Draft lesson.

Cohomological bounds, denominators and truncation adjunctions

Original text public domain (CC0).

Cohomology bounds need not be bounds on the terms of a chosen complex. Smart truncation changes the representative so that the two bounds agree. Applying that change to denominators also controls every derived morphism, including its equality relations. This proves the bounded embeddings and truncation adjunctions without a supply of injective or projective objects.

Let \(\mathcal A\) be any abelian category. Use a fixed ambient Grothendieck universe containing its objects, arrows and the integers as sets, as in Universes and small categories. The Hom sets below are in that ambient universe; they need not belong to a smaller universe in which \(\mathcal A\) was originally locally small. Complexes are cochain complexes, with \(X[r]^n=X^{n+r}\) and differential \((-1)^r d_X\).

We use Derived morphisms through quasi-isomorphism roofs, Sections 1–4 for both fraction calculi and their equality tests. Triangles after quasi-isomorphism localization, Sections 1–6 supplies the derived triangulation, exact Hom sequences and natural short-exact-sequence triangles. Smart truncations and image boundary homotopies, Sections 1–5 proves all four truncations on complexes and homotopy classes, including every boundary homotopy and the specific short exact sequences used here.

1. Four localizations and cohomology detection

The quotient construction also has a form before ordinary degrees are chosen. Let \((\mathcal B,T)\) be an abelian category with an additive translation equivalence. Cohomology in abelian categories with translation, Sections 3–4 proves that cohomology is cohomological on the triangulated homotopy category \(K_c(\mathcal B)\) of square-zero differential objects, and that \(H(TX)\simeq TH(X)\) with the coherent signed translation. Let \(N\) be its full subcategory with \(H(X)=0\). It contains zero and is replete. The displayed shift comparison makes it stable under translation and its inverse. Exactness of the cohomology sequence and its rotations proves triangle two-out-of-three. Finally cohomology sends any retraction equation to a retraction equation; a retract of zero has zero identity. Thus \(N\) is also closed under retracts.

Null systems, roofs and the retract kernel, Sections 1–7 therefore constructs the triangulated quotient \(D_c(\mathcal B)=K_c(\mathcal B)/N\). Its multiplicative arrows are precisely quasi-isomorphisms: the cohomology sequence of a completed arrow, and its adjacent shifts, says its cone is acyclic exactly when its cohomology map is invertible. The quotient inverts those arrows, including the images of differential quasi-isomorphisms before passage to homotopy. The ambient universe must contain the actual denominator diagrams, as specified above. This construction does not assert local smallness in a smaller starting universe. In the rest of the lesson we specialize to ordinary complexes in \(\mathcal A\).

Write \(K^+,K^-,K^b\) for the full homotopy categories of complexes whose terms are respectively bounded below, bounded above, or bounded on both sides. Write \(K\) for all complexes. A complex is acyclic when all its cohomology objects vanish. The acyclic objects in each of these four categories form a thick triangulated subcategory: shifts preserve vanishing; the long exact cone sequence proves closure under cones; and a retract of an acyclic complex induces a retract of zero in every cohomological degree. A retract of zero has identity zero and is a zero object.

Localize each category at its quasi-isomorphisms, obtaining

The constructions in the two localization prerequisites apply within each bounded category. Indeed finite sums, shifts and cones retain each kind of boundedness. The incoming homotopy pullback has terms \(X^n\oplus V^n\oplus Y^{n-1}\); the outgoing homotopy pushout is a cone. Their terms retain the relevant bounds too. The cancellation constructions use shifts and cones of acyclic objects in the same bounded category. Thus the fraction axioms, addition, composition, equality tests and triangle transport are all available there, with the same signed translation. This proves that (1.1) consists of triangulated localizations, rather than assuming that the bounded ones are already full subcategories of \(D\).

Let \(q:C(\mathcal A)\to K(\mathcal A)\) take a chain map to its homotopy class, and let \(Q:K(\mathcal A)\to D(\mathcal A)\) be the localization. The complex localization is their composite \(Qq\); its target is the same \(D\), not another quotient. The identical convention applies to each bounded version. Integer shifts are those fixed in the introduction. A distinguished triangle with first vertex \(X\) has its last arrow into \(X[1]\); an abbreviated last arrow always retains this codomain and the specified positive projection convention.

For \(X,Y\in K\), all three descriptions of derived Hom are available:

In the first index arrows point toward incoming refinements; in the second they point toward outgoing refinements. The third uses both directions, and a representative \(f:U\to V\) represents \((Qt)^{-1}Qf(Qs)^{-1}\). Incoming and outgoing denominator categories in these orientations are filtered by the square and cancellation axioms. For fixed \(s\), the outgoing formula computes \(\operatorname{Hom}_D(U,Y)\). Precomposition by the invertible \(Qs\) identifies this with \(\operatorname{Hom}_D(X,Y)\), compatibly with incoming refinements. Taking the incoming colimit gives the third formula. Equivalently, its finite equality tests reduce to a common refinement in each coordinate. This argument also proves (1.2) in each bounded localization with its own denominators.

Every \(H^n\) inverts quasi-isomorphisms and so descends through each localization. It is cohomological: the long sequence is exact on image cone triangles by cone cohomology, and isomorphisms of triangles preserve that exactness. Consequently

The reverse implication uses the inverted quasi-isomorphism \(0\to X\); the forward implication applies cohomology. An ordinary arrow \(f:X\to Y\) is zero in the localization exactly when some incoming quasi-isomorphism makes it null-homotopic. The outgoing equality test gives the equivalent condition that some outgoing quasi-isomorphism makes it null-homotopic. Finally, an arbitrary derived arrow \(u\) is invertible exactly when every \(H^n(u)\) is invertible. Complete it to a distinguished triangle. Its third vertex has zero cohomology exactly when all these maps are isomorphisms, by the long exact sequence. Apply (1.3) and the exact Hom criterion for an invertible first arrow in a triangle.

2. Truncations control the entire denominator index

Set \(U_aX=\tau_{\le a}X\) and \(L_aX=\tau_{\ge a}X\). Explicitly their boundary terms are \(Z^aX\) and \(X^a/B^aX\), respectively. The other two constructions in the truncation prerequisite are \(V_aX=\widetilde\tau_{\le a}X\), which adds the image term \(B^{a+1}X\), and \(W_aX=\widetilde\tau_{\ge a}X\), which adds \(B^aX\) in degree \(a-1\). Their cohomology formulas show that all four preserve quasi-isomorphisms. They therefore descend, with their natural maps, to

The natural comparisons \(U_a\to V_a\) and \(W_a\to L_a\) are invertible after localization. The original boundary homotopies are part of this descent: the assertion does not assume that a functor on complexes automatically descends through homotopy.

If \(X\) has a lower cohomology bound \(b\), then \(U_aX,V_aX\) have cohomology only in \([b,a]\). If it has an upper cohomology bound \(b\), then \(L_aX,W_aX\) have cohomology only in \([a,b]\). The interval is empty when its lower endpoint exceeds its upper endpoint. Thus cutting the opposite side of a cohomologically bounded-below or bounded-above object gives bounded cohomology.

Proposition 2.1 (bounded denominator comparison). Suppose \(X^n=0\) for \(n>a\). In the first formula of (1.2), one may restrict to denominators \(s:U\to X\) with \(U^n=0\) for \(n>a\). The restriction preserves both representatives and their equality tests. Dually, if \(Y^n=0\) for \(n<a\), the outgoing formula may be restricted to denominators \(t:Y\to V\) with \(V^n=0\) for \(n<a\).

Proof. For any incoming denominator \(s:U\to X\), its source has no cohomology above \(a\). Hence \(j_U:U_aU\to U\) is a quasi-isomorphism. The restricted denominator \(s j_U\) has the asserted term bound. Since \(U_aX=X\), it is also \(U_a(s)\) under that identification.

Consider now a common refinement witnessing equality of two roofs with bounded sources \(U,V\). Apply \(U_a\) to the refinement diagram and to its homotopy equalities. It fixes \(U,V,X\), preserves its denominator conditions and leaves a common refinement with the required bound. Its numerators still have their original target: compose the truncated refinement inclusion with the original numerator. Naturality of \(j\) and the preserved homotopy equations show that the numerators agree on this refinement. Thus the same equality holds using the restricted index.

For the categorical initiality assertion before reversing incoming indices, fix a denominator \(T\to X\). The category of arrows from bounded denominators into it is nonempty by \(U_aT\to T\). Given two objects of that category, take a common denominator refinement over \(X\). Their two composites to \(T\) become equal after a further incoming cancellation for \(T\to X\). Truncate that refinement above \(a\). It still gives both arrows and the equality. This proves connectedness, and hence initiality; after reversal it is cofinality for the filtered colimit. The same argument equalizes parallel comparison arrows. It proves the complete index comparison as well as its concrete roof version.

For the outgoing assertion use \(V\to L_aV\). It is a quasi-isomorphism because \(V\) has the lower cohomology bound of \(Y\), and \(L_aY=Y\). Apply \(L_a\) to common outgoing refinements and their homotopy equations. For connectedness take a common outgoing refinement, equalize the composites from the fixed denominator by outgoing cancellation, and then apply \(L_a\). All arrows now have the lower term bound. This proves the dual cofinality and equality assertion. \(\square\)

The same conclusions hold if the incoming restricted index merely requires an upper bound, or the outgoing restricted index merely requires a lower bound. The stricter indices of Proposition 2.1 are contained in those indices. Any original representative and any finite comparison diagram can be refined into the stricter index, by the constructions just proved. In particular the comparison is stronger than an assertion that a convenient representative happens to exist.

3. Full embeddings and representatives with prescribed bounds

The inclusions of bounded homotopy categories induce fully faithful exact functors

Proof. For two bounded-above complexes use incoming roofs. Every original roof in \(D\) can have its source truncated above a common upper bound of its denominator target, by Proposition 2.1. Thus it is a roof in \(K^-\), proving fullness. Any equality of two such roofs in \(D\) has a common refinement which the same proposition makes bounded above. Hence that equality already holds in \(D^-_{\mathrm{loc}}\), proving faithfulness.

For two bounded-below complexes use outgoing roofs and the lower-bound part of Proposition 2.1. It proves fullness and faithfulness by the same representative and equality arguments. For bounded complexes first work inside \(K^-\), whose localization has just been proved fully faithful in \(D\). Use outgoing roofs in that category. Apply a lower truncation to an outgoing denominator: it retains its existing upper term bound and makes it bounded below. The same operation retains upper bounds on the outgoing common refinements proving equality. This proves fullness and faithfulness of \(D^b_{\mathrm{loc}}\to D^-_{\mathrm{loc}}\). Composing gives the third assertion. All these functors preserve image triangles and the specified shifts, so are exact. \(\square\)

We now identify their images with full subcategories and write them as \(D^+,D^-,D^b\). Their essential images consist respectively of objects with bounded-below, bounded-above, or bounded cohomology. One inclusion follows from term bounds. For the converse a lower cohomology bound gives the quasi-isomorphism \(X\to L_aX\), and an upper one gives \(U_bX\to X\). When both hold, use the zigzag

Both maps are quasi-isomorphisms and the final complex has terms only in \([a,b]\). If \(a>b\) the asserted cohomology vanishes in every degree, so choose the zero complex instead.

More generally, let \(a,b\in\mathbb Z\cup\{-\infty,+\infty\}\), with \(a\le b\), and suppose \(H^nX=0\) outside \([a,b]\). Omit a truncation at an infinite endpoint in (3.2). This gives a representative whose terms are zero outside that same interval. The subcategory defined by either one-sided cohomology bound is triangulated when the bound may vary from object to object; the fixed interval subcategories themselves are full additive subcategories, with their indicated fixed bounds.

For finite \(a\le b\), the two cut orders \(L_aU_bX\) and \(U_bL_aX\) have a canonical isomorphism as complexes. If \(a<b\), their terms and differential factorizations agree, including the quotient at \(a\) and kernel at \(b\). If \(a=b\), the first is \(Z^aX/B^aX\) in degree \(a\), while the second is \(\ker(X^a/B^aX\to X^{a+1})\) in that degree. The canonical cycle/quotient isomorphism identifies them naturally with \(H^aX[-a]\). The isomorphisms commute with maps of complexes by uniqueness of the kernel and quotient factorizations, and hence with every derived arrow after localization.

In particular objects with cohomology only in degree zero are isomorphic to their \(H^0\) in degree zero. The degree-zero embedding of \(\mathcal A\) is fully faithful by the derived-triangle prerequisite, Theorem 6.2. Thus it is an equivalence onto this full subcategory, not merely a correspondence on objects.

4. The common-degree Hom formula and orthogonality

Theorem 4.1. Suppose \(X^n=0\) for \(n>a\) and \(Y^n=0\) for \(n<a\). There are natural isomorphisms

Proof. A chain map can have only its degree-\(a\) component nonzero. That component kills the incoming image \(B^aX\) and has image in \(Z^aY\). Since \(Z^aX=X^a\) and \(B^aY=0\), it factors uniquely as

Every middle arrow defines a chain map by this formula. A degree-minus-one homotopy has components \(X^n\to Y^{n-1}\). If its source can be nonzero then \(n\le a\), so its target is zero. Hence all homotopies are zero and distinct chain maps stay distinct in \(K\).

Use incoming denominators with source terms at most \(a\), as in Proposition 2.1. For each denominator \(s:U\to X\), the same calculation gives \(\operatorname{Hom}_K(U,Y)=\operatorname{Hom}_{\mathcal A}(H^aU,H^aY)\). The arrow \(H^a(s)\) is invertible. The map sending a numerator \(f\) to \(H^a(f)H^a(s)^{-1}\) identifies every group in this diagram with \(\operatorname{Hom}_{\mathcal A}(H^aX,H^aY)\); its transition maps become identities. The nonempty filtered colimit is therefore that same group. This proves (4.1), including surjectivity, injectivity and naturality. \(\square\)

The outgoing denominator computation gives the same specified map. Proposition 2.1 permits target terms at least \(a\). For \(t:Y\to V\) with that bound, chain and homotopy Hom from \(X\) to \(V\) are \(\operatorname{Hom}_{\mathcal A}(H^aX,H^aV)\), by (4.2). Send a numerator \(g\) to \(H^a(t)^{-1}H^a(g)\). The denominator induces an isomorphism on \(H^a\), so this again makes every transition an identity on \(\operatorname{Hom}_{\mathcal A}(H^aX,H^aY)\). The outgoing formula in (1.2) proves its bijectivity. Thus both variance computations agree with the actual cohomology map of the represented derived arrow.

Replacing objects by the representatives of Section 3 gives the same formula whenever \(H^nX=0\) for \(n>a\) and \(H^nY=0\) for \(n<a\). In particular

If the source cohomology vanishes above \(a\) and the target cohomology vanishes below \(a+1\), then

Indeed choose the upper-bounded source and lower-bounded target representatives. Every incoming numerator with the source bound of Proposition 2.1 is zero term by term, because its degree ranges have no overlap. Formula (1.2) proves (4.4).

For objects \(A,B\in\mathcal A\), define \(\operatorname{Ext}^r_{\mathcal A}(A,B)=\operatorname{Hom}_D(A[0],B[r])\) for every integer \(r\). Our shift convention puts \(B[r]\) in degree \(-r\). Hence (4.4) gives negative Ext vanishing, and (4.1) gives naturally

5. Three natural triangles and two adjunctions

The five exact rows of the truncation prerequisite yield the natural triangles

Here every unspecified last arrow is the positive cone projection composed with the inverse cone-to-quotient quasi-isomorphism of the derived-triangle prerequisite. If an ordered cone lists its shifted source first, swap its two summands to use this target-first convention; its positive last projection remains the same map.

For (5.1), use the exact row \(0\to U_aX\to X\to W_{a+1}X\to0\) and the natural quasi-isomorphism \(W_{a+1}X\to L_{a+1}X\). For (5.2), use \(0\to V_{a-1}X\to U_aX\to H^aX[-a]\to0\), and replace its first vertex by the naturally isomorphic \(U_{a-1}X\). For (5.3), use \(0\to H^aX[-a]\to L_aX\to W_{a+1}X\to0\), then the same last comparison. These are actual short exact rows of complexes with their boundary maps checked in that prerequisite. Their triangle naturality holds for chain maps, then for inverse quasi-isomorphisms, and consequently for all derived arrows.

Let \(D^{\le a}\) and \(D^{\ge a}\) denote the full subcategories defined by these fixed cohomology bounds. The inclusion \(D^{\le a}\hookrightarrow D\) has right adjoint \(U_a\), and the inclusion \(D^{\ge a}\hookrightarrow D\) has left adjoint \(L_a\). More precisely, the natural maps give bijections

For the first, apply \(\operatorname{Hom}_D(T,-)\) to (5.1). The two outside groups are Hom into \(L_{a+1}X[-1]\) and into \(L_{a+1}X\), whose cohomology starts at \(a+2\) and \(a+1\), respectively. Both vanish by (4.4). For the second use (5.1) with cutoff \(a-1\), and apply \(\operatorname{Hom}_D(-,T)\). The outside groups have source \(U_{a-1}X\) and \(U_{a-1}X[1]\), with upper cohomology bounds \(a-1\) and \(a-2\). Both vanish. Exactness proves the asserted bijections. The maps defining them are natural, so the bijections are natural in both variables. Their counit and unit are precisely \(U_aX\to X\) and \(X\to L_aX\); uniqueness in (5.4) proves their triangle identities.

The strict-bound notation is \(\tau_{<a}=U_{a-1}\) and \(\tau_{>a}=L_{a+1}\). The cut orders at the same degree both give \(H^aX[-a]\), as proved in Section 3. All these statements hold in the full unbounded derived category as well as in each bounded subcategory when their objects lie there.

6. Two explicit computations

Example 6.1 (a finite window inside an unbounded complex). Over a field \(k\), let \(X^n=k\) for every integer. Put \(d^{-1}=d^0=0\). Below degree \(-1\), put \(d^n=1\) for even \(n\) and zero for odd \(n\). Above degree zero, put \(d^n=1\) for odd \(n\) and zero for even \(n\). Consecutive products are zero, so this is a complex. Every term in either outer tail belongs to a consecutive pair \(k\xrightarrow1k\). Those pairs have zero cohomology. Degree zero has no incoming or outgoing differential, and thus \(H^0X=k\); all other cohomology vanishes.

Although this representative has nonzero terms in every degree, \(L_0U_0X\) is \(k[0]\). The zigzag (3.2) is an isomorphism to it in \(D(k)\). The cut-order calculation gives the same answer naturally, without taking an infinite direct sum or product in the proof. An endomorphism in \(D(k)\) of this object is exactly a scalar on \(H^0X\), by (4.1). The arbitrarily long acyclic tails add no derived endomorphisms.

Example 6.2 (a boundary unseen by cohomology). Consider the exact row of abelian groups

Its derived boundary is \(\delta:\mathbb Z/2\to\mathbb Z[1]\), with the cone prescription of Section 5. Every cohomology map of \(\delta\) is zero: its source has cohomology only in degree zero and its target only in degree minus one. Nevertheless \(\delta\ne0\). If it were zero, the exact Hom sequence of the row triangle, with source \(\mathbb Z/2\), would lift its identity to a derived arrow \(\mathbb Z/2\to\mathbb Z\). By degree-zero full faithfulness this would be an actual group section of reduction. Every group map from \(\mathbb Z/2\) to the torsion-free group \(\mathbb Z\) is zero, so no such section exists.

The same proof gives the general assertion: the boundary of \(0\to A\to E\to B\to0\) is zero exactly when that row splits. A zero boundary lifts \(1_B\) to an actual section by full faithfulness. Conversely a section lifts \(1_B\), so exactness of Hom makes the boundary zero. Thus cohomology detects isomorphisms and zero objects, while it can lose a nonzero arrow.

7. Four graded exercises with complete solutions

Exercise 1 (foundation: check the boundary term). Let \(X\) be \(\mathbb Z\xrightarrow6\mathbb Z\) in degrees \(-1,0\). Compute \(U_{-1}X\), \(L_0X\) and the triangle (5.1) with \(a=-1\). Determine which natural map is a quasi-isomorphism.

Solution. Multiplication by six has zero kernel, so \(U_{-1}X\) is zero. Its image is \(6\mathbb Z\), hence \(L_0X\) is \((\mathbb Z/6)[0]\). The natural map \(X\to L_0X\) is zero in degree \(-1\) and reduction in degree zero. It is a chain map because it kills \(6\mathbb Z\). The cohomology of \(X\) is zero in degree \(-1\) and \(\mathbb Z/6\) in degree zero; reduction induces identity under that quotient identification. It is therefore a quasi-isomorphism. The triangle is \(0\to X\to(\mathbb Z/6)[0]\to0\). The comparison \(0\to X\) is not a quasi-isomorphism because its degree-zero cohomology map is \(0\to\mathbb Z/6\).

Exercise 2 (intermediate: a common cutoff). Over a field let \(X\) be \(k^2\xrightarrow{(u,v)\mapsto u}k\) in degrees \(-1,0\), and let \(Y\) be \(k^2\xrightarrow{(u,v)\mapsto v}k\) in degrees \(0,1\). Compute \(\operatorname{Hom}_D(X,Y)\). Repeat after replacing the first differential by zero.

Solution. In the first case \(H^0X=0\), while \(H^0Y=k\), the first coordinate of \(Y^0\). Both term bounds meet at zero, so Theorem 4.1 gives zero derived Hom. Directly, the only possible chain-map component is \(f^0:k\to k^2\). It must kill the surjective incoming differential of \(X\), hence is zero. If that differential is replaced by zero, then \(H^0X=k\). The component \(f^0\) must have image in \(\ker d_Y^0=k\oplus0\), and so has the form \(t\mapsto(\lambda t,0)\). Every such component gives a chain map. Every homotopy has zero components by the degree bounds, and Proposition 2.1 shows that extra denominators introduce no additional arrows or identifications. Thus the Hom group is naturally \(k\), parametrized by \(\lambda\).

Exercise 3 (advanced: specify the adjunction shifts). Suppose \(T\) has no cohomology above \(a\). Prove directly that its maps to \(X\) factor uniquely through \(U_aX\to X\), and identify the precise lower cohomology bounds of the two tail objects whose Hom groups must vanish. Give the corresponding calculation for a target with no cohomology below \(a\).

Solution. Apply the exact covariant Hom sequence to (5.1). The relevant segment is

The last target has lower bound \(a+1\). Since \(H^n(Z[-1])=H^{n-1}Z\), the first target has lower bound \(a+2\). Orthogonality makes both outside groups zero. The middle arrow is therefore bijective, proving both existence and uniqueness of the factorization. For a target \(T\) with lower bound \(a\), use the triangle \(U_{a-1}X\to X\to L_aX\to U_{a-1}X[1]\). The contravariant Hom segment has outside groups \(\operatorname{Hom}(U_{a-1}X[1],T)\) and \(\operatorname{Hom}(U_{a-1}X,T)\). Their sources have upper bounds \(a-2\) and \(a-1\). Both vanish, so \(\operatorname{Hom}(L_aX,T)\to\operatorname{Hom}(X,T)\) is bijective. These are exactly the adjunction maps and their actual shift bounds.

Exercise 4 (synthesis: splitting a row is a derived test). In any abelian category, prove that a short exact row \(0\to A\xrightarrow iE\xrightarrow pB\to0\) splits if and only if its derived boundary is zero. Prove also that every cohomology map of that boundary is zero. For (6.1), explain why this does not contradict the conservativity statement in Section 1.

Solution. The natural row triangle is \(A\to E\to B\xrightarrow\delta A[1]\). Its covariant Hom sequence with source \(B\) contains \(\operatorname{Hom}_D(B,E)\to\operatorname{Hom}_D(B,B)\to\operatorname{Hom}_D(B,A[1])\). The image of \(1_B\) is \(\delta\). Hence \(\delta=0\) exactly when there is a derived arrow \(s:B\to E\) with \(ps=1_B\). Degree-zero full faithfulness turns this into an actual section. Such a section gives the split isomorphism \(A\oplus B\to E\), \((a,b)\mapsto i(a)+s(b)\): the map \(1_E-sp\) factors uniquely through \(i\), giving its inverse together with \(p\). Conversely a splitting supplies the section and hence makes \(\delta=0\). The objects \(B\) and \(A[1]\) have cohomology in disjoint single degrees, so all \(H^n(\delta)\) are zero. In (6.1) the boundary is nonzero by Example 6.2. Conservativity detects whether a map is invertible; it does not assert that a map with zero cohomology maps is zero. Here the cohomology maps are not isomorphisms between the nonzero source and target cohomology objects, so that invertibility criterion is never applicable.

Freely accessible references

• The Stacks Project Authors, Derived categories, Section 13.11, particularly Lemma 13.11.6, for the bounded localization comparison. The denominator equality tests and the general adjunction arguments used in this lesson are proved above and in the linked course prerequisites.
• The Stacks Project Authors, Truncation of complexes, Section 12.15, for the kernel and quotient conventions. The course truncation prerequisite proves the image-extended constructions and their homotopies as well.