaces splits by lifting a basis of its quotient. Hence tensoring with V preserves it, so G is exact. Its full class of vector spaces is G-injective: identity replacements supply density, and an exact functor preserves cycles, boundaries and their cohomology quotients, so it takes acyclic complexes to acyclic complexes. The image under F of each injective module is a vector space in this class. These are exactly the compatible hypotheses of Section 3, for arbitrary V, including zero and infinite-dimensional V.

The projective resolution of S is \(P^{-1}=R\xrightarrow{t}P^0=R\), augmented to S. Multiplication by t is monic, its cokernel is S, and both terms are free. The normalized one-sided Hom computation of Derived Hom, Proposition 6.1 computes \(R^+F(R)\) by \(\operatorname{Hom}(P,R)\). Explicitly an injective replacement \(R\to I\) gives k-linear chain maps \(\operatorname{Hom}(P,R)\to\operatorname{Hom}(P,I)\leftarrow\operatorname{Hom}(S,I)\). The first is a quasi-isomorphism by exact Hom from each projective term; the second is one by exact Hom into each injective term. The proved finite-diagonal comparison applies, and its maps are k-linear, so this computes the k-vector-valued F, with its normalized unit. The terms of \(\operatorname{Hom}(P,R)\) are R in degrees zero and one, and its differential is \(-t\), from the formula \(d(f)=d_Rf-(-1)^nfd_P\) at n zero. Its kernel is zero and its cokernel is \(R/tR\simeq k\). Thus \(R^qF(R)\) is k for q one and zero otherwise. Formula (3.2) gives \(R^j(GF)(R)=R^{j-1}G(k)\). Exactness makes the latter V for j one and zero for all other integers j. The isomorphism is natural in V and in module maps through the normalized comparison (3.3).

Exercise 4 — A nonzero connecting map between two cohomology degrees (advanced)

Return to \(R=k[\varepsilon]/(\varepsilon^2)\), \(S=R/(\varepsilon)\), and \(G=\operatorname{Hom}_R(S,-)\). Let \(Z=(R\xrightarrow{\varepsilon}R)\) in degrees zero and one. Construct its two-cohomology truncation triangle. Compute \(R^+G(Z)\), and show that its connecting map induces isomorphisms \(\operatorname{Ext}^n_R(S,S)\to\operatorname{Ext}^{n+2}_R(S,S)\) for every \(n\ge0\). Deduce that Z does not split as \(S[0]\oplus S[-1]\).

Solution. Its cohomology is \(\varepsilon R\simeq S\) at zero and \(R/\varepsilon R\simeq S\) at one. The inclusion of \(\varepsilon R\) in the first term and the quotient of the second term give the first two derived maps of the canonical triangle

Indeed this is (4.3) for a zero and b one; the first map induces the chosen isomorphism on \(H^0\), and the second does so on \(H^1\). The last arrow is the specified truncation arrow, hence defines an element of \(\operatorname{Ext}^2_R(S,S)\).

Both terms of Z are injective by (5.2), so the computing-class theorem identifies \(R^+G(Z)\) with G(Z). That complex is k in degrees zero and one with differential zero. Its cohomology is k at these two degrees and zero everywhere else. Applying the same theorem to (5.3) gives \(\operatorname{Ext}^n_R(S,S)=R^nG(S)=k\) for n nonnegative and zero for n negative. The equality with derived Hom is the normalized one-sided Hom theorem in the prerequisite.

Apply \(R^+G\) and cohomology to (6.1), retaining the same connecting arrow as in (4.4). For k zero, the map \(R^0G(S)\to H^0(R^+G(Z))\) is an isomorphism: on the computing complexes it sends a map into S to its inclusion into \(\varepsilon R\), and both spaces are k. At the next degree, the exact sequence reads

The first map is monic because the preceding group is \(R^{-1}G(S)=0\), and its source and target are one-dimensional, so it is an isomorphism. The next map is zero, and exactness makes \(\delta^1: R^0G(S)\to R^2G(S)\) both monic and epic. For \(n\ge1\), use k equal to n plus one in the long exact sequence: the two middle-object groups \(H^{n+1}(R^+G(Z))\) and \(H^{n+2}(R^+G(Z))\) vanish. Thus \(\delta^{n+1}: R^nG(S)\to R^{n+2}G(S)\) is an isomorphism as well. These are the maps induced by the actual \(R^+G(\kappa)\) with its shift comparison. In particular \(\kappa\ne0\), since its image induces a nonzero isomorphism.

If Z were isomorphic to \(S[0]\oplus S[-1]\), additivity and shift compatibility would identify its derived image with \(R^+G(S)\oplus R^+G(S)[-1]\). In every degree at least one the latter has cohomology \(k\oplus k\); the computed derived image of Z has k in degree one and zero in every degree at least two. This contradiction proves that no such splitting exists. The example works over every field, including characteristic two.

Freely accessible references

• The Stacks Project Authors, Truncation of complexes, Section 12.15, for the cochain kernel and quotient conventions.
• The Stacks Project Authors, Right derived functors and injective resolutions, Section 13.20, for the classical enough-injectives setting. Theorem 2.1 above proves the first-cohomology statement directly from universal derivability.
• The Stacks Project Authors, Composition of right derived functors, Section 13.22, for the classical compatible-injective comparison. Sections 3 and 4 above retain arbitrary compatible additive computing classes and construct the two-degree sequence directly from its truncation triangle.