The higher mapping cone axiom

Xu Cheng; Xiaoyan Yang

doi:10.1515/math-2025-0170

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The higher mapping cone axiom

Xu Cheng and Xiaoyan Yang

Published/Copyright: July 7, 2025

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$Open Mathematics$

From the journal Open Mathematics Volume 23 Issue 1

Abstract

The aim of the article is to better understand the higher octahedral axiom of n -angulated categories. We discuss several possible additional axioms for pre- n -angulated categories and prove that they are all equivalent to the higher octahedral axiom.

Keywords: n-angulated category; higher mapping cone

MSC 2010: 18G80; 18E10

1 Introduction

Triangulated categories were introduced independently in algebraic geometry by Verdier [1] and in algebraic topology by Puppe [2]. These constructions have since played a crucial role in representation theory, algebraic geometry, algebraic topology, commutative algebra and even theoretical physics. Let n be an integer greater than or equal to 3. Geiss et al. [3] introduced “higher dimensional” analogues of triangulated categories, called n-angulated categories. The classical triangulated categories are the special case n = 3 . Other examples of n -angulated categories can be found in [3–5]. The fourth axiom for n -angulated categories is a generalization of the octahedral axiom for triangulated categories. Bergh and Thaule [6] discussed the axioms of n -angulated categories systematically and showed that the higher octahedral axiom is equivalent to the higher mapping cone axiom. They also proved that TR3 is redundant under the presence of TR4 and the other axioms also in the higher setting. Lin and Zheng [7] used homotopy cartesian diagrams to give several new equivalent statements of the higher mapping cone axiom.

The aim of this note is to introduce the higher base change, higher cobase change, higher mapping cone axiom and the higher octahedral axiom, and show all these axioms are equivalent in pre- n -angulated categories.

2 Some basic facts of pre- n -angulated categories

This section is devoted to recalling some notions and basic consequences of pre- n -angulated categories. For terminology, we shall follow [3,6].

Let T be an additive category with an automorphism Σ : T → T and n an integer greater than or equal to 3. A sequence of morphisms in T

X • : X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1

is an n - Σ -sequence. Its left and right rotations are the two n - Σ -sequences

X 2 → α 2 X 3 → α 3 … → α n Σ X 1 → ( − 1 ) n Σ α 1 Σ X 2 , Σ − 1 X n → ( − 1 ) n Σ − 1 α n X 1 → α 1 X 2 → α 2 … → α n − 2 X n − 1 → α n − 1 X n ,

respectively. A trivial n- Σ -sequence is a sequence of the form

( T X ) • : X → 1 X → 0 → … → 0 → Σ X

or any of its rotations for X ∈ T . An n - Σ -sequence X • is exact if the induced sequence

Hom T ( − , X • ) : … → Hom T ( − , X 1 ) → … → Hom T ( − , X n ) → Hom T ( − , Σ X 1 ) → …

of representable functors T op → Ab is exact.

A morphism of n - Σ -sequences is given by a sequence of morphisms φ • = ( φ 1 , φ 2 , … , φ n ) such that the following diagram

commutes. It is an isomorphism if φ 1 , … , φ n are all isomorphisms in T , and a weak isomorphism if φ i and φ i + 1 are isomorphisms for some 1 ≤ i ≤ n (with φ n + 1 = Σ φ 1 ).

The category T is pre-n-angulated if there exists a collection N of n - Σ -sequences satisfying the following three axioms:

1. N is closed under direct sums, direct summands and isomorphisms of n - Σ -sequences.
2. For all X ∈ T , the trivial n - Σ -sequence ( T X ) • belongs to N .
3. For each morphism α 1 : X 1 → X 2 in T , there exists an n - Σ -sequence in N whose first morphism is α 1 .
An n - Σ -sequence X • belongs to N if and only if its left rotation belongs to N .
Each commutative diagram with rows in N

can be completed to a morphism of n - Σ -sequences.

In this case, the collection N is a pre- n -angulation of the category T (relative to the automorphism Σ ), and the n - Σ -sequences in N are called n-angles.

Let X • and Y • be two n - Σ -sequences and φ , ψ be two morphisms from X • to Y • . A homotopy Θ from φ to ψ is given by diagonal morphisms Θ i

such that

φ i − ψ i = Θ i α i + β i − 1 Θ i − 1 for i = 2 , 3 , … , n , Σ φ 1 − Σ ψ 1 = Σ Θ 1 Σ α 1 + β n Θ n .

In this case, we say that φ and ψ are homotopic. A morphism homotopic to the zero morphism is called null-homotopic.

Remark 2.1

Let

X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1

be an n - Σ -sequence in N .

One checks that if you change the sign on any n − 1 of α 1 , … , α n , then the resulting n - Σ -sequence is still in N .
It follows from (N2) that the following n - Σ -sequence
Σ ℓ X 1 → ( − 1 ) ℓ n Σ ℓ α 1 Σ ℓ X 2 → ( − 1 ) ℓ n Σ ℓ α 2 … → ( − 1 ) ℓ n Σ ℓ α n − 1 Σ ℓ X n → ( − 1 ) ℓ n Σ ℓ α n Σ ℓ + 1 X 1
is still in N for any ℓ ∈ Z .
All n -angles are exact by [3, Proposition 2.5].

Lemma 2.2

Let X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 ∈ N . Then,

… → Hom T ( Σ X 1 , Y ) → Hom T ( X n , Y ) → … → Hom T ( X 1 , Y ) → …

is exact in Ab for any Y ∈ T .

Proof

This is almost verbatim the dual argument as the corresponding statement for pre-triangulated categories, see for example [8, Section 1.1.10].□

Let C be an additive category and f : A → B a morphism in C . A weak cokernel of f is a morphism g : B → C such that for all C ′ ∈ C the sequence

Hom C ( C , C ′ ) → g * Hom C ( B , C ′ ) → f * Hom C ( A , C ′ )

is exact in Ab. Equivalently, g is a weak cokernel of f if g f = 0 and for each morphism h : B → C ′ with h f = 0 there exists a morphism p : C → C ′ such that h = p g . The concept of weak kernel is defined dually.

Lemma 2.3

Let X 1 → α 1 … → α i − 1 X i be a sequence in T . If α j − 1 is a weak kernel of α j for 2 ≤ j ≤ i − 1 , then there exists an n-angle

X 1 → α 1 … → α i − 1 X i → α i … → α n − 1 X n → α n Σ X 1 .

Proof

This follows directly from (N1) for i = 2 . Assume that 3 ≤ i ≤ n − 2 . Applying (N1) and (N2) for α 2 , there exists an n -angle

X 1 ′ → α 1 ′ X 2 → α 2 X 3 → α 3 … → α n − 1 X n → α n ′ Σ X 1 ′ .

Since α 2 α 1 ′ = 0 , there is φ 1 : X 1 ′ → X 1 such that the following diagram

commutes. By the higher analogue of the 5-Lemma, φ 1 is an isomorphism. We thus obtain the desired n -angle. Assume that i = n − 1 . There exists an n -angle as follows by induction:

X 1 ′ → α 1 ′ X 2 → α 2 … → α n − 2 X n − 1 → α n − 1 X n → α n ′ Σ X 1 ′ .

By analogy with the preceding proof, we obtain the desired n -angle.□

Lemma 2.4

Each commutative diagram with rows in N

can be completed to a morphism of n - Σ -sequences, where 2 ≤ i ≤ n − 1 . In particular, the sequence constructed is homotopic to the one obtained by (N3) from ( φ 1 , φ 2 ) .

Proof

If i = 2 , then we are done. Assume that i > 2 . By (N3), we may choose φ 3 ′ , … , φ n ′ such that ( φ 1 , φ 2 , φ 3 ′ , … , φ n ′ ) is a morphism of n - Σ -sequences. Since Hom T ( X j + 1 , Y j ) → α j ∗ Hom T ( X j , Y j ) → α j − 1 ∗ Hom T ( X j − 1 , Y j ) is exact by Lemma 2.2, there is Θ j : X j + 1 → Y j such that ( φ j − φ j ′ ) − β j − 1 Θ j − 1 = Θ j α j for 3 ≤ j ≤ i , where Θ 2 = 0 . Set φ i + 1 = β i Θ i + φ i + 1 ′ , φ j = φ j ′ for i + 2 ≤ j ≤ n and Θ j = 0 for i + 1 ≤ j ≤ n . Then, ( φ 1 , φ 2 , φ 3 , … , φ n − 1 , φ n ) is a morphism of n - Σ -sequences.□

Next we show that if an n - Σ -sequence looks like it might admit one of these n -angles as a direct summand, then it actually does (loosely speaking).

Lemma 2.5

Suppose we have an n-angle of the form

A ⊕ X 1 → α β γ δ A ⊕ X 2 → a α 2 X 3 → α 3 … → α n − 1 X n → b α n Σ ( A ⊕ X 1 ) .

If α ∈ Aut ( A ) , then it is isomorphic to a direct sum of n-angles

A → 1 A → 0 → … → 0 → Σ A ,

X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 ,

where α 1 = δ − γ α − 1 β .

Proof

Since the composition of maps in an n -angle is zero, we have that a = − α 2 γ α − 1 and ( Σ α ) b + ( Σ β ) α n = 0 . This yields an isomorphism of n - Σ -sequences

where α 1 = δ − γ α − 1 β . The proof is complete.□

Use the same arguments as above, we have the following lemmas.

Lemma 2.6

Suppose we have an n-angle of the form

X 1 → α 1 … → α i − 2 X i − 1 → a α i − 1 A ⊕ X i → α β γ δ A ⊕ X i + 1 → b α i + 1 X i + 2 → α i + 2 … → X n → α n Σ X 1

for i = 2 , … , n − 1 . If α ∈ Aut ( A ) , then it is isomorphic to a direct sum of n-angles

0 → … → 0 → A → 1 A → 0 → … → 0 , X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 ,

where α i = δ − γ α − 1 β for i = 2 , … , n − 1 .

Lemma 2.7

Suppose we have an n-angle of the form

A ⊕ X 1 → a α 1 X 2 → α 2 … → α n − 2 X n − 1 → b α n − 1 Σ A ⊕ X n → α β γ δ Σ ( A ⊕ X 1 ) .

If α ∈ Aut ( A ) , then it is isomorphic to a direct sum of n-angles

A → 0 → … → 0 → Σ A → 1 Σ A , X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 ,

where α n = δ − γ α − 1 β .

3 Additional axioms

The aim of this section is to introduce some possible additional axioms for a pre- n -angulated category, which are inspired by works of Hubery [9]. A pre- n -angulated category ( T , Σ , N ) is called n-angulated if it satisfies any of these extra axioms. We briefly describe the axioms before giving the precise formulation.

(N4) The following commutative diagram with rows in N

can be completed to a morphism of n -angles such that the mapping cone

X 2 ⊕ Y 1 → − α 2 0 φ 2 β 1 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n 0 φ n β n − 1 Σ X 1 ⊕ Y n → − Σ α 1 0 Σ φ 1 β n Σ X 2 ⊕ Σ Y 1

is in N .

Axiom A. The following commutative diagram with rows in N

for 2 ≤ i ≤ n − 1 can be completed to a morphism of n -angles such that the mapping cone

X 2 ⊕ Y 1 → − α 2 0 φ 2 β 1 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n 0 φ n β n − 1 Σ X 1 ⊕ Y n → − Σ α 1 0 Σ φ 1 β n Σ X 2 ⊕ Σ Y 1

is in N .

Axiom B 0 . The following commutative diagram with rows in N

$3$

can be completed to a morphism of n -angles such that the n - Σ -sequence

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n − 1 0 φ n − 1 β n − 2 X n ⊕ Y n − 1 → φ n β n − 1 Y n → ( Σ α 1 ) β n Σ X 2

is in N .

Axioms B 0 is a special case of Axiom (N4) when one of the maps is known to be an isomorphism. Axiom B 0 can be thought of as analogous to the existence of ( n − 2 ) -pushout diagrams and ( n − 2 ) -pullback diagrams in an ( n − 2 ) -abelian category [10].

Axiom B 1 . The following commutative diagram with rows in N

can be completed to a morphism of n -angles such that the n - Σ -sequence

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n − 1 0 φ n − 1 β n − 2 X n ⊕ Y n − 1 → φ n β n − 1 Y n → ( Σ α 1 ) β n Σ X 2

is in N .

Axiom C. Given an n - Σ -sequence in N

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n − 1 0 φ n − 1 β n − 2 X n ⊕ Y n − 1 → φ n β n − 1 Y n → δ Σ X 2 .

If X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 ∈ N or X 1 → β 1 Y 2 → β 2 … → β n − 1 Y n → β n Σ X 1 ∈ N , then there exists a commutative diagram with rows in N

such that δ = ( Σ α 1 ) β n .

Axiom C is a kind of converse to Axiom B 1 and Axiom B 1 ′ , and can be thought of as analogous to the fact that parallel maps in an ( n − 2 ) -pullback/ ( n − 2 ) -pushout diagram, α 2 , β 2 have isomorphic kernels and α n − 1 , β n − 1 have isomorphic cokernels.

Axiom D. For any n - Σ -sequence X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 ∈ N and any morphism φ 2 : X 2 → Y 2 , there exists a commutative diagram of n -angles

$3$

such that γ n θ n = ( Σ α 1 ) β n .

Axiom D ′ . For any n - Σ -sequence Y 1 → β 1 Y 2 → β 2 … → β n − 1 Y n → β n Σ Y 1 ∈ N and any morphism φ n − 1 : X n − 1 → Y n − 1 , there exists a commutative diagram of n -angles

$3$

such that ( Σ ψ 1 ) γ n = α n β n − 1 .

Axiom D and Axiom D ′ can be thought to be “higher dimensional” analogues of cobase change and base change for triangulated categories [11].

4 Equivalence of the additional axioms

In this section, we prove that all possible axioms in the above section are equivalent, which can be applied to explain the higher mapping axiom (N4).

(N4) implies A. Suppose that (N4) holds and that we are given a diagram as in A. We may choose φ 3 ′ , … , φ n ′ such that the mapping cone

(†) X 2 ⊕ Y 1 → − α 2 0 φ 2 β 1 X 3 ⊕ Y 2 → − α 3 0 φ 3 ′ β 2 … → − α n 0 φ n ′ β n − 1 Σ X 1 ⊕ Y n → − Σ α 1 0 Σ φ 1 β n Σ X 2 ⊕ Σ Y 1

is in N . It follows from Lemma 2.4 that ( φ 1 , φ 2 , φ 3 , … , φ n ) and ( φ 1 , φ 2 , φ 3 ′ , … , φ n ′ ) are homotopic. Hence, [4, Lemma 2.1] implies that the mapping cone

(‡) X 2 ⊕ Y 1 → − α 2 0 φ 2 β 1 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n 0 φ n β n − 1 Σ X 1 ⊕ Y n → − Σ α 1 0 Σ φ 1 β n Σ X 2 ⊕ Σ Y 1

is isomorphic to ( † ) , and so ( ‡ ) belongs to N .

A implies (N4). Suppose that A holds and that we are given a diagram as in (N4). Then, there is a morphism ( φ 1 , φ 2 , … φ i , φ i + 1 ′ , … , φ n ′ ) of n -angles by (N3), and so we have a diagram as in A. Hence, A yields φ i + 1 , … , φ n such that the mapping cone

X 2 ⊕ Y 1 → − α 2 0 φ 2 β 1 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n 0 φ n β n − 1 Σ X 1 ⊕ Y n → − Σ α 1 0 Σ φ 1 β n Σ X 2 ⊕ Σ Y 1

is in N .

(N4) implies B 0 . Suppose that (N4) holds and that we are given a diagram as in B 0 . We may choose φ 3 , … , φ n such that the following n - Σ -sequence

X 1 ⊕ X 2 → 0 − α 2 β 1 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n 0 φ n β n − 1 Σ X 1 ⊕ Y n → 1 β n − Σ α 1 0 Σ X 1 ⊕ Σ X 2

is in N . Thus, by Lemma 2.7, this has the following n - Σ -sequence

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n − 1 0 φ n − 1 β n − 2 X n ⊕ Y n − 1 → φ n β n − 1 Y n → ( Σ α 1 ) β n Σ X 2

as a direct summand. Hence, B 0 holds.

B 0 implies (N4). Suppose that B 0 holds and that we are given a diagram as in (N4). We begin by considering an isomorphism of n - Σ -sequences

$4$

Since the top row is in N , so is the bottom row. Applying B 0 , the following commutative diagram with rows in N

can be completed to a morphism of n -angles and such that the n - Σ -sequence

X 2 ⊕ Y 1 → − α 2 0 φ 2 β 1 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − a 0 φ n β n − 1 Σ X 1 ⊕ Y n → − Σ α 1 0 Σ φ 1 β n Σ X 2 ⊕ Σ Y 1

is in N . Note that α n 0 = − 1 0 Σ φ 1 β n − a φ n , and so a = α n . Hence, (N4) holds.

The proof of A ⇔ B 1 is similar to that of (N4) ⇔ B 0 .

B 1 implies C. Suppose that B 1 holds and that we are given an n -angle as in C. Suppose further that we fix the n -angle X 1 → α 1 X 2 → α 2 … → α n − 1 X n → α n Σ X 1 . It is easy to verify that X 2 → − α 2 X 3 → − α 3 … → − α n − 1 X n → − α n Σ X 1 → − Σ α 1 Σ X 2 ∈ N .

Applying B 1 , the following commutative diagram

$4$

can be completed to a morphism of n -angles and such that the n - Σ -sequence

X 3 ⊕ Y 2 → 1 0 α 3 0 − φ 3 − β 2 X 3 ⊕ X 4 ⊕ Y 3 → − α 3 1 0 0 α 4 0 0 − φ 4 − β 3 … → − α n − 2 1 0 0 α n − 1 0 0 − φ n − 1 − β n − 2 X n − 1 ⊕ X n ⊕ Y n − 1 → − α n − 1 1 0 0 − φ n − β n − 1 X n ⊕ Y n → − α n − β n Σ X 1 → 0 − Σ ( φ 2 α 1 ) Σ ( X 3 ⊕ Y 2 )

is in N . Note that δ = ( Σ α 1 ) β n and α n = β n φ n . By Lemma 2.5, this has the n - Σ -sequence

Y 2 → 0 − β 2 X 4 ⊕ Y 3 → 1 0 α 4 0 − φ 4 − β 3 X 4 ⊕ X 5 ⊕ Y 4 → − α 4 1 0 0 α 5 0 0 − φ 5 − β 4 … → − α n − 2 1 0 0 α n − 1 0 0 − φ n − 1 − β n − 2 X n − 1 ⊕ X n ⊕ Y n − 1 → − α n − 1 1 0 0 − φ n − β n − 1 X n ⊕ Y n → − α n − β n Σ X 1 → − Σ ( φ 2 α 1 ) Σ Y 2

as a direct summand. Rotating the above n -angle and using Lemma 2.5 again, we obtain the following n -angle:

Y 3 → 0 − β 3 X 5 ⊕ Y 4 → 1 0 α 5 0 − φ 5 − β 4 … → − α n − 2 1 0 0 α n − 1 0 0 − φ n − 1 − β n − 2 X n − 1 ⊕ X n ⊕ Y n − 1 → − α n − 1 1 0 0 − φ n − β n − 1 X n ⊕ Y n → − α n − β n Σ X 1 → − Σ ( φ 2 α 1 ) Σ Y 2 → ( − 1 ) n + 1 Σ β 2 Σ Y 3 .

Continuing this process, rotating the n - Σ -sequence and setting β 1 = φ 2 α 1 . We obtain the required commutative diagram with rows in N .

Suppose instead we had fixed an n -angle X 1 → β 1 Y 2 → β 2 … → β n − 1 Y n → β n Σ X 1 . We apply the above construction to obtain n -angles X 1 → α ¯ 1 X 2 → α 2 … → α n − 1 X n → α ¯ n Σ X 1 and X 1 → β ¯ 1 Y 2 → β 2 … → β n − 1 Y n → β ¯ n Σ X 1 satisfying Axiom C. By Lemma 2.4, we have the following commutative diagram of n -angles

$4$

By the higher analogue of the 5-Lemma, a ∈ Aut ( X 1 ) . Set α 1 = α ¯ 1 a − 1 and α n = ( Σ a ) α ¯ n . Then, we have a commutative diagram with rows in N :

$4$

We obtain the required commutative diagram of n -angles.

C implies B 1 . Suppose that C holds and that we are given a diagram as in B 1 . Let A ∈ T . Applying the functor Hom T ( A , − ) to the commutative diagram in B 1 , we obtain the following exact sequence of abelian groups:

Hom T ( A , X 2 ) → − α 2 φ 2 ∗ Hom T ( A , X 3 ⊕ Y 2 ) → − α 3 0 φ 3 β 2 ∗ … → − α n − 1 0 φ n − 1 β n − 2 ∗ Hom T ( A , X n ⊕ Y n − 1 ) .

Hence, Lemma 2.3 yields an n -angle:

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n − 1 0 φ n − 1 β n − 2 X n ⊕ Y n − 1 → a b W → c Σ X 2 .

We can apply C to obtain a commutative diagram of n -angles

such that c = ( Σ α 1 ) d . By Lemma 2.4, we have a commutative diagram of n -angles

$4$

Clearly, θ is an isomorphism. Set φ n = θ a . Then, ( 1 , φ 2 , … , φ n − 1 , φ n ) is a morphism of n -angles. Thus, the following diagram is an isomorphism of n - Σ -sequences

$4$

Since the top row is in N , so is the bottom row. Hence, B 1 holds.

B 0 implies D. Suppose that B 0 holds and that we are given an n -angle and a morphism as in D. By (N1), there are two n -angles

X 1 → β 1 Y 2 → β 2 … → β n − 1 Y n → β n Σ X 1 , X 2 → φ 2 Y 2 → γ 2 Z 3 → γ 3 … → γ n − 1 Z n → γ n Σ X 2 ,

where β 1 = φ 2 α 1 . One can check that X 1 → α 1 X 2 → α 2 X 3 → − α 3 … → − α n − 1 X n → ( − 1 ) n + 1 α n Σ X 1 belongs to N . Applying B 0 , the following commutative diagram

$4$

can be completed to a morphism of n -angles and such that the n - Σ -sequence

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → α 3 0 φ 3 β 2 … → α n − 1 0 ( − 1 ) n φ n − 1 β n − 2 X n ⊕ Y n − 1 → ( − 1 ) n + 1 φ n β n − 1 Y n → ( Σ α 1 ) β n Σ X 2

is an n -angle. Again applying B 0 , the following commutative diagram

$4$

can be completed to a morphism of n -angles and such that the mapping cone is in N . Hence, Lemma 2.5 implies that the following n - Σ -sequence

X 3 → α 3 φ 3 X 4 ⊕ Y 3 → − α 4 0 φ 4 − β 3 ψ 4 θ 3 X 5 ⊕ Y 4 ⊕ Z 3 → − α 5 0 0 − φ 5 − β 4 0 ψ 5 θ 4 γ 3 … → − α n − 1 0 0 ( − 1 ) n − 1 φ n − 1 − β n − 2 0 ψ n − 1 θ n − 2 γ n − 3 X n ⊕ Y n − 1 ⊕ Z n − 2 → ( − 1 ) n φ n − β n − 1 0 ψ n θ n − 1 γ n − 2 Y n ⊕ Z n − 1 → θ n γ n − 1 Z n → γ n ′ Σ X 3

is in N , where γ n ′ = ( Σ α 2 ) γ n . Note that γ n θ n = ( Σ α 1 ) β n . Hence, D holds.

D implies B 0 . Suppose that D holds and that we are given a diagram as in B 0 . Then, we have a commutative diagram of n -angles

$4$

One can check that the following n - Σ -sequence

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → α 3 0 φ 3 β 2 … → α n − 1 0 ( − 1 ) n φ n − 1 β n − 2 X n ⊕ Y n − 1 → ( − 1 ) n + 1 φ n β n − 1 Y n → ( Σ α 1 ) β n Σ X 2

is an n -angle. We also have an isomorphism of n - Σ -sequences

$4$

Note that the lower row is an n -angle, it follows from Lemma 2.7 that

X 2 → − α 2 φ 2 X 3 ⊕ Y 2 → − α 3 0 φ 3 β 2 … → − α n − 1 0 φ n − 1 β n − 2 X n ⊕ Y n − 1 → φ n β n − 1 Y n → ( Σ α 1 ) β n Σ X 2

is an n -angle. Hence, B 0 holds.

Acknowledgments

The authors thank the referee for important comments and suggestions on improving this article.

Funding information: Partially supported by Conventional Projects for Graduate Education Reform in the Second Batch of the 14th Five-Year Plan in Zhejiang Province (JGCG2024341).
Author contributions: X.C. wrote the manuscript, X.Y. proposed the research questions. All authors revised the manuscript.
Conflict of interest: The authors declare that they have no conflict of interest.
Data availability statement: No data, models or code were generated or used during the study.

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Received: 2025-01-12

Revised: 2025-05-14

Accepted: 2025-05-16

Published Online: 2025-07-07

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