Draft lesson.

Stabilizing an endofunctor through eventual maps

Original text public domain (CC0).

An endofunctor can forget maps, add copies of objects, or behave like a shift without having an inverse. Stabilization makes it invertible by comparing maps only after enough iterations. The resulting equality is eventual equality, so a faithful embedding of the original category is a conclusion to test, not a premise to assume.

Let \(\mathcal C\) be a locally small category and \(T:\mathcal C\to\mathcal C\) any functor. Fix the universe and Choice conventions of Universes and small categories, Sections 1–2. In particular \(\mathbb N,\mathbb Z\) and countable tagged unions and quotients of small sets are small. Object collections and categories of functors may belong to a larger ambient universe. For target translation equivalences, retain the chosen inverse, coherent powers \(P_n\) and comparisons \(\mu_{r,s}\) of Translations, coherent powers and universal triangle comparisons, Sections 1–2. No faithfulness, additivity or essential surjectivity of \(T\) is assumed in Sections 1–4.

1. Eventual equality supplies the morphisms

For objects \((X,n),(Y,m)\), with \(n,m\in\mathbb Z\), let

The transition from stage \(k\) to stage \(k+1\) is \(f\mapsto T(f)\). Write \([f,k]\) for the class of a stage map. Define

This is an equivalence relation: reflexivity and symmetry are immediate, and two equalities become one equality at the larger of their witnessing stages. No transition map need be injective. The maps from each stage to these classes form a cocone. Every other cocone of functions has the unique factor \([f,k]\mapsto v_k(f)\), because (1.2) identifies precisely the eventual transition images. Thus the class set is the colimit of the stage sets. It is small by the tagged-union and quotient bounds in the universe prerequisite. Allowing negative stages at which both displayed exponents are nonnegative gives the same set: every such stage maps to a nonnegative one, and every equality can be witnessed in a nonnegative tail.

Define \(\operatorname{Stab}(\mathcal C,T)\) to have all pairs \((X,n)\) as objects and these colimits as its Hom sets. For \([f,k]:(X,n)\to(Y,m)\) and \([g,l]:(Y,m)\to(Z,p)\), choose \(r\ge k,l\) and put

The source and target of the composite are \(T^{n+r}X,T^{p+r}Z\), with middle object \(T^{m+r}Y\). Raising \(r\) applies the same further power of \(T\) to the whole composite. Changing either representative makes it equal to the old one at some common later stage, so gives the same composite there. This proves independence of every choice. Three maps can be represented at one stage, where associativity is that of \(\mathcal C\). The identity of \((X,n)\) is \([1_{T^{n+k}X},k]\) at any allowed stage \(k\); its class is independent of \(k\), and (1.3) proves the two identity laws. Consequently this construction is a locally small category.

2. An actual inverse shift and the original-category map

Define a shift and its inverse by

All the indicated stages are allowed. The shifted exponents for \(S[f,k]\) are \(n+k+1,m+k+1\), while those for \(R[f,k]\) at stage \(k+1\) are \(n+k,m+k\). Eventual equality is preserved, so these formulas define maps on classes. Applying them to (1.3) proves functoriality. Both composites on morphisms are \([f,k]\mapsto[T(f),k+1]=[f,k]\), by (1.2). On objects both composites are the identity. Hence \(R=S^{-1}\) strictly on this category.

There is a functor

It preserves identities and composition at stage zero. A natural isomorphism

is represented by \(1_{TX}\) at stage zero; the inverse is represented by the same underlying identity in the reverse direction. The naturality square is represented by \(T(f)\) on both paths.

For every \(a\ge0\), the map

is likewise represented by \(1_{T^aX}\) at stage zero and has the corresponding identity inverse. These maps are the iterates of \(\gamma\): \(\beta_{a+1,X}=\gamma_{T^aX}S(\beta_{a,X})\). The equality follows from the common underlying identity after increasing the stage if necessary.

Every stabilized arrow has a useful exact factorization. For \([f,k]:(X,n)\to(Y,m)\), both \(n+k,m+k\) are nonnegative and

The right side is represented by \(f\) at stage zero between \((X,n+k),(Y,m+k)\). Its transition to stage \(k\) is \(T^k(f)\), which is the representative obtained from the left side by (2.1). Thus (2.5) is an equality of arrows, not only an unspecified comparison. It will control the universal property and all natural transformations.

3. Extending into a translation equivalence

Let \(U:\mathcal A\to\mathcal A\) be an equivalence, let \(F:\mathcal C\to\mathcal A\) be a functor, and choose a natural isomorphism

Use coherent powers \(P_n\) of \(U\), with \(P_k=U^k\) for \(k\ge0\). Define natural isomorphisms \(\rho_a:FT^a\to P_aF\), for nonnegative \(a\), by

Each term has the stated type. Naturality follows inductively from naturality of \(\theta\). For every \(f:T^{n+k}X\to T^{m+k}Y\), that naturality also gives

Indeed substitute (3.2) and cancel the naturality square \(\theta_{T^{m+k}Y}F(Tf)=UF(f)\theta_{T^{n+k}X}\).

Set \(\widehat F(X,n)=P_nFX\). To define \(\widehat F[f,k]\), first form

Use the coherent comparisons to regard it as

Full faithfulness of \(U^k\) gives a unique map

This uses the inverse of a Hom bijection, so it does not replace an equivalence by a literal inverse functor.

For allowed \(k\), coherence and \(n+k\ge0\) give \(\mu_{k+1,n}=U\mu_{k,n}\); the positive-power comparisons on the other side of that coherence square are identities. Equations (3.3)–(3.5) therefore give \(\widetilde q_{k+1}=U\widetilde q_k\). The unique lift in (3.6) is unchanged when the representative is advanced. Equality at a common later stage proves independence under (1.2). Identity representatives give identity maps. For composition, use one common stage in (1.3). The middle factors \(\rho^{-1}\rho\) and \(\mu\mu^{-1}\) cancel, so the lift of the composite is the composite of the lifts. Thus \(\widehat F\) is a functor, and \(\widehat F\iota=F\) on objects and arrows.

Its translation comparison is

To check naturality, compare (3.5) for \([f,k]\) and for \(S[f,k]=[Tf,k]\). Equation (3.3) gives \(q_{S,k}=Uq_k\). The two ways of aligning the powers agree because coherence gives

After applying the faithful functor \(U^k\), these are exactly the naturality equations for (3.7), so they also hold before applying it. At \(n=0\), \(\chi\) is the identity. Formula (3.6) at stage zero for \(\gamma_X\) gives \(\widehat F(\gamma_X)=\theta_X^{-1}\). Consequently the comparison induced on its restriction is precisely the given one:

This proves existence of an extension with the specified translation data.

4. Natural transformations give the full universal property

Pairs \((F,\theta)\) as in (3.1) form a category. Its maps \(a:(F,\theta)\to(G,\vartheta)\) are natural transformations satisfying

Identity transformations and composites satisfy this equation by substitution. The analogous category of translation-compatible functors from \((\operatorname{Stab},S)\) has specified comparisons \(HS\to UH\) and transformations commuting with them.

Theorem 4.1. Restriction along \(\iota\), with comparison induced by \(\gamma^{-1}\), is an equivalence between these two categories of compatible functors.

Proof. Existence is Section 3. For a map satisfying (4.1), define

Induction in (3.2) gives \(P_p(a_X)\rho^F_{p,X}=\rho^G_{p,X}a_{T^pX}\) for every nonnegative \(p\). Combine these equations at \(p=n+k,m+k\) with naturality of \(a\) at the stage map \(f\). Naturality of the power comparisons then gives the naturality equation for (4.2) after applying \(U^k\) to it. Faithfulness of \(U^k\) proves the equation itself. Coherence and naturality of \(\mu_{1,n}\) prove compatibility with (3.7). Thus restriction has the indicated lift of compatible transformations between the canonical extensions.

We prove fullness and faithfulness for arbitrary compatible functors, rather than only the canonical ones. Let \((H,\alpha:HS\to UH)\) be such a functor. Since \((X,n)=S^n\iota X\), iteration of \(\alpha\) gives coherent natural comparisons

Concretely, for every integer \(n\) they satisfy

For positive \(n\) this is recursion. To go downwards, (4.4) prescribes \(U(\kappa_{n,X})\) as the composite \(\mu_{1,n}^{-1}\kappa_{n+1,X}\alpha_{(X,n)}^{-1}\). Full faithfulness of \(U\) gives its unique preimage; the inverse is obtained in the same way, and faithfulness proves both inverse equations. Naturality follows by applying \(U\) to the required square and using the already established next square. The coherence for longer shift blocks follows by induction in (4.4), its downward inverse equations, and coherence of \(\mu\). This gives (4.3) for all integers with the actual chosen comparison data.

If \((K,\delta)\) is another compatible functor and \(a:H\iota\to K\iota\) satisfies the induced equation (4.1), a compatible transformation must have components

This is forced upwards by compatibility and downwards by full faithfulness of \(U\). These components do satisfy the compatibility equation, by (4.4). They are natural on \(\iota(f)\) by naturality of \(a\), and natural on \(\gamma_X\) by (4.1) with the restricted comparisons. They are consequently natural on the iterates \(\beta_{p,X}\) and their inverses. Formula (2.5) proves naturality on \(S^k[f,k]\). Compatibility with the iterated shift comparisons then turns that equation into the image under \(U^k\) of the naturality square for \([f,k]\); its faithfulness proves that square. Thus (4.5) defines the required transformation on every arrow. It is unique because its components were forced. This proves fullness and faithfulness of restriction.

The canonical extension of Section 3 gives every restricted pair. A fully faithful functor which is essentially surjective is an equivalence: choose a preimage with its isomorphism for each object, lift maps through the full-faithful Hom bijections, and use faithfulness for the identity and composition laws. This supplies the asserted equivalence here. \(\square\)

In particular an extension is unique up to a compatible isomorphism after one fixes the isomorphism of its restriction with \((F,\theta)\). That compatible isomorphism is unique when its restriction is prescribed. An unrestricted uniqueness assertion about unspecified natural isomorphisms is not needed and is not made.

5. Additivity and two concrete outcomes

Suppose now that \(\mathcal C\) is additive and \(T\) is additive. Addition on stabilized maps is represented by adding two maps at a common stage. Passing to a later stage preserves addition; choosing another pair of representatives gives the same sum by (1.2). The zero and negative maps are constructed in the same way, so each Hom set is an abelian group. Equation (1.3) is bilinear at a common stage.

The object \((0,0)\) is zero: every \(T^k0\) is zero because \(T\) is additive, and hence every sufficiently late Hom set to or from this object is a singleton. For \((X,n),(Y,m)\), choose \(r\ge\max(0,-n,-m)\) and put

The actual injections and projections in \(\mathcal C\), represented at stage \(r\), satisfy \(p_i i_j=\delta_{ij}\) and \(i_1p_1+i_2p_2=1_B\) in stabilization. These equations give both biproduct universal properties: a map into \(B\) is \(i_1a+i_2b\) and a map out is \(ap_1+bp_2\); composing with the legs proves the required uniqueness. Thus stabilization is additive. The shift, its inverse and \(\iota\) are additive by their formulas. If the target is additive and \(U,F\) are additive, (3.5) is a composite of Hom-group isomorphisms and the group map induced by \(F\). The inverse Hom bijection in (3.6) is also a group isomorphism, so \(\widehat F\) is additive. Theorem 4.1 therefore restricts to the additive compatible-functor categories.

Example 5.1 (a complete loss of original maps). Let \(\mathcal C=\mathsf{Set}\) and \(T\) be the constant singleton functor. After one further transition every stage map becomes the same singleton identity. Conversely every sufficiently late stage Hom set is a singleton, even when an earlier Hom set was empty. Hence every stabilized Hom set is a singleton, and all objects are isomorphic. The two different maps from a singleton to a two-element set have equal images under \(\iota\). Stabilization exists here, but \(\iota\) is not faithful.

Example 5.2 (new maps without loss). Let \(\mathcal C\) have one object \(*\), with endomorphism monoid \((\mathbb N,+)\); composition is addition and its identity is zero. Let \(T(a)=2a\). A class \([a,k]\) between any two labels is identified with \(a/2^k\). Equality of two such fractions is exactly eventual equality after multiplying at a common stage. Every nonnegative dyadic rational occurs at a sufficiently late allowed stage. Composition adds the fractions, and the identity is zero. The original embedding sends \(a\) to the integer \(a\), so it is faithful but is not full. The shift doubles every fraction and its inverse halves it. Maps represented by zero between different labels are mutual isomorphisms, since their composites are the zero identity. This example is an ordinary category; its Hom monoid is not being called an abelian group.

6. Four graded exercises with complete solutions

Exercise 1 (foundation: discard an initial segment). For a sequence of sets \(A_k\) with arbitrary transition functions, prove directly that deleting any finite initial segment gives the same colimit. Do not assume the functions are injective.

Solution. Use the eventual-equality construction of Section 1. Every representative at a deleted stage maps to the first retained stage, so the tail map is surjective. If two retained representatives are equal in the full colimit, their equality has a common witnessing stage. Increase it to at least the first retained stage; that equality witnesses their equality in the tail. Thus the map is injective as well. The two procedures respect the class maps from every retained stage, so the bijection is the canonical colimit comparison. This applies to the nonnegative-stage restriction in (1.1).

Exercise 2 (intermediate: exactly when the original maps survive). Prove that \(\iota\) is faithful exactly when \(T\) is faithful. Prove that \(\iota\) is fully faithful exactly when \(T\) is fully faithful. If \(T\) is an equivalence, show that \(\iota\) is an equivalence as well.

Solution. If \(T\) is faithful, all powers on each Hom set are injective; equality of two stage-zero maps in (1.2) forces equality of the original maps. Conversely \(Tf=Tg\) gives \(\iota f=\iota g\); faithfulness of \(\iota\) then forces \(f=g\). If \(T\) is fully faithful, each transition from a stage-zero Hom set is a bijection, so every class has a unique stage-zero representative. Thus \(\iota\) is fully faithful.

For the other implication, assume \(\iota\) is fully faithful. The natural comparison \(S\iota\simeq\iota T\) and the invertibility of \(S\) identify the map on Hom sets induced by \(T\) with the composite of the bijections induced by \(\iota,S\) and the comparison. This composite is bijective, hence \(T\) is fully faithful. Finally, if \(T\) is an equivalence, choose \(T^{-1}\). For a label \(n\ge0\), (2.4) identifies \((X,n)\) with \(\iota(T^nX)\). For \(n=-a<0\), choose \(Y\) with \(T^aY\simeq X\); the isomorphism \(S^a\iota Y\simeq\iota X\), shifted back by \(S^{-a}\), identifies \(\iota Y\) with \((X,-a)\). Thus every object is in the essential image, and the full-faithful result makes \(\iota\) an equivalence.

Exercise 3 (advanced: doubling creates matrix maps). On abelian groups let \(T(X)=X\oplus X\), with \(T(f)=\operatorname{diag}(f,f)\). Describe \(\operatorname{End}_{\operatorname{Stab}}(\mathbb Z,0)\). Exhibit a new non-scalar automorphism and a nontrivial idempotent, and split that idempotent through an object with a negative label.

Solution. At stage \(k\) the object is \(\mathbb Z^{2^k}\), and its endomorphism ring is \(M_{2^k}(\mathbb Z)\). Transition sends \(A\) to \(\operatorname{diag}(A,A)\), a unital ring embedding. Therefore the stabilized endomorphism ring is their directed colimit, with addition and composition computed at a common matrix stage. The matrix \(P=\left(\begin{smallmatrix}0&1\\1&0\end{smallmatrix}\right)\) at stage one has \(P^2=1\). Its later block-diagonal images contain off-diagonal ones and cannot equal any scalar integer identity. Hence it is an automorphism outside the image of the stage-zero endomorphism ring.

The matrix \(e=\operatorname{diag}(1,0)\) at stage one satisfies \(e^2=e\). It and \(1-e\) remain nonzero at every later stage because the transition is injective. At stage one the projection \((1,0):\mathbb Z^2\to\mathbb Z\) represents a map \(p:(\mathbb Z,0)\to(\mathbb Z,-1)\), and the column \(\binom10\) represents \(j\) in the reverse direction. Both stages are allowed because \(-1+1=0\). Their composites are \(jp=e\) and \(pj=1_{(\mathbb Z,-1)}\). This is the requested actual splitting. The Hom maps introduced by stabilization need not originate as maps at stage zero.

Exercise 4 (synthesis: the chosen comparison affects an extension). In Example 5.2 take as target the one-object category with endomorphism group \((\mathbb Q,+)\) and translation \(U(q)=2q\). Let \(F(a)=ra\), for \(r\in\mathbb Q\), and let the comparison \(\theta\) have component \(c\in\mathbb Q\). Compute \(\widehat F\) on a dyadic map \(q:(*,n)\to(*,m)\). For fixed \(r\), find every compatible natural transformation between the extensions for comparisons \(c,c'\).

Solution. The two functors \(FT\) and \(UF\) agree on maps, and every \(c\) is an invertible natural comparison because the target composition is commutative addition. Formula (3.2) gives \(\rho_a=(2^a-1)c\). For a representative \([b,k]\), with \(q=b/2^k\), (3.4) has value

Applying the inverse of the Hom bijection for \(U^k\), which divides by \(2^k\), gives

This is defined for negative labels as well. For composable dyadic maps, the two middle terms \(2^m c\) cancel, proving composition directly; the zero identity from a label to itself maps to zero. Shifting doubles (6.1), so the extended translation comparison is the identity morphism zero. The map \(\gamma:(*,1)\to(T*,0)\) has fraction zero and image \(-c\); its inverse gives exactly the restricted comparison \(c\), as required by (3.9).

On restriction, a natural transformation has one component \(t\in\mathbb Q\). Equation (4.1) is \(2t+c=c'+t\), so \(t=c'-c\) is forced. Its extension has component \(2^n(c'-c)\) at label \(n\). For a map from label \(n\) to label \(m\), inserting these two components and (6.1) into the naturality equation gives the same value on both sides. Shift compatibility doubles the component, so it holds too. The components are invertible, and no other compatible transformation is possible by their forced restriction and Theorem 4.1. The formula exhibits both dependence on the chosen comparison and the precise uniqueness after that comparison is fixed.

References

• The Stacks Project Authors, Filtered colimits, for the eventual-equality description of morphism-stage colimits. Section 1 proves the particular sequential construction used here.
• The chosen inverse and integer-power comparisons are proved in Translations, coherent powers and universal triangle comparisons, Sections 1–2.