Draft lesson.

Cosyzygies and the bounded injective quotient

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

Removing maps through injectives does not make cosyzygy invertible. Stabilization does: it identifies maps after sufficiently many cosyzygies. We prove that the resulting category is precisely the bounded derived category with bounded injective complexes made zero. The proof computes all morphisms and their equality witnesses, and identifies the complete signed triangles arising from short exact rows.

Conventions and prerequisites

Let \(\mathcal A\) be an abelian category with enough injectives, and let \(\mathcal I\) be its full subcategory of injective objects. Use an ambient universe containing the objects, arrows, integer sequences and choices below. No finite global dimension, projectives or infinite sums in \(\mathcal A\) are assumed. An injective object is one into which a map extends along every monomorphism.

Complexes are cohomological. We use \(X[r]^j=X^{j+r}\), \(d_{X[r]}=(-1)^r d_X\), and

The last arrow of the cone triangle is the positive projection \(C(f)\to X[1]\). Thus the connecting map of a short exact row is the cone projection composed with the inverse of its quotient quasi-isomorphism. On a shifted triangle, keeping its first two arrows positively shifted requires the factor \((-1)^r\) on its last arrow.

We use these complete earlier arguments:

• Stabilizing an endofunctor through eventual maps, Sections 1–5: the actual category, inverse shift, additive structure and compatible functor extension.
• Derived morphisms through quasi-isomorphism roofs, Section 5: the complete bounded-injective homotopy induction and the Hom bijection into such a complex.
• Injective replacements and right derived functors, Sections 1–2: construction of bounded-below replacements and their comparison equivalence.
• Null systems, roofs and the retract kernel, Sections 1–8: both fraction calculi, zero-map factorization, every quotient triangle axiom and the exact comparison of replacements.
• Triangles after quasi-isomorphism localization, Section 6: natural short-exact-row triangles and full faithfulness of objects in degree zero.
• Cohomological bounds, denominators and truncation adjunctions, Sections 1–3: bounded derived full embeddings, including roof equality tests and bounded representatives.
• Right derived tensor and torsion-free quotients, Sections 2–3: the divisible-injective equivalence and enough injectives for the abelian-group example.

1. The additive quotient and actual cosyzygy maps

Let \(\mathcal J(A,B)\) be the maps \(A\to B\) factoring through an injective object. A sum of two such maps factors through the direct sum of their two injectives. This sum is injective: extend its two components separately. Negatives and zero also factor through injectives, and precomposition or postcomposition preserves a factorization. Hence \(\mathcal J\) is an additive two-sided ideal. Define

Objects are unchanged, composition is induced by composition in \(\mathcal A\), and the biproduct equations descend. This is an additive category. An object \(A\) is zero in it exactly when \(1_A\) factors through an injective. Such a factorization exhibits \(A\) as a retract of an injective, so \(A\) is itself injective: extend a map into the larger object and then apply the retraction. Conversely an injective's identity factors through itself.

Choose, for every \(A\), a short exact row

taking \(I_0=0\). Given \(f:A\to B\), injectivity of \(I_B\) extends \(i_Bf\) across \(i_A\) to a map \(a:I_A\to I_B\). There is a unique \(\widetilde f:\Sigma A\to\Sigma B\) with

Two choices \(a,a'\) have difference \(c p_A\), for a map \(c: \Sigma A\to I_B\). Their induced maps have difference \(p_Bc\), which factors through the injective \(I_B\). Thus \(\underline{\widetilde f}\) is independent of the extension.

If \(f=vu\) with \(u:A\to J\), \(v:J\to B\) and \(J\) injective, extend \(u\) to \(\widehat u:I_A\to J\). The allowed choice \(a=i_Bv\widehat u\) induces the zero map because \(p_Bi_B=0\). Consequently (1.3) descends to an endofunctor \(\Sigma\) on \(\underline{\mathcal A}\). Identity extensions, composites of extensions and sums of extensions prove, respectively,

These equalities are in the quotient; arbitrary chosen extensions are not required to be strictly composable in \(\mathcal A\).

Changing all embeddings gives a naturally isomorphic endofunctor. Indeed extend \(i'_A\) across \(i_A\) and conversely, giving \(a_A:I_A\to I'_A\) and \(b_A:I'_A\to I_A\). Their induced cokernel maps \(\theta_A:\Sigma A\to\Sigma' A\) and \(\psi_A\) satisfy \(\psi_A\theta_A=1\), \(\theta_A\psi_A=1\) in the quotient, since \(b_Aa_A-1\) and \(a_Ab_A-1\) kill the respective embeddings. For a map \(f\), the difference of the two composites of comparison extensions kills \(i_A\); its induced cokernel difference therefore factors through \(I'_B\). This proves naturality. The same argument with three choices shows that comparison through the intermediate choice equals direct comparison. In particular the comparison is independent of the extensions and obeys its cocycle law.

2. Eventual maps and an invertible suspension

Write \(\mathcal S=\operatorname{Stab}(\underline{\mathcal A},\Sigma)\). Its objects are \((A,n)\) for integers \(n\), and

with transition \(f\mapsto\Sigma f\). Equality means equality at some common later stage, not equality before any transition. Represent a map by \([\underline f,k]\). Composition and addition are performed at one common stage, and then passed to its class. Independence, associativity, bilinearity and all biproduct equations are proved in the stabilization prerequisite; they apply because of (1.4).

The suspension and its strict inverse are

On objects the inverse lowers the integer by one. Both composites send a morphism to its own next-stage representative and hence are identity. Let \(\iota A=(A,0)\). There are actual isomorphisms

represented by identity maps. They and their inverses have the precise composition relations in the prerequisite. Every arrow \(u=[\underline f,k]:(A,n)\to(B,m)\) satisfies

This factorization will reduce arbitrary arrows to maps between objects of \(\mathcal A\).

3. Bounded injective complexes and their derived quotient

There is a fully faithful exact functor

For completeness the Hom assertion is not an inference from injectivity of individual source terms. A bounded injective target \(J\) is bounded below. For a map from an acyclic complex \(N\) into \(J\), start a homotopy below its lower bound. If its equations hold up to degree \(r-1\), the residual in degree \(r\) kills \(\operatorname{im}d_N^{r-1}=\ker d_N^r\). It therefore factors through \(\operatorname{im}d_N^r\hookrightarrow N^{r+1}\), and injectivity of \(J^r\) extends it to the next homotopy component. Induction makes the whole map null-homotopic. The cone Hom sequence then makes precomposition by every quasi-isomorphism a Hom bijection into \(J\). Roofs and their equality test give

The bounded derived full embedding prerequisite identifies the right side with bounded derived Hom whenever \(X\) is bounded. Thus (3.1) is fully faithful. Finite sums, shifts and cones of bounded injective complexes stay bounded injective, so it preserves the full cone triangles and is exact.

Let \(\mathcal N\) be the replete essential image of (3.1), and put

Full faithfulness and cones make \(\mathcal N\) a full triangulated subcategory. The null-system prerequisite constructs (3.3), both roof calculi, and every triangulated axiom. Its incoming denominators are maps whose cones belong to \(\mathcal N\). It also proves the useful exact criterion: a derived map becomes zero under \(q\) if and only if it factors through an actual object of \(\mathcal N\). We do not replace this statement by an unproved assumption that a null system is already closed under retracts.

Let \(\delta_A:\Sigma A\to A[1]\) be the connecting map of (1.2), with convention (0.1). Naturality of short exact row triangles says

for every extension (1.3). Its image \(q\delta_A\) is invertible, since the middle vertex \(I_A\) becomes zero. Maps through injectives become zero in \(\mathcal Q\), so the stalk functor descends to an additive functor \(F:\underline{\mathcal A}\to\mathcal Q\). Equation (3.4) makes \(q\delta:F\Sigma\to[1]F\) a natural isomorphism.

Define \(\rho_{0,A}=1\), and, before applying \(q\), define

The stabilization construction extends \(F\) to an additive functor

Explicitly its value on a stage map is

All three maps inside the parentheses are between the shifts with indices \(n+k,m+k\). The displayed final shift gives the claimed domain and codomain. Equation (3.4) proves invariance under the next transition; quotienting through injectives and equality at a later stage prove all remaining independence. At a common stage the middle \(\rho\) factors cancel, proving composition and addition. Integer shifts of complexes give the chosen coherent powers, and the translation comparison \(\Phi S\to[1]\Phi\) is identity. In particular

4. Cosyzygy denominators compute all morphisms

Put \(A_r=\Sigma^rA[-r]\), and

Then \(d_{0,A}=1\) and \(d_{r+1,A}=d_{r,A}e_{r,A}\). For \(t\ge r\), write \(e_{r,t,A}\) for the composite from \(A_t\) to \(A_r\), including identity when \(t=r\).

Lemma 4.1 (finite cones with their bounds). Each \(d_{r,A}\) has a cone represented by injective terms only in degrees \(0,\ldots,r-1\). Each \(e_{r,A}\) has a cone isomorphic to \(I_{\Sigma^rA}[-r]\). In particular all these maps are denominators for (3.3).

Proof. Inverse rotation of the triangle of (1.2) gives a triangle starting with \(-\delta_A[-1]\) and third vertex \(I_A\). Replacing the sign on its first vertex gives a completion of \(+\delta_A[-1]\) with the same third object. Shifting this argument gives the assertion for \(e_r\). The identity has zero cone. Suppose a cone \(C_r\) of \(d_r\) has the stated injective representative. The octahedron for \(e_r,d_r\) gives a triangle

Both endpoints are bounded injective complexes, so the last map is represented by a chain map by (3.1). A shifted cone of that map, with the vertex sign appropriate to inverse rotation, represents the middle object. Its terms are the direct sums of the terms of \(I_{\Sigma^rA}[-r]\) and \(C_r\). They lie in degrees \(0,\ldots,r\) and are injective. Triangle comparison identifies this middle object with the prescribed \(C_{r+1}\). Induction proves the first assertion with its exact degree bound. \(\square\)

Lemma 4.2 (the last-degree kernel). For objects \(U,B\), the map

is surjective. When \(r\ge1\), its kernel consists of maps that, in particular, factor through an injective object. For \(r=0\) it is the degree-zero Hom identification and has zero kernel.

Proof. Complete \(d_{r,B}\) with cone \(C_r\) as in Lemma 4.1. By (3.2), \(\operatorname{Hom}_{D^b}(U[-r],C_r)=0\): its source is a stalk in degree \(r\), and every term of the injective target in degree \(r\) and above is zero. The triangle Hom sequence therefore makes

surjective. The source is ordinary Hom by the degree-zero full embedding and shifting, proving surjectivity in (4.3).

If \(d_{r,B}f[-r]=0\), exactness writes \(f[-r]\) as a composite through \(C_r[-1]\). The map into this bounded injective complex has an actual chain representative by (3.2). Its only possible component is \(U\to C_r[-1]^r\), an injective object. The outgoing derived map need not have a chain representative. Apply \(H^r\) instead: since \(r\) is the highest term degree of \(C_r[-1]\), its cohomology is the quotient of \(C_r[-1]^r\) by its incoming image. The induced composite \(f\) is consequently

It factors through the indicated injective term. This proves the needed kernel assertion without treating an injective complex as projective. \(\square\)

Theorem 4.3 (all stabilized Hom groups). The functor \(\Phi\) is fully faithful. In particular there is a natural group isomorphism

Proof of surjectivity. Represent a quotient arrow by \(qg(qs)^{-1}\), where \(s:V\to A\) has bounded injective cone \(J\) and \(g:V\to B\). Choose \(r\) larger than the upper term degree of one bounded injective representative of \(J\). Equation (3.2) gives

The triangle Hom sequence lifts \(d_{r,A}\) to \(a:A_r\to V\) with \(sa=d_{r,A}\). This is a refinement of the incoming denominator; \(a\) is itself a denominator by triangle two-out-of-three. Lemma 4.2 applied to \(U=\Sigma^rA\) writes \(ga=d_{r,B}f[-r]\) for an actual map \(f:\Sigma^rA\to\Sigma^rB\). Thus the given arrow is

which is exactly (3.7) at stage \(r\). Every quotient map has the required stabilized representative.

Proof of injectivity and equality. Suppose a stage map \(f\) has image zero in (4.5). Set \(b=d_{r,B}f[-r]:A_r\to B\). Then \(qb=0\), so the exact zero-factorization criterion in Section 3 writes \(b\) through a bounded injective complex \(J\). Choose \(t\ge r,1\) larger than its upper term bound. Since \(A_t\) is concentrated in degree \(t\), (3.2) makes \(\operatorname{Hom}_{D^b}(A_t,J)=0\). Hence

Iterate actual extensions (1.3) for \(f\), and call the resulting representative \(f_t:\Sigma^tA\to\Sigma^tB\). Equation (3.4) at each step gives the exact derived equality

Lemma 4.2 now says \(f_t\) factors through an injective. Thus \(\Sigma^{t-r}\underline f=0\) in the stable category: the class was eventually zero. If two stage maps have equal images, first advance them to a common stage and apply this argument to their difference. This gives an actual later stable equality witness and proves injectivity of (4.4). It proves all equality tests, not only a supply of convenient roofs.

Finally, for arbitrary pairs \((A,n),(B,m)\), choose \(k\) with \(n+k,m+k\ge0\). The invertible suspension and (2.3) identify their Hom group with that between \(\iota\Sigma^{n+k}A\) and \(\iota\Sigma^{m+k}B\), after applying \(S^k\). The corresponding identifications under \(\Phi\) are shifts and the \(q\rho\) isomorphisms. The already proved stalk-pair bijection therefore gives the Hom bijection for these pairs too. Formula (3.7) shows that these bijections are those of the single functor \(\Phi\), hence preserve composition, sums and both-variable naturality. \(\square\)

5. Every bounded object is a shifted stalk in the quotient

Theorem 5.1. The additive functor \(\Phi\) is an equivalence.

Proof. It remains to prove essential surjectivity. Given \(X\in D^b(\mathcal A)\), the bounded representative and replacement prerequisites give a quasi-isomorphism to a bounded-below injective complex \(I\). Its cohomology is bounded. Choose an integer \(r\) above its highest nonzero cohomology degree and at least its lower term bound. The smart truncation \(E=\tau_{\le r}I\) is bounded and quasi-isomorphic to \(I\). It has injective terms below \(r\), term \(Z^rI=\ker d_I^r\) in degree \(r\), and no higher terms. There is a short exact sequence of complexes

The last complex, with its original degrees and differential truncated to zero after \(r-1\), is bounded injective. Its image under \(q\) is zero. The short-row triangle therefore makes \(q(Z^rI[-r])\to qE\) invertible. Since \(E\simeq X\), this exhibits \(qX\) as the image of \((Z^rI,-r)\). This proves density. Together with Theorem 4.3 it proves equivalence: choose the just exhibited preimages and isomorphisms, and lift maps by the full-faithful Hom bijections; those bijections give the identity, composition, unit and counit equations. No assertion that a bounded object has a bounded injective resolution was used. \(\square\)

6. Signed short-row triangles and all four axioms

For a short exact row \(0\to A\xrightarrow{i}B\xrightarrow{p}C\to0\), extend \(i_A:A\to I_A\) across \(i\) to \(b:B\to I_A\). There is a unique map \(h:C\to\Sigma A\) with

Changing \(b\) changes it by a map \(c p\) into \(I_A\), and changes \(h\) by \(p_Ac\). Thus \(\underline h\) is independent of the extension. Naturality of short-row triangles, applied to the row map \((1_A,b,h)\) into (1.2), gives its connecting map exactly as

Define its stable connecting arrow to be

By (3.8), \(\Phi\partial=q\delta_{(i,p)}\), with a positive sign relative to our specified cone convention. A commuting row map \((a,b',c)\) gives \(\delta'c=a[1]\delta\); full faithfulness of \(\Phi\) gives the full stable triangle morphism, including its last square. Changing choices of embeddings gives the corresponding same statement through the canonical comparisons of Section 1.

Declare a triangle row distinguished if it is isomorphic as a complete triangle to one of

The sign is necessary for positively shifted first and second arrows. For one shift the cone comparison \(C(f[1])\to C(f)[1]\) is \((y,x)\mapsto(y,-x)\). Its last projection square introduces minus sign. Iterating in both directions gives exactly \((-1)^j\), including negative \(j\).

Theorem 6.1. The triangles (6.4) and their complete isomorphs are precisely the triangles whose \(\Phi\) images are distinguished in \(\mathcal Q\). They make \((\mathcal S,S)\) triangulated, and \(\Phi\) is an exact equivalence

Proof of the asserted class of triangles. Equations (6.2)–(6.4) show that every row distinguished triangle has distinguished image. Conversely first construct a row completion of every stable arrow. For an actual representative \(f:U\to V\), the map

is monic, since its second component is monic. Take its cokernel \(C_f\) and the corresponding short exact row. In the stable category projection \(V\oplus I_U\to V\) is an isomorphism with inverse its inclusion: the other composite differs from identity by the projection onto an injective summand. Transport the row triangle along this isomorphism; its first arrow is \(\iota\underline f\). For an arbitrary arrow (2.4), use this completion at its common stage, transport by the two \(\beta\)'s, and shift back by \(-k\) with the sign in (6.4). It is still a row distinguished triangle and has the original first arrow.

Now take any complete triangle with distinguished \(\Phi\) image. Complete its first arrow by the preceding row construction. In \(\mathcal Q\), the triangle morphism axiom compares the two distinguished completions with identities on their first two vertices. Exact Hom sequences make the third comparison invertible. Full faithfulness of \(\Phi\) lifts this comparison and its inverse, including the last square. Thus the proposed triangle is a complete isomorph of the row completion. This proves equality of the two classes before assigning any axioms to the source.

Proof of all axioms. Identity and zero triangles, closure under complete isomorphism, and completion of every arrow follow either from the class identification or from (6.6). Rotation of the image triangle has last arrow \(-[1]\Phi f\). The identity translation comparison and faithfulness make its lift exactly \(-Sf\); inverse rotation is obtained in the same way. For a square between distinguished source triangles, apply \(\Phi\), complete the square in \(\mathcal Q\), and lift the third map by fullness. Faithfulness proves every square, including the connecting square. This proves the full morphism axiom.

For clarity the octahedral assertion includes prescribed completions. Given \(f:X\to Y\), \(g:Y\to Z\), and chosen triangles for \(f,g,gf\), take the prescribed octahedron of their images in \(\mathcal Q\). Lift its arrows \(u:C_f\to C_{gf}\), \(v:C_{gf}\to C_g\) by fullness, and put

where \(i_f:Y\to C_f\) is the second arrow and \(p_g:C_g\to SY\) the last arrow of the triangle for \(g\). Faithfulness lifts the four required face equations exactly:

The image of the fourth row \(C_f\xrightarrow u C_{gf}\xrightarrow v C_g\xrightarrow w S C_f\) is the distinguished fourth row of that octahedron. The class identification makes the lifted row row distinguished. This proves the full octahedral axiom with the given three completions and exact last map (6.7). Thus all four axioms hold. Preservation of complete triangles and the identity translation comparison make \(\Phi\) exact; its quasi-inverse is exact by lifting triangles and the translation comparison along its Hom bijections. \(\square\)

7. Exactly which stalks disappear

Corollary 7.1. For an object \(A\), the following are equivalent: \(qA=0\); \(\iota A=0\) in \(\mathcal S\); some \(\Sigma^r A\) is injective; and \(A\) has finite injective dimension. There need not be one common bound for every object.

Proof. The equivalence identifies the first two conditions. The identity of \(\iota A\) is zero exactly when some transition sends \(1_A\) to zero in \(\underline{\mathcal A}\). Its transition is \(1_{\Sigma^rA}\), which is zero exactly when \(\Sigma^rA\) is injective, by Section 1. The iterated rows (1.2) then give a finite injective resolution

Conversely a finite injective resolution represents \(A\) by a bounded injective complex in \(D^b\), so \(qA=0\). Each direction has been proved without assuming that all objects have finite dimension or that the original stable stalk functor is faithful. \(\square\)

Over the integers every abelian group has a length-one injective resolution: embed it in a divisible group, whose quotient is divisible and hence injective. Therefore this stabilized category and its bounded derived quotient are zero. The unstabilized category need not be zero: \(\mathbb Z\) is not injective. At the opposite extreme, the dual-number example below has nonzero maps at every stage.

8. Four graded exercises with complete solutions

Exercise 1 — A nonzero stable map can disappear (introductory)

For abelian groups choose \(\mathbb Z\hookrightarrow\mathbb Q\), so \(\Sigma\mathbb Z=\mathbb Q/\mathbb Z\). Compute the stable endomorphism group of \(\mathbb Z\), its first cosyzygy transition, and the stabilization of the whole category of abelian groups.

Solution. Ordinary endomorphisms of \(\mathbb Z\) are integer multiplication. An injective abelian group is divisible. The image of a divisible group under any homomorphism is divisible: if \(x=f(y)\), divide \(y\) by a positive integer before applying \(f\). The only divisible subgroup of \(\mathbb Z\) is zero, since a nonzero integer cannot be divided in \(\mathbb Z\) by every positive integer. Consequently every map from an injective group to \(\mathbb Z\) is zero, and every endomorphism through an injective is zero. Hence

The group \(\mathbb Q/\mathbb Z\) is divisible, so it is injective and is a zero object of the stable category. Every first transition from (8.1) is zero. Thus all these nonzero stable endomorphisms become zero after one stage. More generally every group embeds in an injective divisible group \(J\); its quotient is divisible because representatives in \(J\) can be divided before passing to the quotient. It is injective, so every object's first cosyzygy is already zero in the stable category. All sufficiently advanced groups in (2.1) are therefore zero. Every stabilized Hom group is zero, every object has zero identity, and the entire stabilization is the zero category. Theorem 6.1 identifies the bounded derived quotient as zero as well. This explains explicitly why an unproved faithful-embedding premise would fail.

Exercise 2 — Nonzero maps at every stage (intermediate)

Let \(k\) be any field and \(\Lambda=k[t]/(t^2)\). Work in finite-dimensional \(\Lambda\)-modules. Prove there are enough injectives, compute the stable Hom groups between \((k,n)\) and \((k,m)\), and give the complete triangle of \(0\to k\xrightarrow{1\mapsto t}\Lambda\to k\to0\). Include the sign after one positive shift.

Solution. For a finite-dimensional module \(V\), every linear functional \(\ell:V\to k\) gives a module map

Indeed \(e_\ell(tv)=\ell(tv)t=t e_\ell(v)\). Conversely a module map \(e:V\to\Lambda\) is uniquely of this form: take \(\ell\) to be its coefficient of \(t\); \(e(tv)=t e(v)\) forces the constant coefficient to be \(\ell(tv)\). This gives a natural isomorphism \(\operatorname{Hom}_\Lambda(V,\Lambda)\simeq\operatorname{Hom}_k(V,k)\). Linear functionals extend across inclusions of finite-dimensional vector spaces by extending a basis. Thus \(\Lambda\) is injective. For a basis of the linear dual, the map with components (8.2) embeds \(V\) into a finite sum of copies of \(\Lambda\): vanishing of all its \(t\)-coefficients forces \(v=0\). This proves enough injectives.

To identify all injectives, put \(T=t\) on \(V\). Choose a basis \(u_i\) of \(\operatorname{im}T\), lifts \(v_i\) with \(Tv_i=u_i\), and extend the \(u_i\)'s to a basis of \(\ker T\) by \(w_j\). The \(v_i,u_i,w_j\) form a basis: applying \(T\) proves independence of the \(v_i\) coefficients; the kernel basis then proves the rest, and subtracting lifts of \(Tv\) proves spanning. Thus \(V\simeq\Lambda^a\oplus k^b\). The displayed row does not split, because a section of \(\Lambda\to k\) would send \(1\) to an element killed by \(t\), hence to a multiple of \(t\), whose image in \(k\) is zero. Therefore \(k\) is not injective: if its embedding into \(\Lambda\) split by a retraction, the row would split. Retracts of injectives are injective, so an injective \(V\) has \(b=0\). Injectives are exactly the finite free modules.

Every \(k\to\Lambda\) lands in \(kt\), and every \(\Lambda\to k\) kills \(t\). Thus maps \(k\to k\) through any finite free module are zero, and

Choose the displayed embedding at every cosyzygy of \(k\), identifying its quotient with \(k\). A scalar map lifts to the same scalar on \(\Lambda\), so \(\Sigma\) acts as identity on (8.3). Every sufficiently advanced group in (2.1) for \((k,n),(k,m)\) is consequently \(k\) with identity transitions. Its colimit is \(k\), composition multiplies scalars, and the class of \(1\) identifies any two integer labels. The middle term \(\Lambda\) is zero stably. In (6.1) choose \(b=1_\Lambda\), so \(h=1_k\). The complete triangle is

Its last map is the identity scalar under the just specified label identifications. After one shift, keeping the first two maps positively shifted, the last map is \(-S\gamma_k^{-1}\), hence scalar \(-1\). In characteristic two the two scalars coincide; the sign formula is still valid. Since no \(\Sigma^r k=k\) is injective, \(k\) has infinite injective dimension by Corollary 7.1 and its quotient object is nonzero.

Exercise 3 — A cone for an arbitrary stable map (advanced)

Give the three arrows of the distinguished stable triangle completing an actual map \(f:A\to B\), using (6.6). Compute its third object and last map when \(f=0\), and its third object when \(f=1_A\). Explain how it completes a map represented only at stage \(k\).

Solution. Let \(C_f=\operatorname{coker}(f,i_A)\), let \(p:B\oplus I_A\to C_f\) be the quotient, and let \(j_B:B\to B\oplus I_A\) be inclusion. The projection \(b:B\oplus I_A\to I_A\) satisfies \(b(f,i_A)=i_A\). Therefore there is a unique \(h_f:C_f\to\Sigma A\) with \(h_fp=p_Ab\). The required triangle has arrows

Projection and inclusion identify the middle object of its original row with \(B\) stably, and their connecting square is unaffected, since the first and third vertices were not changed. This proves the displayed complete triangle, not just its first arrow.

If \(f=0\), the cokernel is \(B\oplus\Sigma A\), its second arrow is inclusion of \(B\), and \(h_f\) is projection onto \(\Sigma A\). Equation (8.5) is consequently the distinguished split cone of the zero map, with the last projection followed by \(\gamma_A^{-1}\). If \(f=1_A\), the map \((b,z)\mapsto z-i_Ab\) induces an isomorphism \(C_f\simeq I_A\): its kernel is exactly the image of \((1_A,i_A)\), and it is epic with section \(z\mapsto(0,z)\). Thus its third object is zero stably, giving the identity triangle.

For a stage representative \(u=[\underline f,k]:(A,n)\to(B,m)\), first apply this construction to \(f:\Sigma^{n+k}A\to\Sigma^{m+k}B\). Transport the first two vertices by the inverses of \(\beta_{n+k,A},\beta_{m+k,B}\), including the first vertex's suspension in the last square. This completes \(S^ku\) by (2.4). Apply \(S^{-k}\) and multiply its last arrow by \((-1)^k\). That is the signed translate (6.4), and its first arrow is exactly \(u\). This supplies a distinguished completion for every eventual map, with its full last arrow.

Exercise 4 — Changing embeddings and detecting finite dimension (advanced)

Given embeddings \(A\to I\), \(A\to I'\) into injectives, prove the dual Schanuel isomorphism \(\Sigma A\oplus I'\simeq\Sigma' A\oplus I\). Then prove that the identity of \(\iota A\) vanishes exactly when a finite injective resolution exists. Explain why these conclusions assert neither a fixed dimension bound nor that \(\underline{\mathcal A}\to\mathcal S\) is faithful.

Solution. Form the pushout \(E=I\amalg_A I'\). Pushouts of monomorphisms are monic, so it has short exact rows

Their quotient descriptions follow from the pushout property: killing the respective summand leaves precisely the cokernel of the other embedding. Each inclusion has a retraction because its domain is injective. For any short exact row with a retraction \(r:E\to I'\), the kernel \(\ker r\) maps isomorphically onto the cokernel of \(I'\to E\): every element of \(E\) decomposes by \(e=i'r(e)+(e-i'r(e))\), or categorically the maps \(i'r\) and \(1-i'r\) give the biproduct decomposition. Hence the two rows split and give

This is an ordinary object isomorphism, though its splittings need not be unique. The actual canonical stable comparison and its cocycle law were proved separately in Section 1, without choosing a particular splitting of (8.6).

The identity of \(\iota A\) is represented by \(1_{\Sigma^rA}\) at any stage \(r\). It vanishes exactly when a sufficiently advanced such identity factors through an injective. That factorization makes \(\Sigma^rA\) a retract of an injective, hence injective, and the successive chosen rows give the finite resolution (7.1). Conversely a finite injective resolution of \(A\) makes its stalk belong to \(\mathcal N\); its quotient identity is zero. The injectivity proof of Theorem 4.3 gives a finite later stage at which the stable identity vanishes. This proves both directions even if the given finite resolution uses different embeddings.

The needed stage depends on the object and on its finite resolution. Nothing in this argument bounds all objects at one common stage; Exercise 2 even has an object with no finite such stage. Exercise 1 has a nonzero identity in the original stable category whose image in stabilization is zero. Thus the construction proves a quotient equivalence and a precise eventual kernel criterion, while retaining these two limitations.

9. Freely accessible references

The proofs above use the explicit course interfaces listed at the start. Freely accessible primary references for their basic ingredients are the Stacks Project's bounded-injective Hom comparison, Tag 05TG, its extension proof, Tag 013P and uniqueness proof, Tag 013S, and its triangulated quotient construction, Section 13.6. These are references for those ingredients; the cosyzygy Hom calculation and the precise row-triangle equivalence are proved here.