Draft lesson.

Higher extensions, roofs and Yoneda composition

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0). Level: advanced. Prerequisites: abelian categories, cohomological complexes, quasi-isomorphism roofs, smart truncations and the classification of short exact extensions.

A higher extension records several successive failures to split. Its middle objects need not be determined by its derived class, and two nonsplit steps can have zero composite. We construct every positive-degree derived morphism from an exact row, prove the precise equivalence relation on rows, and identify their splicing with composition. A sign calculation makes these statements compatible with the degree-one cone convention.

Conventions and prerequisites

Let \(\mathcal A\) be an abelian category and let \(A,B\) be objects of it. Complexes are cohomological:

Morphisms compose from right to left. Fix an ambient universe in which the localization and its Hom groups exist. The general results below require neither enough injectives nor enough projectives.

Cohomological bounds and truncation adjunctions, Sections 1–4 proves the incoming roof equality test, smart-cut comparisons and heart full faithfulness. Triangles after quasi-isomorphism localization, Sections 2 and 6 proves exact Hom sequences and the positive cone boundary of a short exact row. Short exact extensions and the Baer group, Sections 1–4 proves degree-one classification, its splitting criterion, arbitrary endpoint pullbacks and pushouts, and their exactness by universal properties. We use these complete proofs as prerequisites.

An n-extension of B by A, for \(n\geq1\), is a specified exact row

A morphism of such extensions is a commuting row map with the identity on each endpoint. Define equivalence by the equivalence relation generated by these row maps, allowing either direction. Section 3 will prove the stronger common-refinement description. In degree one a row map is an isomorphism in the middle; that assertion fails in higher degree.

1. The augmented roof and its normalization

Form \(K_e\) by putting \(A\) in degree \(-n\), \(E_{n-1}\) in degree \(1-n\), through \(E_0\) in degree zero, with the row maps as differentials. Define

All unspecified components are zero. Consecutive row maps compose to zero, so this is a complex. The maps are chain maps: \(p\) kills the last incoming differential, while \(f_e\) has no incoming component from degree \(-n-1\). Exactness gives

Thus \(s_e\) is a quasi-isomorphism. Write \(Q\) for passage to the derived category and set

In particular the actual roof for the normalized class has numerator \(\varepsilon_n f_e\). The first four signs are \(+1,-1,-1,+1\). In degree one \(K_e\) is precisely the cone of \(i\), so \(\varepsilon_1=1\) and \(\delta(e)\) is the positive boundary used in the prerequisite. Section 4 will explain the higher signs by an explicit composition roof.

For a commuting row map with arbitrary endpoint maps \(a:A\to A'\) and \(b:B\to B'\), its augmented chain map \(k:K_e\to K_{e'}\) satisfies

Inverting the two denominator maps gives \(\delta(e')b=a[n]\delta(e)\). In particular endpoint-preserving row maps, and therefore equivalent extensions, have the same class. This argument does not require \(k\) itself to be a quasi-isomorphism when the endpoints change.

2. Realizing every derived morphism

Theorem 2.1. For every \(n\geq1\) and every \(\alpha:B\to A[n]\), there is an n-extension with \(\delta(e)=\alpha\).

Proof. Choose a roof \(B\xleftarrow{q}U\xrightarrow{f}A[n]\). Smart upper truncation at zero and then smart lower truncation at \(-n\) give a roof with terms only in \([-n,0]\). The upper cut is a quasi-isomorphism because \(U\) has the cohomology of \(B[0]\). The lower cut is one for the same reason. Its boundary term is the quotient by \(\operatorname{im}d_U^{-n-1}\); the chain-map equations make both numerator and denominator descend through that quotient. Restricting at the upper cut and descending at the lower cut leave the same derived arrow, by the prerequisite's roof comparison.

Denote the resulting complex by \(L\), its numerator component by \(f:L^{-n}\to A\), and its augmentation still by \(q:L^0\to B\). Exactness of its augmented row says that

is exact. In particular \(d\) is monic. Put \(g=\varepsilon_n f\) and form the pushout

The relation is \(jd=ig\). To verify exactness at this modified end, let \(C\) be the cokernel of \(d\). Then \(0\to L^{-n}\xrightarrow d L^{1-n}\to C\to0\) is short exact. The explicit kernel-and-cokernel proof in the degree-one prerequisite, applied to this row and \(g\), proves \(0\to A\xrightarrow i E_{n-1}\to C\to0\). This is an arbitrary pushout of a monomorphism, so no assumption on \(g\) is needed. For \(n=1\), \(C=B\). For \(n\geq2\), exactness of (2.1) identifies \(C\) with the image of the next differential. Composing its epimorphism from \(E_{n-1}\) with that image inclusion, and retaining all the remaining maps in (2.1), gives

At the modified object the kernel is \(A\), since the map factors as its quotient onto \(C\) followed by a monomorphism. At the next object its image is the old image \(C\), hence the old kernel. All subsequent positions retain their previous exactness. This proves exactness of the entire row, including the case \(n=1\).

There is a chain map \(k:L\to K_e\) with components \(g\) in degree \(-n\), \(j\) in degree \(1-n\), and identities in the remaining degrees. The equation \(jd=ig\) checks its first square; the definition through \(C\) checks the next one; the others are unchanged. It obeys

For \(n=1\), the augmentation equation at degree zero is the pushout quotient equation. Both \(L\) and \(K_e\) have only degree-zero cohomology, and (2.4) identifies the map there with the identity of \(B\). Thus \(k\) is a quasi-isomorphism. Equation (2.4) proves equality of the original roof with the normalized extension roof, so \(\delta(e)=\alpha\). \(\square\)

We will also use the same construction with a raw numerator \(g:L^{-n}\to A\), omitting the first multiplication by \(\varepsilon_n\). Its augmented roof then represents the raw class \(Q(g)Q(q)^{-1}\); its normalized class is \(\varepsilon_n\) times that arrow.

3. Equality of classes and the homotopy correction

Theorem 3.1. Two n-extensions have the same normalized class if and only if they are equivalent. When their classes agree, there is a single n-extension mapping to both, with the identity on each endpoint.

Proof. Section 1 proves one direction. For the converse equality of normalized classes implies equality of raw classes, since their common multiplier \(\varepsilon_n\) is invertible. The roof equality test supplies a complex \(U\), quasi-isomorphisms

and equalities \(s_et=s_{e'}t'\) and \(f_et=f_{e'}t'\) in the homotopy category. Smart upper truncation at zero and lower truncation at \(-n\) give the same situation with \(U\) supported in \([-n,0]\). The maps to the augmented complexes descend at the lower cut because those targets have no term below \(-n\). The numerator homotopy descends too: its only potentially relevant component has source degree \(1-n\), which is unaffected by that quotient.

Maps from this \(U\) to \(B[0]\) have no nonzero homotopies: the only possible component would have source \(U^1=0\). Thus their augmentation maps are equal as actual chain maps; call the common map \(q\). Equality of the numerator maps gives an actual map \(h:U^{1-n}\to A\) such that

This is exactly the cochain homotopy equation; the target differential is zero and only this component can contribute.

Apply the raw-numerator construction at the end of Section 2 to \((U,q,g)\), obtaining an extension \(e''\). Its new first middle object is the cokernel of \((g,-d_U^{-n})\). To map it to the first middle object of \(e\), use the pair

This kills the cokernel relation because \(t^{1-n}d_U^{-n}=ig\). It therefore descends. Together with the remaining components of \(t\), the identity of \(A\), and the identity of \(B\), it is a row map \(e''\to e\).

For the map into \(e'\), the necessary pair is instead

Indeed its composite with \(d_U^{-n}\) is \(i'g'+i'hd_U^{-n}=i'g\), by (3.2). The relation is killed again. The added term is killed by the next differential of the row, because that differential kills \(i'\). For \(n=1\), its augmentation is killed for the same reason, \(p'i'=0\). Thus all further squares commute using the unmodified components of \(t'\), and the last square uses the strict equality of augmentations \(q\). This gives \(e''\to e'\) with both endpoint identities.

The resulting common predecessor proves equivalence. Theorem 2.1 supplies every derived class, so these classes give a bijection between extension equivalence classes and \(\operatorname{Ext}^n(B,A)\). The common-predecessor condition is consequently itself an equivalence relation: it is equivalent to equality in this Hom group. No assertion that all its middle arrows are isomorphisms is needed. \(\square\)

The homotopy term in (3.4) matters. Equality of derived morphisms need not provide strictly equal numerators on the first representatives. Replacing homotopy equality by strict equality without this correction would omit precisely the map needed to construct the extension comparison.

4. Splicing and the sign of composition

Let \(u\) be a p-extension of \(C\) by \(A\), and \(v\) a q-extension of \(B\) by \(C\), where \(p,q\geq1\). Write their adjacent ends as \(E_0\xrightarrow{r}C\xrightarrow{j}F_{q-1}\). Their splice \(u\star v\) is the row obtained by replacing these two arrows and the object \(C\) with \(jr:E_0\to F_{q-1}\). It is a \((p+q)\)-extension: its kernel at \(E_0\) is \(\ker r\), by monicity of \(j\), and its image in \(F_{q-1}\) is \(j(C)\), by epicity of \(r\); the other positions are unchanged.

Theorem 4.1. With the normalization of (1.3),

The right-hand side is composition \(B\to C[q]\to A[p+q]\).

Proof. First compute with the raw numerators. The chain map \(s_u[q]:K_u[q]\to C[q]\) is degreewise epic: its only nonzero target term is \(C\) in degree \(-q\), where it is \(r\). Form the termwise pullback of this map along \(f_v:K_v\to C[q]\), and denote the resulting complex by \(P\). Its maps \(\pi:P\to K_u[q]\) and \(\rho:P\to K_v\) satisfy

In degree \(-q\), the pullback of \(E_0\to C\) and \(1_C\) identifies canonically with \(E_0\). Below that degree its terms are those of \(K_u[q]\); above it they are those of \(K_v\). Thus its left-block differentials are \((-1)^q\) times the row maps of \(u\), its seam differential is \(jr\), and its right-block differentials are the row maps of \(v\). This description checks (4.2) as actual chain maps. Multiplying a differential by a sign does not change its image or kernel. Exactness of the spliced row therefore proves that \(P\) has only degree-zero cohomology \(B\); the map \(\rho\) induces the identity there. Hence \(\rho\) is a quasi-isomorphism and (4.2) is a valid composition roof. It gives

Compare this roof with \(K_{u\star v}\), whose row differentials have no inserted signs. There is a chain isomorphism \(J:K_{u\star v}\to P\) which is the identity at \(E_0\) and throughout the right block. At a left term r steps before \(E_0\), its component is \((-1)^{qr}\) times the identity. This satisfies every left differential equation since consecutive factors differ by \((-1)^q\), and the seam has identity factors on both sides. At \(A\), r is p, so

Consequently \(\lambda(u)[q]\lambda(v)=(-1)^{pq}\lambda(u\star v)\). Finally

because \(\binom{p+q}{2}=\binom p2+\binom q2+pq\). Multiplying the raw formula by the two normalization factors proves (4.1). This derives the signs from the actual shifted differentials and the roof comparison. \(\square\)

Splicing is associative on specified rows: both parenthesizations retain the same terms and the same seam composites. Its class is independent of the representatives by (4.1) and Theorem 3.1. The composition is bilinear on extension classes because composition and shift are additive in the derived category; in particular zero in either factor gives zero in the product. This does not say that a zero product has a split constituent.

Corollary 4.2 (degree-one factorization). Every \(\alpha:B\to A[n]\), \(n\geq1\), factors as

For \(n\geq2\), realize \(\alpha\) by Theorem 2.1, put \(C=\ker(E_0\to B)\), and take \(\beta\) to be the class of \(0\to C\to E_0\to B\to0\), with \(\gamma\) the class of the remaining \((n-1)\)-extension. Theorem 4.1 gives exactly (4.6). For \(n=1\), choose \(C=A\), \(\beta=\alpha\), and \(\gamma=1_A\). Thus the positive-degree statement includes its smallest degree without an undefined negative-length extension.

5. Zero classes and the lifting diagram

For \(n\geq2\), decompose an n-extension into its front \((n-1)\)-extension \(u\) of \(C\) by \(A\) and its last short exact row

Theorem 5.1. For \(n\geq2\), the following conditions are equivalent: the class of the n-extension is zero; its front class is the restriction of a class in \(\operatorname{Ext}^{n-1}(E_0,A)\); there is an \((n-1)\)-extension \(w\) of \(E_0\) by \(A\) and an actual commuting row map from the specified front extension \(u\) to \(w\), with endpoint maps \(1_A,j\). They are also equivalent to the existence of a lifted extension whose pullback along j is equivalent to u. All the diagram rows are exact. The actual row-map assertion chooses a suitable representative of the lift; it does not assert that every representative admits that row map.

For an r-extension ending in an object \(Y\), its pullback along \(b:Y'\to Y\) is formed only at its last middle object. The other terms map to the new last middle object by their old map and the zero map to \(Y'\). Its new last short exact piece is the ordinary pullback of \(0\to\ker(F_0\to Y)\to F_0\to Y\to0\). The degree-one prerequisite proves that this pullback is exact, with the same kernel. Therefore the full pulled-back row is exact. Its natural row map and (1.4) prove \(\delta(b^*w)=\delta(w)b\).

Proof of the theorem. Applying \(\operatorname{Hom}_D(-,A[n-1])\) to the distinguished triangle of (5.1) gives the exact segment

The last map sends \(\gamma\) to \(\gamma[1]\delta(v)\), up to the rotation sign, which does not change its kernel. By Theorem 4.1, its value at \(\gamma=\delta(u)\) vanishes exactly when the original extension class vanishes. Exactness of (5.2) is therefore precisely the existence of \(\theta:E_0\to A[n-1]\) with \(\theta j=\delta(u)\). Theorem 2.1 realizes \(\theta\) by \(w\); the pullback formula just proved and Theorem 3.1 say that \(j^*w\) and \(u\) are equivalent.

We now prove the stronger assertion about the fixed front row. Suppose the class is zero. Then \(Q(f_e)=0\). The outgoing equality test in Derived morphisms through quasi-isomorphism roofs, Section 4 supplies a quasi-isomorphism \(a:A[n]\to V\) and a homotopy

Apply smart lower truncation to V at \(-n\). Its boundary quotient kills the term \(d_V^{-n-1}h^{-n}\) in the degree-minus-n equation, and all other surviving homotopy components descend to that quotient or retain their targets. Next apply smart upper truncation at \(-1\). The equation in degree zero is \(d_V^{-1}h^0=0\), so \(h^0:E_0\to V^{-1}\) factors through the new cycle term. Thus the homotopy survives both cuts. Both target cuts are quasi-isomorphisms on the relevant side, because V has cohomology only in degree \(-n\). We obtain an exact row

Here a identifies A with the first kernel: there is no term below \(-n\), and its map on \(H^{-n}\) is an isomorphism. The last map d is epic because the last cohomology vanishes and there is no term above \(-1\).

For \(1\leq r\leq n-1\), put

Write \(q: E_1\twoheadrightarrow C\) for the final quotient of the front row, so its last differential in \(K_e\) is \(jq\). The homotopy equations give

In the middle equation d means the corresponding differential of V. For example \(0=d_Vh^{-r}+h^{-r+1}d_{K_e}^{-r}\), and the alternating factors in (5.5) turn it into the stated commuting square.

Form \(P=V^{-2}\times_{V^{-1}}E_0\) using the epic map d and the map z. Its projection onto \(E_0\) is epic and its kernel is \(\ker d\), by the complete short-exact pullback proof in the degree-one prerequisite. Consequently

is an \((n-1)\)-extension. At \(P\), its incoming image is the kernel \(\ker d\); the earlier exact positions are those of (5.4). When \(n=2\), (5.7) means \(0\to A\to P\to E_0\to0\), with \(A\to P\) induced by \((a,0)\). Equation (5.6) gives the row map from the fixed u: use \(b_r\) for \(r\geq2\), and use \((b_1,jq):E_1\to P\) at its last middle object. Its two pullback components agree by the last equation of (5.6). The next-to-last square has second component zero because consecutive original row maps compose to zero. The first square follows from \(b_{n-1}i=a\); for \(n=2\) it follows from that equation and \(qi=0\). The last square is the projection equation into \(E_0\). Thus all squares commute, with endpoint maps \(1_A,j\).

Conversely, suppose there is any actual row map from u into an \((n-1)\)-extension w ending in \(E_0\). Regard its middle objects as \(V^{-n},\ldots,V^{-2}\), append \(E_0\) in degree \(-1\), and use its row maps as differentials. The first injection \(a:A[n]\to V\) is a quasi-isomorphism by exactness of w. If its middle comparison maps are \(b_r:E_r\to V^{-r-1}\), define

with all other components zero. The first square gives \(h^{1-n}i=a\). At every interior degree the two homotopy summands cancel because adjacent coefficients have opposite signs. At degree minus one they are \((-1)^{n-2}jq\) and \((-1)^{n-1}jq\), so cancel again. At degree zero the target differential vanishes. These checks prove (5.3). Therefore \(Q(f_e)=0\), since \(Q(a)\) is invertible, and the extension class is zero. This proves the actual-diagram equivalence without replacing the fixed front row or assuming enough injectives. \(\square\)

In degree two, the front and lifted extensions have degree one. Their equality on pullback gives an actual endpoint-preserving isomorphism \(u\simeq j^*w\). Composing it with the pullback map gives a diagram with the specified front row itself:

Its middle vertical map is monic and its cokernel is \(B\). Indeed the pullback description identifies it with the inverse image of \(C\subset E_0\); its quotient is the cokernel of \(C\to E_0\), namely \(B\). Thus the two rows together with their middle and right columns form the exact lifting diagram. In higher degree, (5.4)–(5.7) construct the analogous actual row map from the specified front row. If one instead fixes a representative of a lift in advance, Theorem 3.1 guarantees equivalence with its pullback, with a common predecessor, and does not turn that equivalence into termwise isomorphism.

The degree-one endpoint. For \(n=1\), the original row is \(0\to A\xrightarrow{i}E_0\to B\to0\), and its class is zero exactly when it splits, by the degree-one prerequisite. The appropriate degree-zero version of (5.2) is

Vanishing means that \(1_A\) lifts to a retraction \(r:E_0\to A\) with \(ri=1_A\). Such a retraction supplies a section of the quotient: if \(k:B\to E_0\) is the unique factorization of \(1_{E_0}-ir\) through the cokernel, then \(pk=1_B\) because \(p(1-ir)=p\) and \(p\) is epic. Thus (5.10) is exactly the splitting test. Treating its lift as an empty positive-length extension would be incorrect. For example a split row with \(A=B=k\) has \(E_0=k\oplus k\); the supposed exact empty row \(0\to A\to E_0\to0\) would require \(A\simeq E_0\), which is false for these finite-dimensional spaces.

There is also an exact denominator criterion in every positive degree. The class vanishes if and only if some incoming quasi-isomorphism \(t:U\to K_e\) makes \(f_et\) null-homotopic. This is the zero test for roofs, applied to \(Q(f_e)\); the invertible factor \(Q(s_e)\) and the sign \(\varepsilon_n\) do not affect vanishing. Smart cuts may again make \(U\) supported in \([-n,0]\). The homotopy then has the single relevant equation

One cannot generally take \(t=1_{K_e}\). Exercise 4 gives a zero class for which the raw projection is not null-homotopic on that original complex.

6. Four graded exercises with complete solutions

Exercise 1 — Two nonsplit steps can compose to zero (introductory)

Let p be a prime. Splice two copies of \(0\to\mathbb Z/p\to\mathbb Z/p^2\to\mathbb Z/p\to0\), with injection multiplication by p and quotient reduction. Prove both short rows are nonsplit, but their two-extension class is zero by constructing the diagram (5.9).

Solution. An element killed by p in \(\mathbb Z/p^2\) is a multiple of p, and so maps to zero under reduction modulo p. Hence the quotient has no section from \(\mathbb Z/p\), and each row is nonsplit. Put \(W=\mathbb Z/p^3\). The bottom row of (5.9) is

It is exact: the kernel of reduction is generated by \(p^2\), of order p. The top middle vertical map \(\mathbb Z/p^2\to W\) is multiplication by p, and the right vertical map \(\mathbb Z/p\to\mathbb Z/p^2\) is also multiplication by p. Both are well defined and injective. The left square reads \(p\cdot p=p^2\); the right square sends a residue x on either route to px modulo \(p^2\). All endpoint identifications are therefore the required ones. The quotient of \(W\) by the image \(pW\) is \(\mathbb Z/p\), which checks the middle vertical exact row as well. Theorem 5.1 proves that the product, hence the spliced two-extension, has zero class. No constituent splitting was used or obtained. \(\square\)

Exercise 2 — Equivalence is weaker than middle isomorphism (intermediate)

Fix a field k, let \(A=B=k\), and for a finite-dimensional vector space V consider

with maps \(a\mapsto(a,0)\), \((a,v)\mapsto(v,0)\), and \((v,b)\mapsto b\). Check exactness, compare \(e_V\) with \(e_0\), and explain why extension equivalence cannot mean an isomorphism at every middle position.

Solution. The first arrow is monic. The kernel of the middle arrow is \(A\oplus0\), exactly its image. The middle image is \(V\oplus0\), exactly the kernel of the last projection, which is epic. Thus the row is exact. There is an endpoint-preserving row map \(e_V\to e_0\): project \(A\oplus V\) onto A and \(V\oplus B\) onto B. The middle square is zero on both routes. Conversely the inclusions of A and B give a row map \(e_0\to e_V\); its middle square is again zero. Thus \(e_V\) and \(e_0\) are equivalent by explicit row maps. The last short row of \(e_0\) splits, so its class is zero, and hence so is that of \(e_V\). If \(V\ne0\), its middle dimensions are \(1+\dim V\), whereas those of \(e_0\) are 1. Corresponding middle objects are not isomorphic. The bijection of Theorem 3.1 classifies the correct row equivalence, rather than these middle objects separately. \(\square\)

Exercise 3 — All powers over the dual numbers (advanced)

Let \(R=k[t]/(t^2)\) for a field k and let \(S=R/(t)\). Let \(\eta\in\operatorname{Ext}^1_R(S,S)\) be the normalized class of \(0\to S\xrightarrow{t}R\to S\to0\). Compute all positive-degree groups, the raw and normalized classes of the repeated splice, and the powers of \(\eta\), checking the sign independently on a projective resolution.

Solution. Take the bounded-above projective complex P with \(P^j=R\) for \(j\leq0\), zero otherwise, differential multiplication by t for \(j<0\), and augmentation \(P^0\to S\) the quotient. Its consecutive differentials compose to zero. In every negative degree, the kernel and image of multiplication by t are both the ideal \((t)\), so its augmentation is a quasi-isomorphism. Hereditary categories and cohomology splittings, Lemma 3.1 proves that this bounded-above projective complex computes the actual derived Hom, even though it continues indefinitely to the left.

A chain map \(P\to S[n]\) has the single component \(P^{-n}\to S\). Any R-linear map there is allowed because t acts by zero on S. A homotopy changes that component by a map followed by multiplication by t, which is also zero. Thus \(\operatorname{Ext}^n_R(S,S)\simeq k\) for every \(n\geq0\), with quotient \(P^{-n}\to S\) as a specified basis.

The n-fold splice has n middle objects R, injection t, interior differentials t and final quotient to S. It is exact by the same kernel/image calculation. Its augmentation roof is compared to P by identities in degrees \(1-n\) through zero and quotient \(R\to S\) in degree \(-n\); the first chain square says \(t\cdot(\bmod t)=t\). Hence its raw class has coordinate 1 in the preceding basis, and its normalized class has coordinate \(\varepsilon_n\).

For an independent composition check, lift \(\eta\) to the chain map \(u:P\to P[1]\) with \(u^j=(-1)^{j+1}1_R\) for \(j\leq-1\) and zero in degree zero. Below degree minus one the chain-map equation is \(-t u^j=u^{j+1}t\), which the alternating signs satisfy; at degree minus one both sides are zero. Composing with the shifted augmentation gives quotient in degree minus one, so u represents \(\eta\). In the n-fold shifted composite, its degree-minus-n component is the product of the signs at degrees \(-n,-n+1,\ldots,-1\), namely

Thus \(\eta^n\) has coordinate \(\varepsilon_n\), in agreement with Theorem 4.1 and the normalized n-extension. It is nonzero for every n. This also holds in characteristic two, where all signs equal 1. When the characteristic differs from two, \(\eta^2\ne0\) shows that this Ext algebra is not forced to be graded commutative merely because R is a commutative ring. \(\square\)

Exercise 4 — A zero class that needs a new denominator (advanced)

For the two-extension of Exercise 1, prove that \(f_e:K_e\to A[2]\) is not null-homotopic on its original augmented complex. Construct an actual quasi-isomorphism \(q:L\to K_e\) for which \(f_eq=0\) as a chain map. Distinguish zero in the homotopy category from zero in the derived category.

Solution. The augmented complex has \(A=\mathbb Z/p\) in degree minus two, \(E_1=\mathbb Z/p^2\) in degree minus one, and \(E_0=\mathbb Z/p^2\) in degree zero. Its differentials are the injection p and multiplication by p. A homotopy of \(f_e\) to zero would require a map \(h:E_1\to A\) with \(hi=1_A\). Every map to A kills p times any element, so \(hi=0\), a contradiction. Equivalently that retraction would split the first nonsplit short row.

Let L have \(\mathbb Z/p^2\) in degree minus one and \(\mathbb Z/p^3\) in degree zero, with differential multiplication by p, and zero in every other degree. This differential is monic; its cokernel is \(\mathbb Z/p\). Define \(q^{-1}=1_{\mathbb Z/p^2}\) and \(q^0\) to be reduction modulo \(p^2\). The chain square is exactly

The map in degree minus two has zero source, so its chain square is also satisfied. Both L and \(K_e\) have only degree-zero cohomology \(\mathbb Z/p\), and q induces the identity there. Thus q is an actual quasi-isomorphism. Since \(L^{-2}=0\), the composite \(f_eq\) is the zero chain map. Therefore the raw roof, and hence its normalized class, is zero in the derived category, although its original numerator is a nonzero morphism in the homotopy category. This is why the incoming denominator in (5.11) is essential. \(\square\)

Freely accessible references

• The Stacks Project, Ext groups, Section 13.27: general derived Ext, Yoneda extensions, their equivalence and composition. The arguments above spell out the pushout exactness, common-refinement homotopy correction and signed composition under (0.1).
• The Stacks Project, Extensions, Section 12.6: endpoint maps, short exact extensions and their pullback/pushout operations.
• The linked course prerequisites contain the complete localization, truncation, exact Hom and degree-one classification proofs used above.