Draft lesson.

Two cohomology degrees, Ext classes and polynomial models

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

Two complexes can have isomorphic cohomology objects without being isomorphic in the derived category. When only two cohomology degrees are possible, one precise Ext class records the missing information. It also determines which pairs of cohomology maps come from derived morphisms. Compatible pairs always lift, but a second Ext group measures the possible choices of lift.

We prove these assertions in an arbitrary abelian category. We then compute a complete example over a polynomial ring in two variables with any commutative base ring. The construction provides explicit complexes for every class, explicit maps between them, and a classification that retains the zero class when cohomology identifications are forgotten.

Conventions and prerequisites

Fix an ambient universe containing the complexes, localization and Hom groups being used. All rings are commutative and unital unless another convention is stated. We allow the zero ring. Complexes are cohomological:

For a chain map \(f:U\to V\), the cone is target first,

and its distinguished triangle has positive last projection: \(U\to V\to C(f)\to U[1]\). Translation of a distinguished triangle by \(r\) leaves the shifted first two arrows unchanged and multiplies its shifted last arrow by \((-1)^r\).

We use these complete course proofs:

• Cohomological bounds and truncation adjunctions, Sections 1 and 3–5, proves cohomology detection of isomorphisms, heart full faithfulness, orthogonality, bounded embeddings and the natural truncation triangles on every derived arrow.
• Triangles after localization, Sections 2–4 and 6, proves exact Hom sequences, the morphism axiom with arbitrary derived first arrows, signed rotations and short-exact-row triangles.
• Projective complexes compute derived Hom, Lemma 3.1, proves the actual Hom comparison for a bounded-above projective complex. Its Section 1 also proves the vector-space Ext vanishing used in Exercise 1.
• Higher extensions and their normalization, Sections 1–2, fixes the sign of the augmented extension roof and the normalized Yoneda class.
• Functorial cones and derived connecting maps, Sections 2 and 4, proves the signed cone-to-quotient and translation comparisons. We use the same convention here.

Freely accessible references are Stacks, truncation of complexes, Stacks, Ext groups, and Stacks, Koszul regular sequences. The polynomial resolution below is proved directly, including its signs and exactness over a base ring with zero divisors.

Write \(D=D(\mathcal A)\) for the derived category of an abelian category \(\mathcal A\), and define

Negative Ext vanishes and degree-zero Ext is ordinary Hom, by the first prerequisite. None of the general results in Sections 1–5 assumes enough injectives or enough projectives.

1. The canonical class and its precise degree

Fix integers \(a<b\). Suppose \(H^nX=0\) for \(n\notin\{a,b\}\). Either of the two allowed cohomology objects may also be zero. Put

The natural truncation triangle at \(a\) is \(\tau_{\le a}X\to X\to\tau_{\ge a+1}X\to\tau_{\le a}X[1]\). Its first vertex has cohomology only in degree \(a\), and its third vertex only in degree \(b\). The single-degree equivalence proved by smart truncation identifies them naturally with \(U_X\) and \(V_X\), using their actual cohomology identifications. We therefore obtain a specified natural triangle

The maps \(H^a(i_X)\) and \(H^b(p_X)\) are identities under those identifications. Translate the last arrow by \(b\) to identify its group as

We call \(\kappa_X\), with the identification (3), the two-degree class of \(X\). In particular adjacent degrees give an Ext group in degree two, not degree one.

For clarity, the single-degree identifications used here are natural derived identifications rather than chosen splittings of cycles. A complex with cohomology only in degree \(a\) has a canonical truncation zigzag to \(H^a[-a]\); all its maps are quasi-isomorphisms. The zigzag is natural before localization, so it is natural on chain maps and inverse quasi-isomorphisms, and hence on all derived arrows. This is the single-degree construction in Section 3 of the first prerequisite.

The same prerequisite gives two further formulas we will repeatedly use. If source cohomology ends at \(c\) and target cohomology starts at \(c\), then

If the source ends at \(c\) and the target starts at \(c+1\), that Hom group is zero. These statements are proved by bounded denominator representatives and their equality tests, so they control the actual derived maps as well as their existence. Every object considered here is in the fully embedded \(D^b(\mathcal A)\), since its cohomology has a finite bound, even if its initially chosen complex is unbounded.

2. Naturality, translation and the splitting test

For a derived morphism \(f:X\to Y\) between objects with these same two possible degrees, put \(\alpha=H^a(f)\) and \(\beta=H^b(f)\). Naturality of the complete truncation triangle gives

Under (3), its last identity says that pushing \(\kappa_X\) forward along \(\alpha\) equals pulling \(\kappa_Y\) back along \(\beta\). This applies to every derived morphism, not only maps on a chosen complex.

Proposition 2.1. Under the standard cohomology identifications for a shift, the class of \(X[r]\), whose allowed degrees are \(a-r,b-r\), is

After translating to the Ext group in (3), its coordinate is \((-1)^r\) times the coordinate of \(\kappa_X\).

Proof. The kernel and quotient defining smart truncation commute with translation, with the cutoff translated by the same integer. Their first two structural maps commute with the standard termwise shift identifications. For one shift, the isomorphism from the cone of the shifted inclusion to the shifted cone is \((v,u)\mapsto(v,-u)\). It commutes with the inclusion and with the cone-to-quotient map, but changes the positive last projection by a minus sign. Thus the canonical short-row boundary in the shifted truncation triangle is \(-\kappa_X[1]\). This is also the explicit quotient comparison in the fifth prerequisite, Section 4. Iterating gives (6) for positive shifts. Solving the same identity for the inverse shift gives it for negative shifts, and the zero shift is the identity. The Ext identification translates by \(b-r\); together with the preceding shift by \(r\) this is translation by \(b\), so no further sign enters that identification. \(\square\)

Theorem 2.2. The following are equivalent:

1. \(\kappa_X=0\).
2. There is a section \(s:V_X\to X\) of \(p_X\).
3. There is an isomorphism \(U_X\oplus V_X\to X\) inducing the specified identity maps on both cohomology objects.
4. The object \(X\) is isomorphic to \(U_X\oplus V_X\), even without specifying the cohomology identifications.

Proof. Apply \(\operatorname{Hom}_D(V_X,-)\) to (2). Its exact sequence includes

The identity of \(V_X\) lifts precisely when \(\kappa_X=0\), proving the equivalence of 1 and 2. Such a section gives \((i_X,s):U_X\oplus V_X\to X\). On degree \(a\) this is the identity through \(i_X\); on degree \(b\) it is the identity because \(p_Xs=1\). All other cohomology objects vanish. Cohomology detection of isomorphisms makes this arrow invertible, proving 3. Condition 3 implies 4.

The split object \(U_X\oplus V_X\) has class zero: its projection onto \(V_X\) has a section, and consecutive arrows of a distinguished triangle compose to zero, so its last arrow vanishes. If it is isomorphic to \(X\), naturality (5), with invertible endpoint cohomology maps, forces \(\kappa_X=0\). Thus 4 implies 1. \(\square\)

The theorem asserts existence of a splitting. It does not choose one naturally; Section 4 identifies its possible ambiguity.

3. Every compatible pair lifts with both prescribed endpoints

Theorem 3.1. Let \(X,Y\) have cohomology only in the allowed degrees \(a<b\). For arbitrary maps \(\alpha:A_X\to A_Y\), \(\beta:B_X\to B_Y\), there is a derived morphism \(f:X\to Y\) with exactly these two cohomology maps if and only if

Proof. Necessity is (5). For sufficiency, rotate the two triangles (2) twice. They become

and the analogous triangle for \(Y\). Condition (7) is exactly a commuting square between their first arrows, with the prescribed first two components \(\beta[-b]\) and \(\alpha[1-a]\). The derived triangle morphism axiom proved in the second prerequisite completes this square by a map \(g:X[1]\to Y[1]\), respecting its last square as well.

Put \(f=g[-1]\). The two signed equations for \(g\), after canceling the minus signs and translating back, give

Since \(H^a(i_X),H^a(i_Y)\) and \(H^b(p_X),H^b(p_Y)\) are the prescribed identities, these equations give \(H^a(f)=\alpha\) and \(H^b(f)=\beta\). The morphism axiom is valid for arbitrary derived arrows; it does not require replacing the requested cohomology maps by different endpoints, or choosing strict representatives on the original complexes. \(\square\)

In particular, if \(\alpha\) and \(\beta\) are isomorphisms and satisfy (7), any such lift is an isomorphism: it induces isomorphisms on all cohomology, including the zero objects in the other degrees.

4. All choices of lift and the square-zero ambiguity

Put \(d=b-a>0\), and let \(\mathcal C(X,Y)\) be the abelian group of pairs \((\alpha,\beta)\) satisfying (7). It is a subgroup of the direct sum of the two ordinary Hom groups, since the compatibility equation is additive in each entry.

Theorem 4.1. There is a natural short exact sequence of abelian groups

where, viewing \(h\) as a map \(V_X\to U_Y\),

Proof. Theorem 3.1 proves surjectivity onto \(\mathcal C(X,Y)\). The maps (9) are additive. The map \(i_Yhp_X\) has zero degree-\(a\) cohomology because \(H^a(V_X)=0\), and zero degree-\(b\) cohomology because \(H^b(U_Y)=0\). Thus its image is in the kernel of \(H\).

Conversely suppose \(H(f)=0\). Formula (4), at degree \(b\), identifies \(\operatorname{Hom}_D(X,V_Y)\) with \(\operatorname{Hom}(B_X,B_Y)\). It follows that \(p_Yf=0\). The exact Hom sequence of the triangle for \(Y\) supplies \(g:X\to U_Y\) with \(f=i_Yg\). This \(g\) is unique: the preceding group is \(\operatorname{Hom}_D(X,V_Y[-1])=0\), since \(X\) ends in degree \(b\), while \(V_Y[-1]\) starts in degree \(b+1\).

Next \(fi_X=0\), since (4) at degree \(a\) identifies \(\operatorname{Hom}_D(U_X,Y)\) with \(\operatorname{Hom}(A_X,A_Y)\), and its cohomology map is zero. The map \(\operatorname{Hom}_D(U_X,U_Y)\to\operatorname{Hom}_D(U_X,Y)\) induced by \(i_Y\) is an isomorphism under that same identification. Therefore \(gi_X=0\). Apply the contravariant Hom sequence of the triangle for \(X\), with target \(U_Y\). Its relevant part is

Here the labeled maps are precomposition by the indicated arrows. The first group is \(\operatorname{Ext}^{-1}(A_X,A_Y)=0\). Consequently there is a unique \(h:V_X\to U_Y\) with \(g=hp_X\). The identification \(\operatorname{Hom}_D(V_X,U_Y)=\operatorname{Ext}^{b-a}(B_X,A_Y)\) follows by translating by \(b\). This proves exactness at the middle group and injectivity of \(J\), since both factorizations were unique. \(\square\)

For each compatible pair, the set of lifts is therefore a torsor for \(\operatorname{Ext}^d(B_X,A_Y)\): once one lift \(f_0\) is chosen, all lifts are uniquely \(f_0+J(h)\). A torsor here means a set with a free and transitive action of the additive group. There is generally no distinguished choice of \(f_0\).

These formulas also describe naturality of (8). If \(f:Y\to Z\) has lower cohomology map \(\alpha_f\), then

If \(g:W\to X\) has upper cohomology map \(\beta_g\), then

They follow directly from the first two equations in (5), and are the usual pushforward and pullback actions on the Ext group.

Maps with zero cohomology in these two degrees form a square-zero ideal. Indeed, for \(h:V_X\to U_Y\) and \(k:V_Y\to U_Z\),

because consecutive triangle arrows compose to zero. For one object \(X\), its automorphisms inducing the identity on both cohomology objects are exactly

The inverse is \(1_X-J(h)\), and multiplication of these automorphisms corresponds to addition of their parameters by (10). Conversely any such automorphism differs from the identity by a unique \(J(h)\), by (8). The same statement applies to every endomorphism with identity cohomology: it is automatically invertible.

5. Classification with fixed cohomology identifications

Fix \(A,B\in\mathcal A\) and degrees \(a<b\). A marked object is a derived object with cohomology only in these degrees, together with isomorphisms \(H^aX\simeq A\), \(H^bX\simeq B\). A marked isomorphism must induce the identity of \(A\) and \(B\) after these identifications.

Theorem 5.1. The marked isomorphism classes are in bijection with

The class is the canonical truncation class transported by the marks. Forgetting the marks gives the orbits for the action

Proof. Naturality makes the class invariant under marked isomorphisms. If two marked objects have the same class, Theorem 3.1 lifts the pair \((1_A,1_B)\), and the lift is an isomorphism by cohomology detection. This proves injectivity of the proposed classification.

For surjectivity, put \(U=A[-a]\), \(V=B[-b]\), and choose any distinguished triangle for the derived arrow \(-\kappa[-1]:V[-1]\to U\). Its cone object \(X\), after one rotation, fits into

This use of a cone means a choice of a completed triangle; it does not assert a functorial cone for an arbitrary derived arrow. Existence follows from the derived triangle theorem, including arrows given by roofs. Its cohomology sequence shows that \(X\) has no cohomology outside \(a,b\), with \(H^a(i)\) and \(H^b(p)\) isomorphisms. Use these as its marks.

We verify that the last arrow in (14) is precisely its canonical marked class. Formula (4) shows that an arrow \(U\to X\) is determined by its degree-\(a\) map, so \(i\) agrees with the canonical marked truncation inclusion. Likewise an arrow \(X\to V\) is determined by its degree-\(b\) map, so \(p\) agrees with the canonical projection. Complete the identity square on \(U\to X\) to a triangle comparison from (14) to the canonical triangle. Its third component \(c:V\to V\) obeys \(cp=p\). Applying \(H^b\) gives \(H^b(c)=1_B\); full faithfulness on a single degree implies \(c=1_V\). The last square now gives equality of the two last arrows. Thus the class is \(\kappa\), proving surjectivity and the exact marks in (12).

Forgetting marks allows invertible cohomology maps \(u,v\). Equation (7) says exactly that two classes are related by (13). Conversely any such pair of units satisfies (7), lifts by Theorem 3.1, and has an invertible lift. This proves both directions of the orbit classification. \(\square\)

The bijection concerns isomorphism classes. It does not make all marked objects rigid: their marked automorphism groups are given by (11).

6. A signed polynomial resolution over every base ring

Let \(R\) be any commutative ring, put \(\Lambda=R[x,y]\), and let \(M=R\) be a \(\Lambda\)-module by evaluation at \(x=y=0\). Let \(\epsilon:\Lambda\to M\) be that evaluation map. Consider

where

Proposition 6.1. Sequence (15) is exact. Its first three nonzero modules form a free resolution in cohomological degrees \(-2,-1,0\).

Proof. Multiplication by \(x\) in \(R[x,y]\) is injective: it shifts each coefficient of a polynomial from monomial \(x^ry^s\) to \(x^{r+1}y^s\), so a zero product forces every coefficient to be zero. This argument does not cancel an element of \(R\). Thus \(i\) is injective, and \(ji=0\) by commutativity.

Suppose \(j(a,b)=0\), so \(xb=ya\). Set \(x=0\). Then \(y a(0,y)=0\) in \(R[y]\). Multiplication by \(y\) is injective by the same coefficient argument, so \(a(0,y)=0\). The polynomial \(a\) is consequently divisible by \(x\); write \(a=xc\). Substitution gives \(x(b-yc)=0\), and injectivity of multiplication by \(x\) gives \(b=yc\). Hence \((a,b)=i(c)\), proving exactness at \(\Lambda^2\). Uniqueness of \(c\) also follows from that injectivity.

The image of \(j\) is the ideal \((x,y)\): every value is in this ideal, while \(j(0,1)=x\) and \(j(-1,0)=y\). The kernel of \(\epsilon\) is the same ideal, since a polynomial with zero constant coefficient is the sum of its terms divisible by \(x\) and its remaining terms divisible by \(y\). Finally \(\epsilon\) is surjective on constants. This proves every position in (15). All three terms are finite free modules and hence projective, proving the resolution assertion. \(\square\)

Another common ordering uses the column \(c\mapsto(-yc,xc)\) and the row \((a,b)\mapsto xa+yb\). The signed change of coordinates \(L(a,b)=(b,-a)\), together with multiplication by \(-1\) on the leftmost \(\Lambda\) and identity on the next \(\Lambda\), identifies (15) with that resolution: \(Li(c)=-(-yc,xc)\), and \((x,y)L(a,b)=j(a,b)\). Thus the row accompanying the particular column in (16) has the required minus sign.

Proposition 6.2. With the projective coordinates of (15),

Proof. Let \(P\) be the complex in Proposition 6.1. Its augmentation to \(M[0]\) is a quasi-isomorphism. The projective Hom lemma in the third prerequisite proves \(\operatorname{Hom}_D(M,M[n])=\operatorname{Hom}_K(P,M[n])\). The complex \(\operatorname{Hom}_\Lambda(P,M)\) has terms \(M,M^2,M\) in degrees 0, 1, 2. All its differentials are zero, since precomposing with either row or column in (16) multiplies a component by \(x\) or \(y\), both of which act as zero on \(M\). Thus its cohomology gives (17).

Equivalently, for each of degrees 0, 1, 2, the indicated component is an arbitrary chain-map component to \(M[n]\). Its chain condition is automatic for the same reason, and every possible homotopy changes it by zero. In the other degrees there is no common nonzero degree of the source and target complexes. This directly checks both the chain-map and homotopy parts of the computation. In degree two, \(\lambda\in R\) is represented by the sole component

These computations establish Ext in the actual derived category, not a separately defined resolution group. \(\square\)

7. Explicit complexes representing every parameter

For \(\lambda\in R\), define the \(\Lambda\)-module

The indicated set is the image of a module map, hence a submodule. Evaluation \(\epsilon\) is \(\Lambda\)-linear when \(M\) has its specified quotient-module action. Let \([m,v]\) denote the class of \((m,v)\). Its defining relation is \([0,i(c)]=[\lambda\epsilon(c),0]\).

The map \(M\to E_\lambda\), \(m\mapsto[m,0]\), is injective. Indeed if \((m,0)=(-\lambda\epsilon(c),i(c))\), then \(i(c)=0\), so \(c=0\) by Proposition 6.1 and then \(m=0\). The map

is well-defined because \(ji=0\). If its value vanishes, \(v=i(c)\) for a unique \(c\), and \([m,v]=[m+\lambda\epsilon(c),0]\). Its kernel is therefore exactly the copy of \(M\). Its image is \((x,y)\), since the images of \([0,v]\) give every value of \(j\). We have proved the exact row

Define \(X_\lambda\) to be (20) in degrees \(-1,0\). Row (21) gives the fixed marks \(H^{-1}(X_\lambda)=M\), via the injection, and \(H^0(X_\lambda)=M\), via \(\epsilon\). There is no other cohomology.

Proposition 7.1. Under (17), the canonical marked class of \(X_\lambda\) is exactly \(\lambda\in R\).

Proof. Here \(U=M[1]\), \(V=M[0]\). The injection \(U\to X_\lambda\) has cone \(K_\lambda\) with terms

in degrees \(-2,-1,0\), with exactly the maps of (21). Its quotient comparison \(q:K_\lambda\to M[0]\) is \(\epsilon\) in degree zero, and is a quasi-isomorphism by exactness of (21). Its positive last projection \(p:K_\lambda\to M[2]\) is the identity in degree \(-2\). Thus the natural triangle of this short exact row, transported along \(q\), has last arrow \(Qp(Qq)^{-1}\). Its first two maps are the fixed cohomology inclusion and projection for \(X_\lambda\); the canonical-comparison argument of Theorem 5.1 identifies this last arrow with \(\kappa_{X_\lambda}\).

There is an actual chain map \(\ell:P\to K_\lambda\) with components

The square in degree \(-1\) commutes by (20). The square in degree \(-2\) commutes because \([0,i(c)]=[\lambda\epsilon(c),0]\), which is precisely the relation in (19). Also \(q\ell\) is the resolution augmentation \(P\to M[0]\). Therefore

The numerator is (18), so its projective coordinate is \(\lambda\), as claimed. This also proves directly that every parameter in (17) has the explicit representative (20). \(\square\)

There is a useful sign distinction. The augmented raw roof of the two-extension (21) is the same \(Qp(Qq)^{-1}\), hence has coordinate \(\lambda\). The normalized Yoneda class used in the fourth prerequisite multiplies that raw roof by \(\varepsilon_2=-1\). Its coordinate is \(-\lambda\). Thus the truncation class here is the negative of that normalized two-extension class. The chosen projective coordinate and the positive last projection fix the sign; neither changes the vanishing or orbit criteria.

For \(u,v\in R\), a pair of scalar cohomology maps from \(X_\lambda\) to \(X_\mu\) is compatible precisely when

An actual chain map already realizes it:

To check well-definedness, its value on the relation indexed by \(c\) in (19) is \((-u\lambda\epsilon(c),i(vc))\), which is the target relation indexed by \(vc\) exactly when (23) holds. Its differential square commutes because \(j(va,vb)=v j(a,b)\). Its maps on the marked cohomology objects are exactly \(u\) and \(v\). Theorem 4.1 says that all derived lifts of that same pair differ from (24) by a unique element of \(R^2\); these additional lifts need not have a chain-map representative on these particular two-term complexes.

8. Forgetting marks: the complete unit-orbit classification

Every \(\Lambda\)-linear endomorphism of \(M=R\) is multiplication by its value at 1, so \(\operatorname{End}_\Lambda(M)=R\). It is invertible exactly when that value is a unit of \(R\). Theorem 5.1 and Proposition 7.1 therefore prove the following classification.

Theorem 8.1. Objects of \(D(\Lambda)\) with cohomology zero outside degrees \(-1,0\), and with both cohomology objects isomorphic to \(M\), have:

• marked isomorphism classes parametrized by all \(\lambda\in R\);
• unmarked isomorphism classes parametrized by the orbits of \(R^\times\) acting on \(R\) by multiplication.

In particular, \(X_\lambda\simeq X_\mu\) if and only if \(\mu=w\lambda\) for some unit \(w\in R^\times\). The zero orbit represents the split object \(M[1]\oplus M[0]\).

Proof. A marked isomorphism must have \(u=v=1\), so (23) is \(\lambda=\mu\). The general marked classification proves there are no other marked classes. An unmarked isomorphism induces units \(u,v\) satisfying \(\mu v=u\lambda\), and therefore \(\mu=(uv^{-1})\lambda\). Conversely if \(\mu=w\lambda\), take \(u=w,v=1\) in (24). Its cohomology maps are units, so it is an isomorphism in the derived category. Every unit ratio occurs in this way, proving that both endpoint identifications have been accounted for. Finally the splitting criterion identifies the zero class, and multiplication by a unit cannot turn a nonzero element into zero. \(\square\)

For \(R=\mathbb Z\), the only units are \(1,-1\): an integer with an integer inverse has absolute value one. Thus an orbit is \(\{n,-n\}\), with a unique representative \(|n|\), and the isomorphism classes are

The representative 0 must be retained. The number of nonzero classes is infinite.

For a field \(k\), every nonzero element is a unit, and any two nonzero parameters differ by a unit ratio. There are exactly two classes: the zero class and the nonzero class. The same statement applies in characteristic two. If \(R\) is the zero ring, all modules and parameters in this construction are zero, so there is a single orbit and a single object class. Its unit group consists of its identity element \(0=1\), and the orbit formula still applies.

Over a general ring, the precise invariant is the unit-multiplication orbit of the element. The proof does not replace that orbit by a classification of ideals. This matters when zero divisors are present.

9. Four graded exercises with complete solutions

Exercise 1 — Foundation: degrees, shifts and vector spaces

Let \(X\) have cohomology only in degrees 2 and 5. Identify its class group, the class group for \(X[3]\), and the sign relating the two classes. When \(\mathcal A\) is the category of vector spaces over a field, prove that \(X\) splits and that any pair of its cohomology maps to another such object has a unique derived lift.

Solution. Here \(a=2,b=5,d=3\), so

The shift \(X[3]\) has its two possible cohomology degrees \(-1,2\), with the same respective cohomology objects under the standard shift identifications. Its gap is still 3, and its class belongs to the same degree-four Ext group. Proposition 2.1 makes its coordinate \(-\kappa_X\), because \((-1)^3=-1\).

In vector spaces every short exact sequence splits: choose a basis of the quotient, lift its basis vectors and extend linearly to a section. Section 1 of the third prerequisite uses those splittings degree by degree to contract any complex onto its cohomology. It proves \(K\simeq D\) and vanishing of every nonzero-degree Ext group; it applies to arbitrary vector-space dimensions. Thus this degree-four class is zero. Theorem 2.2 gives

For a second object \(Y\) with the same allowed degrees, its class also vanishes, so every pair \((\alpha,\beta)\) is compatible. Theorem 3.1 supplies a lift. Its ambiguity in (8) is \(\operatorname{Ext}^3(H^5X,H^2Y)=0\), proving uniqueness of the derived arrow with those prescribed maps. This uniqueness does not say that a chosen chain-level splitting of a representative complex is unique; it identifies the resulting derived morphism.

Exercise 2 — Intermediate: nonunique lifts on a split object

Let \(k\) be a field, \(\Lambda=k[x,y]\), \(M=k\) with \(x,y\) acting by zero, and \(X=M[1]\oplus M[0]\). Compute \(\operatorname{End}_D(X)\), its composition law and its automorphisms. Identify all automorphisms that preserve the two cohomology identifications.

Solution. The four biproduct Hom groups are

The first two are degree-zero full faithfulness, the third is (17), and the last is negative Ext vanishing. In the displayed summand order, each endomorphism is uniquely

For another such map \(f_{a',b',h'}\), biproduct composition gives

It is invertible exactly when \(a,b\ne0\), either by cohomology detection or by the explicit inverse

The automorphisms preserving both marks are \(f_{1,1,h}\), with law \(h'+h\) and inverse \(-h\). They form the additive group \(k^2\), so preserving both cohomology maps does not force a derived morphism to be the identity. On the naive split stalk complex, an ordinary chain map has no component from degree zero to degree minus one. The nonzero off-diagonal derived maps are instead represented after projective replacement or by roofs, as the resolution calculation (17) guarantees.

Exercise 3 — Advanced: construct the prescribed lift on the models

For any base ring \(R\), any \(\lambda,\mu,u,v\in R\), determine exactly when formula (24) defines a chain map \(X_\lambda\to X_\mu\). Prove that its cohomology maps are the requested scalars. Over \(\mathbb Z\), decide whether \(X_2\) and \(X_{-2}\) are isomorphic, construct an isomorphism with its actual endpoints, and decide whether they are isomorphic with both marks fixed.

Solution. The relation \((-\lambda\epsilon(c),i(c))\) maps to

In the target quotient this class equals \([ (\mu v-u\lambda)\epsilon(c),0]\), by relation (19) for \(\mu\). The copy of \(M\) is injective in \(E_\mu\), so this vanishes for all \(c\) if and only if \(\mu v-u\lambda=0\); necessity already follows by taking \(c=1\). This proves exactly the well-definedness condition (23), even if \(R\) has zero divisors. The differential square is \(j(v(a,b))=v j(a,b)\), proving the chain-map condition.

On the kernel copy of \(M\) in degree minus one, \([m,0]\) maps to \([um,0]\), so its lower cohomology map is \(u\). On the quotient of \(\Lambda\) by \((x,y)\) in degree zero, multiplication by \(v\) induces the scalar \(v\). Thus both requested endpoints are realized exactly. Necessity of the same relation for any derived lift also follows from Theorem 3.1 and Proposition 7.1; it is not merely a restriction of this particular chain-map formula.

For \(\lambda=2,\mu=-2\) over \(\mathbb Z\), choose \(u=-1,v=1\). Then \(u\lambda=\mu v=-2\). The map is

The same formula in the opposite direction is well-defined and is its inverse already as a chain map. It induces \(-1\) on \(H^{-1}\) and \(1\) on \(H^0\). Thus the two objects are unmarked isomorphic. With both marks fixed, compatibility would require \(u=v=1\) and hence \(2=-2\) in \(\mathbb Z\), which is false. They are not marked isomorphic. Their canonical classes are distinct; their normalized Yoneda classes have the opposite coordinates, so the conclusion is unchanged by keeping that separate convention.

Exercise 4 — Synthesis: enumerate orbits and marked automorphisms

Enumerate the unmarked object classes for \(R=\mathbb Z\), for a field, and for \(R=\mathbb Z/8\). In the last case distinguish the parameters 2 and 4, and decide whether 2 and 6 give the same class. For every \(\lambda\) over every base ring, compute the group of automorphisms of \(X_\lambda\) preserving both cohomology marks. Include the zero parameter.

Solution. Theorem 8.1 reduces the classification to multiplication by units. Over \(\mathbb Z\), the orbits are \(\{n,-n\}\), with one representative \(n\ge0\). Over a field there are two orbits, \(\{0\}\) and its nonzero elements. The zero parameter is the split object in both cases.

In \(\mathbb Z/8\), the units are \(1,3,5,7\). Each odd residue is a unit: \(1^2,3^2,5^2,7^2\) are all 1 modulo 8. An even residue cannot be a unit because its product with any residue is even and cannot equal 1. The four orbits are therefore

Multiplying 2 by the four units gives only 2 or 6, while multiplying 4 by any odd residue gives 4. Thus \(X_2\) and \(X_4\) are not isomorphic, and \(X_2\) and \(X_6\) are isomorphic. An explicit isomorphism for the latter uses \(u=3,v=1\) in (24), since \(3\cdot2=6\) and 3 is a unit. The zero orbit is separate because multiplying a nonzero parameter by an invertible element cannot annihilate it.

For arbitrary \(R\) and \(\lambda\), the class has adjacent cohomology degrees, so the marked automorphism formula (11) uses \(\operatorname{Ext}^1_\Lambda(M,M)=R^2\). Every such automorphism is uniquely

with product corresponding to addition in \(R^2\) and inverse given by \(-h\). The square-zero identity (10) proves both formulas. Conversely (8) shows that any endomorphism inducing the two identity maps has exactly this form. This includes \(\lambda=0\), the split object; its cohomology marks still permit these off-diagonal derived automorphisms. Over the zero ring, \(R^2=0\), and the group consists only of the identity of the zero object.