Draft lesson.

Triangles after quasi-isomorphism localization

Original text public domain (CC0).

A derived morphism can require a replacement of its source. A triangle must accommodate that replacement, and an octahedron must accommodate two composable replacements at once. We prove that the roof category has a triangulated structure, with all the required comparisons. We then construct the triangles of short exact sequences and explain what they remember beyond cohomology.

Let \(\mathcal A\) be an abelian category in an ambient Grothendieck universe in which its objects and arrows form sets and \(\mathbb Z\) is small. Use the homotopy category \(K=K(\mathcal A)\), its quasi-isomorphisms \(S\), and its ordinary additive localization \(Q: K\to D\). Derived morphisms through quasi-isomorphism roofs proves both Ore squares and cancellations, the complete associative roof category, incoming filtered colimits, outgoing representatives and equality tests. No localization theorem for triangles is assumed from it.

Chain homotopies, mapping cones and triangles, Sections 2–6, proves the triangulated structure on \(K\), both exact Hom sequences, the explicit octahedron and cone cohomology exactness. Its convention is \(C_f^n=Y^n\oplus X^{n+1}\), with differential \((d_Yy+fx,-d_Xx)\) and positive last projection \(p_f: C_f\to X[1]\). Exact complexes, signs and lifting covers, Sections 3–4, supplies epimorphic covering lifts and the five-term test. Coordinates below mean biproduct matrices. When a cycle or boundary in a general abelian category needs a lift, we explicitly allow an epimorphic change of test object; projective objects are not assumed.

1. Shift and the proposed triangles

Define shift on a roof by

Shifts preserve quasi-isomorphisms and common refinements, so this is well defined. A shifted Ore square is an Ore square; the composition formula therefore proves preservation of composition. Common-denominator addition proves additivity. The inverse is shift by \(-r\), and successive shifts add their indices with the differential signs already checked in \(K\). Thus \(D\) has the additive auto-equivalence \([1]\), and \(Q(X[r])=(QX)[r]\), also on arrows.

Call a triangle in \(D\) distinguished if it is isomorphic, as a triangle including its last map, to the image of a distinguished triangle of \(K\). Equivalently it is isomorphic to an image cone triangle

The equivalence follows because the preceding lesson proves that every distinguished triangle of \(K\) is isomorphic to a cone triangle. This definition alone is not yet the triangulated localization theorem. We verify its axioms below.

2. Exact Hom sequences before the morphism axiom

We first establish exactness directly from the roof descriptions. This avoids using an unproved triangulated structure to compare third vertices.

For a fixed target \(T\), let \(O_T\) have objects \(t: T\to V\) in \(S\). An arrow from \(t\) to \(u: T\to W\) is \(a: V\to W\) with \(at=u\) in \(K\). Two-out-of-three puts every such \(a\) in \(S\).

Lemma 2.1. The category \(O_T\) is filtered, and naturally in \(X\),

The representative \((t,f)\) on the right is sent to \((Qt)^{-1}Qf\).

Proof. Identity is an object. Apply the outgoing Ore square to two denominators from \(T\). It produces a common outgoing denominator; the two comparison arrows are in \(S\), one by the square condition and the other by two-out-of-three. If \(at=bt\), outgoing cancellation produces \(c\in S\) with \(ca=cb\). Its composite with the target denominator is still a denominator. These prove all three filtered conditions.

Every derived arrow has an outgoing representative, giving surjectivity in (2.1). For injectivity take two representatives and move them to a common denominator \(u: T\to W\). If their images in \(D\) agree, invertibility of \(Qu\) says that the two resulting ordinary arrows \(X\to W\) agree after applying \(Q\). The outgoing equality test gives a further denominator \(c: W\to W'\) that makes them equal in \(K\). They are consequently equal in the filtered colimit. Conversely such an equality plainly gives the same derived arrow. Finite common refinements show that addition of representatives is preserved, so this is an abelian-group isomorphism. Precomposition in \(X\) commutes with every transition, proving naturality. \(\square\)

Lemma 2.2. Applying either \(\operatorname{Hom}_D(T,-)\) or \(\operatorname{Hom}_D(-,T)\) to a distinguished triangle gives an exact sequence of abelian groups continued in both directions by shifts and rotations.

Proof. For an image triangle, the first sequence is the filtered incoming-denominator colimit of the exact sequences \(\operatorname{Hom}_K(U,-)\), indexed by denominators \(U\to T\). The second is, by Lemma 2.1, the filtered outgoing-denominator colimit of the exact sequences \(\operatorname{Hom}_K(-,V)\), indexed by \(T\to V\). Naturality of the Hom sequences makes each a diagram of complexes of abelian groups.

Here is the exactness step for either diagram. Represent a class in a middle group at one index. If its next image is zero in the colimit, the filtered equality relation supplies a later index at which that image is already zero. Exactness at that index supplies a preceding-group preimage. It represents a preimage of the original class in the colimit. Every preceding image is killed by the next map because that is true at each index. This proves both containments at every position. The signs of rotations change no kernel or image. Finally an isomorphism of triangles gives an isomorphism of these sequences, proving the assertion for every distinguished triangle. \(\square\)

Lemma 2.3. In a morphism between distinguished triangles, if its first two components are isomorphisms, so is its third component.

Proof. Apply \(\operatorname{Hom}_D(T,-)\). In five successive exact positions, with the third-vertex group in the middle, the four outside maps are isomorphisms: they are the first and second components and their shifts. The abelian-group five-term test makes the middle map an isomorphism. If \(c: C\to C'\) is that component, surjectivity with \(T=C'\) gives \(e\) with \(ce=1\). Injectivity with \(T=C\) gives \(ec=1\), since \(c(ec-1)=0\). Thus \(c\) is invertible. Rotation gives the assertion when any other two corresponding vertices are isomorphic. \(\square\)

3. Clear a square and complete its third vertex

Lemma 3.1. Suppose \(f_0: X_0\to Y_0\) and \(f_1: X_1\to Y_1\) are arrows of \(K\), and \(\alpha: QX_0\to QX_1\), \(\beta: QY_0\to QY_1\) satisfy \(\beta Qf_0=Qf_1\alpha\). There is an arrow \(f': X'\to Y'\) of \(K\) and two commuting squares in \(K\),

Proof. Choose an incoming roof for \(\beta\), with denominator \(y: Y'\to Y_0\) and numerator \(b\). Incoming Ore for \(f_0\) and \(y\) gives \(x_0: X''\to X_0\) in \(S\) and \(k: X''\to Y'\) with \(f_0x_0=yk\). Represent \(\alpha Qx_0\) by a roof with denominator \(z: X'''\to X''\) in \(S\) and numerator \(a_0: X'''\to X_1\). Put \(x_*=x_0z\) and \(f_*=kz\). We then have

The last equation uses the given derived square and the first two equations. The incoming equality test supplies \(c: X'\to X'''\) in \(S\) with \((bf_*-f_1a_0)c=0\) in \(K\). Set \(x=x_*c\), \(f'=f_*c\), and \(a=a_0c\). These satisfy every equation in (3.1). Notice that only \(x,y\) must be denominators; the given \(\alpha,\beta\) can be arbitrary derived maps. \(\square\)

Theorem 3.2 (triangle morphisms). A commuting square between the first arrows of two distinguished triangles in \(D\) extends to a morphism of triangles.

Proof. Transport through their defining triangle isomorphisms to reduce to images of cone triangles for \(f_0,f_1\). Lemma 3.1 gives the intermediate cone triangle for \(f'\). The cone completion in \(K\) supplies triangle morphisms to the two original cone triangles, with components

The first two components of the first morphism are quasi-isomorphisms. Apply the cone cohomology exact sequences in \(K\): in five successive positions every outside comparison is an isomorphism, so each \(H^n(c_0)\) is an isomorphism. Thus \(c_0\in S\). Its image is invertible, and

completes the derived square. Indeed the composite of the second image triangle morphism with the inverse of the first has first two components exactly \(\alpha,\beta\) by (3.1), and third component (3.4). Its last square also commutes, because both original triangle morphisms respect the positive last cone projections. Transport back to the given triangles. No uniqueness of \(\gamma\) is asserted. \(\square\)

4. The four axioms, including prescribed octahedra

Theorem 4.1. The distinguished triangles of Section 1 make \(D\) a triangulated category, and \(Q\) is an exact functor with its strict shift identification.

Proof of the first three axioms. Closure under triangle isomorphisms is part of the definition. The identity triangle is an image identity triangle of \(K\). Given a roof \(\varphi=Qf(Qs)^{-1}: QX\to QY\), where \(s: U\to X\) is in \(S\), transport the image cone triangle of \(f: U\to Y\) along the source isomorphism \(Qs\). Its arrows are

This proves triangle existence for every derived arrow. Rotation commutes with \(Q\), including its minus sign: the rotated last map is \(-\varphi[1]\). Both directions of rotation hold because they hold in \(K\) and commute with triangle isomorphisms. The same argument gives shifted triangles with last map multiplied by \((-1)^r\). The morphism axiom is Theorem 3.2.

Proof of the octahedron axiom. Take arbitrary composable derived arrows \(\varphi: QX\to QY\), \(\psi: QY\to QZ\), with any prescribed distinguished triangles for \(\varphi,\psi,\psi\varphi\). Write incoming roofs

Incoming Ore gives \(c: W\to U\) in \(S\) and \(d: W\to V\) with \(ac=td\). The actual composable pair in \(K\) is now \(W\xrightarrow dV\xrightarrow bZ\). Its vertices compare to the original vertices by the isomorphisms \(Q(sc),Qt,1_{QZ}\). The equations \(Qt\,Qd=\varphi Q(sc)\) and \(Qb=\psi Qt\) show that this is the required simultaneous replacement, and their composite gives \(Q(bd)=\psi\varphi Q(sc)\).

Use the explicit octahedron in \(K\) for \(d,b\). Its third vertices are \(C_d,C_{bd},C_b\), and its fourth row is the distinguished triangle \(C_d\xrightarrow uC_{bd}\xrightarrow vC_b\xrightarrow wC_d[1]\), with every face checked in the prerequisite. Apply \(Q\). For each of the three prescribed triangles, Theorem 3.2 extends the corresponding pair of vertex isomorphisms to a triangle comparison. Lemma 2.3 makes its third component invertible. Denote these components by \(\lambda_1: QC_d\to A\), \(\lambda_2: QC_{bd}\to B\), \(\lambda_3: QC_b\to C\). Define

Write the prescribed three rows as \(QX\xrightarrow\varphi QY\xrightarrow{i_\varphi}A\xrightarrow{p_\varphi}QX[1]\), \(QY\xrightarrow\psi QZ\xrightarrow{i_\psi}C\xrightarrow{p_\psi}QY[1]\), and \(QX\xrightarrow{\psi\varphi}QZ\xrightarrow{i_h}B\xrightarrow{p_h}QX[1]\). Transport of the checked faces gives precisely

The fourth row \(A\xrightarrow UB\xrightarrow VC\xrightarrow WA[1]\) is distinguished, since the three \(\lambda_i\) give an isomorphism from the image fourth row. Its consecutive composites vanish, as they did in the original octahedron. Every other face likewise commutes after conjugating its endpoints by the same vertex isomorphisms. Thus the octahedron exists for the prescribed triangles, including triangles whose initial arrows have no chain-map representative on the original vertices. This proves the fourth axiom. Finally images of \(K\)-triangles are distinguished by definition, so \(Q\) is exact. \(\square\)

5. Universal exact functors and the acyclic kernel

For completeness, one elementary consequence of the axioms will also apply to arbitrary triangulated target categories. In a distinguished triangle \(A\xrightarrow fB\xrightarrow gC\xrightarrow hA[1]\), all consecutive composites are zero. To prove \(gf=0\), complete the square from the identity triangle \(A\xrightarrow1A\to0\) with first two vertical maps \(1,f\); its third map from zero forces the equality. Rotate to get the other composites.

If \(k: T\to B\) satisfies \(gk=0\), compare the rotated identity triangle \(T\to0\to T[1]\xrightarrow{-1}T[1]\) to the rotated triangle \(B\xrightarrow gC\xrightarrow hA[1]\xrightarrow{-f[1]}B[1]\), with first two components \(k,0\). The morphism axiom supplies \(l: T[1]\to A[1]\). Its last square says \((-f[1])l=-k[1]\); shifting back gives \(f l[-1]=k\). Together with the zero-composite assertion and rotations this proves exactness of all covariant Hom sequences in any triangulated category.

It follows that \(f\) in such a triangle is invertible if and only if \(C\simeq0\). If \(C\) is zero, the Hom sequence makes \(\operatorname{Hom}(T,f)\) bijective for every \(T\), and the inverse is obtained from the identity of \(B\) and checked by injectivity at \(A\). If \(f\) is invertible, the sequence

forces the middle group to vanish. Taking \(T=C\) gives \(1_C=0\), which makes \(C\) a zero object. The same Hom argument and the five-term test also prove the two-vertex isomorphism criterion in an arbitrary triangulated category.

Theorem 5.1. For any triangulated category \(\mathcal B\), restriction along \(Q\) gives an equivalence between exact functors \(D\to\mathcal B\) and exact functors \(K\to\mathcal B\) that annihilate all acyclic complexes. Morphisms are natural transformations compatible with the specified shift isomorphisms.

Proof. Let \(F\) be exact. If it annihilates every acyclic complex, then for \(s\in S\) its cone is acyclic, so the exact image cone triangle and the criterion just proved make \(F(s)\) invertible. Conversely, if \(F\) inverts \(S\), then for an acyclic \(N\) the quasi-isomorphism \(0\to N\) makes \(F(N)\simeq F(0)=0\), since \(F\) is additive. Thus the two conditions agree.

The ordinary localization theorem gives the unique descended functor

It is additive: compute a sum with one common denominator, then apply additivity of \(F\) and distributivity of composition. Its shift isomorphism has the same object components as the specified isomorphism \(F(X[1])\simeq F(X)[1]\). Naturality for a roof follows from naturality for its numerator and denominator, followed by inversion of the denominator; coherence with successive shifts is the same objectwise coherence as before descent. Every distinguished triangle is an isomorphic image triangle, and \(F\) preserves the latter, including the last map with its shift identification. Hence \(\overline F\) is exact.

The ordinary localization theorem extends every natural transformation uniquely with unchanged object components. Its compatibility with shift is an equality on those components, so holds after descent exactly when it held before descent. Restriction is therefore fully faithful and has the stated image. \(\square\)

Let \(N\) be the full subcategory of acyclic complexes in \(K\). It is closed under isomorphisms, both shifts and cones of its arrows: cone cohomology exactness makes the third vertex acyclic when the first two are. It is also closed under any direct summand that exists in \(K\). Indeed a retraction and section induce such maps on each cohomology object, and a retract of zero has identity zero. Thus \(N\) is thick. The zero-detection theorem for roofs says that it is exactly the object kernel of \(Q\), and the cone criterion says that \(S\) consists exactly of arrows with cone in \(N\). In the sense of Theorem 5.1, \(D\) is the Verdier quotient \(K/N\). This assertion constructs the quotient and its exact universal property; it does not presume resolution existence.

The triangulated structure is unique with the specified shift and exact \(Q\). To check uniqueness, any arrow admits (4.1), which is distinguished in any such structure. Any other distinguished triangle for that arrow admits a comparison from (4.1) with identity on its first two vertices, by the morphism axiom of that structure. The general two-vertex criterion just proved makes the third comparison invertible. Thus it is isomorphic to (4.1), and so belongs to the triangles of Section 1. The reverse inclusion is forced by exactness of \(Q\).

6. Exact sequences and objects in degree zero

Theorem 6.1. A termwise short exact sequence of complexes \(0\to X\xrightarrow iY\xrightarrow pZ\to0\) gives a distinguished triangle naturally for commuting morphisms of short exact sequences.

Proof. Define the chain map

It is a chain map because \(pi=0\) and \(p\) commutes with differentials. We verify that it is a quasi-isomorphism. For surjectivity on cohomology, represent a class in \(H^nZ\) by a cycle after an epimorphic change of test object. Lift it to \(y\) in \(Y^n\), using the epimorphism \(p^n\) and another covering lift. Then \(d_Yy\) lies in the kernel of \(p^{n+1}\), so is uniquely \(ix\). Its next differential gives \(i d_Xx=0\), hence \(d_Xx=0\). Thus \((y,-x)\) is a cycle in \(C_i\) that maps to the given cycle. These covering lifts prove that \(H^n(q)\) is epic.

For injectivity, represent a cohomology class by a cycle \((y,x)\) whose image is zero. After a covering lift its image is a boundary \(py=d_Zz'\). Lift \(z'\) to \(y'\) in degree \(n-1\), after a further covering change. Subtract the cone boundary \(D_i(y',0)=(d_Yy',0)\). The remaining first coordinate is killed by \(p\), so equals \(iu\) for a unique \(u\). The cycle equation gives \(i(d_Xu+x)=0\), hence \(x=-d_Xu\). The remaining pair \((iu,-d_Xu)\) is the boundary \(D_i(0,u)\). Thus the class vanishes after an epimorphic covering change. Canceling that epimorphism proves that the kernel of \(H^n(q)\) is zero. This proves monicity, and hence invertibility, in every degree.

Transport the cone triangle along \(Qq\). Its resulting arrows are

For a commuting morphism of short exact sequences with components \(a,b,c\), the strict cone map \(r(y,x)=(by,ax)\) obeys \(q'r=cq\) and \(p_{i'}r=a[1]p_i\). Inverting the quasi-isomorphisms \(q,q'\) gives \(\delta'Qc=Qa[1]\delta\). The other two squares were already given. This proves naturality, including the connecting map. \(\square\)

Every cohomology functor on \(K\) descends to \(D\). Applied to a distinguished triangle its long sequence is exact: for image triangles this is cone cohomology exactness, and triangle isomorphisms preserve exactness. In particular (6.2) gives the natural cohomology sequence of a short exact sequence of complexes.

Theorem 6.2. The functor sending an object of \(\mathcal A\) to its complex in degree zero is fully faithful in \(D\):

Proof. Represent a derived arrow by \(A[0]\xleftarrow sP\xrightarrow fB[0]\). Define \(r=H^0(f)H^0(s)^{-1}\). This is a morphism \(A\to B\). Form \(P'\) with \(P'^n=P^n\) for \(n<0\), \(P'^0=Z^0P\), and \(P'^n=0\) for \(n>0\). Its degree-minus-one differential lands in \(Z^0P\), so it is a complex with an inclusion \(j: P'\to P\). It has the same cohomology as \(P\) in nonpositive degrees, and its positive cohomology is zero. The latter is also zero for \(P\), because \(s\) is a quasi-isomorphism to \(A[0]\). Hence \(j\in S\).

The map \(f-rs\) has zero induced map on \(H^0P\). Restricted to \(Z^0P\), it factors through the quotient onto \(H^0P\), because both \(f\) and \(s\) kill the degree-zero boundaries. Its restriction is therefore zero. All its other components have zero target. Consequently \((f-rs)j=0\) as a chain map, and the roof equality test gives \(Qf=Qr\,Qs\). The original derived arrow is \(Qr\). Conversely its cohomology map is \(r\), proving uniqueness. Composition and addition are preserved by this identification because cohomology preserves both. Thus (6.3) is natural and is a full-faithfulness statement without enough injectives or projectives. \(\square\)

7. Two worked examples

Example 7.1 (an extension invisible on cohomology maps). In abelian groups take the short exact sequence \(0\to\mathbb Z\xrightarrow n\mathbb Z\to\mathbb Z/n\to0\), with \(n>1\). Its triangle is

The cone in (6.1) is \(P_n=(\mathbb Z\xrightarrow n\mathbb Z)\) in degrees minus one and zero, and its map to \(\mathbb Z/n\) is reduction. The last positive cone projection is identity in degree minus one. Thus \(\delta_n\) is represented by the roof with that denominator and numerator one. The bounded projective test proved in the roof lesson computes \(\operatorname{Hom}_D((\mathbb Z/n)[0],\mathbb Z[1])=\mathbb Z/n\), with homotopies changing the integer numerator by a multiple of \(n\). Hence \(\delta_n\) is nonzero.

Every \(H^j(\delta_n)\) is nevertheless zero, since its source and target cohomology are in different degrees. Cohomology detects invertibility of derived maps: for a roof \(Qf(Qs)^{-1}\), if all its cohomology maps are invertible, all \(H^j(f)\) are invertible, so \(f\in S\) and the roof is invertible. It does not detect zero morphisms. The triangle retains the nonsplit extension that the individual cohomology maps of \(\delta_n\) lose.

Example 7.2 (a new ambiguity in triangle completion). Put \(M=(\mathbb Z/2)[0]\), \(Y=\mathbb Z[2]\), and take the zero arrow \(M\to Y\). Its cone is \(C=Y\oplus M[1]\), with the summand inclusion \(j\) and positive projection \(p\). Let \(a=\delta_2[1]: M[1]\to Y\), nonzero because shifting is an equivalence. In the derived category,

The biproduct identities verify their two inverse products, and \(cj=j\), \(pc=p\). Thus both \((1_M,1_Y,1_C)\) and \((1_M,1_Y,c)\) complete the same identity square to an automorphism of this distinguished triangle. They differ because projecting \(c-1\) from the second summand to the first gives \(a\ne0\). No chain map between these two off-diagonal summands represents \(a\): their nonzero terms are in degrees minus one and minus two. Thus even the ambiguity in a cone comparison can include a morphism introduced by localization.

8. Four graded exercises with complete solutions

Exercise 1 (foundation: the split triangle). For arbitrary complexes \(X,Y\), prove that \(X\to X\oplus Y\to Y\to X[1]\), with summand inclusion, summand projection and zero last map, is distinguished already in \(K\), hence in \(D\). Supply the actual cone comparison and homotopy.

Solution. The cone of the inclusion has coordinates \((x,y,x')\) and differential \((d_Xx+x',d_Yy,-d_Xx')\). Projection \(r(x,y,x')=y\) and inclusion \(i(y)=(0,y,0)\) are chain maps, with \(ri=1\). The family \(H(x,y,x')=(0,0,x)\), of degree minus one, obeys

Hence \(r\) is invertible in \(K\). Its composite with the cone inclusion of \(X\oplus Y\) is the required projection onto \(Y\). The cone last projection preceded by \(i\) is zero. This is an isomorphism of cone triangles with the stated split triangle. Applying \(Q\) proves the derived assertion.

Exercise 2 (intermediate: cone of a derived extension map). Calculate a distinguished triangle starting with \(\delta_2: (\mathbb Z/2)[0]\to\mathbb Z[1]\), including the next two maps and their signs. Compare it to twice rotating (7.1).

Solution. Write \(P=(\mathbb Z\xrightarrow2\mathbb Z)\) and let \(q: P\to(\mathbb Z/2)[0]\) be reduction. The numerator \(a: P\to\mathbb Z[1]\) is identity in degree minus one. Its cone has \(C_a^{-2}=\mathbb Z\), \(C_a^{-1}=\mathbb Z\oplus\mathbb Z\), with differential

The map \(R: C_a\to\mathbb Z[1]\) in degree minus one is \(R(u,v)=2u+v\). It is a chain map, its kernel in that degree is precisely the image of (8.2), and (8.2) is injective. It is surjective via \((0,v)\), so \(R\) is a quasi-isomorphism. On the cone inclusion of \(\mathbb Z[1]\), \(R\) is multiplication by two. The last projection followed by \(q[1]\) is reduction of \(v\); since \(2u+v\) has the same reduction, it is reduction after \(R\). Transporting the source by \(Qq\) and the third vertex by \(QR\) gives

Twice rotating (7.1) gives next maps \(-2\) and minus reduction. Multiplication by minus one on its third vertex changes both to the signs in (8.3), while leaving its first arrow unchanged. This verifies the triangle isomorphism and the positive-projection convention.

Exercise 3 (advanced: splitting an ordinary extension). For a short exact sequence \(0\to A\xrightarrow iB\xrightarrow pC\to0\) in \(\mathcal A\), prove that its connecting arrow \(C[0]\to A[1]\) is zero exactly when the sequence splits in \(\mathcal A\). In that case exhibit the triangle isomorphism to Exercise 1.

Solution. If the connecting map is zero, Lemma 2.2 applied to \(\operatorname{Hom}_D(C[0],-)\) says that \(1_C\) lifts through \(Qp\) to a map \(s: C[0]\to B[0]\) in \(D\). Theorem 6.2 makes it an ordinary map \(C\to B\), and faithfulness gives \(ps=1_C\) in \(\mathcal A\). Conversely an ordinary section gives \(\delta=\delta p s=0\), using the consecutive-zero property of the distinguished triangle.

For a section \(s\), the map \(1_B-sp\) is killed by \(p\), so factors uniquely as \(ir\). Monicity of \(i\) gives \(ri=1_A\) and \(rs=0\). Thus \((i,s): A\oplus C\to B\) and \((r,p)\) are inverse, by expanding their biproduct composites. Under this isomorphism the first two arrows are summand inclusion and projection, and the connecting arrow is zero. Together with identity on \(A,C\) this is the required isomorphism to the split triangle. This criterion used no injective resolution.

Exercise 4 (synthesis: termwise splitting does not split complexes). Let \(f=2: \mathbb Z[0]\to\mathbb Z[0]\). The cone gives a termwise split sequence \(0\to\mathbb Z[0]\xrightarrow{j_f}C_f\xrightarrow{p_f}\mathbb Z[1]\to0\). Compute its connecting arrow \(\mathbb Z[1]\to\mathbb Z[1]\), and prove that the sequence has no chain-map section despite splitting in every degree.

Solution. More generally, for \(f: X\to Y\), the cone of \(j_f\) has coordinates \((y,x,y')\) and differential \((d_Yy+fx+y',-d_Xx,-d_Yy')\). Its comparison to \(X[1]\) is \(q(y,x,y')=x\), exactly (6.1) for this sequence. The chain section \(i(x)=(0,x,-fx)\) and homotopy \(H(y,x,y')=(0,0,y)\) satisfy \(qi=1\) and \(DH+HD=1-iq\), as checked directly in the rotation proof. Its last cone projection is \(y'\), so the connecting arrow is \(-f[1]\). For the given \(f\), it is therefore \(-2\).

This is nonzero: shift and Theorem 6.2 identify \(\operatorname{End}_D(\mathbb Z[1])\) with \(\operatorname{End}_{\mathsf{Ab}}(\mathbb Z)=\mathbb Z\). A chain section would yield a derived section of \(Qp_f\), forcing the connecting map to vanish by its consecutive-zero property. That contradicts \(-2\ne0\). Yet each term is the biproduct \(C_f^n=Y^n\oplus X^{n+1}\), so its sequence splits degree by degree. The cross term \(f\) obstructs compatibility of these sections with the differential.

References

• The Stacks Project Authors, Localization of triangulated categories, Section 13.5, and The canonical delta-functor, Section 13.12, for further reading. The denominator clearing, axioms and exact-sequence constructions used here are proved above.
• Pierre Schapira, An Introduction to Categories and Homological Algebra, Sections 6.4 and 7.1, for the localization and derived-category framework.