Draft lesson.

Chain homotopies, mapping cones and triangles

Original text public domain (CC0).

A homotopy records why two maps induce the same cohomology map. A mapping cone records the failure of a map to be a homotopy equivalence, and, in an abelian category, its failure to be a quasi-isomorphism. We construct both records and prove the triangulated axioms, including the octahedron, by actual maps.

Let \(\mathcal A\) be a locally small additive category. We use cohomological complexes indexed by \(\mathbb Z\). Only finite biproducts in \(\mathcal A\) are needed. Assume \(\mathcal A\) is abelian in Section 6. Exact complexes, signs and lifting covers proves the cohomology descriptions, covering lifts and five-term test used below; Formal linear combinations and finite sums supplies the biproduct calculus. All products of Hom groups below are products in abelian groups, not products in \(\mathcal A\).

1. Homotopies and the Hom complex

A chain map \(a: X\to Y\) has components \(a^n: X^n\to Y^n\) satisfying \(d_Ya=ad_X\). Composition and addition are componentwise. The category of these complexes and maps is \(C(\mathcal A)\). Its zero object and finite biproducts are componentwise. For an integer \(r\), define

The differential squares to zero, the shifted map is a chain map, and successive shifts add their indices, including their signs. Thus shifting is an invertible additive functor.

A homotopy from \(b\) to \(a\) is a family \(s^n: X^n\to Y^{n-1}\) such that

Zero, negation and addition of these families prove reflexivity, symmetry and transitivity. Precomposing with a chain map \(u\), or postcomposing with a chain map \(v\), changes the homotopy to \(su\), or \(vs\). Hence null-homotopic maps form an additive ideal. The quotient is the additive homotopy category \(K(\mathcal A)\). An isomorphism there is precisely a chain map with a two-sided inverse up to homotopy. Shifting a homotopy by \(r\) uses \((-1)^r s^{n+r}\); substitution into (1.2) proves that shifting descends to this quotient.

For two complexes form the complex of abelian groups

Expanding the next differential gives two terms containing a squared differential, both zero, and two terms \(-(-1)^m d_Ytd_X\) and \(-(-1)^{m+1}d_Ytd_X\), which cancel. Thus \(\partial^2=0\). For composable homogeneous families \(t,u\), expanding the two products gives

In degree zero the cycles are the chain maps and the boundaries are (1.2). In degree \(m\), a cycle is a chain map \(X\to Y[m]\), by (1.1). A boundary \(\partial s=d_Ys+(-1)^m sd_X\) is its homotopy using \((-1)^m s\). Consequently

The identifications are natural: chain-map composition preserves cycles and boundaries by (1.4).

2. A cone with all structural maps

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

Here and below a formula in coordinates means the corresponding biproduct matrix; it requires no elements in the objects of \(\mathcal A\). Squaring (2.1) gives \((d_Y^2y+(d_Yf-fd_X)x,d_X^2x)=(0,0)\). There are chain maps

Their composite is zero. The composite \(j_ff\) is null-homotopic by \(x\mapsto(0,x)\) from \(X^n\) to \(C_f^{n-1}\). The composite \(f[1]p_f\) is null-homotopic by \((y,x)\mapsto y\) from \(C_f^n\) to \(Y[1]^{n-1}\): its differential expression is \(-d_Yy+d_Yy+fx=fx\).

A triangle is three maps \(X\to Y\to Z\to X[1]\) in \(K(\mathcal A)\). Call it distinguished if it is isomorphic, with all four maps commuting, to

This fixes our cone and connecting-map convention. For a strict square \(bf=f'a\), the map \((y,x)\mapsto(by,ax)\) is a chain map between cones. It respects identities and composition. For a square commuting only up to homotopy a further component is necessary; Section 4 supplies it. There is no assertion that an unchosen homotopy gives a unique cone map.

3. Rotation, with its minus sign

Lemma 3.1. Rotating a distinguished triangle gives the distinguished triangle \(Y\to Z\to X[1]\xrightarrow{-f[1]}Y[1]\). Distinguished triangles are also closed under the shifted convention whose last map is \((-1)^r p_f[r]\).

Proof. Compute the cone of \(j_f\) in coordinates \((y,x,y')\). Its differential is

The maps \(r(y,x,y')=x\) to \(X[1]\) and \(i(x)=(0,x,-fx)\) in the other direction are chain maps, and \(ri=1\). For \(H(y,x,y')=(0,0,y)\), of degree minus one, direct substitution gives

Thus \(r\) is an isomorphism in the homotopy category. The cone inclusion of \(C_f\) followed by \(r\) is \(p_f\). The last cone projection followed after \(i\) is \(-f[1]\). These are the required triangle comparisons.

The chain isomorphism \(C_{f[r]}\to C_f[r]\) sends \((y,x)\) to \((y,(-1)^r x)\). Its diagonal and off-diagonal entries commute with the two differentials by (1.1). It preserves the inclusion and changes the projection by \((-1)^r\), proving the shift assertion. Three rotations give arrows \(-f[1],-j_f[1],-p_f[1]\). Multiplying the middle object by minus one identifies this triangle with the shifted triangle just proved. Thus if a rotation is distinguished, two further rotations and an inverse shift show that the original triangle is distinguished. The rotation condition holds in both directions. \(\square\)

4. Exact Hom sequences and completing a square

Lemma 4.1. For every complex \(T\), applying either \(\operatorname{Hom}_{K}(T,-)\) or \(\operatorname{Hom}_{K}(-,T)\) to a distinguished triangle gives an exact sequence of abelian groups, continued in both directions by shifts and rotations.

Proof. We first check the covariant sequence for (2.3). Suppose \(b: T\to Y\) has \(j_fb\) null-homotopic by \((s,t): T^n\to Y^{n-1}\oplus X^n\). The second component says \(d_Xt=td_T\), and the first says \(b-ft=d_Ys+sd_T\). Thus \([b]=[f][t]\). This proves exactness at \(\operatorname{Hom}_K(T,Y)\).

Suppose \(c: T\to C_f\) has \(p_fc=-d_Xh+hd_T\). Subtract from \(c\) the null-homotopic map defined by \((0,h): T^n\to C_f^{n-1}\). The remaining map has second component zero, hence is \(j_fb\) for a chain map \(b: T\to Y\). This proves exactness at the cone position. Finally, if \(a: T\to X[1]\) has \(f[1]a=-d_Yh+hd_T\), then \((h,a): T\to C_f\) is a chain map and projects to \(a\). This proves the next position. The reverse containments follow from the null-homotopies after (2.2). Shifting these three checks covers every position; multiplying a map by minus one does not change its kernel or image.

For the contravariant sequence, a map \(b: Y\to T\) with \(bf=d_Th+hd_X\) extends to the chain map

Indeed, the two expressions for its differential differ by \((bf-d_Th-hd_X)x\). This proves exactness at \(\operatorname{Hom}_K(Y,T)\). If a chain map \(c: C_f\to T\) restricts on \(Y\) to \(d_Ts+sd_Y\), subtract the homotopy defined by \((y,x)\mapsto sy\). The remainder vanishes on \(Y\), so factors as \(ap_f\); its chain condition makes \(a: X[1]\to T\) a chain map. This proves the cone position. Apply these same two checks to the rotations of Lemma 3.1 to cover every position. Isomorphic triangles have isomorphic Hom sequences, so the assertions hold for every distinguished triangle. \(\square\)

Lemma 4.2 (completion). If \([b][f]=[f'][a]\), a square between the first maps of cone triangles extends to a triangle morphism. If \([a]\) and \([b]\) are invertible, its cone component is invertible.

Proof. Choose a homotopy \(bf-f'a=d_{Y'}h+hd_X\). Define

The upper components of \(cD_f\) and \(D_{f'}c\) agree because \(bf-hd_X=d_{Y'}h+f'a\); the lower ones agree because \(a\) is a chain map. The inclusion and projection squares commute strictly. The first square commutes in \(K\) by the chosen homotopy.

When the endpoint maps are invertible, Lemma 4.1 gives a commuting diagram of five exact Hom-group positions, with the cone position in the middle and four outside isomorphisms. The five-term result proved in the prerequisite makes \(\operatorname{Hom}_K(T,c)\) invertible for every \(T\). Taking \(T=C_{f'}\) gives \(e\) with \(ce=1\); injectivity for \(T=C_f\) gives \(ec=1\), since \(c(ec-1)=0\). Thus \([c]\) is invertible.

Transport the construction through the isomorphisms defining distinguished triangles to complete any square between such triangles. It also follows that a distinguished triangle whose first arrow is \([f]\) is isomorphic to the particular cone triangle of any chosen chain representative \(f\), with identity comparisons on \(X,Y\): complete the square comparing its cone presentation to \(f\), then use the endpoint-isomorphism result. \(\square\)

The identity cone is contractible: \(H(y,x)=(0,y)\) gives \(D_1H+HD_1=1\). Hence \(X\xrightarrow1X\to0\to X[1]\) is distinguished. Every arrow in \(K\) has a chain representative and therefore a cone triangle. Together with isomorphism closure, Lemmas 3.1 and 4.2 prove the identity, existence, rotation and morphism axioms. Only the octahedron remains.

5. An explicit octahedron

Take chain maps \(X\xrightarrow fY\xrightarrow gZ\), and write \(h=gf\). Set

The formulas commute with the cone differentials: the upper cross terms are \(gf\), and the lower entries use the chain equations for \(f,g\); for \(w\) both differentials give \((-d_Yy,0)\). To prove that \(C_f\xrightarrow uC_h\xrightarrow vC_g\xrightarrow wC_f[1]\) is distinguished, compute its first cone. In degree \(n\), its four coordinates lie in \(Z^n,X^{n+1},Y^{n+1},X^{n+2}\), and

Define chain maps

For \(r\), the second component of \(rD\) is \(-d_Yy-ft+f(-d_Xx+t)=-d_Y(y+fx)\); the first is \(d_Zz+g(y+fx)\). For \(i\), substitution into (5.2) gives \((d_Zz+gy,0,-d_Yy,0)\). Thus both are chain maps and \(ri=1\). The degree-minus-one map \(H(z,x,y,t)=(0,0,0,x)\) satisfies

This proves the homotopy equivalence. The cone inclusion followed by \(r\) is \(v\), and the cone projection preceded by \(i\) is \(w\). Hence this is an isomorphism of triangles to the cone triangle of \(u\).

The octahedron contains the four distinguished triangles

Their faces obey

These are literal coordinate identities. For example, the two paths in the last identity send \((z,y)\) to \((y,0)\). The consecutive composites in the fourth triangle vanish in \(K\): for \(vu\) use the homotopy \((y,x)\mapsto(0,y)\); for \(wv\) use \((z,x)\mapsto(0,-x)\) valued in \(C_f[1]^{n-1}\); and for \(u[1]w\) use \((z,y)\mapsto(z,0)\) valued in \(C_h[1]^{n-1}\). Their differential expressions are respectively \((gy,fx)\), \((fx,0)\), and \((gy,0)\).

Thus the maps and all the required faces of the octahedral axiom are specified. For triangles initially given by other cone presentations, use the identity-on-endpoints comparisons of Lemma 4.2 and transport these maps. This proves the axiom for arbitrary composable arrows and their distinguished triangles. Therefore \(K(\mathcal A)\), with shift (1.1) and triangles (2.3), is a triangulated category.

Corollary 5.1. A chain map \(f\) is a homotopy equivalence if and only if \(C_f\) is contractible.

Proof. Contractibility means that the identity is null-homotopic, equivalently the complex is a zero object in \(K\). If \(C_f\) is zero, the Hom sequences of Lemma 4.1 make \(\operatorname{Hom}_K(T,f)\) bijective for all \(T\). The right-inverse and injectivity argument in Lemma 4.2 produces a two-sided inverse. Conversely, if \([f]\) is invertible, those sequences give \(\operatorname{Hom}_K(T,C_f)=0\) for all \(T\). Take \(T=C_f\) to make its identity zero. \(\square\)

6. Cohomology and quasi-isomorphisms

Assume \(\mathcal A\) is abelian. Put \(Z^nX=\ker d_X^n\), \(B^nX=\operatorname{im}d_X^{n-1}\), and \(H^nX=Z^nX/B^nX\). The incoming image is contained in the outgoing kernel because \(d_X^2=0\); its inclusion is monic. These are the actual kernel and cokernel constructions in Section 2 of the prerequisite. A chain map restricts to cycles, maps boundaries into boundaries, and induces a map on the quotient. Uniqueness of these factorizations proves functoriality. For (1.2), restriction to cycles makes \(a-b=d_Ys\), which lands in boundaries. Thus cohomology factors through \(K(\mathcal A)\), and

Multiplication of a differential by minus one preserves its kernel and image, giving this identification with the structural quotient maps.

Theorem 6.1. A cone has the natural long exact sequence

Proof. The null-homotopies in Section 2 show that successive composites are zero. We verify equality of images and kernels at the three different kinds of positions.

We may write a class as a cycle after an epimorphic change of its test object: the quotient \(Z^n\twoheadrightarrow H^n\) pulls back to an epimorphism. If a cycle has zero class, it factors through \(B^n\); pulling back \(X^{n-1}\twoheadrightarrow B^n\) supplies, after a further epimorphic cover, a preimage under the differential. Finite successive covers compose to an epimorphism. These statements and the exactness criterion for generalized maps are proved in Section 3 of the prerequisite. Thus the following coordinate chases prove categorical exactness, without enough projectives or an embedding theorem.

At \(H^nY\), let a cycle \(y\) become a boundary \((y,0)=D_f(v,u)\) in the cone, after those covers. The lower coordinate says \(d_Xu=0\), and the upper says \(y=d_Yv+fu\). Its class is the image of \([u]\). Conversely, a cycle \(u\) has \((fu,0)=D_f(0,u)\), so its image is killed by the next map.

At \(H^nC_f\), take a cycle \((y,x)\) whose projected class is zero. After a cover, write \(x=d_Xu\). Subtract the cone boundary \(D_f(0,-u)=(-fu,d_Xu)\). The remaining cycle is \((y+fu,0)\), the image of a cycle in \(Y\). Conversely, any cycle with second coordinate zero projects to zero.

At \(H^{n+1}X\), take a cycle \(x\) whose image under \(f\) is a boundary, say \(fx=d_Yv\) after a cover. Then \((-v,x)\) is a cone cycle and projects to \(x\). Conversely, a cone cycle obeys \(fx=-d_Yy\), so its projected class is killed by \(H^{n+1}f\).

Each construction produces a lift of every kernel test map after an epimorphic cover, which is precisely the cited exactness criterion. Shifting the index covers the entire sequence. The maps are the functorial quotient maps already constructed; for a triangle morphism, all the connecting squares commute in \(K\) and therefore on cohomology. This proves naturality. \(\square\)

A quasi-isomorphism is a chain map inducing isomorphisms on all cohomology objects. Exactness of (6.2) proves that \(f\) is a quasi-isomorphism exactly when \(C_f\) is acyclic. In one direction two adjacent cohomology maps of \(f\) being invertible force the intervening cone cohomology to vanish. In the other, vanishing of the two adjacent cone terms makes \(H^nf\) both monic and epic, hence invertible. A homotopy equivalence is consequently a quasi-isomorphism. The converse fails, as the next example shows.

7. Two worked examples

Example 7.1 (an exact complex with a surviving identity). In abelian groups take the complex \(A\) with \(A^0=\mathbb Z\), \(A^1=\mathbb Z\), \(A^2=\mathbb Z/2\), differentials multiplication by two and reduction modulo two, and zero elsewhere. At degree zero multiplication by two has zero kernel. At degree one the kernel of reduction is \(2\mathbb Z\), its incoming image. At degree two reduction is surjective. Thus all cohomology vanishes. A contracting homotopy would, in degree two, give a section \(s^2: \mathbb Z/2\to\mathbb Z\) of reduction. Every such homomorphism is zero because \(\mathbb Z\) has no nonzero element killed by two. No section exists. The complex is acyclic but not contractible; its identity survives in \(K\), and \(0\to A\) is a quasi-isomorphism without a homotopy inverse.

Example 7.2 (the cone of an integer). Regard \(\mathbb Z\) as a complex concentrated in degree zero. For multiplication by \(m\), its cone has \(\mathbb Z\) in degrees minus one and zero, with differential \(m\). Hence

For nonzero \(m\) the first group is zero; for \(m=0\) both are \(\mathbb Z\). The cone is contractible precisely when \(m=1\) or \(m=-1\), with contracting map from degree zero to minus one multiplication by \(m^{-1}\). Indeed a contraction in those two degrees requires \(ms=sm=1\). This calculates separately the differential, its cohomology and its homotopy obstruction.

8. Four graded exercises with complete solutions

Exercise 1 (foundation: a shifted Hom class). Take \(X=\mathbb Z\) in degree zero and \(Y\) with \(Y^{-1}=Y^0=\mathbb Z\), differential multiplication by three, and zero elsewhere. Compute \(H^{-1},H^0\) of the Hom complex and the corresponding Hom groups in (1.5). Identify the homotopy represented by its boundary.

Solution. A family in (1.3) has just its component on \(X^0\). The Hom complex is therefore \(\mathbb Z\xrightarrow3\mathbb Z\) in degrees minus one and zero. Its cohomology is zero in degree minus one and \(\mathbb Z/3\) in degree zero. A chain map \(X\to Y\) is an arbitrary integer into \(Y^0\). Two such maps differ by a homotopy exactly when their difference is three times the integer \(s^0: X^0\to Y^{-1}\). Thus \(\operatorname{Hom}_K(X,Y)=\mathbb Z/3\). A chain map \(X\to Y[-1]\) has its degree-zero component in \(Y^{-1}\), killed by the shifted differential \(-3\), so it is zero. This gives \(\operatorname{Hom}_K(X,Y[-1])=0\), including the shift sign and the boundary homotopy.

Exercise 2 (intermediate: different completions of one square). Over a field \(k\), take \(X=k\) in degree zero, \(Y=k\) in degree minus one, and \(f=0: X\to Y\). Complete the square whose endpoint maps are both identities using the homotopies zero and \(h^0=1_k: X^0\to Y^{-1}\). Are the resulting cone maps equal in \(K\)?

Solution. Both sides of the square are zero, so both families are homotopies of its difference: all differentials vanish. The cone \(C_f\) is \(k\oplus k\) in degree minus one and has zero differential. Formula (4.2) gives respectively

Both commute with the inclusion and projection and are invertible, the second with inverse having upper-right entry minus one. They are distinct. A homotopy on this cone has zero differential expression, so it cannot identify distinct maps. They remain distinct in \(K\). A completion exists, but its unspecified homotopy does not determine a unique map.

Exercise 3 (advanced: bounded projectives). Call an object projective when every map from it lifts through every epimorphism. Show that an acyclic bounded-above complex of projective objects in an abelian category is contractible. No existence of projectives elsewhere in the category is assumed.

Solution. Choose \(N\) above which the complex \(P\) is zero. Acyclicity gives exact sequences \(0\to Z^nP\to P^n\to Z^{n+1}P\to0\). Start with \(Z^{N+1}P=0\). Descending in \(n\), if \(Z^{n+1}P\) is projective, its identity lifts through \(P^n\twoheadrightarrow Z^{n+1}P\), giving a section. The kernel is then a direct summand, so \(P^n\simeq Z^nP\oplus Z^{n+1}P\). A direct summand of a projective object is projective: include it, lift the resulting map from the projective object, and restrict the lift to the summand. This proves the descending induction, for each finite distance from \(N\), over all integers below it. Choose the sections and these identifications.

The differential is \((u,v)\mapsto(v,0)\) in the next biproduct, since its quotient component identifies with the next cycle inclusion. Define \(s^n(u,v)=(0,u)\) in \(P^{n-1}\). Then \(ds(u,v)=(u,0)\) and \(sd(u,v)=(0,v)\); their sum is the identity. This is a contracting homotopy. Example 7.1 has a nonprojective top object and therefore does not satisfy these hypotheses.

Exercise 4 (synthesis: an arithmetic octahedron). Apply Section 5 to multiplication by two followed by multiplication by three on \(\mathbb Z\) in degree zero. Compute \(u,v,w\), the degree-zero cohomology sequence of the fourth triangle, and its short exactness directly.

Solution. The three cones are the two-term complexes with differentials \(2,3,6\). The map \(u: C_2\to C_6\) is identity in degree minus one and multiplication by three in degree zero; indeed \(6\cdot1=3\cdot2\). The map \(v: C_6\to C_3\) is multiplication by two in degree minus one and identity in degree zero; indeed \(3\cdot2=1\cdot6\). The map \(w: C_3\to C_2[1]\) is identity from the degree-minus-one group to the first component \(\mathbb Z\) of the shifted cone, and zero on degree zero. The target differential kills that first component, so it is a chain map. The composite \(vu\) is null-homotopic by the map from degree zero of \(C_2\) to degree minus one of \(C_3\) equal to the identity: the two differential expressions are multiplication by two and three at the respective degrees, exactly the composite components.

The fourth triangle gives

There is no degree-minus-one cohomology of \(C_3\), and no degree-one cohomology of \(C_2\), which supplies the zeros. The first map sends the two classes to \(0,3\), so is injective. The second is reduction modulo three, so is surjective, and its kernel is exactly \(\{0,3\}\). This checks every position of the sequence and every degree of its maps.

References

• Pierre Schapira, An Introduction to Categories and Homological Algebra, Sections 4.2–4.4, 5.5 and 6.1–6.3, for further reading on complexes and triangles.
• The Stacks Project Authors, Cones and termwise split sequences, Distinguished triangles in the homotopy category, and The definition of a triangulated category. Their termwise-split connecting-map convention uses a minus sign on the last cone projection. Our convention is fixed in (2.3); rotation and the full octahedron are proved above with that convention.