Draft lesson.

Smart truncations and image boundary homotopies

Original text public domain (CC0).

A truncation can keep the cohomology on one side of a chosen degree while changing the boundary object. Kernels and cokernels give two familiar constructions. Keeping an additional image term gives two more constructions which fit directly into short exact sequences. All four preserve homotopies, but their boundary homotopies have different types.

Let \(\mathcal A\) be any abelian category, with the size conventions of Universes and small categories. We use cochain complexes and the signed shifts of Graded complexes, contravariant signs and stupid truncations, Section 1. Cohomology in abelian categories with translation, Sections 3 and 5 supplies the complete cycle, boundary, quotient and homotopy-invariance interfaces. No projective or injective objects are assumed.

1. Four constructions and their natural maps

For a complex \(X\), write

Use the inclusions \(z_X^k:Z_X^k\to X^k\), \(i_X^k:B_X^k\to X^k\), the quotient \(q_X^k:X^k\to Q_X^k\), and the image factorization

Square zero gives a unique monomorphism \(B_X^k\to Z_X^k\) whose composite with \(z_X^k\) is \(i_X^k\). It also gives \(d_X^{k+1}i_X^{k+1}=0\): cancel the epimorphism \(e_X^k\) in the square-zero equation. Kernels and images will always mean these subobjects with their specified inclusions.

For a fixed integer \(n\), set \(U_nX=\tau_{\le n}X\), \(V_nX=\widetilde\tau_{\le n}X\), \(W_nX=\widetilde\tau_{\ge n}X\), and \(L_nX=\tau_{\ge n}X\). Their terms are

The boundary differential of \(U_nX\) is the unique \(\widehat d_X^{n-1}:X^{n-1}\to Z_X^n\) with \(z_X^n\widehat d_X^{n-1}=d_X^{n-1}\). For \(V_nX\), it is \(e_X^n:X^n\to B_X^{n+1}\). For \(W_nX\), it is \(i_X^n:B_X^n\to X^n\), from degree \(n-1\) to degree \(n\). For \(L_nX\), it is the unique \(\overline d_X^n:Q_X^n\to X^{n+1}\) with \(\overline d_X^nq_X^n=d_X^n\). All other retained differentials are the original ones, and differentials incident to a zero term are zero.

These are complexes. At a kernel boundary, cancel \(z_X^n\) to see that \(\widehat d_X^{n-1}d_X^{n-2}=0\). At an image boundary, cancel the image inclusion in \(e_X^nd_X^{n-1}=0\), or use \(d_X^ni_X^n=0\) at the other image boundary. At a quotient boundary, cancel \(q_X^n\) in \(d_X^{n+1}\overline d_X^nq_X^n=0\). Thus every changed boundary has square zero in \(\mathcal A\).

A chain map \(f:X\to Y\) induces \(f_Z^k,f_B^k,f_Q^k\), characterized by

The image map also satisfies \(f_B^{k+1}e_X^k=e_Y^kf^k\), by cancelling \(i_Y^{k+1}\). Use these induced maps at the changed terms and \(f^k\) elsewhere. Equations (1.4), and cancellation of the boundary monomorphisms or epimorphisms, prove all the chain-map equations. The same uniqueness proves identity and composition laws. Factoring sums through kernels, images and cokernels proves additivity. Therefore (1.3) defines four additive functors on actual complexes.

There is a natural chain of complex maps

The first is the identity below \(n\), \(z_X^n\) in degree \(n\), and zero above. The second is the identity through \(n\), \(i_X^{n+1}\) in degree \(n+1\), and zero above. The third is zero below \(n-1\), \(e_X^{n-1}\) in degree \(n-1\), and the identity from \(n\) onward. The fourth is zero in degree \(n-1\), \(q_X^n\) in degree \(n\), and the identity above it. The changed chain equations are respectively \(e_X^nz_X^n=0\), (1.2), (1.2), and \(q_X^ni_X^n=0\) together with \(\overline d_X^nq_X^n=d_X^n\). Their remaining equations are the original ones or zero. Equations (1.4) prove naturality of every map. The display denotes a chain of maps; it does not assert that these five objects form an exact sequence.

2. Five short exact sequences, with their actual endpoints

The complex \(H_X^n[-n]\) has its sole nonzero term \(H_X^n\) in degree \(n\). We have natural short exact sequences of complexes

Here the ordered cone is \(C_f^k=A^{k+1}\oplus D^k\), with differential \(\begin{pmatrix}-d_A&0\\f&d_D\end{pmatrix}\), for \(f:A\to D\), as in the graded prerequisite. In particular the last quotient is \(B_X^{n+1}\xrightarrow{1}B_X^{n+1}\) in degrees \(n,n+1\), with the positive identity differential.

Proof of (2.1). Below degree \(n\), the inclusion is the identity and the quotient is zero. In degree \(n\), it is the natural sequence \(0\to B_X^n\to Z_X^n\to H_X^n\to0\). Above it all three terms are zero. The boundary square commutes because both ways from \(X^{n-1}\) to \(X^n\) are \(d_X^{n-1}\). The projection is a chain map because the image of \(\widehat d_X^{n-1}\) is \(B_X^n\), which dies in \(H_X^n\). This proves degreewise exactness and all differential compatibilities.

Proof of (2.2). The differential induces an epimorphism \(\epsilon_X^n:Q_X^n\to B_X^{n+1}\) with \(\epsilon_X^nq_X^n=e_X^n\). It is epic because its composite with \(q_X^n\) is epic. Its kernel is \(H_X^n\), by the cycle/quotient identification in the cohomology prerequisite. Use that kernel inclusion in degree \(n\), and use \(\epsilon_X^n\) followed by the identity in degrees above \(n\) for the quotient map. At degree \(n\), the quotient boundary equation is \(i_X^{n+1}\epsilon_X^n=\overline d_X^n\), obtained by precomposing with \(q_X^n\). The differential kills its kernel. At degrees above \(n\), exactness is \(0\to0\to X^k\xrightarrow{1}X^k\to0\); below it every term is zero. This proves the assertion.

Proof of (2.3). The inclusion is the identity below \(n\), \(z_X^n\) at \(n\), and zero above. Its quotient is zero below \(n\), \(X^n/Z_X^n\) at \(n\), and \(X^k\) above. The coimage/image isomorphism identifies \(X^n/Z_X^n\) with \(B_X^{n+1}\), and the quotient map there becomes \(e_X^n\). The induced differential from this term is \(i_X^{n+1}\), since its composite with \(e_X^n\) is \(d_X^n\). Thus the quotient, with its actual differential, is precisely \(W_{n+1}X\). This proves the sequence as complexes, not only as graded objects.

Proof of (2.4). The inclusion is the identity below \(n\), \(i_X^n\) at \(n\), and zero above. The quotient at \(n\) is \(Q_X^n=X^n/B_X^n\), and its differential is \(\overline d_X^n\). Below \(n\) the quotient is zero; above it it agrees with \(X\). The image factorization (1.2) checks the incoming boundary of the inclusion, and \(q_X^ni_X^n=0\) checks the incoming boundary of the quotient. Hence the quotient is \(L_nX\) with the stated maps.

Proof of (2.5). The inclusion is the first map of (1.5). Its quotient is zero below \(n\), \(X^n/Z_X^n\simeq B_X^{n+1}\) at \(n\), \(B_X^{n+1}\) at \(n+1\), and zero above. Since the differential before quotienting is \(e_X^n\), the coimage/image comparison makes its quotient differential the identity. That is the displayed identity cone with exactly the shift \([-n-1]\). Its quotient map is \(e_X^n\) at \(n\), the identity at \(n+1\), and zero elsewhere. All boundary equations have now been checked.

Kernels, image factorizations and quotient maps in each proof are natural under (1.4); their uniquely induced arrows are therefore natural under chain maps as well. Exactness of complexes is degreewise by the abelian-category construction in the cohomology prerequisite, so these degreewise arguments establish all five short exact sequences in \(C(\mathcal A)\).

3. Cohomology at each boundary and the comparison quasi-isomorphisms

Proposition 3.1. The maps in (1.5) induce natural identifications

Proof. For \(U_nX\), degrees strictly below \(n-1\) have the original incoming and outgoing maps. At degree \(n-1\), replacing the target of \(d_X^{n-1}\) by its factor into \(Z_X^n\) leaves its kernel unchanged, because \(z_X^n\) is monic. The incoming image is unchanged. At degree \(n\), the cycles are all of \(Z_X^n\) and the boundaries are \(B_X^n\), giving \(H_X^n\). Degrees above \(n\) have zero terms.

For \(V_nX\), the only original outgoing differential replaced is \(d_X^n\) by \(e_X^n\). Their kernels agree because \(i_X^{n+1}\) is monic. Therefore degree \(n\) gives \(Z_X^n/B_X^n\). In degree \(n+1\), the cycles are \(B_X^{n+1}\) and the incoming differential is epic onto them, so cohomology is zero. All lower degrees are unchanged and all higher ones are zero. The map \(U_nX\to V_nX\) is the identity on the resulting cycle quotient even at \(n\).

For \(W_nX\), the lowest retained differential \(i_X^n\) is monic, so degree \(n-1\) has zero cohomology. At degree \(n\), the outgoing kernel is \(Z_X^n\) and the incoming image is \(B_X^n\), giving \(H_X^n\). Degrees above it have their original incoming images and outgoing kernels. For \(L_nX\), the cycles in degree \(n\) are \(\ker\overline d_X^n\simeq H_X^n\), with no incoming term. In degree \(n+1\), the image of \(\overline d_X^n\) is the original \(B_X^{n+1}\), since \(\overline d_X^nq_X^n=d_X^n\) and \(q_X^n\) is epic. Higher degrees are unchanged. The map \(W_nX\to L_nX\) gives the canonical cycle/quotient isomorphism at \(n\) and the identity on cohomology above it. This proves every position in (3.1). Naturality follows from (1.4). \(\square\)

It follows that \(U_nX\to V_nX\) and \(W_nX\to L_nX\) are quasi-isomorphisms. Each of the four functors sends any quasi-isomorphism to a quasi-isomorphism, by applying (3.1) to the given map. The maps \(U_nX\to X\) and \(X\to L_nX\) are quasi-isomorphisms when the cohomology omitted on the corresponding side is zero.

The quotient in (2.5) is contractible: its only nonzero homotopy component is the identity from degree \(n+1\) back to degree \(n\). This verifies \(1=dh+hd\) in both degrees. It does not make the short exact sequence split in each degree. A quasi-isomorphism in this paragraph consequently need not be an isomorphism in \(K(\mathcal A)\); Exercise 3 gives an explicit example.

4. All four actual boundary homotopies

Let \(f:X\to Y\) be a null-homotopic chain map. Choose maps \(h^k:X^k\to Y^{k-1}\) satisfying

Restriction to \(Z_X^k\) kills the second term, and the first factors through \(B_Y^k\). Therefore \(H^k(f)=0\), as proved in the cohomology prerequisite. For two homotopic maps, apply this to their difference to obtain equality of their cohomology maps. We now construct an actual homotopy for each truncated map.

Kernel truncation. For \(U_n(f)\), put

These maps have targets \((U_nY)^{j-1}\). Away from \(n-1,n\), the equation is (4.1) or an equation of zero maps. At \(n-1\), the changed second term is \(h^nz_X^n\widehat d_X^{n-1}=h^nd_X^{n-1}\), so (4.1) gives the equation. At \(n\), postcompose the desired equation with \(z_Y^n\). Its left side becomes \(f^nz_X^n\), and its right side becomes \(d_Y^{n-1}h^nz_X^n\); the missing second term of (4.1) is zero because \(d_X^nz_X^n=0\). Cancel the monomorphism \(z_Y^n\). Thus \(U_n(f)=d_{U_nY}k+kd_{U_nX}\) in every degree.

Image-extended upper truncation. For \(V_n(f)\), put

At \(n\), the changed second term is \(h^{n+1}i_X^{n+1}e_X^n=h^{n+1}d_X^n\), giving (4.1). At \(n+1\), postcompose with \(i_Y^{n+1}\). The equation becomes \(f^{n+1}i_X^{n+1}=d_Y^nh^{n+1}i_X^{n+1}\), which is (4.1) because \(d_X^{n+1}i_X^{n+1}=0\). Cancel \(i_Y^{n+1}\). All other degrees retain (4.1) or are zero. This homotopy uses a restriction of its source at the added image term.

Cokernel truncation. For \(L_n(f)\), take

At \(n\), precompose the desired equation with \(q_X^n\). The left side becomes \(q_Y^nf^n\). On the right it becomes \(q_Y^nh^{n+1}\overline d_X^nq_X^n=q_Y^nh^{n+1}d_X^n\). These agree by (4.1), since \(q_Y^nd_Y^{n-1}=0\). Cancel the epimorphism \(q_X^n\). At \(n+1\), the changed first term is \(\overline d_Y^nq_Y^nh^{n+1}=d_Y^nh^{n+1}\), giving (4.1). The remaining degrees are unchanged or zero. This homotopy uses a quotient of its target at the first nonzero backward map.

Image-extended lower truncation. For \(W_n(f)\), the formula is

Here \(k^n:X^n\to B_Y^n=(W_nY)^{n-1}\). Its first factor \(h^n\) has target \(Y^{n-1}\), and the second \(e_Y^{n-1}\) has target \(B_Y^n\). Thus the old \(h^n\) is not being regarded as a map into the image. At degree \(n\), the changed first term is \(i_Y^ne_Y^{n-1}h^n=d_Y^{n-1}h^n\), giving (4.1). At \(n-1\), the required equality is

Postcompose with \(i_Y^n\). The left side becomes \(f^ni_X^n\), whereas the right side becomes \(d_Y^{n-1}h^ni_X^n\). They agree by (4.1) and \(d_X^ni_X^n=0\). Cancel the monomorphism \(i_Y^n\), proving (4.6). Above \(n\), both terms are the original ones; below \(n-1\), every term is zero. This proves the last homotopy without a lift or a splitting.

5. Descent, support bounds and what the comparisons imply

The formulas in Section 4 prove that each additive truncation functor carries the null-homotopy ideal into the corresponding null-homotopy ideal. If two chain maps differ by a null homotopy, their images under any truncation do too. Thus taking their homotopy classes gives well-defined additive functors

Identities and composition are inherited from the actual functors in Section 1; the induced maps in (1.5) remain natural after quotienting Hom groups. The support bounds are explicit: \(U_n\) vanishes above \(n\), \(V_n\) above \(n+1\), \(W_n\) below \(n-1\), and \(L_n\) below \(n\). A zero original term has zero cycles, image and quotient when the corresponding adjacent map requires it, so an existing bound on the other side remains a bound after these constructions. In particular each functor sends bounded complexes to bounded complexes. The homotopies just constructed have their source and target in exactly these full support subcategories, so the quotient categories in (5.1) are the stated ones.

A quasi-isomorphism is a chain map inducing an isomorphism on every \(H^k\). This property is unchanged upon replacing its chain representative by a homotopic one. Formula (3.1) and the four homotopies establish the same quasi-isomorphism comparisons in the corresponding homotopy categories. They make no claim that truncation is exact on every short exact sequence, or that the quasi-isomorphisms are already invertible there. The five specific exact sequences were proved individually in Section 2.

6. Four graded exercises with complete solutions

Exercise 1 (introductory: boundary calculation). Over \(\mathbb Q\), let \(X\) have terms \(\mathbb Q,\mathbb Q^3,\mathbb Q\) in degrees \(0,1,2\), with differentials \(a\mapsto(a,0,0)\) and \((a,b,c)\mapsto c\). Compute all four truncations at \(n=1\), their cohomology and their two comparison maps.

Solution. The square is zero. In degree \(1\), \(B_X^1=\mathbb Q e_1\), \(Z_X^1=\mathbb Q e_1\oplus\mathbb Q e_2\), and \(Q_X^1=\mathbb Q^3/\mathbb Q e_1\). Also \(B_X^2=\mathbb Q\). Hence \(U_1X\) is \(\mathbb Q\to\mathbb Q e_1\oplus\mathbb Q e_2\) in degrees \(0,1\), with \(a\mapsto ae_1\). The complex \(V_1X\) is the original three-term complex. The complex \(W_1X\) is \(\mathbb Q e_1\hookrightarrow\mathbb Q^3\to\mathbb Q\) in degrees \(0,1,2\), with the last map the third coordinate. The complex \(L_1X\) is \(\mathbb Q^3/\mathbb Q e_1\to\mathbb Q\) in degrees \(1,2\), sending the class of \((a,b,c)\) to \(c\).

All four have their sole nonzero cohomology \(\mathbb Q e_2\) in degree \(1\). The map \(U_1X\to V_1X\) is the identity in degree \(0\), the inclusion of the first two coordinate directions in degree \(1\), and zero in degree \(2\). The map \(W_1X\to L_1X\) is zero in degree \(0\), the quotient by \(\mathbb Q e_1\) in degree \(1\), and the identity in degree \(2\). Each induces the identity on the displayed cohomology generator. This checks the boundary kernels and images, not only the dimensions.

Exercise 2 (intermediate: a homotopy which must change its target). Let \(X=(\mathbb Z\xrightarrow{2}\mathbb Z)\) in degrees \(0,1\), and let \(f=2\,1_X\). Give null homotopies after all four truncations at \(1\). Explain why the lower image boundary cannot use the original homotopy unchanged.

Solution. The homotopy \(h^1=1_{\mathbb Z}:X^1\to X^0\), with all other components zero, gives \(dh+hd=2\,1_X\). Both \(U_1X\) and \(V_1X\) are \(X\), since \(Z_X^1=\mathbb Z\) and \(B_X^2=0\); their homotopy is the original one. The complex \(L_1X\) is \(\mathbb Z/2\mathbb Z\) in degree \(1\), and its map induced by \(f\) is zero, so the zero homotopy works.

The remaining complex is \(W_1X=(2\mathbb Z\hookrightarrow\mathbb Z)\) in degrees \(0,1\). The induced \(f\) multiplies both terms by \(2\). Here \(e_X^0:\mathbb Z\to2\mathbb Z\) sends \(a\) to \(2a\). Formula (4.5) gives \(k^1:\mathbb Z\to2\mathbb Z\), \(a\mapsto2a\). In degree \(0\), its composite with the inclusion sends \(b\in2\mathbb Z\) to \(2b\); in degree \(1\), the inclusion composed with \(k^1\) sends \(a\) to \(2a\). These are exactly the two homotopy equations. The original identity \(\mathbb Z\to\mathbb Z\) does not even have image contained in \(2\mathbb Z\), so treating it as a map into the added image term would be undefined.

Exercise 3 (advanced: a contractible quotient without a homotopy equivalence). Let \(X=(\mathbb Z\twoheadrightarrow\mathbb Z/2\mathbb Z)\) in degrees \(0,1\), with its quotient differential. Show that \(U_0X\to V_0X\) is a quasi-isomorphism with contractible quotient but is not invertible in \(K(\mathbf{Ab})\).

Solution. We have \(U_0X=2\mathbb Z\) concentrated in degree \(0\), while \(V_0X=X\), since \(B_X^1=\mathbb Z/2\mathbb Z\). The comparison \(u\) is the inclusion \(2\mathbb Z\hookrightarrow\mathbb Z\) in degree \(0\). Both cohomologies are \(2\mathbb Z\) in degree \(0\) and zero elsewhere, and \(u\) induces the identity on them. Its quotient is \(\mathbb Z/2\mathbb Z\xrightarrow{1}\mathbb Z/2\mathbb Z\) in degrees \(0,1\); the identity backward map contracts it.

If \([u]\) had an inverse represented by \(g:X\to U_0X\), then \(gu\) would be homotopic to \(1_{U_0X}\). A homotopy between maps of a complex concentrated in one degree has zero boundary, so \(g^0|_{2\mathbb Z}=1_{2\mathbb Z}\) would hold strictly. In particular \(g^0(2)=2\), forcing \(g^0(1)=1\) as an equality of integers. But its target is \(2\mathbb Z\), a contradiction. Thus no such inverse exists. In fact the two objects cannot be isomorphic in \(K\) by another map either: any chain map \(2\mathbb Z[0]\to X\) inducing a cohomology isomorphism must send the generator \(2\) to \(2\) or \(-2\), so it is \(u\) or \(-u\). An isomorphism in \(K\) induces a cohomology isomorphism, and invertibility of either of these maps would imply invertibility of \(u\). The degree-zero sequence \(0\to2\mathbb Z\to\mathbb Z\to\mathbb Z/2\mathbb Z\to0\) does not split; indeed a section would send the nonzero element of order two to an element of order two in \(\mathbb Z\), where none exists.

Exercise 4 (advanced: one surviving cohomology degree). Suppose \(H_X^k=0\) for every \(k\ne n\). Construct a canonical span of quasi-isomorphisms between \(X\) and \(H_X^n[-n]\), and determine whether this forces an isomorphism in the homotopy category.

Solution. The span is

The left map is the natural inclusion from (1.5). It induces the identity on cohomology at and below \(n\) by (3.1); above \(n\) both cohomologies are zero by hypothesis. The right map is the quotient in (2.1), namely \(Z_X^n\twoheadrightarrow H_X^n\) in degree \(n\) and zero elsewhere. It is a chain map because this quotient kills the incoming image \(B_X^n\). It induces the identity on \(H_X^n\), while all other cohomology objects of source and target are zero. Both maps are therefore quasi-isomorphisms, with no chosen splitting or injective resolution. The span is natural for maps of complexes with this cohomology condition, by the naturality already established. It need not give an isomorphism in \(K(\mathcal A)\): the complex and single surviving degree in Exercise 3 satisfy the hypothesis and its inclusion has no homotopy inverse.

References

• The Stacks Project, Section 12.15, Truncation of complexes, for the canonical kernel and cokernel constructions and their cohomology descriptions. The image-extended constructions, all five exact sequences and every boundary homotopy are proved above.
• Graded complexes, contravariant signs and stupid truncations, Section 1, for signed degrees, support and the ordered cone convention.
• Cohomology in abelian categories with translation, Sections 3 and 5, for the natural cycle/quotient identities and homotopy-invariant cohomology in any abelian category.