The remaining part of this section is devoted to proving Theorem 4.46 and Corollary 4.49.

**Lemma 4.50.** Let $\Lambda$ be any Artin algebra. Let $X \xrightarrow{a} Y \xrightarrow{b} Z$ be an almost-split conflation in $K^{[1,0]}(\mathrm{proj}\,\Lambda)$ such that $Z$ is not in add($\Sigma\Lambda$). Denote by $F$ the functor $\mathrm{Hom}_{K^b}(\Lambda, -) : K^{[1,0]}(\mathrm{proj}\,\Lambda) \to \mathrm{mod}\,\Lambda$. Then there is an almost-split sequence in mod $\Lambda$:

$$0 \to FX \to FY \to FZ \to 0$$

*Proof.* The almost-split conflation induces an exact sequence

$$F\Sigma^{-1}Y \to F\Sigma^{-1}Z \to FX \to FY \to FZ \to F\Sigma X.$$

Because $X$ is concentrated in nonpositive degrees, $F\Sigma X = 0$. Let us verify that the map $F\Sigma^{-1}Y \to F\Sigma^{-1}Z$ is surjective. Let $\Lambda \xrightarrow{f} \Sigma^{-1}Z$ be a morphism in $K^b(\mathrm{proj}\,\Lambda)$. As $Y \xrightarrow{b} Z$ is right almost-split, and $Z$ is indecomposable not in add($\Sigma\Lambda$), the morphism $\Sigma f$ factors through $b$. This implies the required surjectivity, and we have a short exact sequence

$$0 \to FX \xrightarrow{Fa} FY \xrightarrow{Fb} FZ \to 0.$$

Because $\mathrm{mod}\,\Lambda$ and $K^b(\mathrm{proj}\,\Lambda)$ are Krull-Schmidt, it is immediate to check that $Fb$ (resp., $Fa$) inherits from $b$ (resp., $a$) the property of being right (resp., left) almost-split. $\square$

**Lemma 4.51.** Let $\Lambda$ be an Artin algebra, let $\tau X \to E \to X$ be an Auslander-Reiten conflation in $K^{[1,0]}(\mathrm{proj}\,\Lambda)$ and denote $FX = M$. We make the following assumptions.

(1) $X \notin \mathrm{add}(\Sigma\Lambda)$.

(2) $M$ and $\tau M$ are $\tau$-rigid.

(3) $\mathrm{Hom}_{\Lambda}(M, \tau^2 M) = 0$.

(4) $\tau M$ is a brick.

Then $\tau X \to E \to X$ is a mutation conflation.

*Proof.* First note that the results [8, Lemmas 3.4 and 3.5], even though stated for finite-dimensional algebras, hold with the same proofs for any Artin algebra. For convenience of the reader, we recall those two results: Let $Y, Z \in K^{[1,0]}(\mathrm{proj}\,\Lambda)$ be indecomposable and not in add($\Sigma\Lambda$), and let $P \in \mathrm{proj}\,\Lambda$. Then $\mathrm{Hom}_{K^b}(Y, \Sigma Z) = 0$ if and only if $\mathrm{Hom}_{\Lambda}(FZ, \tau FY) = 0$ and $\mathrm{Hom}_{K^b}(P, Y) = 0$ if and only if $\mathrm{Hom}_{\Lambda}(P, FY) = 0$.

Note also that Lemma 4.50 implies that $F\tau X$ is isomorphic to $\tau M$.

By assumption, $X$ is not in add($\Sigma\Lambda$). The same is true of $\tau X$ because objects in add($\Sigma\Lambda$) are injective in $K^{[1,0]}(\mathrm{proj}\,\Lambda)$ and an Auslander-Reiten conflation does not split. It thus follows from [8, Lemma 3.4] that $X$ and $\tau X$ are rigid in $K^{[1,0]}(\mathrm{proj}\,\Lambda)$. We have to prove that both $E \oplus X$ and $E \oplus \tau X$ are rigid.

(1) $\mathrm{Hom}_{K^b}(E, \Sigma X) = 0$.

The Auslander-Reiten conflation induces an exact sequence

$$\mathrm{Hom}_{K^b}(X, \Sigma X) \to \mathrm{Hom}_{K^b}(E, \Sigma X) \to \mathrm{Hom}_{K^b}(\tau X, \Sigma X).$$
As already remarked, $X$ is rigid so that the left-most term vanishes. By assumption

$$ \mathrm{Hom}_{\Lambda}(FX, \tau F \tau X) = \mathrm{Hom}_{\Lambda}(M, \tau^2 M) = 0, $$

so that the right-most term also vanishes by [8, Lemma 3.4].

$$ (2) \mathrm{Hom}_{K^b}(\tau X, \Sigma E) = 0. $$

As $\Lambda$ is projective in $K^{[-1,0]}(\mathrm{proj} \Lambda)$, the sequence $\tau M \to FE \to M$ is short exact. It induces a sequence

$$ \mathrm{Hom}_{\Lambda}(M, \tau^2 M) \to \mathrm{Hom}_{\Lambda}(FE, \tau^2 M) \to \mathrm{Hom}_{\Lambda}(\tau M, \tau^2 M) $$

from which we deduce that $\mathrm{Hom}_{\Lambda}(FE, \tau^2 M) = 0$. This implies $\mathrm{Hom}_{K^b}(\tau X, \Sigma E) = 0$ by [8, Lemma 3.4].

$$ (3) \mathrm{Hom}_{K^b}(X, \Sigma E) = 0. $$

Let us show that $\mathrm{Hom}_{\Lambda}(FE, \tau M) = 0$. Let $FE \xrightarrow{f} \tau M$ be any morphism, and consider the short exact sequence $0 \to \tau M \xrightarrow{a} FE \xrightarrow{b} M \to 0$ induced by the Auslander-Reiten conflation. As $\tau M$ is a brick, the composition $f$ is zero. The morphism $f$ then factors through the cokernel $b$, which implies $f = 0$ because $M$ is $\tau$-rigid.

$$ (4) \mathrm{Hom}_{K^b}(E, \Sigma E) = 0. $$

The Auslander-Reiten conflation induces an exact sequence

$$ \mathrm{Hom}_{K^b}(X, \Sigma E) \to \mathrm{Hom}_{K^b}(E, \Sigma E) \to \mathrm{Hom}_{K^b}(\tau X, \Sigma E). $$

We can conclude by (2) and (3).

$$ (5) \mathrm{Hom}_{K^b}(E, \Sigma \tau X) = 0. $$

We have an exact sequence

$$ \mathrm{Hom}_{K^b}(E, E) \to \mathrm{Hom}_{K^b}(E, X) \to \mathrm{Hom}_{K^b}(E, \Sigma \tau X) \to \mathrm{Hom}_{K^b}(E, \Sigma E). $$

The first map is surjective because $E \to X$ is right almost-split (note that $E$ cannot contain any summand isomorphic to $X$). We can conclude by (4). □

**Lemma 4.52.** Let $\Lambda$ be an Artin algebra, and let $Y \to E \to X$ be a conflation in $K^{[-1,0]}(\mathrm{proj} \Lambda)$. Then:

$$ (1) \text{ If } \mathrm{Hom}_{K^b}(Y, \Sigma E) = 0, \text{ we have } \mathrm{Hom}_{\Lambda}(FX, \tau FY) = 0. $$

$$ (2) \text{ If moreover } X \notin \mathrm{add}(\Sigma \Lambda) \text{ and the conflation is almost-split, then } \mathrm{Hom}_{\Lambda}(FX, \tau^2 FX) = 0. $$

*Proof.* By Lemma 4.50, (2) follows from (1). Let us prove (1). The conflation induces an exact sequence $\mathrm{Hom}_{K^b}(Y, \Sigma E) \to \mathrm{Hom}_{K^b}(Y, \Sigma X) \to \mathrm{Hom}_{K^b}(Y, \Sigma^2, Y)$. The complex $Y$ being two-term, we have $\mathrm{Hom}_{K^b}(Y, \Sigma^2, Y) = 0$. Assuming that $\mathrm{Hom}_{K^b}(Y, \Sigma E) = 0$, we deduce that $\mathrm{Hom}_{K^b}(Y, \Sigma X) = 0$, and hence that $\mathrm{Hom}_{\Lambda}(FX, \tau FY) = 0$ by [8, Lemma 3.4]. □
**Lemma 4.53.** Let $\Lambda$ be an Artin algebra whose indecomposable injective modules are $\tau$-rigid, and let $P$ be an indecomposable projective $\Lambda$-module. Then the Auslander-Reiten conflation $Y \to E \to \Sigma P$ in $K^{[-1,0]}$(proj $\Lambda$) is a mutation conflation if FY is a brick.

*Proof.* In this proof, we implicitly identify the projective indecomposable $\Lambda$-module $P$ with the complex $P$ concentrated in degree 0.

We check that the conflation $Y \to E \to \Sigma P$ is a mutation conflation.

(0) As $P$ is concentrated in only 1 degree, it is rigid in $K^{[-1,0]}$(proj $\Lambda$) and so is $\Sigma P$. Moreover, FY is indecomposable and injective. By assumption, it is $\tau$-rigid, hence $Y$ is rigid in $K^{[-1,0]}$(proj $\Lambda$) by [8, Lemma 3.4].

(1) We have $\text{Hom}_{K^b}(E, \Sigma^2 P) = 0$ because $E$ is concentrated in degrees $-1$ and $0$.

(2) $\text{Hom}_{K^b}(Y, \Sigma E) = 0$: There is an exact sequence

$$ \text{Hom}_{K^b}(Y, \Sigma Y) \to \text{Hom}_{K^b}(Y, \Sigma E) \to \text{Hom}_{K^b}(Y, \Sigma^2 P), $$

where the first term vanishes by (0) and the last term vanishes because $Y$ is in degrees $-1, 0$ while $\Sigma^2 P$ is concentrated in degree $-2$.

(3) $\text{Hom}_{K^b}(E, \Sigma Y) = 0$: Consider the exact sequence

$$ \text{Hom}_{K^b}(\Sigma Y, \Sigma Y) \to \text{Hom}_{K^b}(\Sigma P, \Sigma Y) \to \text{Hom}_{K^b}(E, \Sigma Y) \to \text{Hom}_{K^b}(Y, \Sigma Y). $$

Its last term vanishes by (0). Let us prove that its first morphism is surjective. Composition with $\alpha$ and $F\alpha$ yields a commutative square

<figure><img src="image_35.png" /></figure>

where the vertical maps are surjective. As $\Lambda$ is rigid and $Y$ is two-term, we can apply [52, Proposition 6.2] to the triangulated category $K^b$(proj $\Lambda$) (or Proposition 4.6 to the extriangulated $K^{[-1,0]}$(proj $\Lambda$)). As a consequence, the right-most vertical map is bijective. Because FY is a brick, $\text{End}_\Lambda(FY)$ is a division ring. Moreover, FY is the indecomposable injective with simple socle the simple top of $P$, so that the $\text{End}_\Lambda(FY)$-module $\text{Hom}_\Lambda(FP, FY)$ is 1-dimensional over $\text{End}_\Lambda(FY)$. It follows that the vertical map $(F\alpha)^*$ is either zero or bijective. But $\alpha$ is nonzero with domain in add $\Lambda$ so that $F\alpha$ is nonzero; and $(F\alpha)^*(1_{FY}) = F\alpha$ so that $(F\alpha)^*$ is nonzero. By commutativity, this implies the surjectivity of the top vertical morphism $\alpha^*$. As the shift functor is an automorphism, the morphism $\text{Hom}_{K^b}(\Sigma Y, \Sigma Y) \to \text{Hom}_{K^b}(\Sigma P, \Sigma Y)$ is also surjective, and $\text{Hom}_{K^b}(E, \Sigma Y) = 0$.

(4) $\text{Hom}_{K^b}(E, \Sigma E) = 0$: In the exact sequence

$$ \text{Hom}_{K^b}(E, \Sigma Y) \to \text{Hom}_{K^b}(E, \Sigma E) \to \text{Hom}_{K^b}(E, \Sigma^2 P), $$

the first term is zero by (3) and the last term vanishes because $P$ is concentrated in degree 0.

(5) $\text{Hom}_{K^b}(\Sigma P, \Sigma E) = 0$: Let $P \xrightarrow{f} E$ be any morphism.
<figure><img src="image_36.png" /></figure>

As $P$ is rigid, $f$ factors through the inflation $\beta$ in the conflation $Y \to E \to \Sigma P$, and there is some morphism $g$ such that $f = \beta g$. By (3), the morphism $g$ factors through $P \xrightarrow{\alpha} Y$: there is some $h : Y \to Y$ such that $g = h\alpha$. We then have $f = \beta g = \beta(g - \alpha) = \beta(h - 1)\alpha$. As $Y$ is endo-local, the morphism $h$ or $h-1$ is an automorphism, and we may assume that $h$ is an automorphism. The morphism $\beta h$ is not a section (otherwise $\beta$ would be a section) and $\beta$ is left-almost split: there is an endomorphism $h'$ of $E$ such that $\beta h = h'\beta$. Therefore, we have $f = \beta h\alpha = h'\beta\alpha = 0$. $\square$

**Proof of Theorem 4.46.** The theorem follows from the slightly more general statements of Lemmas 4.51 to 4.53. $\square$

**Lemma 4.54.** Let $\Lambda$ be a brick and 2-acyclic gentle algebra. Then, for any indecomposable $\Lambda$-module $M$ we have $\text{Hom}_{\Lambda}(M, \tau^2 M) = 0$.

*Proof.* Let $\Lambda = \mathbb{K}Q/I$ be any gentle algebra of finite representation type, and let $M$ be an indecomposable $\Lambda$-module. Write $\sigma$ for the string corresponding to $\tau M$. Then the walks (i.e., the maximal strings in the blossoming bound quiver of $(Q, I)$) associated with $M$ and $\tau M$ are, respectively, $\omega_M = \mathbf{\wedge}\sigma\mathbf{\wedge} = p^{-1}\alpha\sigma\beta^{-1}q$ and $\omega_{\tau M} = \mathbf{\wedge}\sigma\check{\mathbf{\wedge}} = p'\alpha'^{-1}\sigma\beta'q'^{-1}$, where $p, p', q, q'$ are paths and $\alpha, \alpha', \beta, \beta'$ are arrows in the blossoming quiver.

Assume that $\text{Hom}_{\Lambda}(M, \tau^2 M) \neq 0$. By [81, Theorem 2.46] or [12, Theorem 5.1], the walk $\omega_M$ kisses the walk $\omega_{\tau M}$: there is a substring $\rho$ of $\omega_M$ and $\omega_{\tau M}$ that is strictly on top of $\omega_M$ and strictly at the bottom of $\omega_{\tau M}$. Remark that $\rho$ cannot start nor end at some endpoint of $\sigma$ (because $\rho$ is a top substring of $\omega_M$).

If the substring $\rho$ of $\omega_M$ is contained in $\sigma$, by the previous remark it is strictly below $\sigma$. This implies that $\omega_M$ kisses itself, hence that $M$ is not $\tau$-rigid. Similarly, if $\rho$ is contained in $\sigma$ as a substring of $\omega_{\tau M}$, then $\tau M$ is not $\tau$-rigid. In both cases, the brick-$\tau$-rigid correspondence of [30] implies that $\Lambda$ is not a brick algebra.

Note that $\rho$ cannot contain $\sigma$, because $\sigma$ is the only common substring of $\omega_M$ and $\omega_{\tau M}$ that contains $\sigma$. Assume now that $\rho$, viewed either as a substring of $\omega_M$ or of $\omega_{\tau M}$ is not contained in $\sigma$. As $\rho$ is strictly on top of $\omega_M$, it contains the source of $\alpha$ or the source of $\beta$. Similarly, $\rho$ contains the target of $\alpha'$ or the target of $\beta'$. Because the target of $\alpha$ is the source of $\alpha'$ (and the target of $\beta$ is the source of $\beta'$), this implies that $Q$ contains either a nonoriented cycle with at most one relation, in which case $\Lambda$ is not a brick algebra, or a 2-cycle, in which case $\Lambda$ is not 2-acyclic. $\square$

**Proof of Corollary 4.49.** The unique exchange relation property follows from Proposition 4.44 (2). By Lemma 4.54, the assumptions of Theorem 4.46 are satisfied. We can thus apply Theorem 4.13 and its corollary, which shows that there is the same number of extremal exchange pairs of the support $\tau$-tilting fan (i.e., the nonkissing fan) as there are meshes in the Auslander-Reiten quiver of $\Lambda$. Hence, the type cone of the support $\tau$-tilting fan is simplicial. $\square$
**4.7** | **Application to other examples of categories**

In this section, we give two more examples of families of categories to which Theorem 4.13 and Corollary 4.14 apply.

**4.7.1** | **2-Calabi–Yau triangulated categories**

Let $\mathcal{C}$ be a triangulated category with shift functor $\Sigma$. We assume moreover that $\mathcal{C}$ is $\mathbb{K}$-linear, Krull–Schmidt, Hom-finite, 2-Calabi–Yau, with a cluster-tilting object $T$. Then $\mathcal{C}$ satisfies all the assumptions of Subsection 4.1 for Subsection 4.3 but, possibly, assumption (1).

We fix this issue by considering a relative extriangulated structure. Let $\mathbb{E}_T$ be the additive sub-bifunctor of $\mathcal{C}(-, \Sigma)$ formed by those morphisms that factor through add $\Sigma T$. Then $(\mathcal{C}, \mathbb{E}_T)$ becomes extriangulated by either [47, Propositions 3.16 or 3.19] (this can also be checked directly).
The triangles of $\mathcal{C}$ are precisely the triangles $X \xrightarrow{f} Y \xrightarrow{g} Z \xrightarrow{h}$, for which the morphism $h$ factors through add $\Sigma T$. By construction, Condition (1) is now satisfied. Moreover, the triangles appearing in Conditions (2) and (3) are extrangles for this relative structure, which shows that those two conditions are still satisfied for the extriangulated structure $(\mathcal{C}, \mathbb{E}_T)$.

*Remark 4.55.* The index of an object $X$ in the triangulated category $\mathcal{C}$, coincide with its class in the Grothendieck group of $\mathcal{C}$ for the relative extriangulated structure given by $\mathbb{E}_T$.

*Remark 4.56.* The main results of Subsection 3.2 can thus be viewed as specific cases of the results of Section 4: Theorem 3.8 and Corollary 3.9 are consequences of Theorem 4.13 and Corollary 4.14.

**4.7.2** | **Objects presented by a subcategory**

We now describe a setup satisfying the assumptions of Subsection 4.1 that, at the same time, subsumes all examples given so far, while being easily checked in practice.

Let $(\mathcal{E}, \mathbb{E}, \mathfrak{s})$ be an extriangulated category (for example, a small exact category or a triangulated category), and let $\mathcal{T}$ be an essentially small subcategory of $\mathcal{E}$.

For any $X, Y \in \mathcal{E}$, let $\mathbb{E}_{\mathcal{T}}(X, Y)$ be the subset of $\mathbb{E}(X, Y)$ consisting of those $\delta$ that satisfy $f^*\delta = 0$, for any $T \in \mathcal{T}$ and any $T \xrightarrow{f} X$. It is easily seen that $\mathbb{E}_{\mathcal{T}}$ is an additive subfunctor of $\mathbb{E}$. Consider $(\mathcal{E}, \mathbb{E}_{\mathcal{T}}, \mathfrak{s}_{\mathcal{T}})$ endowed with the restriction of the additive realization $\mathfrak{s}$. As the deflations in $(\mathcal{E}, \mathbb{E}_{\mathcal{T}}, \mathfrak{s}_{\mathcal{T}})$ are precisely the deflations in $(\mathcal{E}, \mathbb{E}, \mathfrak{s})$ that are $\mathcal{T}$-epic, they are closed under composition. We can thus apply [47, Proposition 3.16] to obtain that $(\mathcal{E}, \mathbb{E}_{\mathcal{T}}, \mathfrak{s}_{\mathcal{T}})$ is an extriangulated category. With that relative structure, the objects in $\mathcal{T}$ become projective (alternatively, one can directly apply [47, Proposition 3.19]). Define $\mathcal{C}$ to be the full subcategory of $\mathcal{E}$ whose objects $X$ admit an $\mathfrak{s}_{\mathcal{T}}$-conflation $T_1 \twoheadrightarrow T_0 \twoheadrightarrow X$.

**Lemma 4.57.** The full subcategory $\mathcal{C}$ is extension-closed in $(\mathcal{E}, \mathbb{E}_{\mathcal{T}}, \mathfrak{s}_{\mathcal{T}})$, and thus inherits an extriangulated structure.

*Proof.* This follows from the proof of Lemma 4.38. □
**Notation 4.58.** Write $\mathcal{T}^{\perp}$ for the full subcategory of $\mathcal{E}$ whose objects are those objects $X \in \mathcal{E}$ that satisfy $\mathcal{E}(T, X) = 0$, for any $T \in \mathcal{T}$.

**Proposition 4.59.** Assume that, for any $T \in \mathcal{T}$, there is an $\mathfrak{s}$-conflation $T \to 0 \to X_T$ in $\mathcal{E}$ with $X_T \in \mathcal{T}^{\perp}$. Then the extriangulated category $\mathcal{C}$ satisfies the assumptions (1) to (3) of Subsection 4.1.

*Proof.* For any $T \in \mathcal{T}$ the existence of an extrtriangle $0 \to T \to T \xrightarrow{0} T$ shows that $\mathcal{T}$ is a full subcategory of $\mathcal{C}$. By Lemma 4.57, the category $\mathcal{C}$ is extriangulated and, by definition of this extriangulated structure, the objects in $\mathcal{T}$ are projective in $\mathcal{C}$. Assumption (3) is satisfied by the definition of $\mathcal{C}$, and assumption (2) is equivalent to that of the proposition. $\square$

*Remark 4.60.* The categories considered in Sections 4.6 and 4.7.1 all fall into the setting of Proposition 4.59.

## 4.8 | Examples

We conclude with a couple of examples for two associative algebras.

**Example 4.61 (A gentle algebra).** Consider the algebra $\Lambda$ given the path algebra of the quiver

$$1 \xrightarrow{\alpha} 2 \xrightarrow{\beta} 3 \xrightarrow{\gamma} 4$$

modulo the relation $\gamma\beta\alpha = 0$. The Auslander-Reiten quiver of the extriangulated category $K^{[-1,0]}(\text{proj } \Lambda)$ is depicted below.

<figure><img src="image_37.png" /></figure>

The Auslander-Reiten quiver of the module category is obtained by taking the quotient by the ideal generated by the shifted projective modules.

<figure><img src="image_38.png" /></figure>

The conditions of Theorem 4.46 are satisfied.
**Example 4.62** (Preprojective algebra of type $A_3$). Let $\Lambda$ be the preprojective algebra of type $A_3$: it is the path algebra of the quiver

$$1 \xrightarrow[\bar{\alpha}]{\alpha} 2 \xrightarrow[\bar{\beta}]{\beta} 3$$

modulo the relations $\alpha\bar{\alpha} = 0$, $\bar{\beta}\beta = 0$ and $\bar{\alpha}\alpha + \beta\bar{\beta} = 0$. The Auslander-Reiten quiver of the extriangulated category $K^{[-1,0]}(\text{proj } \Lambda)$ is depicted below. Note that it is periodic.

<figure><img src="image_39.png" /></figure>

The Auslander-Reiten quiver of the module category $\text{mod}(\Lambda)$ can be obtained by applying the functor $\text{Hom}(\Lambda, -)$.

<figure><img src="image_40.png" /></figure>

One can check directly that the conditions of Theorem 4.46 is not satisfied. For example, the module $\Lambda^{(2)}$ is $\tau$-rigid because $\tau(\Lambda^{(2)}) = \Lambda^{(3)}$, while $\text{Hom}_{\Lambda}(\Lambda^{(2)}, \tau^2(\Lambda^{(2)})) = \text{Hom}_{\Lambda}(\Lambda^{(2)}, \Lambda^{(1)}) \neq 0$.

## ACKNOWLEDGEMENTS

We thank H. Thomas for comments on a preliminary version of the paper and for informing us about the upcoming extended version of [17]. We are grateful to C. Stump for reporting a typo in Example 2.18. We also thank an anonymous referee for relevant comments. Yann Palu, Vincent Pilaud, and Pierre-Guy Plamondon were partially supported by the French ANR Grant SC3A (15 CE40 0004 01). Arnau Padrol and Vincent Pilaud were partially supported by the French ANR Grant CAPPS (17 CE40 0018).

## JOURNAL INFORMATION

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