Consistent pairs of $\mathfrak{s}$-torsion pairs in extriangulated categories with negative first extensions

1.   Introduction

Let $\mathcal{K}$ be an abelian category or a triangulated category and $\mathcal{X}, \mathcal{Y}$ be two subcategories. We denote by $\mathcal{X}\ast\mathcal{Y}$ the subcategory which consists of objects $Z$ in $\mathcal{K}$ such that there is a short exact sequence

with $X\in\mathcal{X}$ and $Y\in\mathcal{Y}$ or there is a triangle

with $X\in\mathcal{X}$ and $Y\in\mathcal{Y}$, where $\Sigma$ is the shift functor. It is called the extension subcategory of $\mathcal{Y}$ by $\mathcal{X}$. This is a classic research object and has been used extensively in the representation theory of algebra. Gentle and Todorov ^[6] proved that in an abelian category with enough projective objects, the extension subcategory of two covariantly finite subcategories is covariantly finite. Chen ^[5] proved a triangulated version of Gentle-Todorov's result. Jorgensen and Kato ^[10] gave some necessary and sufficient conditions for $\mathcal{X}\ast\mathcal{Y}$ to be triangulated for a pair of triangulated subcategories $\mathcal{X}$ and $\mathcal{Y}$. Yoshizawa considered the extension subcategory when the two given subcategories are Serre in ^[14].

In order to discuss cotorsion pairs in a more general case, Nakaoka and Palu ^[12] introduced the notion of extriangulated categories. In particular, exact categories and extension-closed subcategories of triangulated categories are typical examples of extriangulated categories. Hence, many results told on exact categories and triangulated categories can be unified in the same framework ^[7,11,13,15]. However, it is worth mentioning that there exist numerous examples of extriangulated categories that are neither triangulated nor exact categories (see ^[8,12,15]). Recently, Adachi, Enomoto and Tsukamoto ^[1] introduced the notion of extriangulated categories with negative first extensions (that is, an additive bifunctors $\mathbb{E}^{-1}$ satisfying certain conditions) and $\mathfrak{s}$-torsion pairs as a general framework for the studies of $t$-structures on triangulated categories and torsion pairs in abelian categories. Let $\mathcal{K}$ be an extriangulated category with a negative first extension. According to ^[1], a pair $(\mathcal{T}, \mathcal{F})$ of subcategories in $\mathcal{K}$ is called an $\mathfrak{s}$-torsion pair if it is a torsion pair in the usual sense and if $\mathbb{E}^{-1}(\mathcal{T}, \mathcal{F}) = 0$ holds. In this case, we also call $\mathcal{T}$ (respectively, $\mathcal{F}$) a torsion class (respectively, torsion-free class), and they are mutually determined. Exact categories and triangulated categories naturally admit negative first extension structures, then torsion pairs in exact categories and $t$-structures on triangulated categories are exactly $\mathfrak{s}$-torsion pairs ^[1]. Let $s_{i} = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ for any $i = 1, 2$. We have the following natural question: Is the extension subcategory $\mathcal{T}_{1}\ast\mathcal{T}_{2}$ of $\mathcal{T}_{2}$ by $\mathcal{T}_{1}$ a torsion class? Bondal provided some sufficient condition for the intersection of two $t$-structures on a triangulated category to be a $t$-structure in terms of consistent pairs ^[4]. In this paper, we extend the notion of the consistent pair to the extriangulated setup, then give a necessary and sufficient condition for $\mathcal{T}_{1}\ast\mathcal{T}_{2}$ to be a torsion class. More specifically, we show that $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ is an $\mathfrak{s}$-torsion pair if and only if $(s_{1}, s_{2})$ is an upper consistent pair, which generalizes Bondal's result [4, Proposition 6]. We also give some conditions under which the operations of $\ast$ and intersection satisfy the distributive laws. For a $\tau$-rigid module $M = M_{1}\oplus M_{2}$ over a finite-dimensional algebra, we denote by Fac$M$ the subcategory of the module category which consists of all factor modules of finite direct sums of copies of $M$. As an application of the main theorem, we show that Fac$M$ is equal to the extension subcategory of Fac$M_{2}$ by Fac$M_{1}$ (see Proposition 4.4 for details).

We include some notations here. Throughout this paper, we assume that every category is skeletally small that is, the isomorphism classes of objects form a set, and $\mathcal{K}$ denotes an additive category. The symbol $\mathcal{K}^{\rm op}$ denotes the opposite category of $\mathcal{K}$. When we say that $\mathcal{D}$ is a subcategory of $\mathcal{K}$, we always mean that $\mathcal{D}$ is a full subcategory which is closed under isomorphisms. For a collection $\mathcal{X}$ of objects in $\mathcal{K}$, we define $\mathcal{X}^{\perp}: = \{ C\in \mathcal{K}|{\rm Hom}_{\mathcal{K}}(\mathcal{X}, C) = 0\}$ and $^{\perp}\mathcal{X}: = \{ C\in \mathcal{K}|{\rm Hom}_{\mathcal{K}}(C, \mathcal{X}) = 0\}$.

2.   Preliminaries

We briefly recall some definitions and some needed properties of extriangulated categories from ^[12] and $\mathfrak{s}$-torsion pairs from ^[1].

Let $\mathcal{K}$ be an additive category equipped with an additive bifunctor

where Ab is the category of abelian groups. For any objects $A, C\in \mathcal{K}$, an element $\delta\in \mathbb{E}(C, A)$ is called an $\mathbb{E}$-extension. Let $\mathfrak{s}$ be a correspondence which associates an equivalence class

to any $\mathbb{E}$-extension $\delta\in \mathbb{E}(C, A)$. Such $\mathfrak{s}$ is called a realization of $\mathbb{E}$ if it makes the diagrams in [12, Definition 2.9] commutative. A triplet $(\mathcal{C}, \mathbb{E}, \mathfrak{s})$ is called an extriangulated category if it satisfies the following conditions.

● (ET1) $\mathbb{E}:\mathcal{K}^{\rm op}\times\mathcal{K}\rightarrow {\rm Ab}$ is an additive bifunctor.

● (ET2) $\mathfrak{s}$ is an additive realization of $\mathbb{E}$.

● (ET3) Let $\delta\in \mathbb{E}(C, A)$ and $\delta^{'}\in\mathbb{E}(C^{'}, A^{'})$ be any pair of E-extensions, realized as

For any commutative square

in $\mathcal{K}$, there exists a morphism $(a, c):\delta\rightarrow \delta^{'}$ satisfying $cy = y^{'}b$.

● (ET3)$^{\rm op}$ Dual of (ET3).

● (ET4) Let $\delta\in\mathbb{E}(C, A)$ and $\delta^{'}\in\mathbb{E}(F, B)$ be $\mathbb{E}$-extensions realized by

respectively, then there exists an object $E\in\mathcal{ C}$ and a commutative diagram

where all rows and columns are $\mathbb{E}$-extensions in $\mathcal{K}$.

● (ET4)$^{\rm op}$ Dual of (ET4).

Definition 2.1. (^[12]) Let $\mathcal{K}$ be an extriangulated category.

(1) A sequence $A\stackrel{f}{\longrightarrow} B\stackrel{g}{\longrightarrow} C$ in $\mathcal{K}$ is called a conflation if it realizes some $\mathbb{E}$-extension $\delta\in\mathbb{E}(C, A)$. The pair $(A\stackrel{f}{\longrightarrow} B\stackrel{g}{\longrightarrow} C, \delta)$ is called an $\mathbb{E}$-triangle (or $\mathfrak{s}$-conflation), and it is written in the following way:

(2) Let $A\stackrel{x}{\rightarrow } B\stackrel{y}{\rightarrow } C\stackrel{\delta}{\dashrightarrow}$ and $A^{'}\stackrel{x^{'}}{\rightarrow } B^{'}\stackrel{y^{'}}{\rightarrow } C^{'}\stackrel{\delta^{'}}{\dashrightarrow}$ be any pair of $\mathbb{E}$-triangles. If a triplet $(a, b, c)$ realizes $(a, c):\delta\rightarrow \delta^{'}$, then we write it as

and call $(a, b, c)$ a morphism of $\mathbb{E}$-triangles.

(3) Let $\mathcal{X}$ and $\mathcal{Y}$ be subcategories of $\mathcal{K}$ and let $\mathcal{X}\ast\mathcal{Y}$ denote the subcategory of $\mathcal{K}$ consisting of $M\in\mathcal{K}$ which admits an $\mathbb{E}$-triangle $X\rightarrow M\rightarrow Y\dashrightarrow$ in $\mathcal{K}$ with $X\in\mathcal{X}$ and $Y\in\mathcal{Y}$. The subcategory $\mathcal{X}$ is said to be extension-closed (or equivalently, closed under extensions) if $\mathcal{X}\ast\mathcal{X}\subseteq\mathcal{X}$. Note that the operation $\ast$ on subcategories is associative by (ET4) and (ET4)$^{\rm op}$.

Exact categories and extension-closed subcategories of triangulated categories are extriangulated categories. The extension-closed subcategories of an extriangulated category are again extriangulated categories (see [12, Remark 2.18]). The following proposition gives a criterion for an extension subcategory of two given extension-closed subcategories to be extension-closed.

Proposition 2.2. Let $\mathcal{X}$ and $\mathcal{Y}$ be extension-closed subcategories of an extriangulated category, then the following conditions are equivalent.

$({\rm{1}})$ $\mathcal{X}\ast\mathcal{Y}$ is an extension-closed subcategory.

$({\rm{2}})$ $\mathcal{Y}\ast\mathcal{X}\subseteq\mathcal{X}\ast\mathcal{Y}$.

Proof. (1) implies (2). For any $C\in\mathcal{Y}\ast\mathcal{X}$, there exists an $\mathbb{E}$-triangle

such that $Y\in\mathcal{Y}$ and $X\in\mathcal{X}$. Note that $Y$ is in $\mathcal{Y}\subseteq\mathcal{X}\ast\mathcal{Y}$ and $X$ is in $\mathcal{X}\subseteq\mathcal{X}\ast\mathcal{Y}$. Since $\mathcal{X}\ast\mathcal{Y}$ is closed under extensions by the assumption (1), then the $\mathbb{E}$-triangle above implies that $C\in\mathcal{X}\ast\mathcal{Y}$.

(2) implies (1). Since the operation $\ast$ is associative and $\mathcal{Y}\ast\mathcal{X}\subseteq\mathcal{X}\ast\mathcal{Y}$, we have the following formulas

Hence $\mathcal{X}\ast\mathcal{Y}$ is closed under extensions. □

The following is a basic property of extriangulated categories.

Lemma 2.3. ([12, Proposition 3.3]) Let $\mathcal{K}$ be an extriangulated category. For any $\mathbb{E}$-triangle $A\rightarrow B\rightarrow C\dashrightarrow$, we have the following exact sequences.

Definition 2.4. (^[1]) Let $\mathcal{K}$ be an extriangulated category. A negative first extension structure on $\mathcal{K}$ consists of the following data.

● (NE1) $\mathbb{E}^{-1}:\mathcal{K}^{\rm op}\times\mathcal{K}\rightarrow {\rm Ab}$ is an additive bifunctor.

● (NE2) For each $\delta\in \mathbb{E}(C, A)$, there exist two natural transformations

such that for each $\mathbb{E}$-triangle $A\stackrel{f}{\longrightarrow} B\stackrel{g}{\longrightarrow} C\dashrightarrow$ and each $W\in\mathcal{K}$, two sequences

are exact.

Thus, $(\mathcal{K}, \mathbb{E}, \mathfrak{s}, \mathbb{E}^{-1})$ is called an extriangulated category with a negative first extension.

In what follows, we assume that $\mathcal{K}$ is an extriangulated category with a negative first extension $\mathbb{E}^{-1}$.

Definition 2.5. (^[1]) A pair $(\mathcal{T}, \mathcal{F})$ of subcategories of $\mathcal{K}$ is called an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ if it satisfies the following three conditions.

● (STP1) $\mathcal{K} = \mathcal{T}\ast\mathcal{F}$.

● (STP2) ${\rm Hom}_{\mathcal{K}}(\mathcal{T}, \mathcal{F}) = 0$.

● (STP3) $\mathbb{E}^{-1}(\mathcal{T}, \mathcal{F}) = 0$.

In this case, $\mathcal{T}$ (respectively, $\mathcal{F}$) is called a torsion class (respectively, torsion-free class) in $\mathcal{K}$.

Let $\mathcal{D}$ be a triangulated category with the shift functor $\Sigma$. By regarding triangulated category $\mathcal{D}$ as the extriangulated category with the negative first extension $\mathbb{E}^{-1}(C, A) = {\rm Hom}_{\mathcal{D}}(C, \Sigma^{-1}A)$ for all $A, C\in\mathcal{D}$, then $t$-structures on $\mathcal{D}$ are exactly $\mathfrak{s}$-torsion pairs in $\mathcal{D}$. By regarding an exact category $\mathcal{E}$ as the extriangulated category with the negative first extension $\mathbb{E}^{-1} = 0$, then torsion pairs in the exact category $\mathcal{E}$ are exactly $\mathfrak{s}$-torsion pairs in $\mathcal{E}$ ^[1].

The following proposition shows that torsion class and torsion-free class in $\mathcal{K}$ are mutually determined.

Lemma 2.6. ([1, Proposition 3.2]) Let $(\mathcal{T}, \mathcal{F})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$, then the following statements hold.

$({\rm{1}})$ $\mathcal{T }^{\perp} = \mathcal{F}$.

$({\rm{2}})$ $^{\perp}\mathcal{F} = \mathcal{T }$.

In particular, $\mathcal{T}$ and $\mathcal{F}$ are extension-closed subcategories which are closed under direct summands.

Lemma 2.7. ([1, Proposition 3.7]) Let $(\mathcal{T}, \mathcal{F})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$. For each $C\in \mathcal{K}$, there uniquely exists an $\mathbb{E}$-triangle

such that $T_{C}\in\mathcal{T}$ and $F^{C}\in\mathcal{F}$ (up to isomorphism of $\mathbb{E}$-triangles).

Let $(\mathcal{T}, \mathcal{F})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$. According to Lemma 2.7, $(\mathcal{T}, \mathcal{F})$ gives functors ${\rm t}: \mathcal{K}\rightarrow \mathcal{T}$ and ${\rm f}: \mathcal{K}\rightarrow \mathcal{F}$ and natural transformations ${\rm t}\rightarrow {\rm id}_{\mathcal{C}}\rightarrow {\rm f}$ such that there uniquely exists an $\mathbb{E}$-triangle

for each $C\in\mathcal{K}$. We call the functors $({\rm t}, {\rm f})$ the torsion functors of the $\mathfrak{s}$-torsion pair $(\mathcal{T}, \mathcal{F})$. An $\mathbb{E}$-triangle given as (2.1) is called a canonical $\mathbb{E}$-triangle with respect to the torsion pair $(\mathcal{T}, \mathcal{F})$.

3.   Main results

The concept of the consistent pair of $t$-structures on a triangulated category has been introduced by Bondal in ^[4]. We extend this notion to an extriangulated category with a negative first extension.

Definition 3.1. Let $\mathcal{K}$ be an extriangulated category with a negative first extension. Let $s_{i}: = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ with the torsion functors $({\rm t}_{i}, {\rm f}_{i})$ for any $i = 1, 2$. If $\mathcal{F}_{1}$ is stable under the functor ${\rm f}_{2}$ that is, ${\rm f}_{2}\mathcal{F}_{1}\subset\mathcal{F}_{1}$, then the pair $(s_{1}, s_{2})$ is called an upper consistent pair of $\mathfrak{s}$-torsion pairs. If $\mathcal{T}_{2}$ is stable under the functor ${\rm t}_{1}$ that is, ${\rm t}_{1}\mathcal{T}_{2}\subset\mathcal{T}_{2}$, then the pair $(s_{1}, s_{2})$ is called a lower consistent pair of $\mathfrak{s}$-torsion pairs. A pair is said to be consistent if it is upper or lower consistent.

There exists inconsistent pairs of $\mathfrak{s}$-torsion pairs [4, Subsection 1.2]. The following lemma provides a necessary condition.

Lemma 3.2. Let $s_{i}: = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pairs in $\mathcal{K}$ for any $i = 1, 2$.

$({\rm{1}})$ Suppose that $\mathbb{E}(\mathcal{T}_{1}, \mathcal{T}_{2}) = 0$, then $(s_{1}, s_{2})$ is upper consistent.

$({\rm{2}})$ Suppose that $\mathbb{E}(\mathcal{F}_{1}, \mathcal{F}_{2}) = 0$, then $(s_{1}, s_{2})$ is lower consistent.

Proof. We only prove (1); the proof of (2) is similar. Denote by $({\rm t}_{i}, {\rm f}_{i})$ the torsion functors of $(\mathcal{T}_{i}, \mathcal{F}_{i})$, $i = 1, 2$. For any $C\in\mathcal{F}_{1}$, take the canonical $\mathbb{E}$-triangle

with respect to the $\mathfrak{s}$-torsion pair $(\mathcal{T}_{2}, \mathcal{F}_{2})$. Applying ${\rm Hom}_{\mathcal{K}}(\mathcal{T}_{1}, -)$ to the preceding $\mathbb{E}$-triangle deduces an exact sequence

By assumption, $\mathbb{E}(\mathcal{T}_{1}, {\rm t}_{2}C) = 0$, so ${\rm Hom}_{\mathcal{K}}(\mathcal{T}_{1}, {\rm f}_{2}C) = 0$. It follows from Lemma 2.6 that ${\rm f}_{2}C\in\mathcal{F}_{1}$. Therefore, $(s_{1}, s_{2})$ is upper consistent. □

Lemma 3.3. Let $A\stackrel{f}{\longrightarrow} B\stackrel{g}{\longrightarrow} C\dashrightarrow$ be an $\mathbb{E}$-triangle, then the following assertions hold.

$({\rm{1}})$ If $f = 0$, then $g$ is a split monomorphism.

$({\rm{2)}}$ If $g = 0$, then $f$ is a split epimorphism.

Proof. We only prove (1); the proof of (2) is similar. First there exists an $\mathbb{E}$-triangle $0\rightarrow B\stackrel{{\rm id}_{B}}{\longrightarrow} B\dashrightarrow$, then we consider the following diagram

where the left square is commutative. By (ET3) there exists a morphism $h:C\rightarrow B$ such that the following diagram

commutates. Hence $hg = {\rm id}_{B}$. □

Now we can show the main theorem of this paper.

Theorem 3.4. Let $\mathcal{K}$ be an extriangulated category with a negative first extension. Let $s_{i}: = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ for any $i = 1, 2$, then

$({\rm{1}})$ $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ if and only if $(s_{1}, s_{2})$ is upper consistent.

$({\rm{2}})$ $(\mathcal{T}_{1}\cap\mathcal{T}_{2}, \mathcal{F}_{1}\ast\mathcal{F}_{2})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ if and only if $(s_{1}, s_{2})$ is lower consistent.

Proof. We only prove (1); the proof of (2) is similar. Denote by $({\rm t}_{i}, {\rm f}_{i})$ the torsion functors of $(\mathcal{T}_{i}, \mathcal{F}_{i})$, $i = 1, 2$. We first show the necessity. Assume that $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$, then for any $C\in\mathcal{F}_{1}$, there exists a canonical $\mathbb{E}$-triangle

with respect to $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ such that $T\in\mathcal{T}_{1}\ast\mathcal{T}_{2}$ and $F\in\mathcal{F}_{1}\cap\mathcal{F}_{2}\subset\mathcal{F}_{2}$. It suffices to prove that $T$ is in $\mathcal{T}_{2}$. Indeed, if it is done, then the $\mathbb{E}$-triangle (3.1) is also a canonical $\mathbb{E}$-triangle with respect to the $\mathfrak{s}$-torsion pair $(\mathcal{T}_{2}, \mathcal{F}_{2})$ by Lemma 2.7. Therefore ${\rm f}_{2}C = F\in\mathcal{F}_{1}\cap\mathcal{F}_{2}\subset\mathcal{F}_{1}$, so $(s_{1}, s_{2})$ is upper consistent. Since $T\in\mathcal{T}_{1}\ast\mathcal{T}_{2}$, there exists an $\mathbb{E}$-triangle

such that $T_{1}\in\mathcal{T}_{1}$ and $T_{2}\in\mathcal{T}_{2}$. Applying ${\rm Hom}_{\mathcal{K}}(T_{1}, -)$ to the $\mathbb{E}$-triangle (3.1), we have an exact sequence

Since $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ is an $\mathfrak{s}$-torsion pair and $T_{1}\in\mathcal{T}_{1}\subset\mathcal{T}_{1}\ast\mathcal{T}_{2}$, we have $\mathbb{E}^{-1}(T_{1}, F) = 0$. Since $(\mathcal{T}_{1}, \mathcal{F}_{1})$ is an $\mathfrak{s}$-torsion pair, we have ${\rm Hom}_{\mathcal{K}}(T_{1}, C) = 0$, so $a\in{\rm Hom}_{\mathcal{K}}(T_{1}, T) = 0$. By Lemma 3.3, $b$ is a split monomorphism and $T$ is a direct summand of $T_{2}$. It followes from Lemma 2.6 that $T\in\mathcal{T}_{2}$. Hence, $(s_{1}, s_{2})$ is an upper consistent pair.

Let $(s_{1}, s_{2})$ be an upper consistent pair. To prove the sufficiency, we verify the conditions (STP1), (STP2) and (STP3) of the definition of $\mathfrak{s}$-torsion pair individually.

(STP1) Let $C\in\mathcal{K}$. There exists a canonical $\mathbb{E}$-triangle

with respect to the $\mathfrak{s}$-torsion pair $(\mathcal{T}_{1}, \mathcal{F}_{1})$. Similarly, since $(\mathcal{T}_{2}, \mathcal{F}_{2})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$, there exists a canonical $\mathbb{E}$-triangle

On the basis of the assumption that $(s_{1}, s_{2})$ is upper consistent, we have ${\rm f_{2}}{\rm f_{1}}C\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$. Applying (ET4)$^{\rm op}$ to (3.2) and (3.3) induces a commutative diagram

where all rows and columns are $\mathbb{E}$-triangles. In the top horizontal $\mathbb{E}$-triangle, since ${\rm t_{1}}C\in \mathcal{T}_{1}$ and ${\rm t_{2}}{\rm f_{1}}C\in\mathcal{T}_{2}$, then $A\in \mathcal{T}_{1}\ast\mathcal{T}_{2}$. Thus we obtain $C\in(\mathcal{T}_{1}\ast\mathcal{T}_{2})\ast(\mathcal{F}_{1}\cap\mathcal{F}_{2})$ by the left vertical $\mathbb{E}$-triangle $A\rightarrow C\rightarrow {\rm f_{2}}{\rm f_{1}}C\dashrightarrow$, meaning that $\mathcal{K} = (\mathcal{T}_{1}\ast\mathcal{T}_{2})\ast(\mathcal{F}_{1}\cap\mathcal{F}_{2})$.

(STP2) Since $(\mathcal{T}_{1}, \mathcal{F}_{1})$ and $(\mathcal{T}_{2}, \mathcal{F}_{2})$ are $\mathfrak{s}$-torsion pairs, we obtain that $\mathcal{C}(\mathcal{T}_{i}, \mathcal{F}_{1}\cap\mathcal{F}_{2}) = 0$, for any $i = 1, 2$. This implies that $\mathcal{C}(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2}) = 0$ by Lemma 2.3.

(STP3) Since $(\mathcal{T}_{1}, \mathcal{F}_{1})$ and $(\mathcal{T}_{2}, \mathcal{F}_{2})$ are $\mathfrak{s}$-torsion pairs, we obtain that $\mathbb{E}^{-1}(\mathcal{T}_{i}, \mathcal{F}_{1}\cap\mathcal{F}_{2}) = 0$, for any $i = 1, 2$. This implies that $\mathbb{E}^{-1}(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2}) = 0$ by (NE2). This completes the proof of the sufficiency. □

Remark 3.5. (1) Let $s_{1}: = (\mathcal{T}, \mathcal{F})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ and $s_{2}: = (\mathcal{K}, 0)$ be a trivial $\mathfrak{s}$-torsion pair. Since $(\mathcal{T}\ast\mathcal{K}, \mathcal{F}\cap 0) = (\mathcal{K}, 0)$, $(\mathcal{T}\cap\mathcal{K}, \mathcal{F}\ast 0) = (\mathcal{T}, \mathcal{F})$, by Theorem 3.4, $(s_{1}, s_{2})$ is simultaneously lower and upper consistent.

(2) Let $\mathcal{K}$ be a triangulated category, then Theorem 3.4 recovers [4, Proposition 6].

When $\mathcal{K}$ is a triangulated category, the following corollary is an analogue of [9, Proposition 2.4] which plays a crucial role in the study of $n$-cluster tilting subcategories.

Corollary 3.6. Let $(\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ for any $i\geq 1$.

$({\rm{1}})$ Suppose that $\mathbb{E}(\mathcal{T}_{i}, \mathcal{T}_{j}) = 0$, for any $i < j$. Put

then $(\mathcal{X}_{n}, \mathcal{Y}_{n})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$.

$({\rm{2}})$ Suppose that $\mathbb{E}(\mathcal{F}_{i}, \mathcal{F}_{j}) = 0$ for any $i < j$. Put

then $(\mathcal{X}_{n}, \mathcal{Y}_{n})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$.

Proof. We only prove (1); the proof of (2) is similar. We show by induction on $n$. The case $n = 1$ is obvious. Assume that the assertion is true for $n-1$ that is, $(\mathcal{X}_{n-1}, \mathcal{Y}_{n-1})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$. Since $\mathbb{E}(\mathcal{T}_{i}, \mathcal{T}_{n}) = 0$ for any $i < n$, by Lemma 2.3 we obtain that

It follows from Lemma 3.2 that $((\mathcal{X}_{n-1}, \mathcal{Y}_{n-1}), (\mathcal{T}_{n}, \mathcal{F}_{n}))$ is upper consistent. By Theorem 3.4, $(\mathcal{X}_{n-1}\ast\mathcal{T}_{n}, \mathcal{Y}_{n-1}\cap\mathcal{F}_{n}) = (\mathcal{X}_{n}, \mathcal{Y}_{n})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$. □

The following lemma is useful in constructing consistent pairs.

Lemma 3.7. Let $(\mathcal{T}, \mathcal{F})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$. Let $A\stackrel{f}{\longrightarrow} B\stackrel{g}{\longrightarrow} C\dashrightarrow$ be an $\mathbb{E}$-triangle, then the following assertions hold.

$({\rm{1}})$ If $B, C\in\mathcal{F}$, then $A\in\mathcal{F}$.

$({\rm{2}})$ If $A, B\in\mathcal{T}$, then $C\in\mathcal{T}$.

Proof. We only prove (1); the proof of (2) is similar. Applying ${\rm Hom}_{\mathcal{K}}(\mathcal{T}, -)$ to the given $\mathbb{E}$-triangle, we have an exact sequence

Since the left hand side and right hand side vanish, we obtain ${\rm Hom}_{\mathcal{K}}(\mathcal{T}, A) = 0$. It follows from Lemma 2.6 that $A\in\mathcal{F}$. □

Next we will construct some new consistent pairs by using the known ones, which enables us to determine the new $\mathfrak{s}$-torsion pairs. Meanwhile, the conditions under which the operations of $\ast$ and intersection satisfy the distributive laws are given.

Corollary 3.8. Let $s_{i}: = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ for any $i = 1, 2, 3$. Suppose that $(s_{1}, s_{2})$, $(s_{1}, s_{3})$ are upper consistent and $(s_{2}, s_{3})$ is lower consistent, then the following assertions hold.

$({\rm{1}})$ $((\mathcal{T}_{1}, \mathcal{F}_{1}), (\mathcal{T}_{2}\cap\mathcal{T}_{3}, \mathcal{F}_{2}\ast\mathcal{F}_{3}))$ is upper consistent.

$({\rm{2}})$ $((\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2}), (\mathcal{T}_{1}\ast\mathcal{T}_{3}, \mathcal{F}_{1}\cap\mathcal{F}_{3}))$ is lower consistent.

$({\rm{3}})$ $\mathcal{T}_{1}\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3}) = (\mathcal{T}_{1}\ast\mathcal{T}_{2})\cap(\mathcal{T}_{1}\ast\mathcal{T}_{3})$, and $\mathcal{F}_{1}\cap(\mathcal{F}_{2}\ast\mathcal{F}_{3}) = (\mathcal{F}_{1}\cap\mathcal{F}_{2})\ast(\mathcal{F}_{1}\cap\mathcal{F}_{3})$.

Proof. Denote by $({\rm t}_{i}, {\rm f}_{i})$ the torsion functors of $(\mathcal{T}_{i}, \mathcal{F}_{i})$ for any $i = 1, 2, 3$.

(1) By Theorem 3.4, $(\mathcal{T}_{2}\cap\mathcal{T}_{3}, \mathcal{F}_{2}\ast\mathcal{F}_{3})$ is an $\mathfrak{s}$-torsion pair in $\mathcal{K}$. Denote by $({\rm t}_{23}, {\rm f}_{23})$ the torsion functors of $(\mathcal{T}_{2}\cap\mathcal{T}_{3}, \mathcal{F}_{2}\ast\mathcal{F}_{3})$. We need to prove that ${\rm f}_{23}\mathcal{F}_{1}\subset\mathcal{F}_{1}$. Let $C\in\mathcal{F}_{1}$. Since $(\mathcal{T}_{2}, \mathcal{F}_{2})$ and $(\mathcal{T}_{3}, \mathcal{F}_{3})$ are $\mathfrak{s}$-torsion pairs, we have the following two canonical $\mathbb{E}$-triangles

Since $(s_{2}, s_{3})$ is lower consistent and ${\rm t}_{3}C\in\mathcal{T}_{3}$, we have ${\rm t}_{2}{\rm t}_{3}C\in\mathcal{T}_{2}\cap\mathcal{T}_{3}$. Applying (ET4) to (3.4) and (3.5) induces a commutative diagram

where all rows and columns are $\mathbb{E}$-triangles. Thus there exists an $\mathbb{E}$-triangle

such that $A\in\mathcal{F}_{2}\ast\mathcal{F}_{3}$, so the $\mathbb{E}$-triangle (3.6) is a canonical $\mathbb{E}$-triangle with respect to the $\mathfrak{s}$-torsion pair $(\mathcal{T}_{2}\cap\mathcal{T}_{3}, \mathcal{F}_{2}\ast\mathcal{F}_{3})$. Since $(s_{1}, s_{3})$ is upper consistent and $C\in\mathcal{F}_{1}$, we have ${\rm f}_{3}C\in\mathcal{F}_{1}$, then it follows from the $\mathbb{E}$-triangle ${\rm t}_{3}C\rightarrow C\rightarrow {\rm f}_{3}C\dashrightarrow$ and Lemma 3.7 that ${\rm t}_{3}C\in\mathcal{F}_{1}$. Since $(s_{1}, s_{2})$ is upper consistent, we have ${\rm f}_{2}{\rm t}_{3}C\in\mathcal{F}_{1}$. Since the torsion-free class $\mathcal{F}_{1}$ is closed under extensions, we obtain $A\in\mathcal{F}_{1}$ by the $\mathbb{E}$-triangle ${\rm f}_{2}{\rm t}_{3}C\rightarrow A\rightarrow {\rm f}_{3}C\dashrightarrow$ and the canonical $\mathbb{E}$-triangle (3.6) implies that ${\rm f}_{23}C = A\in\mathcal{F}_{1}$. Therefore, $((\mathcal{T}_{1}, \mathcal{F}_{1}), (\mathcal{T}_{2}\cap\mathcal{T}_{3}, \mathcal{F}_{2}\ast\mathcal{F}_{3}))$ is upper consistent.

(2) By Theorem 3.4, $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ and $(\mathcal{T}_{1}\ast\mathcal{T}_{3}, \mathcal{F}_{1}\cap\mathcal{F}_{3})$ are $\mathfrak{s}$-torsion pairs in $\mathcal{K}$. Denote by $({\rm t}_{12}, {\rm f}_{12})$ the torsion functors of $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$. To complete the proof, we need to show that ${\rm t}_{12}(\mathcal{T}_{1}\ast\mathcal{T}_{3})\subset \mathcal{T}_{1}\ast\mathcal{T}_{3}$. Let $C\in\mathcal{T}_{1}\ast\mathcal{T}_{3}$, then there exists an $\mathbb{E}$-triangle

such that $T_{1}\in\mathcal{T}_{1}$ and $T_{3}\in\mathcal{T}_{3}$. Since $(\mathcal{T}_{1}, \mathcal{F}_{1})$ and $(\mathcal{T}_{3}, \mathcal{F}_{3})$ are $\mathfrak{s}$-torsion pairs, we have the following two canonical $\mathbb{E}$-triangles

Applying ${\rm Hom}_{\mathcal{K}}(-, {\rm f}_{3}{\rm f}_{1}T_{3})$ to the $\mathbb{E}$-triangle (3.8), we have an exact sequence

Notice that $(s_{1}, s_{3})$ is upper consistent and ${\rm f}_{1}T_{3}\in \mathcal{F}_{1}$, and we have ${\rm f}_{3}{\rm f}_{1}T_{3}\in\mathcal{F}_{1}\cap \mathcal{F}_{3}$. Since $(\mathcal{T}_{1}, \mathcal{F}_{1})$ is an $\mathfrak{s}$-torsion pair and ${\rm t}_{1}T_{3}\in\mathcal{T}_{1}$, ${\rm f}_{3}{\rm f}_{1}T_{3}\in\mathcal{F}_{1}$, we have $\mathbb{E}^{-1}({\rm t}_{1}T_{3}, {\rm f}_{3}{\rm f}_{1}T_{3}) = 0$. Since $(\mathcal{T}_{3}, \mathcal{F}_{3})$ is an $\mathfrak{s}$-torsion pair and $T_{3}\in\mathcal{T}_{3}$, ${\rm f}_{3}{\rm f}_{1}T_{3}\in\mathcal{F}_{3}$, we have ${\rm Hom}_{\mathcal{K}}(T_{3}, {\rm f}_{3}{\rm f}_{1}T_{3}) = 0$ and $b\in{\rm Hom}_{\mathcal{K}}({\rm f}_{1}T_{3}, {\rm f}_{3}{\rm f}_{1}T_{3}) = 0$. It follows from Lemma 3.3 that $a$ is a split epimorphism, so ${\rm f}_{1}T_{3}$ is a direct summand of ${\rm t}_{3}{\rm f}_{1}T_{3}$. By Lemma 2.6, ${\rm f}_{1}T_{3}\in\mathcal{T}_{3}$. Consider the canonical $\mathbb{E}$-triangle

with respect to the $\mathfrak{s}$-torsion pair $(\mathcal{T}_{2}, \mathcal{F}_{2})$. Since $(s_{2}, s_{3})$ is lower consistent and ${\rm f}_{1}T_{3}\in\mathcal{T}_{3}$, we have ${\rm t}_{2}{\rm f}_{1}T_{3}\in\mathcal{T}_{2}\cap\mathcal{T}_{3}$. Since $(s_{1}, s_{2})$ is upper consistent and ${\rm f}_{1}T_{3}\in\mathcal{F}_{1}$, we have

Applying (ET4)$^{\rm op}$ to (3.8) and (3.10) induces a commutative diagram

where all rows and columns are $\mathbb{E}$-triangles. Thus there exists an $\mathbb{E}$-triangle

such that $A\in\mathcal{T}_{1}\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3})$. Moreover, applying (ET4)$^{\rm op}$ to (3.7) and (3.12) induces a commutative diagram

where all rows and columns are $\mathbb{E}$-triangles, which gives an $\mathbb{E}$-triangle

Since the torsion class $\mathcal{T}_{1}$ is closed under extensions, we obtain $B\in\mathcal{T}_{1}\ast(\mathcal{T}_{1}\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3})) = \mathcal{T}_{1}\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3})$ by the $\mathbb{E}$-triangle $T_{1}\rightarrow B\rightarrow A\dashrightarrow$. It follows from Lemma 2.7 that the canonical $\mathbb{E}$-triangle uniquely exists. Since $B\in\mathcal{T}_{1}\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3})\subset\mathcal{T}_{1}\ast\mathcal{T}_{2}$ and ${\rm f}_{2}{\rm f}_{1}T_{3}\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$ by (3.11), the $\mathbb{E}$-triangle (3.13) is a canonical $\mathbb{E}$-triangle with respect to the $\mathfrak{s}$-torsion pair $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$, so ${\rm t}_{12}C = B\in\mathcal{T}_{1}\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3})\subset\mathcal{T}_{1}\ast\mathcal{T}_{3}$. Therefore, $((\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2}), (\mathcal{T}_{1}\ast\mathcal{T}_{3}, \mathcal{F}_{1}\cap\mathcal{F}_{3}))$ is lower consistent.

(3) By Theorem 3.4 and the results of (1) and (2), we have the following two $\mathfrak{s}$-torsion pairs

Hence it suffices to prove that $\mathcal{F}_{1}\cap(\mathcal{F}_{2}\ast\mathcal{F}_{3}) = (\mathcal{F}_{1}\cap\mathcal{F}_{2})\ast (\mathcal{F}_{1}\cap\mathcal{F}_{3})$. It is easy to check that $(\mathcal{F}_{1}\cap\mathcal{F}_{2})\ast (\mathcal{F}_{1}\cap\mathcal{F}_{3})\subset\mathcal{F}_{1}\cap(\mathcal{F}_{2}\ast\mathcal{F}_{3})$. Conversely, let $C\in\mathcal{F}_{1}\cap(\mathcal{F}_{2}\ast\mathcal{F}_{3})$, then we have the following three $\mathbb{E}$-triangles

where $F_{2}\in\mathcal{F}_{2}$, $F_{3}\in\mathcal{F}_{3}$, (3.15) and (3.16) are canonical. Since $(s_{2}, s_{3})$ is lower consistent and ${\rm t}_{3}C\in\mathcal{T}_{3}$, we have ${\rm t}_{2}{\rm t}_{3}C\in\mathcal{T}_{2}\cap\mathcal{T}_{3}$. Applying ${\rm Hom}_{\mathcal{K}}({\rm t}_{2}{\rm t}_{3}C, -)$ to the $\mathbb{E}$-triangle (3.14), we have an exact sequence

so ${\rm Hom}_{\mathcal{K}}({\rm t}_{2}{\rm t}_{3}C, C) = 0$. Applying ${\rm Hom}_{\mathcal{K}}({\rm t}_{2}{\rm t}_{3}C, -)$ to the $\mathbb{E}$-triangle (3.15), we have an exact sequence

Since $(\mathcal{T}_{3}, \mathcal{F}_{3})$ is an $\mathfrak{s}$-torsion pair and ${\rm t}_{2}{\rm t}_{3}C\in\mathcal{T}_{3}$, ${\rm f}_{3}C\in\mathcal{F}_{3}$, we have $\mathbb{E}^{-1}({\rm t}_{2}{\rm t}_{3}C, {\rm f}_{3}C) = 0$ and $a\in{\rm Hom}_{\mathcal{K}}({\rm t}_{2}{\rm t}_{3}C, {\rm t}_{3}C) = 0$. By Lemma 3.3, $b$ is a split monomorphism, then ${\rm t}_{3}C$ is a direct summand of ${\rm f}_{2}{\rm t}_{3}C$. It follows from Lemma 2.6 that ${\rm t}_{3}C\in\mathcal{F}_{2}$. Since $(s_{1}, s_{3})$ is upper consistent and $C\in\mathcal{F}_{1}$, we have ${\rm f}_{3}C\in\mathcal{F}_{1}$. By Lemma 3.7, using the $\mathbb{E}$-triangle (3.15), we obtain ${\rm t}_{3}C\in\mathcal{F}_{1}$. Thus ${\rm t}_{3}C\in\mathcal{F}_{1}\cap\mathcal{F}_{2}$ and ${\rm f}_{3}C\in\mathcal{F}_{1}\cap\mathcal{F}_{3}$, which implies that $C\in(\mathcal{F}_{1}\cap\mathcal{F}_{2})\ast(\mathcal{F}_{1}\cap\mathcal{F}_{3})$. Therefore, $\mathcal{F}_{1}\cap(\mathcal{F}_{2}\ast\mathcal{F}_{3})\subset(\mathcal{F}_{1}\cap\mathcal{F}_{2})\ast (\mathcal{F}_{1}\cap\mathcal{F}_{3})$. Thus we complete the proof. □

Dually, we have the following conclusion.

Corollary 3.9. Let $s_{i}: = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair in $\mathcal{K}$ for any $i = 1, 2, 3$. Suppose that $(s_{1}, s_{3})$, $(s_{2}, s_{3})$ are lower consistent and $(s_{1}, s_{2})$ is upper consistent, then the following assertions hold.

$({\rm{1}})$ $((\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2}), (\mathcal{T}_{3}, \mathcal{F}_{3}))$ is lower consistent.

$({\rm{2}})$ $((\mathcal{T}_{1}\cap\mathcal{T}_{3}, \mathcal{F}_{1}\ast\mathcal{F}_{3}), (\mathcal{T}_{2}\cap\mathcal{T}_{3}, \mathcal{F}_{2}\ast\mathcal{F}_{3}))$ is upper consistent.

$({\rm{3}})$ $(\mathcal{T}_{1}\ast\mathcal{T}_{2})\cap\mathcal{T}_{3} = (\mathcal{T}_{1}\cap\mathcal{T}_{3})\ast(\mathcal{T}_{2}\cap\mathcal{T}_{3})$, and $(\mathcal{F}_{1}\cap\mathcal{F}_{2})\ast\mathcal{F}_{3} = (\mathcal{F}_{1}\ast\mathcal{F}_{3})\cap(\mathcal{F}_{2}\ast\mathcal{F}_{3})$.

4.   Application

In this section, we apply our main theorem to the $\tau$-tilting theory which was introduced by Adachi, Iyama and Reiten in ^[2]. Let $\Lambda$ be a finite-dimensional algebra over an algebraically closed field and ${\rm mod} \Lambda$ the category of finitely generated left $\Lambda$-modules. If $M$ is a $\Lambda$-module, we denote by Fac$M$ the subcategory of ${\rm mod}\Lambda$ which consists of all factor modules of finite direct sums of copies of $M$; the subcategory ${\rm Sub}M$ is defined dually. We denote the number of pairwise nonisomorphic indecomposable summands of $M$ by $|M|$, then $|\Lambda|$ equals the rank of the Grothendieck group of ${\rm mod} \Lambda$.

Definition 4.1. (^[2]) Let $M$ be a $\Lambda$-module.

(1) $M$ is called $\tau$-rigid if ${\rm Hom}_{\Lambda}(M, \tau M) = 0$, where $\tau$ is the Auslander-Reiten translation.

(2) $M$ is called $\tau$-tilting if $M$ is $\tau$-rigid and $|M| = |\Lambda|$.

Let $M$ be a $\tau$-rigid $\Lambda$-module. It is well known that there exist two distinguished torsion pairs in ${\rm mod} \Lambda$, namely

which satisfy ${\rm Fac} M\subseteq{^{\perp}(\tau M)}$ and ${\rm Sub}\tau M\subseteq M^{\perp}$. We have the following characterization of an arbitrary module being $\tau$-tilting.

Lemma 4.2. Let $M$ be a $\Lambda$-module, then $M$ is $\tau$-tilting if and only if $({\rm Fac}M, {\rm Sub}\tau M)$ is a torsion pair in ${\rm mod} \Lambda$.

Proof. According to [2, Theorem 2.12], a $\tau$-rigid module $M$ is $\tau$-tilting if and only if ${\rm Fac} M = {^{\perp}(\tau M)}$.

If $M$ is $\tau$-tilting, then $({\rm Fac}M, {\rm Sub}\tau M) = ({^{\perp}(\tau M)}, {\rm Sub}\tau M)$ is a torsion pair. Conversely, assume that $({\rm Fac}M, {\rm Sub}\tau M)$ is a torsion pair. Since $M\in{\rm Fac}M$, $\tau M\in{\rm Sub}\tau M$, we have ${\rm Hom}_{\Lambda}(M, \tau M) = 0$, which implies that $M$ is $\tau$-rigid. Notice that a torsion-free class and the corresponding torsion class are determined by each other, and $({\rm Fac}M, {\rm Sub}\tau M)$, $(^{\perp}(\tau M), {\rm Sub}\tau M)$ are torsion pairs. We have ${\rm Fac} M = {^{\perp}(\tau M)}$, so $M$ is $\tau$-tilting. □

Lemma 4.3. ([3, Propositions 5.8, 5.6]) Let $M$ and $N$ be two $\Lambda$-modules, then the following conditions are equivalent.

$({\rm{1}})$ ${\rm Hom}_{\Lambda}(N, \tau M) = 0$.

$({\rm{2}})$ ${\rm Ext}^{1}_{\Lambda}(M, {\rm Fac}N) = 0$.

$({\rm{3}})$ ${\rm Ext}^{1}_{\Lambda}({\rm Sub}\tau M, \tau N) = 0$.

Proposition 4.4. Let $M = M_{1}\oplus M_{2}$ be a $\tau$-rigid $\Lambda$-module, then the following assertions hold.

$({\rm{1}})$ ${\rm Fac}M = {\rm Fac}M_{1}\ast{\rm Fac}M_{2}$.

$({\rm{2}})$ ${\rm Sub}\tau M = {\rm Sub}\tau M_{1}\ast{\rm Sub}\tau M_{2}$.

Proof. We only prove (1); the proof of (2) is similar. First, we prove that $(({\rm Fac} M_{1}, M_{1}^{\perp}), ({\rm Fac} M_{2}, M_{2}^{\perp}))$ is lower consistent. Let $X$ be a module in $M_{1}^{\perp}$, then there exists an exact sequence

with respect to the torsion pair $({\rm Fac} M_{2}, M_{2}^{\perp})$ such that $P\in{\rm Fac} M_{2}$ and $Q\in M_{2}^{\perp}$. Applying the functor ${\rm Hom}_{\Lambda}(M_{1}, -)$ to the exact sequence, we have an exact sequence

Since $X\in M_{1}^{\perp}$, the left hand side vanishes. Since ${\rm Fac} M_{2}\subseteq{\rm Fac} (M_{1}\oplus M_{2})$, we have that $P$ also belongs to ${\rm Fac} (M_{1}\oplus M_{2})$. Note that $M_{1}\oplus M_{2}$ is a $\tau$-rigid module, then it follows from Lemma 4.3 that ${\rm Ext}_{\Lambda}^{1}(M_{1}\oplus M_{2}, P) = 0$, so ${\rm Ext}_{\Lambda}^{1}(M_{1}, P) = 0$. Thus ${\rm Hom}_{\Lambda}(M_{1}, Q) = 0$, which implies that $Q$ is in $M_{1}^{\perp}$. Hence $(({\rm Fac} M_{1}, M_{1}^{\perp}), ({\rm Fac} M_{2}, M_{2}^{\perp}))$ is upper consistent, then by Theorem 3.4, $({\rm Fac}M_{1}\ast{\rm Fac}M_{2}, M_{1}^{\perp}\cap M_{2}^{\perp})$ is a torsion pair in ${\rm mod}\Lambda$. Moreover, since

the torsion pairs $({\rm Fac}M_{1}\ast{\rm Fac}M_{2}, M_{1}^{\perp}\cap M_{2}^{\perp})$ and $({\rm Fac} M, M^{\perp})$ share the same torsion-free class, and we have the assertion. □

Let $\mathcal{X}, \mathcal{Y}$ be two subcategories of ${\rm mod} \Lambda$. Generally speaking, $\mathcal{X}\ast\mathcal{Y}\neq \mathcal{Y}\ast \mathcal{X}$. Proposition 4.4 tells us that ${\rm Fac}M_{1}\ast{\rm Fac}M_{2} = {\rm Fac}M_{2}\ast{\rm Fac}M_{1}$ and ${\rm Sub}\tau M_{1}\ast{\rm Sub}\tau M_{2} = {\rm Sub}\tau M_{2}\ast{\rm Sub}\tau M_{1}$ when $M_{1}\oplus M_{2}$ is a $\tau$-rigid $\Lambda$-module. By Proposition 4.4, we can easily get the following conclusion.

Corollary 4.5. Let $M = \bigoplus_{i = 1}^{n}M_{i}$ be a $\tau$-rigid $\Lambda$-module, then the following assertions hold.

$({\rm{1}})$ ${\rm Fac}M = {\rm Fac}M_{1}\ast{\rm Fac}M_{2}\ast\cdots\ast{\rm Fac}M_{n}$.

$({\rm{2}})$ ${\rm Sub}\tau M = {\rm Sub}\tau M_{1}\ast{\rm Sub}\tau M_{2}\ast\cdots\ast{\rm Sub}\tau M_{n}$.

As a byproduct, we have the following characterization of direct sum of $\Lambda$-modules being $\tau$-tilting.

Corollary 4.6. Let $M = \bigoplus_{i = 1}^{n}M_{i}$ be a $\Lambda$-module, then the following conditions are equivalent.

$({\rm{1}})$ $M$ is $\tau$-tilting.

$({\rm{2}})$ $({\rm Fac}M_{1}\ast{\rm Fac}M_{2}\ast\cdots\ast{\rm Fac}M_{n}, {\rm Sub}\tau M_{1}\ast{\rm Sub}\tau M_{2}\ast\cdots\ast{\rm Sub}\tau M_{n})$ is a torsion pair in ${\rm mod} \Lambda$.

Proof. (1) implies (2). If $M$ is a $\tau$-tilting module, then $({\rm Fac}M, {\rm Sub}\tau M)$ is a torsion pair in ${\rm mod} \Lambda$ by Lemma 4.2, then the statement follows from Corollary 4.5.

(2) implies (1). Since $({\rm Fac}M_{1}\ast{\rm Fac}M_{2}\ast\cdots\ast{\rm Fac}M_{n}, {\rm Sub}\tau M_{1}\ast{\rm Sub}\tau M_{2}\ast\cdots\ast{\rm Sub}\tau M_{n})$ is a torsion pair, $M_{i}\in{\rm Fac}M_{1}\ast{\rm Fac}M_{2}\ast\cdots\ast{\rm Fac}M_{n}$ and $\tau M_{j}\in{\rm Sub}\tau M_{1}\ast{\rm Sub}\tau M_{2}\ast\cdots\ast{\rm Sub}\tau M_{n}$, and we have ${\rm Hom}_{\Lambda}(M_{i}, \tau M_{j}) = 0$ for any $1\leq i, j\leq n$. This implies that $M$ is $\tau$-rigid. By Corollary 4.5, we have that $({\rm Fac}M, {\rm Sub}\tau M)$ is a torsion pair in ${\rm mod} \Lambda$. Thus $M$ is $\tau$-tilting by Lemma 4.2. □

5.   Conclusions

We introduced the notion of a consistent pair of $\mathfrak{s}$-torsion pairs in an extriangulated category with a negative first extension. Let $s_{i}: = (\mathcal{T}_{i}, \mathcal{F}_{i})$ be an $\mathfrak{s}$-torsion pair, for any $i = 1, 2$. We showed that $(\mathcal{T}_{1}\ast\mathcal{T}_{2}, \mathcal{F}_{1}\cap\mathcal{F}_{2})$ (respectively, $(\mathcal{T}_{1}\cap\mathcal{T}_{2}, \mathcal{F}_{1}\ast\mathcal{F}_{2})$) is an $\mathfrak{s}$-torsion pair if and only if $(s_{1}, s_{2})$ is an upper (respectively, lower) consistent pair, which generalizes [4, Proposition 6]. Let $M = M_{1}\oplus M_{2}$ be a $\tau$-rigid module over a finite-dimensional algebra. As an application of the main theorem, we proved that ${\rm Fac}M = {\rm Fac}M_{1}\ast{\rm Fac}M_{2}$, where Fac$M$ is the category of all factor modules of finite direct sums of copies of $M$.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was supported by Fujian Province Nature Science Foundation of China (2020J01364), Fujian Province education and research projects for young and middle-aged teachers (JAT190739) and Doctoral Research Launch Project of Longyan University (LB20202003). The authors would like to thank the anonymous reviewers for their comments and suggestions.

Conflict of interest

The authors declare no conflicts of interest.