Draft lesson.

Finite cohomology splittings and derived kernels

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

A finite list of cohomology objects does not always determine a derived object. We give explicit Ext conditions under which it does, prove the splitting with its cohomology identifications, and describe its possible nonuniqueness. We then derive the kernel functor on the category of arrows. A functorial replacement by epic arrows produces a mapping fiber, even in an abelian category with no enough-injectives hypothesis. A concrete nonsplit derived kernel connects the two constructions.

Conventions and prerequisites

Work in an ambient universe containing the abelian category \(\mathcal A\), its complexes, localizations and Hom groups. Complexes are cohomological. Our conventions are

The cone triangle has positive final projection to \(X[1]\). Its shifted triangle uses a minus sign on the last map for one shift. Define \(\operatorname{Ext}^q(B,A)=\operatorname{Hom}_{D(\mathcal A)}(B,A[q])\).

We use the following complete course proofs.

• Cohomological bounds and truncation adjunctions, Sections 1 and 3–5: cohomology detects isomorphisms, bounded embeddings, single-degree identifications, negative Ext and natural truncation triangles.
• Triangles after localization, Sections 2–4 and 6: exact Hom sequences, the morphism axiom on derived arrows, signed triangles, and complete short-exact-row cohomology sequences.
• Two cohomology degrees and Ext classes, Sections 1–5: canonical two-degree classes, splitting with actual endpoints, and changes of cohomology identifications.
• Relative replacements, Section 2, especially Lemma 2.2: the full localization equivalence from outgoing class replacements, including equality of roofs.
• Universal derived functors, Sections 1–4: the absolute universal property, the meaning of an adapted class, and the finite-prefix exactness test.
• Localizing triangulated functors, Sections 1–3: the exact localization through an adapted triangulated family with its unit and shift comparison.
• Projective complexes and derived Hom, Lemma 3.1: bounded-above projective complexes compute actual derived Hom. This is used only for the explicit module calculation in Exercise 3.

Freely accessible primary references are Stacks, the finite cohomology splitting lemma, Stacks, derived functors on derived categories, and Stacks, Ext groups. The constructions below retain their own explicit conventions and proofs.

1. Finite splitting under precise Ext vanishings

Let \(X\in D(\mathcal A)\) have \(H^nX=0\) outside a finite interval \([a,b]\). Write \(A_i=H^iX\), allowing some of these objects to be zero.

Theorem 1.1. Suppose

There is an isomorphism

whose degree-\(i\) cohomology map is the specified identity on \(A_i\). The sum is finite. No enough-injectives, enough-projectives or global dimension assumption is needed.

Proof. Finite cohomology support places \(X\) in the fully embedded bounded derived category. If the interval has one degree, the natural smart-truncation zigzag identifies \(X\) with \(A_a[-a]\) with its cohomology identity. If every cohomology object is zero, cohomology detection identifies \(X\) with zero. These also handle an empty support.

Proceed by induction on \(b-a\). Put \(L=\tau_{\le b-1}X\) and \(V=A_b[-b]\). The canonical truncation triangle is

The cohomology of \(L\) is \(A_i\) for \(i<b\), with the actual truncation identifications. Its pairwise conditions are a subset of (1). The induction therefore supplies a marked isomorphism \(S=\bigoplus_{i=a}^{b-1}A_i[-i]\simeq L\). Finite biproducts and translation give

The equality uses only finite sums of Hom groups. Consequently \(\delta=0\).

The exact sequence obtained by applying \(\operatorname{Hom}_D(V,-)\) to (3) contains

Thus \(1_V\) lifts to a section \(s:V\to X\) of \(p\). The arrow \((i,s):L\oplus V\to X\) has the identity cohomology map in degrees below \(b\), through \(i\), and in degree \(b\), because \(ps=1\). All remaining cohomology objects vanish. It is an isomorphism by cohomology detection. Compose it with the marked induction isomorphism on \(S\) to obtain (2), preserving every mark. The induction terminates after finitely many cutoffs, including cutoffs where \(A_i=0\). \(\square\)

Corollary 1.2. If \(\operatorname{Ext}^q(A_j,A_i)=0\) for every \(q\ge2\) and \(i<j\), then (2) holds with the same marks.

Proof. Each degree \(j-i+1\) in (1) is at least two, so the hypotheses imply every required vanishing. Apply Theorem 1.1. \(\square\)

These are sufficient conditions. They do not assert necessity: a split object can have nonzero groups in (1), with its particular boundary maps nevertheless zero.

2. The ambiguity of finite splittings

Choose a finite ordered set of allowed degrees \(s_1<\cdots<s_r\), with \(r\ge1\), and put \(S_X=\bigoplus_i A_i[-s_i]\), \(S_Y=\bigoplus_i B_i[-s_i]\). Zero summands are allowed. Here \(A_i\) and \(B_i\) denote the cohomology objects in those degrees, rather than their degree labels.

Proposition 2.1. A derived arrow \(S_X\to S_Y\) is uniquely a finite matrix whose target-row \(i\), source-column \(j\) entry lies in

Entries below the diagonal vanish. The diagonal entries are exactly the cohomology maps. Composition is matrix composition using actual derived composition of the shifted arrows.

Proof. Biproduct injections and projections give a mutually inverse pair of constructions between an arrow and its matrix: recover it as the finite sum of injection–entry–projection composites. Translating an entry by \(s_j\) proves (5). For \(i>j\), its Ext degree is negative, so it is zero. A diagonal entry has Ext degree zero and is its ordinary cohomology arrow by heart full faithfulness. An off-diagonal entry has source and target stalks in different degrees and therefore has zero cohomology in every degree. Composing the two finite decompositions gives the matrix law by the biproduct retraction and zero equations. This uses actual shifted arrows, so it imposes no additional unannounced Yoneda sign convention. \(\square\)

For one split object \(S\), let \(N\subset\operatorname{End}_D(S)\) be the subgroup of arrows inducing zero cohomology. It is the group of strictly upper triangular matrices in (5), and is an ideal.

Proposition 2.2. We have \(N^r=0\). The automorphisms inducing the identity on every cohomology object are exactly \(1+n\), \(n\in N\), with inverse

The group need not be additive or abelian. Marked splittings (2), once one exists, form a torsor for this group. In particular the marked splitting is unique precisely when every group \(\operatorname{Ext}^{s_j-s_i}(A_j,A_i)\), \(i<j\), is zero.

Proof. A nonzero term in a product of \(r\) strictly upper triangular matrices would require \(r+1\) strictly increasing indices in a set of size \(r\). There are none, so every such product is zero. Multiplying the finite geometric sum in (6) on either side by \(1+n\) leaves \(1+(-1)^{r-1}n^r=1\). Conversely an endomorphism with identity cohomology has diagonal identity and differs from it by a unique \(n\in N\). Its inverse is therefore (6).

For two marked isomorphisms \(\Phi,\Psi:S\to X\), the map \(\Phi^{-1}\Psi\) is a unique automorphism of \(S\) with identity cohomology. Conversely composing \(\Phi\) with any such automorphism gives a marked splitting. This is a free and transitive right action, which is the stated meaning of torsor. The matrix decomposition identifies \(N\) with the direct sum of all off-diagonal groups, so \(N=0\) is exactly their simultaneous vanishing. If \(N\ne0\), a nonzero entry gives a distinct \(1+n\), hence a distinct splitting. If the entire support is empty, the object is zero and its unique arrow is its identity; this is the separate trivial case. \(\square\)

For two objects already split as in (2), every prescribed family of cohomology maps lifts: take the diagonal matrix and transport it through the chosen splittings. All lifts of that family differ by the off-diagonal additive group from (5). Choices of the two splittings can change this parametrization. In particular a finite splitting existence theorem does not by itself give a natural splitting functor.

3. The abelian arrow category and its kernel functor

An object of \(\operatorname{Arr}(\mathcal A)\) is an arrow \(f:A\to B\). A morphism to \(g:C\to D\) is a pair \((u,v)\) such that \(vf=gu\). Composition is componentwise, as are addition, zero and finite biproducts.

Proposition 3.1. The arrow category is abelian. Kernels, cokernels, images and coimages are computed at the two endpoints. A short exact sequence in it is exactly a pair of short exact endpoint sequences with their commuting vertical arrows.

Proof. For a square \((u,v):f\to g\), the composite \(\ker u\to A\xrightarrow fB\xrightarrow vD\) is zero because it equals \(gu\) on that kernel. Hence \(f\) induces an arrow \(\ker u\to\ker v\). Its square into \(f\) is the kernel of \((u,v)\): a square killed by both components factors uniquely through the two endpoint kernels, and their induced factors commute by cancellation of the monic endpoint inclusion. This proves the full universal property.

Dually, \(g\) induces \(\operatorname{coker}u\to\operatorname{coker}v\), and the quotient square has the cokernel universal property. A square is monic exactly when its kernel object is zero, hence exactly when both components are monic. The dual criterion holds for epimorphisms. The canonical coimage-to-image square is the two endpoint coimage-to-image maps. Each is an isomorphism in \(\mathcal A\); their inverses still commute by composing the square with those inverses. Thus it is an isomorphism in the arrow category. The category has finite biproducts, all kernels and cokernels, and the abelian comparison is invertible, so it is abelian. Componentwise computation of images and kernels now proves the short exact sequence assertion. \(\square\)

Define \(F:\operatorname{Arr}(\mathcal A)\to\mathcal A\) by \(F(f)=\ker f\). A square acts on its kernels by the uniquely induced map.

Proposition 3.2. This functor is additive and left exact.

Proof. If \(k_f:\ker f\to A\) is the inclusion, the induced arrow for \((u,v)\) is determined by \(k_gF(u,v)=u k_f\). This equation, monicity and addition prove additivity and compatibility with composition. For an arrow-category kernel \(h:\ker u\to\ker v\) of a square \((u,v):f\to g\), its ordinary kernel represents arrows \(t:T\to A\) satisfying both \(ft=0\) and \(ut=0\). The kernel of \(F(u,v):\ker f\to\ker g\) represents exactly the same pairs of equations. Their compatible universal isomorphism identifies \(F(h)\) with that kernel. Thus \(F\) preserves kernels, and additivity makes it left exact. \(\square\)

For later use, a short exact row \(0\to f'\to f\to f''\to0\) in the arrow category has a natural exact sequence

To prove this with no elementwise surjectivity assumption on arbitrary abelian objects, put each arrow in the ordinary two-term complex \([A\xrightarrow f B]\) in degrees 0 and 1. Proposition 3.1 gives a termwise short exact row of these complexes. The complete categorical cone-to-quotient proof in the triangle prerequisite, Section 6, supplies its long cohomology sequence. The two cohomology objects of each complex are its kernel and cokernel, and all others vanish. This is precisely (7), including its natural maps. The connecting map uses our positive cone projection; Exercise 4 computes its sign in an explicit row.

4. Epic arrows form a full adapted class

Let \(\mathcal E\) be the full subcategory of arrows that are epimorphisms in \(\mathcal A\). It contains zero, is invariant under isomorphism, and is closed under finite biproducts.

Proposition 4.1. Every arrow embeds functorially into an object of \(\mathcal E\), by the short exact sequence of arrows

The first square has components \(a\mapsto(a,0)\) and \(1_B\). The quotient square has components \((a,b)\mapsto b\) and \(B\to0\).

Proof. The first square commutes, both its components are monic, and its endpoint quotients are \(B\) and zero. The induced quotient arrow is therefore \(B\to0\), giving (8). The middle arrow is split epic via \(b\mapsto(0,b)\); the last arrow is epic because every arrow to zero is epic. For a square \((u,v):f\to g\), the middle morphism is \((u\oplus v,v)\) and the final morphism is \((v,0)\). They commute with all maps in (8) and respect identity and composition. This proves functoriality and cogeneration. \(\square\)

Proposition 4.2. In a short exact row of arrows, an epic middle arrow has an epic quotient arrow. If the first arrow is epic, the induced map from the middle kernel to the quotient kernel is epic. Consequently every short exact row with all its arrows in \(\mathcal E\) remains short exact under \(F\).

Proof. Write the quotient square as \((p_A,p_B):f\to f''\), with both components epic. If \(f\) is epic, then \(f''p_A=p_Bf\) is epic; canceling on its right shows that \(f''\) is epic. For the second assertion, \(\operatorname{coker}f'=0\), so exactness of (7) makes \(\ker f\to\ker f''\) epic. Left exactness supplies its other two positions. These are categorical epimorphisms, not presumed surjections on every Hom group. \(\square\)

This verifies the finite-prefix criterion too. If an arrow \(f'\) has a finite exact prefix \(0\to E_0\to\cdots\to E_r\to f'\to0\) with \(E_i\in\mathcal E\), its final epimorphism exhibits \(f'\) as a quotient of the epic arrow \(E_r\), so \(f'\) is epic. A short exact row \(0\to f'\to f\to f''\to0\), with \(f\in\mathcal E\), therefore has exact kernels by Proposition 4.2. Cogeneration and this test give the full required adapted class. We will now prove the stronger complex-level statements directly.

5. An explicit epic replacement of every complex of arrows

A complex of arrows is exactly a chain map \(f:X\to Y\): its degree-\(n\) arrow is \(f^n:X^n\to Y^n\), and the arrow-complex differential is the commuting square \((d_X^n,d_Y^n)\). Its cohomology and quasi-isomorphisms are componentwise by Proposition 3.1.

Define a complex \(I(f)\) of epic arrows by

Its domain differential and codomain differential are

Coordinates in a biproduct stand for the corresponding component maps; no literal element interpretation in \(\mathcal A\) is required.

Theorem 5.1. The maps

form a functorial degreewise monic quasi-isomorphism. This replacement works for unbounded, bounded-below, bounded-above and bounded complexes. If both endpoint complexes are zero outside \([a,b]\), \(I(f)\) is zero outside \([a,b+1]\).

Proof. Squaring (10) gives zero: its third coordinate is \(d_Yb-d_Y(b-d_Yc)=0\), and its first two coordinates are the squares of their differentials. Its square with the arrow in (9) commutes because \(f d_Xx+d_Yb=d_Y(fx+b)\). The arrow is epic via the identity \(Y^n\) summand. Formula (11) is a commuting square of chain maps and is monic at both endpoints.

Its quotient complex of arrows has codomain zero and domain

It is contractible with degree-minus-one map \(h(b,c)=(c,0)\). Indeed \(dh(b,c)=(d_Yc,c)\) and \(hd(b,c)=(b-d_Yc,0)\), whose sum is \((b,c)\). The termwise short exact quotient row and its cohomology sequence imply that \(t_f\) is a quasi-isomorphism. The bound statement follows directly from the three summand degrees in (9); all sums are finite, including in the unbounded case.

A square of chain maps \((u,v):f\to g\) acts on the domain of \(I\) as \((u,v,v)\) and on its codomain as \(v\); the last component is taken in the one-lower degree. It commutes with (10) and (11), and these formulas preserve composition. A homotopy of squares with components \((h_X,h_Y)\) is sent on domains to \((h_X,h_Y,-h_Y)\), and on codomains to \(h_Y\), with the corresponding degrees. Its third homotopy equation follows by cancellation of \(h_Yb\), leaving \((d_Yh_Y+h_Yd_Y)c\); the first two are the given homotopy equations. The homotopy square itself commutes, since \(g h_X=h_Y f\) in the appropriate source and target degrees. Thus \(I\) also defines a functor on homotopy categories. \(\square\)

Proposition 5.2. If a complex of arrows is termwise epic and is acyclic at both endpoints, its kernel complex is acyclic. Moreover \(F\) sends every quasi-isomorphism between termwise epic complexes to a quasi-isomorphism.

Proof. For a termwise epic chain map \(f:X\to Y\), the sequence of complexes \(0\to\ker f\to X\to Y\to0\) is termwise short exact. If \(X,Y\) are acyclic, its long cohomology sequence makes \(\ker f\) acyclic. For a quasi-isomorphism between two such complexes, its cone is again termwise epic: each term arrow is a finite biproduct of two epic arrows. It is acyclic in the arrow category. Applying the first assertion gives an acyclic kernel cone. Additivity identifies this with the cone of the induced kernel map, so that map is a quasi-isomorphism. \(\square\)

The full homotopy subcategory of epic complexes is closed under both shifts and cones. Theorem 5.1 supplies outgoing replacements into it, and Proposition 5.2 annihilates its acyclic objects after \(F\). These are precisely the adaptation conditions. In particular \(\mathcal E\) is kernel-injective in the bounded-below sense, although its arrows need not be injective objects of the arrow category. The same construction applies without a bound.

6. The signed mapping fiber and every structural map

For a chain map \(f:X\to Y\), define

The equation \(d_T^2=0\) follows from \(fd_X=d_Yf\). A square \((u,v)\) acts as \((x,y)\mapsto(ux,vy)\). This is a strict functor on complexes of arrows.

Proposition 6.1. There is a natural chain isomorphism

The natural map \(\eta_f:\ker f\to T(f)\) is \(k\mapsto(k,0)\).

Proof. The kernel equation in (9) forces the middle coordinate to be \(-fx\). This can also be stated by the kernel universal property: the displayed map from \(X^n\oplus Y^{n-1}\) is monic onto that kernel, with inverse the first and negative-third projections. Applying (10) to (14) gives \((d_Xx,-f d_Xx,-fx+d_Yy)\), which is (14) applied to (13). Thus it is a chain isomorphism. The kernel of (11), followed by \(\theta_f^{-1}\), is exactly \(\eta_f\). All formulas commute with squares, proving naturality. \(\square\)

Proposition 6.2. The mapping fiber induces an exact functor on homotopy categories, with translation comparison

It sends componentwise quasi-isomorphisms to quasi-isomorphisms.

Proof. If a homotopy of squares has components \((h_X,h_Y)\), its mapping-fiber homotopy is \((h_X,-h_Y)\). The lower-left component of \(d_Th+h d_T\) is \(f'h_X-h_Yf=0\) by the homotopy-square condition. Its two diagonal components are the given homotopy differences. Hence \(T\) descends to the homotopy category.

In the shifted source of (15) the differential is \((-d_Xx,fx+d_Yy)\); in the shifted target it is \((-d_Xx,-fx+d_Yy)\). The negative second coordinate in (15) makes these commute. This comparison is natural, and iterating it gives the coherent comparison for every integer shift.

For a square of chain maps \((u,v):f\to g\), its arrow-category cone has endpoint cones \(C(u),C(v)\), with arrow \(\operatorname{diag}(g,f[1])\). Write a term of its fiber in the order \((x',x,y',y)\), with \(x'\in X'^n,x\in X^{n+1},y'\in Y'^{n-1},y\in Y^n\). Its differential has components

The map

is an isomorphism of complexes. In the target order its differential is \[ (d_{X'}x'+ux,\ gx'-d_{Y'}y'+vy,\ -d_Xx,\ -fx+d_Yy); \] substituting \(-y\) proves the equality with (17) applied to (16). It identifies the inclusion from \(T(g)\) on the nose. The final projection on the left first gives \((x,y)\in T(f[1])\); after (15) this is \((x,-y)\), exactly the target cone projection. Therefore the first map, inclusion and signed last map all agree. This proves preservation of cone triangles and hence exactness.

Finally (13) has the degreewise split short exact sequence

A square that is a quasi-isomorphism at \(X,Y\) induces a map of these rows with the two outside cohomology maps invertible. The exact cohomology sequence and the five-term argument make every middle cohomology map invertible. Thus \(T\) inverts exactly the required quasi-isomorphisms for descent. \(\square\)

For comparison with the fixed cone convention, \(T(f)\simeq C(f)[-1]\) by \((x,y)\mapsto(-y,x)\), in the target-first order. Direct substitution in the negative shifted cone differential gives (13).

7. The universal derived kernel, including its unit

Let \(Q\) denote the arrow-category localization and \(Q_A\) the target localization. Apply the same notation in any one of the unbounded, bounded-below, bounded-above or bounded variants. Proposition 6.2 gives an exact functor

Its unit from the termwise kernel is \(Q_A\eta_f\).

Theorem 7.1. Formula (19) is the universal right derived kernel. For every further functor \(L:D^*(\mathcal A)\to\mathcal B\) and every functor \(H:D^*(\operatorname{Arr}(\mathcal A))\to\mathcal B\), composition with its unit gives a bijection

No additivity, exactness or colimit preservation is imposed on \(L,H\).

Proof. First, \(\eta_e\) is a quasi-isomorphism for every termwise epic complex \(e:X\to Y\). Indeed the quotient of its inclusion \(\ker e\to T(e)\) identifies, via \(X/\ker e\simeq Y\), with (12). Its differential is \((d_Yb,b-d_Yc)\), and the same contraction proves it acyclic. Thus the unit is invertible on epic complexes.

Write \(R=R\ker\), \(G=Q_A\ker\), and regard \(\eta:G\to RQ\) as a natural transformation on the homotopy category. Let \(\zeta:LG\to HQ\) be given. For \(t_f:f\to I(f)\) of (11), define

Every inverse exists: \(t_f\) is a quasi-isomorphism and \(I(f)\) is termwise epic. Naturality of \(t,\eta,\zeta\) makes (21) commute with every homotopy-category arrow. Because both \(LR\) and \(H\) already invert denominators, the same commuting equation holds for their inverses and for all derived roofs. Thus it is a natural transformation on the derived category.

Naturality of the unit gives \(R(Qt_f)\eta_f=\eta_{I(f)}G(t_f)\). Insert this in (21), then use naturality of \(\zeta\); the result is \(\mu_{Qf}L\eta_f=\zeta_f\). Conversely any transformation with this property has, at the epic complex \(I(f)\), the forced component \(\zeta_{I(f)}(L\eta_{I(f)})^{-1}\). Naturality at the invertible \(Qt_f\) forces precisely (21) at \(f\). This proves both directions and uniqueness in (20), for arbitrary \(L\). It is the absolute universal property, not merely a formula for object values. The bounds from Theorem 5.1 and (13) ensure that every object and map used lies in the chosen variant. \(\square\)

Equivalently, the localization of the full homotopy subcategory of epic complexes is exactly \(D^*(\operatorname{Arr}(\mathcal A))\). For clarity, density is (11); any outgoing roof between epic complexes can have its middle object replaced by \(I\), giving a roof inside that full subcategory. For two equal roofs, replace the common equality-refinement object as well; the equality remains a homotopy equality between epic complexes, so equality holds in their localization. This proves fullness and faithfulness by the actual fraction tests, as in the relative-replacement prerequisite. On this localization the functor is ordinary termwise kernel, by Proposition 5.2 and the invertible unit. The displayed mapping fiber supplies the exact shift and cone comparisons rather than leaving them implicit in an equivalence of categories.

8. Ordinary arrows: kernel, cokernel and acyclicity

For \(f:A\to B\) concentrated in arrow degree zero, (13) is the complex

Consequently

Proof. Substituting stalk complexes in (13) leaves exactly the two terms and differential in (22). Their cycles and boundaries give (23). The unit is the ordinary kernel inclusion in degree zero, so its degree-zero map is the canonical identity on \(\ker f\). A square \((u,v)\) acts on (22) by \(u\) in degree zero and \(v\) in degree one. Thus both identifications are natural for every square. \(\square\)

The full class of ordinary right-acyclic arrows is exactly \(\mathcal E\). Indeed \(R^1\ker(f)=0\) exactly when \(f\) is epic, and all other positive degrees already vanish. For such arrows the unit is a quasi-isomorphism by its cohomology calculation; conversely an invertible unit forces that cokernel to vanish. This is an adapted class assertion, not a claim that every epic arrow is an injective object.

The two possible cohomology objects in (22) need not split. Theorem 1.1 gives a sufficient splitting condition \(\operatorname{Ext}^2(\operatorname{coker}f,\ker f)=0\). More precisely the canonical class lies in this group and vanishes exactly when the derived kernel splits. Exercise 3 computes a nonzero class, so the distinction persists even though every \(R^n\ker\) above degree one is zero.

9. Four graded exercises with complete solutions

Exercise 1 — Foundation: the exact degrees and all splitting choices

Let \(X\) have cohomology only in degrees \(0,2,5\), denoted \(A,B,C\). List the three groups whose vanishing suffices in Theorem 1.1. After a marked splitting, describe every marked automorphism and its multiplication and inverse. State exactly when the marked splitting is unique.

Solution. The required groups are \(\operatorname{Ext}^3(B,A)\), \(\operatorname{Ext}^6(C,A)\), and \(\operatorname{Ext}^4(C,B)\). Their vanishing supplies \(S=A\oplus B[-2]\oplus C[-5]\simeq X\) with all three identity marks. In this summand order a marked automorphism is uniquely

where \(h\in\operatorname{Ext}^2(B,A)\), \(k\in\operatorname{Ext}^3(C,B)\), and \(l\in\operatorname{Ext}^5(C,A)\). These entries are actual maps between the displayed shifted objects. Matrix multiplication gives

The last product is composition \(C[-5]\xrightarrow kB[-2]\xrightarrow{h'}A\); under Ext coordinates it is the corresponding translated composition. The inverse is \(U(-h,-k,hk-l)\), as multiplying on either side checks. Thus the parameters form a group with a possible cross term, not automatically their additive direct sum. The marked splitting choices form its torsor. They are unique precisely when the three ambiguity groups \(\operatorname{Ext}^2(B,A)\), \(\operatorname{Ext}^3(C,B)\), \(\operatorname{Ext}^5(C,A)\) vanish. Their degrees differ by one from the splitting-existence groups, as the formulas show.

Exercise 2 — Intermediate: an adapted arrow need not be injective

In abelian groups, compute the derived kernel and unit for \(f:\mathbb Z\to\mathbb Z/m\), with \(m\ge2\), the quotient map. Show that the epic arrow \((\mathbb Z\to0)\) is not an injective object of the arrow category. Explain why this does not contradict adaptation.

Solution. Formula (22) is \([\mathbb Z\to\mathbb Z/m]\) in degrees 0 and 1. Its degree-zero cohomology is \(m\mathbb Z\), its degree-one cohomology is zero, and the unit is the inclusion \(m\mathbb Z[0]\to[\mathbb Z\to\mathbb Z/m]\). This is a quasi-isomorphism by the computed cohomology; multiplication by \(m\) identifies its sole cohomology group with \(\mathbb Z\). The assertion includes the actual kernel inclusion, rather than choosing a new endpoint map.

There is a monomorphism \((\mathbb Z\to0)\hookrightarrow(\mathbb Q\to0)\), given by the usual inclusion and the identity of zero. If the first arrow object were injective, its identity would extend through this monomorphism. The domain component would be a group map \(r:\mathbb Q\to\mathbb Z\) restricting to the identity of \(\mathbb Z\). But \(r(1)=n r(1/n)\) for every positive integer \(n\), so \(r(1)\) is divisible by every \(n\) in \(\mathbb Z\), forcing it to be zero, not one. Such an extension is impossible. Thus this epic arrow is not an injective object.

Adaptation requires outgoing replacements and acyclicity of kernel complexes on acyclic epic complexes. Those were proved in Section 5; it never asserts extension of all arrows through arbitrary monomorphisms. The example therefore meets the adapted-class definition and fails the different injective-object definition.

Exercise 3 — Advanced: a nonsplit derived kernel

Let \(k\) be a nonzero field, \(R=k[t]/(t^2)\), and \(M=k=R/(t)\). Compute \(R\ker(f)\) for \(f:R\xrightarrow{t}R\). Give its two cohomology identifications and calculate its canonical two-degree class, with the fixed positive cone convention. Decide whether it is the sum of its cohomology stalks.

Solution. Write each element of \(R\) uniquely as \(a+bt\). Multiplication by \(t\) sends it to \(at\), so its kernel and image are both \(tR\). The cokernel is \(M\). Thus \(Z=R\ker(f)=[R\xrightarrow{t}R]\) in degrees 0 and 1 has marks \(H^0Z=M\) via \(c\mapsto ct\), and \(H^1Z=M\) via quotient evaluation \(\epsilon:R\to M\). Its class lies in \(\operatorname{Ext}^2_R(M,M)\).

Use the free resolution \(P\) with \(P^n=R\) for every \(n\le0\), zero for \(n>0\), every negative differential multiplication by \(t\), and augmentation \(\epsilon\) in degree zero. The equation \(t^2=0\), the just computed kernel and image, and \(\ker\epsilon=tR\) prove exactness in every augmented position. Its terms are projective and it is bounded above. The projective Hom lemma computes actual derived Ext by \(\operatorname{Hom}(P,M)\). All its differentials are zero because \(t\) kills \(M\); in particular \(\operatorname{Ext}^2_R(M,M)=k\), with coordinate \(c\) represented by \(c\epsilon:P^{-2}\to M[2]^{-2}\).

The injection \(M[0]\to Z\) has cone \(K\) with terms \(M,R,R\) in degrees \(-1,0,1\), whose differentials are the injection \(c\mapsto ct\) and multiplication by \(t\). Its quotient comparison \(q:K\to M[-1]\) is \(\epsilon\) in degree 1; exactness makes it a quasi-isomorphism. The positive projection \(p:K\to M[1]\) is identity in degree minus one. The first two transported triangle maps are precisely the marked inclusion and cohomology projection of \(Z\), so its canonical class is \(Qp(Qq)^{-1}\), by the canonical comparison in the two-degree prerequisite.

Translate this arrow by 1 to its degree-two Ext coordinate. The differentials of \(K[1]\), in degrees \(-2,-1,0\), are the negative injection and \(-t\). There is a chain map \(\ell:P\to K[1]\) with components \(\ell^0=1_R\), \(\ell^{-1}=-1_R\), \(\ell^{-2}=\epsilon\), and zero in lower degrees. Its two nontrivial squares say \((-t)(-1)=t\) and \((-t)\epsilon=(-1)t\), where the first \(t\) in the second equation is the injection \(M\to R\). The lower remaining square is \(\epsilon t=0\). Thus all chain equations hold. Moreover \(q[1]\ell\) is the augmentation, and \(p[1]\ell\) has component \(\epsilon\) in degree minus two. The class of \(Z\) therefore has coordinate \(1\in k\), which is nonzero.

The two-degree splitting criterion now proves \(Z\not\simeq M[0]\oplus M[-1]\), even without fixed marks. This example has no higher right-kernel cohomology, but it has a nonzero degree-two obstruction to splitting. Those are distinct assertions. The coordinate just computed is the truncation boundary; it does not silently replace it by a differently normalized Yoneda class.

Exercise 4 — Synthesis: exactness and the sign of the connecting map

Consider the short exact row of arrows in abelian groups

with the evident identity endpoint maps. Compute the full six-term sequence (7), including its connecting sign in our cone convention. Use the row to show that the ordinary kernel functor is not exact, and verify that the derived kernel nevertheless gives its required distinguished triangle.

Solution. At domain endpoints the short row is \(0\to0\to\mathbb Z\xrightarrow1\mathbb Z\to0\); at codomain endpoints it is \(0\to\mathbb Z\xrightarrow1\mathbb Z\to0\to0\). Thus it is short exact in the arrow category. The two-term images are \(P'=[0\to\mathbb Z]\), \(P=[\mathbb Z\xrightarrow1\mathbb Z]\), and \(P''=[\mathbb Z\to0]\), all in degrees 0 and 1.

The cone of the inclusion \(P'\to P\) has \(\mathbb Z\oplus\mathbb Z\) in degree zero, \(\mathbb Z\) in degree one, and differential \((z,w)\mapsto z+w\). Its quotient comparison to \(P''\) is \((z,w)\mapsto z\) in degree zero, and its positive final projection to \(P'[1]\) is \((z,w)\mapsto w\). A degree-zero cycle satisfies \(w=-z\), so the comparison is an isomorphism on its sole cohomology, and the transported last projection is \(-1:\mathbb Z\to\mathbb Z\). The exact sequence is therefore

The map from the middle ordinary kernel, zero, to the final ordinary kernel, \(\mathbb Z\), is not epic. This proves the failure of ordinary exactness. The derived kernel images form the short-exact-complex triangle just computed: \(P\) is contractible, and its last map identifies \(P''\simeq\mathbb Z[0]\) with \(P'[1]\simeq\mathbb Z[0]\) by \(-1\). The cone comparison and its positive projection prove distinguishedness and its actual connecting sign. This also checks the exact functor structure of Sections 6–7 on a row whose ordinary kernels fail to be exact.