Draft lesson.

Graded complexes, contravariant signs and stupid truncations

Original text public domain (CC0).

Index conventions control the degree of a cone, the sign of a dual differential and the last arrow of a truncation triangle. We keep those conventions explicit. Stupid truncation works on actual complexes, yet can destroy a contracting homotopy; its triangles must therefore be constructed before passing to homotopy classes.

Let \(\mathcal C\) be a locally small additive category, with the universe and Choice conventions of Universes and small categories. Kernels and cohomology in \(\mathcal C\) are not required. Chain homotopies, mapping cones and triangles, Sections 1–2 proves the ordinary chain-map, Hom-complex and cone interfaces. Differential objects, curvature and cone homotopies, Theorem 5.1 and Section 7 proves the full triangulated structure and the cone comparison for a split underlying sequence. We use its positive last-projection convention and cone order \(X[1]\oplus Y\).

1. Grading, support and the ordinary homotopy category

The category \(\operatorname{Gr}(\mathcal C)\) has integer families \((X^n)_{n\in\mathbb Z}\) as objects and degreewise maps as morphisms. Composition and addition are degreewise; the zero family and degreewise finite biproducts give an additive category. Its Hom group is a product of the given Hom groups in abelian groups. This product does not ask for infinite products of objects of \(\mathcal C\). The shift \((TX)^n=X^{n+1}\) has the strict inverse \((T^{-1}X)^n=X^{n-1}\).

A differential object in this graded category is precisely a sequence of maps \(d_X^n:X^n\to X^{n+1}\). It is a complex exactly when \(d_X^{n+1}d_X^n=0\) for every integer \(n\). A differential morphism is precisely a family \(f^n\) satisfying \(d_Y^nf^n=f^{n+1}d_X^n\). Thus

The second identification includes its actual maps: an abstract homotopy \(h:X\to T^{-1}Y\) has components \(h^n:X^n\to Y^{n-1}\), and its boundary is

These are exactly the chain homotopies in the linked prerequisite, so the two quotient Hom groups and their compositions agree. For every integer \(r\), the translation formula is

Its square is zero by the square-zero equation for \(X\). Its chain-map equation follows by applying the same scalar to both sides. Two successive shifts have the component \(X^{n+r+s}\) and sign \((-1)^{r+s}\), including negative integers; hence they agree with the combined shift. Formula (1.2) descends under the homotopy \((-1)^r h^{n+r}\), so (1.3) also acts on \(K\).

Write \(C^{[a,b]}(\mathcal C)\) for the full subcategory with \(X^n=0\) outside \(a\le n\le b\), allowing infinite endpoints. Vanishing means that the term is a zero object; chosen zero representatives may be used. Write \(C^{\ge a}\) and \(C^{\le b}\) for the corresponding one-sided intervals. These are additive. Bounded means that some finite \(a,b\) work; bounded below means that some \(a\) works, and bounded above means that some \(b\) works. Their categories are \(C^b,C^+,C^-\), respectively, and \(C^{\mathrm{ub}}=C\). They are the unions of those full support subcategories, with the inherited morphisms. The original category embeds fully faithfully as degree-zero complexes: every underlying map is a chain map, and the only potentially nonzero component is in degree zero.

Use the same support conditions to define \(K^b,K^+,K^-\) and the full additive categories \(K^{[a,b]},K^{\ge a},K^{\le b}\). For a chain map \(f:X\to Y\), our cone has

The category \(K\) is triangulated by the cited general differential-object theorem. If \(X,Y\) are bounded below by \(a_X,a_Y\), their cone is bounded below by \(\min(a_X-1,a_Y)\). If they are bounded above by \(b_X,b_Y\), its upper bound is \(\max(b_X-1,b_Y)\). Both shifts preserve each of the three qualitative boundedness conditions. Therefore \(K^b,K^+,K^-\) are full triangulated subcategories: the cone completion and all its maps remain in the stated subcategory, and the inherited axioms and their transported comparisons restrict there. A fixed interval is different: shifting moves \([a,b]\) to \([a-1,b-1]\). If there is a nonzero object, the complex concentrated at an endpoint shows that a finite interval subcategory is generally not shift-stable.

A distinguished triangle \(X\to Y\to Z\xrightarrow h X[1]\) is sometimes abbreviated \(X\to Y\to Z\xrightarrow{+1}\). The \(+1\) records the target of the omitted last map; \(h\) remains part of the triangle. The abbreviation does not discard the comparison involving that map.

If \(F:\mathcal C\to\mathcal D\) is additive, its degreewise action gives a complex functor, because \(F(d^{n+1})F(d^n)=F(0)=0\). It sends (1.2) to the corresponding homotopy equation, and commutes with every signed shift by additivity. Thus it descends to \(K\). The degreewise biproduct comparison gives \(F(C_f)\simeq C_{Ff}\), preserving the inclusion and positive projection; all four matrix entries are the images of those in (1.4). This is also the specialization of the general translation-compatible functor in the prerequisite to \(\operatorname{Gr}(F)\). Consequently its induced functor is triangulated, including the full last-arrow comparison. Boundedness conditions are preserved because an additive functor sends zero objects to zero.

2. Small coproducts and two cone calculations

Suppose \(\mathcal C\) admits coproducts indexed by a fixed small set \(I\). For a family of complexes \((X_i)\), define \(M^n=\bigoplus_{i\in I}X_i^n\). The unique map \(d_M^n\) whose restriction to each summand is \(X_i^n\xrightarrow{d_i^n}X_i^{n+1}\to M^{n+1}\) squares to zero: its square restricts to \(d_i^{n+1}d_i^n=0\) on each injection. The injections are chain maps.

Every family of chain maps \(X_i\to Y\) gives a unique degreewise coproduct map \(M\to Y\); its chain equation is checked on each injection. Hence \(M\) is the actual coproduct in \(C(\mathcal C)\). It is still a coproduct in \(K(\mathcal C)\). For existence there, choose a representative of each prescribed homotopy class and assemble those chain maps. For uniqueness, if a difference restricts to a null-homotopic map on every \(X_i\), choose those homotopies and assemble their degreewise maps \(M^n\to Y^{n-1}\). Formula (1.2) gives the difference on every injection, so on \(M\). The small-index Choice convention permits these choices. This proves both existence and uniqueness in the quotient and shows that \(C\to K\) preserves these coproducts. No compactness or exactness of coproducts was used. The empty family is the zero complex.

For a map \(f:A\to B\) between degree-zero objects, (1.4) has \(A\) in degree minus one, \(B\) in degree zero and differential \(f\). For a map between complexes \(X^0\xrightarrow{d_X}X^1\) and \(Y^0\xrightarrow{d_Y}Y^1\), with components \(f^0,f^1\), the cone has

in degrees \(-1,0,1\). The composite is \(-f^1d_X+d_Yf^0=0\), exactly the chain-map equation.

For any three-object complex \(L\xrightarrow fM\xrightarrow gN\), its contraction is equivalent to an actual splitting of the short sequence. A contraction has its only possibly nonzero maps \(k:M\to L\), \(h:N\to M\), and the three homotopy equations are

Conversely these equations supply the contraction. They also give the usual biproduct splitting: as in the prerequisite, \(kh=0\) follows from (2.2), and \((f,h)\) and \((k,g)\) are inverse maps \(L\oplus N\rightleftarrows M\). In an abelian category this proves exactness as well, rather than assuming it. Indeed \(f\) is monic because it has a left inverse, and \(g\) is epic because it has a right inverse. If \(ga=0\), then \(a=fka\), the unique factorization through \(f\). If \(bf=0\), then \(b=bhg\), the unique factorization through \(g\). Thus \(f\) and \(g\) are the actual kernel and cokernel maps. The criterion already holds in an arbitrary additive category whenever the displayed splitting equations are meant.

3. The Hom complex keeps source and target visible

For complexes \(X,Y\), put

The target of this last component is \(Y^{k+p+1}\). The linked chain-homotopy lesson, Section 1, proves by expansion that \(\partial^2=0\): the two squared differentials vanish and the two mixed terms have opposite signs. That same proof gives the graded composition rule and the full natural identification \(H^p\operatorname{Hom}^{\bullet}(X,Y)=\operatorname{Hom}_K(X,Y[p])\). It assumes only that \(\mathcal C\) is additive; the products in (3.1) are products of abelian groups.

For clarity, the conventional expression \(\operatorname{Hom}(X,Y)\) always has \(X\) as source and \(Y\) as target here. Its domain is \(C(\mathcal C)^{\mathrm{op}}\times C(\mathcal C)\). A target-first convention uses

These are the same construction with the argument order stated explicitly. For chain maps \(a:X'\to X\), \(b:Y\to Y'\), the action on a family is \(f\mapsto bfa\). Applying (3.1) and the two chain-map equations gives \(\partial(bfa)=b(\partial f)a\), so it is a complex functor. It descends to homotopy categories as well. If \(a-a'=\partial s\), the homotopy between the two precomposition maps sends a degree-\(p\) family \(f\) to \((-1)^pfs\). The graded composition rule gives \(\partial((-1)^pfs)+(-1)^{p+1}(\partial f)s=f\partial s\). If \(b-b'=\partial t\), use \(f\mapsto tf\); the same rule gives \(\partial(tf)+t\partial f=(\partial t)f\). These are the required homotopies of Hom-complex maps. The target-first notation (3.2) therefore also defines a bifunctor \(K(\mathcal C)\times K(\mathcal C)^{\mathrm{op}}\to K(\mathsf{Ab})\).

In degree zero define the subgroups and quotient

Equation (3.1) says \(f\in Z^0\) exactly when \(d_Y^kf^k=f^{k+1}d_X^k\), the actual chain-map condition. At degree minus one its boundary is \(d_Y^{k-1}h^k+h^{k+1}d_X^k\), exactly (1.2). Since the full product describes all such families, its image is precisely the subgroup of null-homotopic chain maps, rather than a smaller finite-support subgroup. Thus

Precomposition and postcomposition preserve the cycle and boundary formulas, so all three identifications are natural in the stated two variances.

4. Reversing degrees requires a signed comparison

Let \(G:\mathcal C^{\mathrm{op}}\to\mathcal D\) be additive. Define

The square of the differential is \((-1)^{2n+1}G(d_X^{-n-1}d_X^{-n-2})=0\), with composition reversed inside \(G\). The chain-map equation follows from \(f^{-n}d_X^{-n-1}=d_Y^{-n-1}f^{-n-1}\). Contravariance reverses the order of composite maps, and preserves identities and addition, so this is an additive functor \(C(\mathcal C)^{\mathrm{op}}\to C(\mathcal D)\).

If \(f-g\) is the boundary of \(h\), the homotopy between \(Gf\) and \(Gg\) has components

Its first boundary term has coefficient \((-1)^{n-1}(-1)^{n-1}=1\) and equals \(G(h^{-n+1}d_X^{-n})\). Its second has coefficient \((-1)^n(-1)^n=1\) and equals \(G(d_Y^{-n-1}h^{-n})\). Their sum is \(G(f^{-n}-g^{-n})\). Thus the functor descends to homotopy categories.

There is a natural chain isomorphism

Both sides have the indicated term. Their differentials are respectively \((-1)^{n+1}G(d_X^{-n})\) and \((-1)^nG(d_X^{-n})\). Multiplying the first by \(\theta^{n+1}\), or multiplying the second by \(\theta^n\), gives the same map \(-G(d_X^{-n})\). The inverse uses the same scalar signs. Every chain map commutes with these scalar components, proving naturality. The target is the inverse shift of \(G(X)\), not its positive shift.

Degree reversal also preserves the stated support bounds with their signs reversed: support \([a,b]\) goes to \([-b,-a]\), so bounded below goes to bounded above, and vice versa. For homological notation with a differential decreasing the subscript, setting \(X^n=X_{-n}\) and \(d_X^n=d^X_{-n}\) identifies it with the cohomological convention; this relabelling has no additional differential sign.

5. Degree weights and evaluation

On a graded object define \(\xi_X^n=n1_{X^n}\). Integer multiplication in Hom groups commutes with every graded map, so this is a natural endomorphism of the identity functor on \(\operatorname{Gr}(\mathcal C)\). Its two translated components satisfy

For a complex, \(d_X:X\to X[1]\) is a chain map: its two chain-equation composites are \(-d_X^{n+1}d_X^n\) and \(d_X^{n+1}d_X^n\), both zero. Its homotopy to zero is

The boundary is \(-n d_X^n+(n+1)d_X^n=d_X^n\). In particular the underlying differential, viewed as this shifted chain map, is zero in \(K\); it need not be zero as an actual map.

For evaluation specialize to abelian groups. Recall the elementary tensor construction: \(A\otimes B\) is the free abelian group on pairs \((a,b)\), modulo the subgroup generated by the two bilinearity relations. A bilinear map into an abelian group kills those relations, so extends uniquely from the pairs and factors uniquely through the quotient. This proves its bilinear universal property. Simple tensors generate the quotient; maps induced by maps of the two factors therefore preserve identities and composition by their values on those generators. Direct sums of abelian groups are their finite-support families, with their actual inclusion universal property.

Let \(H=\operatorname{Hom}^{\bullet}(X,Y)\). The tensor complex has

The two terms are bilinear on each summand, so define the displayed map by the tensor and direct-sum universal properties. Its square has the two zero terms \(\partial^2f\otimes x,f\otimes d_X^2x\) and mixed terms with signs \((-1)^{p+1},(-1)^p\); these cancel. Checking simple tensors checks the whole map. If \(X\) is supported in a finite interval, each diagonal has only finitely many nonzero summands, so no infinite tensor-diagonal choice is involved.

Define evaluation on that summand by

The product projection \(H^p\to\operatorname{Hom}(X^q,Y^{q+p})\) and the bilinear evaluation give an actual homomorphism from its tensor factor. For bounded \(X\), decomposing the product in (3.1) as its finite biproduct makes this the given evaluation on the matched source index and zero on every unmatched source index. The chain equation on a generating tensor is

The two precomposition terms cancel with exactly the signs in (3.1). Thus \(\varepsilon:H\otimes X\to Y\) is a chain map. This proves the bounded-source assertion with all matched and unmatched components. In abelian groups the same construction also works for an unbounded \(X\): direct sums exist, their elements have finite summand support, and (5.4) uses only one product projection at a time. The same generator calculation proves its chain equation without an infinite summation.

6. Stupid truncations are functors on actual complexes

For \(a\in\mathbb Z\), define \(\sigma^{\ge a}X\) to have term \(X^n\) for \(n\ge a\) and zero otherwise, with the original differential when both degrees are retained and zero at the cut. Define \(\sigma^{\le a}X\) in the same way for \(n\le a\). The square-zero and chain-map equations hold at retained positions by the original equations, and at the boundary because the relevant component is zero. Applying the same cuts to chain maps preserves identities, composition and addition. Thus these are additive functors on \(C(\mathcal C)\), landing in \(C^+\) and \(C^-\), respectively. Set \(\sigma^{>a}=\sigma^{\ge a+1}\) and \(\sigma^{<a}=\sigma^{\le a-1}\).

They need not preserve homotopies. Take a nonzero object \(A\) and the two-term identity complex in degrees \(a-1,a\). It is contractible by the identity homotopy from degree \(a\) to degree \(a-1\). Its truncation \(\sigma^{\ge a}\) is \(A\) concentrated in degree \(a\), whose identity cannot be a boundary: every possible adjacent homotopy component is zero. A functor on \(K\) agreeing with this truncation on actual complexes would send the original isomorphism to zero to an isomorphism of this nonzero object to zero, which is impossible. This proves the claimed failure on the quotient. An upper truncation has the analogous obstruction.

7. Three complete truncation triangles and their boundary maps

The general split-underlying-sequence calculation in Differential objects, curvature and cone homotopies, Section 7 applies in \(\operatorname{Gr}(\mathcal C)\). For a degreewise split sequence of complexes \(A\xrightarrow iB\xrightarrow pQ\), choose the underlying degreewise projection \(k:B\to A\) and section \(h:Q\to B\). That proof supplies an actual cone homotopy equivalence \(C_i\to Q\), its inverse and contraction difference. The resulting entire distinguished triangle has last map

This uses the underlying degree shift on \(k\), with the signed differential on the translated complex. The proof does not require the splitting maps \(k,h\) themselves to be chain maps.

First use the degreewise direct-sum decomposition

The inclusion and projection are chain maps by their boundary definitions. Their underlying splitting maps are the evident degreewise projections and inclusions. Equation (7.1) has its only potentially nonzero component

Hence \(\sigma^{>a}X\to X\to\sigma^{\le a}X\xrightarrow\delta(\sigma^{>a}X)[1]\) is distinguished. The chain equation of \(\delta\) at either boundary is also immediate from \(d_X^{a+1}d_X^a=d_X^ad_X^{a-1}=0\).

Second use

Here \(X^a[-a]\) means the original object \(X^a\), first concentrated in degree zero and then shifted by \(-a\); its sole nonzero term is in degree \(a\). The sequence is degreewise split with chain inclusion and projection. Its last arrow again has the only component (7.3), since (7.1) uses the same outgoing boundary differential. This proves the second entire distinguished triangle.

For the third triangle, let \(f:X\to Y\) be a chain map and set

At degree \(n\), the cone has the two summands \(X^{n+1},Y^n\). The inclusion from \(A\) retains the first exactly when \(n\ge a\), and the second exactly when \(n\ge a+1\). The projection onto \(Q\) retains the first exactly when \(n\le a-1\), and the second exactly when \(n\le a\). Thus these summands give a degreewise split sequence. Inclusion and projection are the diagonal strict-square cone maps of (1.4), induced by the actual truncation inclusions and projections on \(X,Y\); their chain equations follow from those strict squares. Therefore the split-sequence theorem gives the distinguished triangle

To see every boundary term, (7.1) gives

and all other components are zero. The target differential from its degree \(a-1\) to degree \(a\) is the negative of the restricted cone differential, so applying it to \(\delta^{a-1}\) gives \((0,-f^{a+1}d_X^a)\). Applying \(\delta^a\) after the quotient differential gives \((0,-d_Y^af^a)\); these agree by the chain-map equation for \(f\). The adjacent boundary equations use \(d_X^2=d_Y^2=0\). This verifies the signs independently of the abstract split-sequence comparison. It also shows why a cone of a truncated map is not generally the truncation of the original cone: the first cone summand already has its degree shifted.

8. Four graded exercises with complete solutions

Exercise 1 (foundation: a two-term arithmetic cone). Let \(X^0=X^1=\mathbb Z\), differential two, and \(Y^0=Y^1=\mathbb Z\), differential three, with zero elsewhere. Take \(f^0=2,f^1=3\). Check the chain map, compute its whole cone and compute its cohomology.

Solution. The chain equation is \(3\cdot2=3\cdot2\). Formula (2.1) gives the cone

in degrees \(-1,0,1\). The composite is zero. The first map is injective, so \(H^{-1}=0\). The middle kernel is \(\{(s,-s):s\in\mathbb Z\}\), and the incoming image consists of its even multiples; hence \(H^0=\mathbb Z/2\). The final cokernel is \(\mathbb Z/3\), giving \(H^1\). All other terms and cohomology groups are zero. Both differentials, the shifted degree and the two signs are thus explicit.

Exercise 2 (intermediate: an unbounded coproduct). In \(K(\mathsf{Ab})\), form the coproduct of \(\mathbb Z[i]\) for \(i\ge0\). Describe it and prove that this family has no coproduct in \(K^b(\mathsf{Ab})\).

Solution. Each summand is \(\mathbb Z\) in degree \(-i\) with zero differential. The termwise sum has \(\mathbb Z\) in each degree \(n\le0\), zero in positive degrees and zero differential. Section 2 proves that its actual injections make it the coproduct in \(K\). It is unbounded below.

Suppose a bounded complex \(B\) were a coproduct in \(K^b\). For a sufficiently large \(i\), \(B^{-i}=0\), so every chain map \(\mathbb Z[i]\to B\) is zero. Its purported coproduct injection is therefore zero in \(K^b\). The family of maps to \(\mathbb Z[i]\) which is its identity on that summand and zero on every other summand would require a map from \(B\) whose composite with that zero injection is the nonzero identity. This is impossible. The identity is nonzero because a zero-differential one-term complex has no nonzero homotopy boundary. Thus no alternative bounded object can supply the coproduct.

Exercise 3 (advanced: inspect the dual shift sign). Apply \(G=\operatorname{Hom}_{\mathbb Z}(-,\mathbb Z)\) to the two-term complex \(X^0=X^1=\mathbb Z\) with differential \(m\). Compute \(GX\), \(G(X[1])\), \((GX)[-1]\) and the comparison (4.3). Give the null homotopy of \(d_X:X\to X[1]\).

Solution. The complex \(GX\) has \(\mathbb Z\) in degrees \(-1,0\), with differential \(-m\). The complex \(G(X[1])\) has \(\mathbb Z\) in degrees \(0,1\), with differential \(-m\), because \(X[1]\) has differential \(-m\). The inverse shift \((GX)[-1]\) has those same degrees and differential \(+m\). The comparison is minus the identity in degree zero and the identity in degree one. Its chain equation is \(1\cdot(-m)=m\cdot(-1)\), so it is the required chain isomorphism. The unsigned identity would fail for a nonzero \(m\) with \(2m\ne0\).

For the differential itself as a shifted chain map, use (5.2): the only nonzero homotopy component is \(h^1=1_{\mathbb Z}:X^1\to X[1]^0\), and \(h^0=0\). Its boundary in degree zero is \(-m\cdot0+1\cdot m=m\), and in degree one is zero. This is exactly \(d_X\).

Exercise 4 (synthesis: truncating a cone at the wrong boundary). Let \(a=0\) and \(f\) be the identity on \(\mathbb Z\) concentrated in degree one. Compute the three terms in (7.6). Compare \(C_{\sigma^{>0}f}\) with \(\sigma^{>0}(C_f)\).

Solution. Both source and target are unchanged by \(\sigma^{>0}\), and both are killed by \(\sigma^{\le0}\). Hence \(C_{\sigma^{>0}f}=C_f\) and \(C_{\sigma^{\le0}f}=0\). The cone \(C_f\) has \(\mathbb Z\) in degrees zero and one with identity differential, so it is contractible. Triangle (7.6) is \(C_f\xrightarrow1C_f\to0\to C_f[1]\), with all the indicated cone maps and zero last arrow.

But \(\sigma^{>0}(C_f)\) keeps only its degree-one copy of \(\mathbb Z\), with zero differential. Its identity is not null-homotopic, so it is not isomorphic in \(K\) to the contractible cone. Thus those two candidate first terms differ even in the homotopy category. The cause is the shifted \(X^{n+1}\) summand in (1.4), exactly the boundary distinction described before (7.7).

References

• Pierre Schapira, An Introduction to Categories and Homological Algebra, Sections 4.2–4.4, for ordinary complexes, their signs and the Hom construction.
• The Stacks Project Authors, Cones and termwise split sequences and Distinguished triangles in the homotopy category, for the additive-category interfaces. The positive last projection and its resulting negative split-sequence connecting map used here are proved in the linked differential-object lesson.