In this paper, let (A,C,B) be a recollement of extriangulated categories. We construct a silting object and a tilting object from the two end terms of a recollement. We also show that the reverse direction holds under natural assumptions. Moreover, we show that our gluing preserves cotorsion pairs.
Citation: Zhen Zhang, Shance Wang. Silting objects and recollements of extriangulated categories[J]. AIMS Mathematics, 2024, 9(9): 24796-24809. doi: 10.3934/math.20241207
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In this paper, let (A,C,B) be a recollement of extriangulated categories. We construct a silting object and a tilting object from the two end terms of a recollement. We also show that the reverse direction holds under natural assumptions. Moreover, we show that our gluing preserves cotorsion pairs.
Belinson et al. [1] introduced the notions of recollements of abelian and triangulated categories, which were in connection with derived categories of sheaves on topological spaces, with the idea that one triangulated category may be glued together from two others. The recollements of abelian and triangulated categories are closely related, and they play an important role in algebraic geometry and representation theory; see, for instance, [2,3,4].
Gluing techniques concerning the recollement of triangulated or abelian categories have been investigated by more and more experts and scholars. For example, the glued cotorsion pairs [5], the glued torsion pairs [6], and so on (e.g., [7,8]). In particular, Liu et al. [9] presented explicit constructions of the gluing of silting objects for a recollement of triangulated categories.
Nakaoka and Palu [10] introduced an extriangulated category that extracts properties from triangulated categories and exact categories. In particular, exact categories and triangulated categories are extriangulated categories. There are a lot of examples of extriangulated categories, which are neither exact nor triangulated categories; see [10,11]. In the work of [12], Wang et al. defined the recollement of extriangulated categories, which generalized recollements of abelian categories and triangulated categories.
Motivated by the applications of gluing techniques in recollements, we want to consider gluing a silting object and a tilting object from two other silting objects and tilting objects in a recollement of extriangulated categories.
In this paper, let (A,C,B) be a recollement of extriangulated categories. Our first main result describes how to glue together silting objects σ in A and ω in B to obtain a silting object ρ in C; see Theorem 3.3. In the reverse direction, our second main result gives sufficient conditions for a silting object ρ of C, relative to the functors involved in the recollement, to induce silting objects in A and B; see Theorem 3.5, which partially extends the results of [9, Theorem 3.1], [13, Theorem 2.10] and [14, Theorem 2.5] in the framework of extriangulated categories. Our third main result constructs a glued tilting object from tilting objects in A and in B; see Theorems 4.2 and 4.3, which extend the result in [15, Theorem 3.5] to extriangulated categories.
This section briefly recalls some definitions and basic properties of extriangulated categories from [10,11]. We omit some details here, but we refer the readers to [10].
Let C be an additive category. All subcategories considered are full additive subcategories closed under isomorphisms. Let
E:Cop×C→Ab |
be a biadditive functor, where Ab is an abelian category. For any pair of objects A,C∈C, an element δ∈E(C,A) is called an E-extension.
Let s be a correspondence that associates an equivalence class
s(δ)=[Ax→By→C] |
to any E-extension δ∈E(C,A). This s is called a realization of E if it makes the diagrams in [10, Definition 2.9] commutative.
We call a triplet (C,E,s) an extriangulated category if it satisfies:
(1) E:Cop×C→Ab is a biadditive functor.
(2) s is an additive realization of E.
(3) E and s satisfy the compatibility conditions (ET3), (ET3)op, (ET4), and (ET4)op in [10, Definition 2.12].
Remark 2.1. By [10, Example 2.13 and Proposition 3.22], triangulated categories and exact categories (with a condition concerning the smallness) are typical examples of extriangulated categories.
For an extriangulated category C, we use the following notations in [10,11]:
● A sequence Aa→Bb→Cδ⇢ is called an E-triangle.
● Let Aa→Bb→Cδ⇢ be an E-triangle. A is called the CoCone of b and denoted by CoCone(B,C); C is called the Cone of a and denoted by Cone(A,B).
● An object P is called projective if, for any E-triangle Ax→By→Cδ⇢ and any morphism c∈C(P,C), there exists a morphism b∈C(P,B), satisfying y∘b=c. The class of all the projective objects in C is denoted by PC. An injective object can be dually defined, and the class of all the injective objects in C is denoted by IC.
● We say that C has enough projective objects if, for any object C∈C, there exists an E-triangle Kx→Py→Cδ⇢ satisfying P∈PC. Dually, we can define C as having enough injective objects.
Let X, Y be subcategories of C.
● Denote Cone(X,Y) by the subcategory of C consisting of M∈C, which admits an E-triangle X→Y→M⇢ in C, with X∈X and Y∈Y. We say that X is closed under Cones if Cone(X,X)⊆X.
● Denote CoCone(X,Y) by the subcategory of C consisting of M∈C, which admits an E-triangle M→X→Y⇢ in C, with X∈X and Y∈Y. We say that X is closed under CoCones if CoCone(X,X)⊆X.
● We denote
ΩX=CoCone(PC,X). |
And Ω is called the syzygy of X. Dually, we define the cosyzygy of X by
ΣX=Cone(X,PC), |
see [16, Definition 4.2, and Proposition 4.3].
For any subcategory X of C, put
Ω0X=X, |
for k≥1, define ΩkX inductively by
ΩkX=Ω(Ωk−1X)=CoCone(PC,Ωk−1X). |
ΩkX is called the k-th syzygy of X. Dually the k-th cosyzygy ΣkX by
ΣkX=Cone(Σk−1,IC) |
for k≥1 can be defined.
For any k>0 and any objects X,Y, Liu and Nakaoka [16, Proposition 5.2] defined the higher extension groups in an extriangulated category having enough projective and injective objects as
Ek+1(X,Y)=E(ΩkX,Y)=E(X,ΣkY). |
They showed the following result:
Lemma 2.2. [16, Proposition 5.2] Let Af→Bg→Cδ⇢ be an E-triangle. For any object X∈C, there are long exact sequences
C(X,A)C(X,f)→C(X,B)C(X,g)→C(X,C)(δ♯)X→E(X,A)f∗→E(X,B)g∗→E(X,C)→E2(X,A)→⋯ |
and
C(C,X)C(g,X)→C(B,X)C(f,X)→C(A,X)(δ♯)X→E(C,X)g∗→E(B,X)f∗→E(A,X)→E2(C,X)→⋯. |
We define
X⊥={Y∈C∣Ek(X,Y)=0,∀k≥1}. |
Dually, we define
⊥X={Y∈C∣Ek(Y,X)=0,∀k≥1}. |
Definition 2.3. [17, Definition 3.1] Let C be an extriangulated category, X and Y subcategories of C. We call a pair (X,Y) a hereditary cotorsion pair if:
(1) Both X and Y are closed under direct summands;
(2) Ek(X,Y)=0 for any k≥1 and X∈X and Y∈Y;
(3) C=Cone(Y,X);
(4) C=CoCone(Y,X).
We recall the concepts and basic properties of recollements of extriangulated categories from [12].
Definition 2.4. [12, Definition 3.1] Let A, C, and B be three extriangulated categories. A recollement of C relative to A and B, denoted by (A,C,B), is a diagram
![]() |
(2.1) |
given by two exact functors i∗, j∗, two right exact functors i∗, j!, and two left exact functors i!, j∗, which satisfies the following conditions:
(R1) (i∗,i∗,i!) and (j!,j∗,j∗) are adjoint triples.
(R2) Im i∗=Ker j∗.
(R3) i∗, j!, j∗ are fully faithful.
(R4) For each C∈C, there exists a left exact EC-triangle sequence
i∗i!C→C→j∗j∗C→i∗A |
with A∈A.
(R5) For each C∈C, there exists a right exact EC-triangle sequence
i∗A1→j!j∗C→C→i∗i∗C |
with A1∈A.
Remark 2.5. (1) If the categories A, B, and C are triangulated, then Definition 2.4 coincides with the definition of a recollement of triangulated categories [1].
(2) If the categories A, B, and C are abelian, then Definition 2.4 coincides with the definition of a recollement of abelian categories [3,4,6].
Lemma 2.6. [12, Lemma 3.3] Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4.
(1) All the natural transformations i∗i∗⇒IdA, IdA⇒i!i∗, IdB⇒j∗j!, j∗j∗⇒IdB are natural isomorphisms.
(2) i∗j!=0=i!j∗.
(3) i∗ preserves projective objects, and i! preserves injective objects.
(4) j! preserves projective objects, and j∗ preserves injective objects.
(5) If i! (resp. j∗) is exact, then i∗ (resp. j∗) preserves projective objects.
(6) If i∗ (resp. j!) is exact, then i∗ (resp. j∗) preserves injective objects.
(7) If C has enough projectives, then A has enough projectives; if C has enough injectives, then A has enough injectives.
(8) If C has enough projectives and j∗ is exact, then B has enough projectives; if C has enough injectives and j! is exact, then B has enough injectives.
(9) If i∗ is exact, then j! is exact.
(10) If i! is exact, then j∗ is exact.
The following results will be used frequently in this paper; their proof is very similar to [18, Proposition 2.8], we omit it.
Lemma 2.7. Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4.
(1) If A has enough projectives and i! is exact, then
EkC(i∗X,Y)≅EkA(X,i!Y) |
for any k≥1, X∈A and Y∈C.
(2) If A has enough injectives and i∗ is exact, then
EkA(i∗X,Y)≅EkC(X,i∗Y) |
for any k≥1, X∈C and Y∈A.
(3) If B has enough projectives and j! is exact, then
EkC(j!Z,Y)≅EkB(Z,j∗Y) |
for any k≥1, Z∈B and Y∈C.
In this section, we always assume A,C and B are three extriangulated categories with enough projective and injective objects.
A subcategory X of C is called thick if it is closed under extensions, Cones, CoCones, and direct summands. Let thickX denote the smallest thick subcategory containing X. When X is a single object M, then thick(M)=thick (add M).
Definition 3.1. An object σ in an extriangulated category C is silting if
(1) Ek(σ,σ)=0 for all k≥1;
(2) thick(σ)=C.
For an object σ, add σ denotes the closure of σ under finite direct sums and summands. Clearly, when σ is a silting object, then add σ is a silting subcategory in the sense of [17, Definition 5.1]. When C is a well generated triangulated category, Definition 3.1 coincides with [13, Definition 1.11].
Example 3.2. Let R be any ring, and let P∧(R) denote the category of right R-modules of finite projective dimension. Since P∧(R) is closed under extensions, it is an extriangulated category. We can easily check that R is a silting object of P∧(R).
For any subcategory X of C, let
X∨0=X∧0=X. |
We denote the full subcategories X∨i and X∧i for i>0 inductively by
X∨i=CoCone(X,X∨i−1), X∧i=Cone(X∧i−1,X). |
Put
X∨:=⋃i≥0X∨i and X∧:=⋃i≥0X∧i. |
Proposition 3.3. Assume that C admits a recollement relative to A and B as (2.1) in Definition 2.4 and σ, ω are silting objects in A and B, respectively. If i∗ is exact, then
thick(i∗σ⊕j!ω)=C. |
Proof. Let C∈C. As i∗ is exact, there exists an E-triangle by [12, Proposition 3.4],
(1) j!j∗C→C→i∗i∗C⇢. |
Clearly, i∗C∈A and j∗C∈B. By [17, Theorem 5.7], there exist two hereditary cotorsion pairs ((add σ)∨, (add σ)∧) in A and ((add ω)∨, (add ω)∧) in B, respectively. So we get two E-triangles associated with the above two hereditary cotorsion pairs:
(2) N→M→i∗C⇢ and (3) V→U→j∗C⇢, |
where N∈(add σ)∧, M∈(add σ)∨ and V∈(add ω)∧, U∈(add ω)∨.
Since i∗ is exact, i∗(add σ)∧⊆ (add i∗σ)∧ and i∗(add σ)∨⊆ (add i∗σ)∨. So i∗i∗C∈thick(i∗σ) by the E-triangle (2). Similarly, since i∗ is exact, j! is exact. So j!j∗C∈thick(j!ω) by the E-triangle (3). Hence C∈thick(j!ω⊕i∗σ) by the E-triangle (1). So
C=thick(i∗σ⊕j!ω). |
This completes the proof.
Theorem 3.4. Let (A,C,B) be a recollement as (2.1) in Definition 2.4. Assume σ and ω are silting objects in A and B, respectively. If
EkC(i∗σ,j!w)=0 |
for any k≥1 and i∗ is exact, then j!w⊕i∗σ is a silting object in C.
Proof. Since i∗ is exact, j! is exact by Lemma 2.6(9). So
EkC(j!ω,j!ω)≅EkB(ω,j∗j!ω)≅EkB(ω,ω)=0 |
and
EkC(j!ω,i∗σ)≅EkB(ω,j∗i∗σ)=0 |
for any k≥1 by Lemmas 2.6(1) and 2.7(3). Since i∗ is exact, i∗ preserves injective objects. So
EkC(i∗σ,i∗σ)≅EkA(i∗i∗σ,σ)≅EkA(σ,σ)=0 |
by Lemmas 2.6(1) and 2.7(2). So for any k≥1, we get
EkC(i∗σ⊕j!ω,i∗σ⊕j!ω)=EkC(i∗σ,i∗σ)⊕EkC(i∗σ,j!ω)⊕EkC(j!ω,i∗σ)⊕EkC(j!ω,j!ω)=0. |
Hence j!w⊕i∗σ is a silting object in C by Proposition 3.2 and Definition 3.1.
Let σ and ω be silting objects; then there exist hereditary cotorsion pairs ((addσ)∨, (addσ)∧) and ((add ω)∨, (add ω)∧) in A and B by [17, Theorem 5.7], respectively. Define
U:={C∈C∣i∗C∈(addσ)∨,j∗C∈(addω)∨} |
and
V:={C∈C∣i!C∈(addσ)∧,j∗C∈(addω)∧}. |
By [12, Theorem 4.4], (U,V) is a glued hereditary cotorsion pair in C.
On the other hand, by Theorem 3.3, we obtain a silting object
ρ=j!ω⊕i∗σ |
in C. So there exists a hereditary cotorsion pair ((add ρ)∨, (add ρ)∧) in C.
The following result indicates that our gluing preserves the cotorsion pair:
Theorem 3.5. In the situation above, let
ρ=j!ω⊕i∗σ. |
Then
(U,V)=((addρ)∨,(addρ)∧). |
Proof. Since σ and ω are silting objects,
(addσ)∧=σ⊥ |
and
(addω)∧=ω⊥ |
by [17, Lemma 4.11]. By Theorem 3.3, ρ is a silting object in C. So
thick(ρ)=C. |
Hence
ρ⊥=(addρ)⊥=(addρ)∧ |
by [17, Lemma 4.11]. So C∈V if and only if i!C∈σ⊥ and j∗C∈ω⊥. By Lemma 2.7, the latter holds if and only if C∈(i∗σ)⊥ and C∈(j!ω)⊥ if and only if
C∈ρ⊥=(addρ)∧. |
So
(U,V)=((addρ)∨,(addρ)∧). |
For the converse of Theorem 3.3, we have
Theorem 3.6. Let (A,C,B) be a recollement as (2.1) in Definition 1.4. Assume that ρ is a silting object in C and i∗ is exact,
(1) If i∗i∗ρ∈ add ρ, then i∗ρ is a silting object in A.
(2) If j!j∗ρ∈ add ρ, then j∗ρ is a silting object in B.
In this case, ρ is a glued silting object with respect to two silting objects, σ in A and ω in B.
Proof. (1) Since i∗ is exact,
EkA(i∗ρ,i∗ρ)≅EkC(ρ,i∗i∗ρ) |
for any k≥1 by Lemma 2.11. Since i∗i∗ρ∈ add ρ,
EkC(ρ,i∗i∗ρ)=0 |
by Definition 3.1. So
EkA(i∗ρ,i∗ρ)=0. |
For any A∈A, then i∗A∈C. Since ρ is a silting object in C, there exists a hereditary cotorsion pair ((add ρ)∨, (add ρ)∧) induced by ρ. So there exists an E-triangle V1→U1→i∗A⇢ with U1∈(add ρ)∨ and V1∈(add ρ)∧. Since i∗ is exact, we get another E-triangle: i∗V1→i∗U1→i∗i∗A⇢, Clearly, i∗V1∈(i∗(add ρ))∧⊆thick(i∗ρ) and i∗U1∈(i∗(add ρ))∨⊆thick(i∗ρ). Also
i∗i∗A≅A, |
so A∈thick(i∗ρ) and
thick(i∗ρ)=A. |
Hence, i∗ρ is a silting object in A.
(2) Since i∗ is exact, j∗ preserves injective objects by Lemma 2.10(6). So
EkB(j∗ρ,j∗ρ)≅EkC(j!j∗ρ,ρ) |
for any k≥1, by Lemma 2.7. Since j!j∗ρ∈ add ρ,
EkB(j∗ρ,j∗ρ)=0. |
For any B∈B, using the same proof method, we can also obtain
thick(j∗ρ)=B. |
Hence, j∗ρ is a silting object in B.
Denote i∗ρ by σ and j∗ρ by ω. Then σ is a silting object in A, and ω is a silting object in B. Moreover, i∗σ∈ add ρ and j!ω∈add ρ. So
Ek≥1C(i∗σ,j!ω)=0. |
By Theorem 3.3, j!ω⊕i∗σ is a silting object. Since i∗ is exact, there exists an EC-triangle j!ω→ρ→i∗σ⇢ by [12, Proposition 3.4]. So
ρ≅j!ω⊕i∗σ |
and ρ glued from σ and ω.
By applying Theorem 3.5 to triangulated categories and using the fact that any triangulated category can be viewed as an extriangulated category, we get the following result:
Corollary 3.7. When (A,C,B) is a recollement of triangulated categories, assume that i∗ is exact. The silting object ρ glued from two silting objects σ in A and ω in B if and only if i∗i∗ρ∈ add ρ and j!j∗ρ∈ add ρ, which is a special case of [13, Theorem 2.26].
In this section, we restrict the gluing procedures to tilting objects. We always assume that A,C,B are three extriangulated categories with enough projective and injective objects.
The notion of tilting objects in an extriangulated category was introduced in [19, Definition 5.2], and we recall it.
Definition 4.1. Let T be an object in an extriangulated category C. It is called tilting if
(1) T∈(PC)∧;
(2) Ek(T,T)=0, for all k≥1;
(3) PC⊆ (add T)∨.
Clearly, if T is a tilting object, then
thick(T)=thick(PC). |
By [17, Proposition 4.10],
thick(PC)=(PC)∧ |
and by [19, Lemma 5.4], any tilting object of C is silting of (PC)∧ and they coincide when every object in C has finite projective dimension by [19, Theorem 5.6].
Theorem 4.2. Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4 and i∗, i! exact. Assume that T1 and T2 are tilting objects in A and B, respectively, such that
EkC(i∗T1,j!T2)=0 |
for all k≥1. Then there exists a glued tilting object ρ with respect to T1 and T2 in C.
Proof. By [19, Lemma 5.4], T1 and T2 are silting objects of (PA)∧ and (PB)∧, respectively, and we have
thick(T1)=thick(PA)=(PA)∧ |
and
thick(T2)=thick(PB)=(PB)∧. |
Moreover, there exist two cotorsion pairs ((add T1)∨, (add T1)∧) in (PA)∧ and ((add T2)∨, (add T2)∧) in (PB)∧, respectively. Denote j!T2⊕i∗T1 by ρ. Since i∗ is exact, for any k≥1,
EkC(i∗T1,i∗T1)≅EkA(T1,T1)=0, EkC(j!T2,j!T2)≅EkB(T2,T2)=0 |
and
EkC(j!T2,i∗T1)≅EkB(T2,j∗i∗T1)=0 |
by Lemmas 2.6 and 2.7.
So
EkC(ρ,ρ)=0, |
and ρ is self-orthogonal. Since i∗ is exact, j! is exact. Since i! is exact, i∗ preserves projective objects. So
ρ=j!T2⊕i∗T1∈(PC)∧ |
by [19, Lemma 5.1].
Since i∗ is exact, for any C∈(PC)∧, we have the following E-triangle by [12, Proposition 3.4],
(1) j!j∗C→C→i∗i∗C⇢. |
Since both i∗ and j∗ preserves projective objects, i∗C∈(PA)∧ and j∗C∈(PB)∧. So we get two E-triangles associated with the above two cotorsion pairs:
(2) N→M→i∗C⇢ and (3) V→U→j∗C⇢, |
where N∈(add T1)∧, M∈(add T1)∨ and V∈(add T2)∧, U∈(add T2)∨.
Since i∗ is exact, i∗i∗C∈thick(i∗T1) by the E-triangle (2). Similarly, j!j∗C∈thick(j!T2) by the E-triangle (3). Hence C∈thick(j!T2⊕i∗T1) by the E-triangle (1). So
thick(ρ)=(PC)∧ |
and thus ρ is a silting object in (PC)∧. Hence ((add ρ)∨, (add ρ)∧) is a cotorsion pair in (PC)∧. Since PC⊆(PC)∧, for any M∈PC, there exists an E-triangle V→U→M⇢, where U∈ (add ρ)∨ and V∈ (add ρ)∧. Since M is a project object, the E-triangle splits. So M is a direct summand of U, and thus PC⊆ (add ρ)∨. So ρ is a tilting object in C by Definition 4.1.
For the converse direction,
Theorem 4.3. Let (A,C,B) be a recollement of extriangulated categories as (2.1) in Definition 2.4 and i∗, i! are exact. Assume that ρ is a tilting object in C.
(1) If i∗i∗ρ∈ add ρ, then i∗ρ is a tilting object in A.
(2) If j!j∗ρ∈ add ρ, then j∗ρ is a tilting object in B.
In this case, ρ glued from two tilting objects in A and B.
Proof. By Theorem 3.5, both i∗ρ and j∗ρ are self-orthogonal. Since i∗ is exact and i∗ preserves projective objects,
i∗ρ∈(i∗PC)∧⊆(PA)∧. |
Since i! is exact, j∗ preserves projective objects. So
j∗ρ∈(j∗PC)∧⊆(PB)∧. |
On the other hand, since ρ is a tilting object in C, PC⊆(addρ)∨. For any M∈PA, then i∗M∈PC. So
M≅i∗i∗M∈(addi∗ρ)∨. |
For any N∈PB, then j!N∈PC. So
N≅j∗j!N∈(addj∗ρ)∨. |
Hence, i∗ρ is a tilting object in A, and j∗ρ is a tilting object in B, by Definition 4.1.
Since i∗ is exact, we have an E-triangle by [12, Proposition 3.4],
(1) j!j∗ρ→ρ→i∗i∗ρ⇢. |
Since i∗i∗ρ∈ add ρ and j!j∗ρ∈ add ρ,
EkC(i∗i∗ρ,j!j∗ρ)=0. |
So
ρ≅i∗i∗ρ⊕j!j∗ρ |
and thus ρ glued from two tilting objects i∗ρ in A and j∗ρ in B.
Corollary 4.4. (1) If we let C be a triangulated category that is K-linear over a field K, B a localizing subcategory of C, and
A=ker(C(A,−)). |
Then Theorem 4.2 is exactly [14, Theorem 2.5] when viewing triangulated categories as extriangulated categories.
(2) Let Λ1, Λ, and Λ2 be artin algebras,
A=modΛ1, C=modΛ andB=modΛ2. |
Then Theorem 4.2 is [15, Theorem 3.5].
In this section, we present the conclusions of the paper.
Let (A,C,B) be a recollement as (2.1) in Definition 2.4. Our first main result describes how to glue together silting objects σ in A and ω in B to obtain a silting object ρ in C; see Theorem 3.3.
Theorem 5.1. Assume σ and ω are silting objects in A and B, respectively. If
EkC(i∗σ,j!w)=0 |
for any k≥1 and i∗ is exact, then j!w⊕i∗σ is a silting object in C.
In the reverse direction, our second main result gives sufficient conditions for a silting object ρ of C, relative to the functors involved in the recollement, to induce silting objects in A and B; see Theorem 3.5, which partially extends the results of [9, Theorem 3.1], [13, Theorem 2.10] and [14, Theorem 2.5] in the framework of extriangulated categories.
Theorem 5.2. Assume that ρ is a silting object in C and i∗ is exact,
(1) If i∗i∗ρ∈ add ρ, then i∗ρ is a silting object in A.
(2) If j!j∗ρ∈ add ρ, then j∗ρ is a silting object in B.
In this case, ρ glued from two silting objects, σ in A and ω in B.
Our third main result constructs a glued tilting object from tilting objects in A and in B; see Theorems 4.2 and 4.3, which extend the result in [15, Theorem 3.5] to extriangulated categories.
Theorem 5.3. Let i∗ and i! be exact. Assume that T1 and T2 are tilting objects in A and B, respectively, such that
EkC(i∗T1,j!T2)=0 |
for all k≥1. Then there exists a glued tilting object ρ with respect to T1 and T2 in C.
Theorem 5.4. Let i∗, i! be exact. Assume that ρ is a tilting object in C.
(1) If i∗i∗ρ∈ add ρ, then i∗ρ is a tilting object in A.
(2) If j!j∗ρ∈ add ρ, then j∗ρ is a tilting object in B.
In this case, ρ glued from two tilting objects in A and B.
Zhen Zhang: contributed the creative ideals and proof techniques for this paper; Shance Wang: consulted the relevant background of the paper and composed the article, encompassing the structure of the article and the modification of grammar. All authors have read and agreed to the published version of the manuscript.
The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.
This research was supported by the National Natural Science Foundation of China (No. 12101344) and Shan Dong Provincial Natural Science Foundation of China (No. ZR2015PA001).
The authors declared that they have no conflicts of interest.
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