Draft lesson.

Functorial cones and derived connecting maps

Written by GPT-6.1 Sol (OpenAI), October 2026. Public domain (CC0).

A cone of a morphism in a derived category usually cannot be chosen functorially from that morphism alone. A natural transformation of additive functors contains more information: its components are actual chain maps after applying the functors to a complex. This additional structure gives a cone functor, including its maps on morphisms, its homotopies and its signed comparison with translation. Computing on injective complexes then produces a natural cone construction on a bounded-below derived category.

The same construction explains a useful phenomenon. A sequence of functors can be exact on injectives even when its value on an arbitrary object is not a short exact sequence. Its derived values nevertheless form a natural distinguished triangle. We will construct every arrow of that triangle, including the connecting map, and compute its sign in two examples.

Prerequisites and conventions

Work in fixed nested universes large enough for the categories and the resolution choices under discussion. Either choose a quasi-inverse on a set of representatives in the larger universe, or use individual injective replacements and the unique comparison maps proved below. No functorial choice of resolutions in the original category is an assumption.

We use the following previously proved interfaces:

• Injective replacements and the bounded-below null-homotopy theorem, Sections 2 and 4: an abelian category with enough injectives has bounded-below injective replacements, and every map from an acyclic complex to a bounded-below complex of injectives is null-homotopic.
• Universal derived functors, Sections 1, 4 and 5: any additive functor is exact on homotopy categories, and its bounded-below right derived functor is computed on injective complexes. Left exactness of the original functor is unnecessary for this statement.
• Triangles after localization, Sections 2 and 6: the positive-projection cone convention gives distinguished triangles after localization at quasi-isomorphisms.
• Cohomological bounds and truncations, Sections 3–5: bounded-below complexes and their shifts form the corresponding derived subcategory.
• Divisible groups and injective embeddings, Section 4: divisible abelian groups are injective, and abelian groups have enough injectives.
• Finite dual-number modules, Exercise 2 and its solution: finite-dimensional modules over the dual numbers have enough injectives, and the regular module is injective.

The free reference Stacks, injective resolution functors treats the equivalence used here. Stacks, derived functors on derived categories gives the general derived-functor setting, and Stacks, the canonical delta-functor discusses the cone-to-quotient comparison. Our proofs below spell out the cone and connecting-map conventions rather than importing a sign from a different convention.

Complexes are cohomological. For a complex \(X\),

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

Its structural maps are \(j_f(y)=(y,0)\) and \(p_f(y,x)=x\). Our distinguished cone triangle is

In an arbitrary additive category, coordinate expressions such as (1) denote biproduct matrices. They do not assume that the objects have elements. A homotopy from \(v\) to \(u\) is a family \(h^n:X^n\to Y^{n-1}\) satisfying

1. An actual cone functor and its homotopies

Let \(\mathcal A,\mathcal B\) be abelian categories, let \(F,G:\mathcal A\to\mathcal B\) be additive functors, and let \(\alpha:F\Rightarrow G\) be a natural transformation. For a complex \(I\), applying either functor termwise gives a complex, since an additive functor preserves zero morphisms and compositions. Define

with differential

The upper-right entry of \(D_I^{n+1}D_I^n\) is \(Gd_I^{n+1}\alpha_{I^{n+1}}-\alpha_{I^{n+2}}Fd_I^{n+1}=0\) by naturality; all other entries vanish because \(d_I^2=0\). Thus (4) is a differential.

For a chain map \(u:I\to J\), put

The chain-map equation for (5) follows from the chain-map equations for \(u\) on the diagonal, and from \(Gu^{n+1}\alpha_{I^{n+1}}=\alpha_{J^{n+1}}Fu^{n+1}\) off the diagonal. Identity maps and compositions are respected entry by entry. Addition of chain maps is also respected. Consequently \(H_\alpha\) is an additive functor on complexes. It sends a bounded-below complex to a bounded-below complex; its lowest possible nonzero degree is at most one earlier.

Proposition 1.1. Homotopic chain maps induce homotopic maps under \(H_\alpha\). More precisely, if (3) holds for \(u,v:I\to J\), the induced homotopy is

Proof. Multiply the biproduct matrices. The diagonal entries of \(D_JS+SD_I\), in degree \(n\), are

These are \(G(u^n-v^n)\) and \(F(u^{n+1}-v^{n+1})\). The upper-right entry is

again by naturality, now for the degree-lowering map \(h^{n+1}:I^{n+1}\to J^n\). The lower-left entry is zero. Therefore \(H_\alpha(u)-H_\alpha(v)=D_JS+SD_I\), which is the required homotopy equation. \(\square\)

Let \(\mathcal I\) denote the full additive subcategory of injectives in \(\mathcal A\). Finite biproducts of injectives are injective: each component of a map into the biproduct extends along a monomorphism, and the extended components combine. In particular, shifts and cones of maps between complexes in \(C^+(\mathcal I)\) remain in \(C^+(\mathcal I)\). Equations (4)–(6) give an actual functor

There are natural chain maps

Both commute with (5), so they descend along with the functor. Naturality here includes the actual structural arrows, not merely the isomorphism class of a cone object.

2. Translation and the double-cone comparison

The functor (7) has a signed translation comparison

Indeed, the source and target differentials, in degree \(n\), are respectively

Conjugating by \(\operatorname{diag}(1,-1)\) transforms the first into the second. The same diagonal matrix is its inverse, and it commutes with (5), proving naturality. Iteration yields the comparison for an integral shift \(r\), with diagonal entries \(1\) and \((-1)^r\).

The structural maps satisfy

Thus the translated cone triangle has the expected negative last arrow. The minus sign in (10) is part of the construction. When the target functor \(F[1]\) is viewed as an exact functor, its standard translation comparison is negative the termwise identification; with that comparison, \(p:H_\alpha\Rightarrow F[1]\) is compatible with translation.

Proposition 2.1. The functor \(H_\alpha\), with comparison (9), is exact on the homotopy categories in (7).

Proof. It suffices to compare its value on a standard cone triangle with the cone triangle of its value on the first map. Let \(u:I\to J\). Use the order

Its differential sends the tuple \((b,c,a,e)\) to

The omitted superscripts are determined by these summands; \(\alpha_J\) acts on \(FJ^{n+1}\), and \(\alpha_I\) on \(FI^{n+2}\). Formula (11) uses the canonical biproduct identifications of additive functors.

On the other hand,

Define

This is an invertible signed permutation. The target differential applied to its value is

which is exactly \(\Theta_u\) applied to (11). Hence (12) is a chain isomorphism. It is natural in strict commutative squares of chain maps because the permutation and sign commute with all diagonal maps.

We also check the arrows of the triangle. Its middle inclusion becomes

Its last arrow becomes

The first arrow is already \(H_\alpha(u)\). Thus (12) identifies the entire image triangle, with its translation comparison, with the standard cone triangle of \(H_\alpha(u)\). Every distinguished triangle in a homotopy category is isomorphic to a standard cone triangle, so the functor is exact. \(\square\)

The calculation is useful beyond this proof: omitting the sign on the last coordinate of (12) would preserve neither the differential nor the last structural arrow.

3. Descent to derived objects and all comparison maps

Assume from now on that \(\mathcal A\) has enough injectives. There is no enough-injectives assumption on \(\mathcal B\).

Lemma 3.1. The localization functor restricts to an exact equivalence

Proof. The injective replacement theorem supplies, for every bounded-below complex \(X\), a quasi-isomorphism \(\rho_X:X\to I_X\) with \(I_X\) bounded below and termwise injective. The replacement-density theorem identifies the derived category with the localization of \(K^+(\mathcal I)\) at its quasi-isomorphisms. These are the full conclusions of Sections 2 and 4 of the first prerequisite, including their maps-on-morphisms statements.

In this subcategory a quasi-isomorphism \(u:I\to J\) has an acyclic cone, still termwise injective and bounded below. Apply the bounded-below null-homotopy theorem to the identity map of that cone. The cone is contractible. In the homotopy category its cone triangle then shows that \(u\) is invertible: the exact Hom sequences give an inverse, or equivalently the zero cone criterion in a triangulated category does so. Thus every morphism inverted by the indicated localization was already invertible. The localization is \(K^+(\mathcal I)\) itself, proving (13), including full faithfulness. Exactness follows because the inclusion and localization preserve standard cone triangles. \(\square\)

Choose an exact quasi-inverse \(S\) to (13), with its comparison to \(E\). Let \(Q_{\mathcal B}:K^+(\mathcal B)\to D^+(\mathcal B)\) be localization. Define

The first two are the universal bounded-below right derived functors proved in the second prerequisite. This remains true for additive functors which are not left exact. The functor \(\mathbf RH_\alpha\) is exact by Proposition 2.1. For each \(X\), (14) can be computed by any injective replacement \(\rho:X\to I\).

Theorem 3.2. The natural transformation \(\alpha\) has a derived transformation \(\mathbf R\alpha:\mathbf RF\Rightarrow\mathbf RG\). There are natural transformations

such that for every \(X\in D^+(\mathcal A)\),

is distinguished. All arrows in (15) are natural in every derived morphism. Different injective replacements give canonically isomorphic whole triangles; these comparison isomorphisms obey the identity, composition and cocycle laws.

Proof. On the model \(I\), the first arrow is \(Q_{\mathcal B}(\alpha_I)\), and the other arrows are (8). Applying localization to the actual cone triangle of \(\alpha_I\) proves distinguishedness. The comparison (9) supplies the translation identification used in its last arrow.

Here is the full maps-on-morphisms argument, independently of a particular choice of \(S\). Let \(f:X\to Y\) be any derived arrow, and choose replacements \(\rho_X:X\to I\) and \(\rho_Y:Y\to J\). Full faithfulness in (13) gives exactly one homotopy class of chain maps \(u:I\to J\) satisfying

Choose any chain-map representative of this class. Naturality of \(\alpha\) and (5), (8) gives the three strict identities

They yield a morphism of the complete triangles. A different representative of the class in (16) has the same image under \(F\) and \(G\), since additive functors carry the homotopy equation to a homotopy equation. It has the same image under \(H_\alpha\) by Proposition 1.1. Thus (17) is independent of the representative. Uniqueness in (16) gives identity and composition laws. This includes arrows originally described by roofs or inverses of quasi-isomorphisms, because (16) starts with an arbitrary arrow in the localized category.

If \(\rho:X\to I\) and \(\rho':X\to J\) are two replacements, the same argument gives a unique homotopy class \(c:I\to J\) with \(E(c)Q(\rho)=Q(\rho')\). The class in the opposite direction is its inverse by uniqueness. Applying \(F,G,H_\alpha\) gives isomorphisms of all three objects commuting with all arrows in (15). For three replacements the composite of the first two comparisons and the direct comparison satisfy the same defining equation, so they are equal. The comparison for identical replacements is the identity. This proves the promised canonical comparison and cocycle laws.

Finally, the derived transformation agrees with the universal derived-functor units. For a bounded-below complex \(X\), those units are represented by \(F\rho_X\) and \(G\rho_X\). The equality

is ordinary naturality before localization. Hence \(\mathbf R\alpha\,\eta_F=\eta_G\,Q_{\mathcal B}K(\alpha)\). The universal characterization identifies this transformation with the usual derived transformation. \(\square\)

The construction proves a specific functorial cone statement. It does not assert that the category of arbitrary functors into a triangulated category is itself triangulated, or that a cone of an arbitrary derived arrow is determined functorially without the chain-level information in \(\alpha\).

4. A row exact on injectives and its connecting map

Let \(F',F,F'':\mathcal A\to\mathcal B\) be additive functors with natural maps

Assume that for every injective \(J\in\mathcal A\),

is exact in \(\mathcal B\). We do not assume exactness of (19) on arbitrary objects, or that the functors are left exact.

For an injective complex \(I\), abbreviate \(A=F'I\), \(B=FI\), \(C=F''I\). The termwise sequence \(0\to A\to B\to C\to0\) is a short exact sequence of complexes. Define

This is a chain map: applying \(q_I\) to the cone differential gives \(\beta(db+\alpha a)=d\beta b\). It is epic in every degree by (19).

Lemma 4.1. The map (20) is a quasi-isomorphism. Its kernel is canonically the contractible complex \(C(1_A)\).

Proof. In degree \(n\), \(\alpha^n\) identifies \(A^n\) with \(\ker\beta^n\). Thus the kernel of (20) is identified with \(A^n\oplus A^{n+1}\) by

Its differential is \((x,a)\mapsto(dx+a,-da)\), because \(d\alpha x+\alpha a=\alpha(dx+a)\). This is exactly \(C(1_A)\), including its differential. The maps

satisfy

Here is a categorical verification that a degreewise epic chain map with this contractible kernel is a quasi-isomorphism; it avoids assuming global element lifts in \(\mathcal B\). For any cycle test map \(z:T\to Z^nC\), pull back the epimorphism \(q_I^n\). Over the resulting epimorphic cover \(T'\to T\), there is a lift \(b:T'\to C(\alpha_I)^n\). Its differential factors uniquely through the kernel as \(db=\kappa k\), with \(k:T'\to C(1_A)^{n+1}\). The equation \(d^2b=0\) and monicity of \(\kappa\) show \(dk=0\). From (21), \(dhk=k\). Therefore \(b-\kappa hk\) is a cycle mapping to \(z\) over this cover. This proves epimorphy on cycles locally along epimorphisms, and hence epimorphy on cohomology. Formally, the cokernel of the induced cohomology map vanishes because every test map into it becomes zero after an epimorphic cover.

For injectivity on cohomology, start with a test map into its kernel. Lift it through the epimorphism from cycles to cohomology after an epimorphic pullback cover. It is then represented by a cycle \(e\) in \(C(\alpha_I)^n\), and \(q_Ie\) is a boundary. Lift a primitive of this boundary after a second epimorphic cover, and lift that primitive through \(q_I^{n-1}\) after a third. Write the final lift as \(b\). Then \(e-db=\kappa k\) for a cycle \(k\) in \(C(1_A)^n\). Equation (21) gives

The original cohomology test map therefore vanishes after an epimorphic cover, so it vanishes before the cover as well. This proves monomorphy on cohomology. Both arguments use only epimorphic pullbacks, kernels, images and the displayed contraction, all available in an abelian category. Hence (20) is a quasi-isomorphism. \(\square\)

Theorem 4.2. Under (19), there is a natural transformation

and a pointwise distinguished natural triangle

It is natural on every arrow of \(D^+(\mathcal A)\), canonically independent of all injective replacements, and natural under commuting morphisms of rows (18) satisfying (19).

Proof. Apply Theorem 3.2 to \(\alpha:F'\Rightarrow F\). In the model \(I\), Lemma 4.1 makes \(Q_{\mathcal B}q_I\) invertible. Define

Since \(q_Ij_I=\beta_I\) and \(\delta_Iq_I=p_I\) after localization, transporting the entire cone triangle along \(q_I\) gives (23). This proves distinguishedness with exactly the arrows in its statement.

For a chain map \(u:I\to J\), naturality of \(\alpha\) and \(\beta\) gives

Inverting \(q_I,q_J\) in the first equality and using the second yields

Homotopic representatives give identical maps by Proposition 1.1 and by the additive homotopy equations for the three functors. Equation (16) then proves (26) on all derived arrows. The unique comparison classes between any two replacements commute with \(\alpha,\beta,q,p\) by the same strict identities. Thus they identify the whole triangles (23), with the identity and cocycle laws already proved.

For the last assertion, consider another row \(\widetilde F'\to\widetilde F\to\widetilde F''\) and natural transformations \(a,b,c\) from the first row to the second satisfying \(b\alpha=\widetilde\alpha a\) and \(c\beta=\widetilde\beta b\). On \(I\), the matrix \(\operatorname{diag}(b_I,a_I[1])\) is a chain map between the cones. Its composites with the two quotient maps agree with \(c_Iq_I\), and its composites with the two projections agree with \(a_I[1]p_I\). Inverting the quotient maps gives \(\widetilde\delta\,\mathbf Rc=\mathbf Ra[1]\delta\). The other two squares are the original commuting squares after derivation. Hence the complete natural triangle is natural under morphisms of rows. \(\square\)

Under the canonical termwise shift identifications, (9), (20) satisfy

Consequently the connecting map obeys

This is the translation sign required by the last arrow in a distinguished triangle.

One can also read the sign directly on cohomology. In a module category, lift a cycle \(z\in C^n\) to \(b\in B^n\) and write \(db=\alpha a\). Then \(a\in A^{n+1}\) is a cycle, since \(\alpha\) is monic and \(\alpha da=d^2b=0\). The cone cycle over \(z\) is \((b,-a)\), so (24) gives

A second lift \(b+\alpha x\) changes \(a\) to \(a+dx\), so this cohomology class is independent of the lift. If \(z\) is a boundary, lift its primitive to \(B^{n-1}\); the resulting lift of \(z\) has zero differential, showing that (28) vanishes on boundaries. In an arbitrary abelian category these statements apply to test maps after the epimorphic pullbacks in Lemma 4.1; the identities descend because the covers are epic. Thus (28) specifies the same morphism without assuming a concrete element description.

With our positive last projection in (2), this is the negative of the lift-and-differentiate boundary \([z]\mapsto[a]\). Either convention yields the same kernels and images in the long exact cohomology sequence, but their actual arrows have opposite signs. Formula (24) fixes the convention unambiguously.

5. Four graded exercises with complete solutions

Exercise 1 — Foundation: detect the homotopy sign

In rational vector spaces, let \(I\) be \(\mathbb Q\xrightarrow{1}\mathbb Q\) in degrees 0 and 1. Take \(F=G=\mathrm{Id}\), \(\alpha=1\), \(u=1_I\) and \(v=0\). Use the homotopy \(h^1=1\), with all other components zero. Compute \(H_\alpha(I)\) and the induced contraction. Show explicitly that using a positive second block in (6) fails.

Solution. Every rational vector space is injective: choose a basis of a subspace, extend it to a basis of the ambient space, and extend a linear map by assigning zero on the new basis vectors. Thus this example also lies in the injective-complex setting.

The cone has terms \(\mathbb Q,\mathbb Q^2,\mathbb Q\) in degrees \(-1,0,1\), with differentials

The original equation \(1_I=dh+hd\) holds in each of degrees 0 and 1. Formula (6) gives

and all other components zero. In degree \(-1\), \(s^0d^{-1}(a)=a\). In degree 0,

whose sum is \((b,a)\). In degree 1, \(d^0s^1(b)=b\). Therefore \(ds+sd=1\).

The incorrect choice \(s^0(b,a)=a\) instead gives

which is not \((b,a)\), for example when \((b,a)=(0,1)\). Over a field of characteristic two this test would not detect the mistake; the rational example does.

Exercise 2 — Intermediate: zero and identity transformations

For arbitrary additive functors \(F,G\), identify \(\mathbf RH_0\) and all arrows of (15) when \(\alpha=0\). When \(F=G\) and \(\alpha=1_F\), give a natural contraction and identify (15). Explain why the translation comparison still carries a sign in the zero case.

Solution. With \(\alpha=0\), (4) is block diagonal, so there is a strict natural isomorphism

Consequently \(\mathbf RH_0=\mathbf RG\oplus\mathbf RF[1]\). The triangle is

It is the cone triangle of zero, pointwise. The map on an arrow is the direct sum of the two derived maps. Despite the zero off-diagonal differential, (9) is still \((b,a)\mapsto(b,-a)\); this is the comparison for this exact cone functor. It makes its last projection transform with the minus sign (10), as required for translated triangles. The direct sum decomposition alone does not prescribe that translation comparison.

For \(\alpha=1_F\), the cone differential is \((b,a)\mapsto(db+a,-da)\). Define \(s^n(b,a)=(0,b)\). Then

so \(ds+sd=1\). This contraction is natural in \(I\), because it is built from biproduct identities. The cone is zero in the homotopy category, and hence \(\mathbf RH_{1_F}=0\). Triangle (15) is naturally isomorphic to

The first arrow is the identity before localization, so its identification does not require a separate choice of cone isomorphism.

Exercise 3 — Advanced: a row exact only on injectives

Fix an integer \(m\ge2\), and put \(T_m(M)=M[m]=\operatorname{Hom}(\mathbb Z/m,M)\) on abelian groups. Use the natural inclusion \(T_m\to\mathrm{Id}\) and multiplication by \(m\) on \(\mathrm{Id}\). Prove exactness on injectives, compute \(\mathbf RT_m(\mathbb Z)\), and compute the actual connecting arrow \(\delta_{\mathbb Z}:\mathbb Z\to\mathbf RT_m(\mathbb Z)[1]\) in convention (2).

Solution. An injective abelian group \(J\) is divisible. Indeed, for \(x\in J\), send \(m\in m\mathbb Z\) to \(x\), and extend this map to \(\mathbb Z\) by injectivity. The image of 1 is an \(m\)-th part of \(x\). Thus

is exact. At \(\mathbb Z\), however, the multiplication map is not epic. This example therefore needs precisely the injective hypothesis of Theorem 4.2, rather than exactness of the underived row everywhere.

Use the resolution

with zero terms elsewhere. Both terms are divisible and hence injective by the fifth prerequisite. Its cohomology is \(\mathbb Z\) in degree 0 and zero in degree 1. Applying \(T_m\) gives zero in degree 0 and \((\mathbb Q/\mathbb Z)[m]\) in degree 1. Identify this last group with \(\mathbb Z/m\) by

This is injective because \(a/m\in\mathbb Z\) exactly when \(m\mid a\), and surjective because any \(m\)-torsion class has a rational representative whose product with \(m\) is an integer. Therefore \(\mathbf RT_m(\mathbb Z)\simeq(\mathbb Z/m)[-1]\).

For this model the cone in (20) has \(\mathbb Q\oplus\mathbb Z/m\) in degree 0 and \(\mathbb Q/\mathbb Z\) in degree 1, with

Its quotient map to \(I\) is \((q,\bar a)\mapsto mq\) in degree 0 and \(r+\mathbb Z\mapsto mr+\mathbb Z\) in degree 1. Its positive last projection is \((q,\bar a)\mapsto\bar a\) in degree 0. For \(z\in\mathbb Z\), the tuple

is a cone cycle, its image under the quotient map is \(z\), and its last projection is \(-\bar z\). These tuples give an actual homomorphism \(\mathbb Z[0]\to C(\alpha_I)\). Its composite with the quotient map is the augmentation \(\mathbb Z[0]\to I\), so it is a quasi-isomorphism by Lemma 4.1 and the two-out-of-three property. It therefore represents the inverse quotient arrow on this source in the derived category. Formula (24) now gives

The identity functor derives to itself, because it already preserves quasi-isomorphisms and its unit is the resolution quasi-isomorphism. Thus the resulting triangle is

At \(\mathbb Z\), its cohomology sequence contains \(\mathbb Z\xrightarrow{m}\mathbb Z\xrightarrow{-\mathrm{red}}\mathbb Z/m\to0\), which is exact. The connecting map is nonzero; for \(m=3\) its value at 1 is 2, so its sign can be observed.

Exercise 4 — Synthesis: nonzero connecting maps in every nonnegative degree

Let \(R=k[t]/(t^2)\), \(S=R/(t)\), and let \(\mathcal A=\mathcal B\) be the category of finite-dimensional \(R\)-modules. Put \(T(M)=M[t]=\operatorname{Hom}_R(S,M)\). Show that the natural row

is exact on injectives. Compute \(\mathbf RT(S)\) and every cohomology component of its connecting map \(\delta_S:\mathbf RT(S)\to\mathbf RT(S)[1]\).

Solution. The sixth prerequisite proves that \(R\) is injective and that every finite module embeds in a finite direct sum of copies of \(R\), so this abelian category has enough injectives. Apply \(\operatorname{Hom}_R(-,J)\), for an injective finite module \(J\), to

Injectivity makes restriction onto \(\operatorname{Hom}_R(S,J)\) surjective: any map from the submodule \((t)\) of \(R\) extends across \(R\). Under the identifications \(\operatorname{Hom}_R(R,J)=J\) and \(\operatorname{Hom}_R(S,J)=J[t]\), restriction sends \(x\) to \(tx\). Its kernel is \(J[t]\), and the first arrow is its inclusion. This proves exactness on injectives.

An injective resolution of \(S\) is

with the first \(R\) in degree 0. In each copy of \(R\), both the image and the kernel of multiplication by \(t\) are \((t)\). The augmentation identifies \(S\) with that kernel in degree 0. Thus this is a quasi-isomorphic injective replacement. Applying \(T\) gives \((t)\simeq S\) in every degree \(n\ge0\), with zero differentials. Consequently

Write \(A=TI\) using the identification \(\bar a\leftrightarrow at\). The cone of \(A\to I\) has \(S\) in degree \(-1\) and \(R\oplus S\) in every degree \(n\ge0\). Its differentials are

The quotient map (20) in a nonnegative degree is \(q^n(r,\bar a)=\bar r\in S\), because the map \(I\to TI\) is multiplication by \(t\). The positive last projection is \(p^n(r,\bar a)=\bar a\in A^{n+1}\).

For any class \(\bar x\in S\), use its scalar representative \(x\in k\). The tuple \((x,-\bar x)\) is a cone cycle, since \(t(x-x)=0\). It maps under \(q^n\) to \(\bar x\), and under \(p^n\) to \(-\bar x\). Since \(q\) is a quasi-isomorphism by Lemma 4.1, this proves

These are computations with cycle representatives. The assignment \(\bar x\mapsto x\in R\) is not an \(R\)-linear splitting of \(R\to S\): multiplication by \(t\) acts trivially on \(S\) but not on its scalar lift in \(R\). No chain-level section of \(q\) is claimed or needed.

The underlying distinguished triangle is

since the identity functor derives to itself. In positive degrees its exact cohomology sequence forces the displayed connecting maps to be isomorphisms; the cone calculation determines their precise sign. In degree 0 the inclusion \(T(S)=S\to S\) is an isomorphism and multiplication by \(t\) is zero, so the same exact sequence also permits and forces the connecting isomorphism. In characteristic two \(-1_S=1_S\), and it remains nonzero.