# Support $\tau_{2}$-tilting and 2-torsion pairs

Jordan McMahon  unaffiliated<EMAIL_ADDRESS>
###### Abstract.

The theory of $\tau$-tilting was introduced by Adachi–Iyama–Reiten; one of the
main results is a bijection between support $\tau$-tilting modules and torsion
classes. We are able to generalise this result in the context of the higher
Auslander–Reiten theory of Iyama. For a finite-dimensional algebra $A$ with
2-cluster-tilting subcategory $\mathcal{C}\subseteq\mathrm{mod}A$, we are able
to find a correspondence between support $\tau_{2}$-tilting $A$-modules and
torsion pairs in $\mathcal{C}$ satisfying an additional functorial finiteness
condition.

##  1\. Introduction

Support-tilting modules were first studied by Ringel [31] and connected to
torsion classes and cluster algebras by Ingalls–Thomas [14]. Since support-
tilting modules are able to capture the behaviour of clusters, this led to
further study. Significantly, Adachi, Iyama and Reiten [1] introduced support
$\tau$-tilting modules, and were able to find a bijection, for an finite-
dimensional algebra $A$, between support $\tau$-tilting $A$-modules and
functorially-finite torsion classes in $\mathrm{mod}A$. See also [18] for a
survey of $\tau$-tilting theory, which has seen much activity in recent years
[4], [8], [9], [17], [19], [28], [29] as well as its generalisations such as
in silting theory [2], [3], [7]. A natural question to ask is whether similar
results are true in the context of higher Auslander–Reiten theory, as
introduced by Iyama in [15], [16], and an active area of research [10], [20],
[22], [23], [24], [30]. As the name suggests, higher Auslander–Reiten theory
has connections to higher-dimensional geometry and topology [11], [12], [25],
[27], [32] and is hence a natural generalisation. The result we find is the
following:

######  Theorem 1.1.

Let $A$ be a finite-dimensional algebra with $2$-cluster-tilting subcategory
$\mathcal{C}\subseteq\mathrm{mod}A$. Then there is a correspondence between
support $\tau_{2}$-tilting $A$-modules and 2-functorially finite torsion pairs
in $\mathcal{C}$.

##  2\. Background and Notation

Consider a finite-dimensional algebra
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$
over a field $K$, and fix a positive integer $d$. We will assume that
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$
is of the form $KQ/I$, where $KQ$ is the path algebra over some quiver $Q$ and
$I$ is an admissible ideal of $KQ$. An
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$-module
will mean a finitely-generated left
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$-module;
by
$\mathrm{mod}{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$
we denote the category of
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$-modules.
The functor $D=\mathrm{Hom}_{K}(-,K)$ defines a duality. Let $\mathrm{add}M$
be the full subcategory of
$\mathrm{mod}{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$
composed of all
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$-modules
isomorphic to direct summands of finite direct sums of copies of $M$. For an
$A$-module $M$, the right annihilator of $M$ is the two-sided ideal
$\mathrm{ann}M:=\\{a\in A|aM=0\\};$ for a class of modules $\mathcal{X}$ we
set $\mathrm{ann}\mathcal{X}:=\\{a\in A|aX=0\ \forall\ X\in\mathcal{X}\\}.$
For an $A$-module $M$, define $\mathrm{Sub}M:=\\{N\in\mathrm{mod}A|\exists\
\text{injection}\ N\hookrightarrow M\\};$ for a class of modules $\mathcal{X}$
we set $\mathrm{Sub}\mathcal{X}:=\\{N\in\mathcal{X}|\exists X\in\mathcal{X},\
\text{injection}\ \ N\hookrightarrow X\\}.$ Define $\mathrm{Fac}M$ and
$\mathrm{Fac}\mathcal{X}$ dually.

###  2.1. Higher cluster-tilting subcategories

Define $\tau_{d}:=\tau\Omega^{d-1}$ to be the _$d$ -Auslander–Reiten
translation_ and
$\tau_{d}^{-}:={\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}\tau^{-}\Omega^{-(d-1)}}$
to be the _inverse $d$-Auslander–Reiten translation_. A subcategory
$\mathcal{C}$ of $\mathrm{mod}A$ is _precovering_ or _contravariantly finite_
if for any $M\in\mathrm{mod}\Lambda$ there is an object $C_{M}\in\mathcal{C}$
and a morphism $f:C_{M}\rightarrow M$ such that $\mathrm{Hom}(C,-)$ is exact
on the sequence

${C_{M}}$${M}$${0}$

for all $C\in\mathcal{C}$. The module $C_{M}$ is said to be a _right
$\mathcal{C}$-approximation_. The dual notion of precovering is
_preenveloping_ or _covariantly finite_. A subcategory $\mathcal{C}$ that is
both precovering and preenveloping is called _functorially finite_.

######  Definition 2.1.

[16, Definition 2.2] For a finite-dimensional algebra
${\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}$,
a subcategory
$\mathcal{C}\subseteq\mathrm{mod}({\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A})$
is a _$d$ -cluster-tilting subcategory_ if it satisfies the following
conditions:

$\displaystyle\mathcal{C}$
$\displaystyle=\\{X\in\mathcal{C}\mid\mathrm{Ext}^{i}_{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}}(M,X)=0\
\forall\ 0<i<d\\}.$ $\displaystyle\mathcal{C}$
$\displaystyle=\\{X\in\mathcal{C}\mid\mathrm{Ext}^{i}_{{\color[rgb]{0,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{0,0,0}\pgfsys@color@gray@stroke{0}\pgfsys@color@gray@fill{0}A}}(X,M)=0\
\forall\ 0<i<d\\}.$

A _right $\mathcal{C}$-resolution_ is a sequence

${\cdots}$${C_{1}}$${C_{0}}$${M}$${0}$

with $C_{i}\in\mathcal{C}$ for each $i$, and which becomes exact under
$\mathrm{Hom}_{A}(C,-)$ for each $C\in\mathcal{C}$. Define a _left
$\mathcal{C}$-resolution_ dually.

######  Theorem 2.2.

[16, Theorem 3.6.1] Let $\mathcal{C}\subseteq\mathrm{mod}A$ be a $d$-cluster-
tilting subcategory. Then

  1. (1)

Any $M\in\mathrm{mod}A$ has a right $\mathcal{C}$-resolution

${0}$${C_{d-1}}$${\cdots}$${C_{1}}$${C_{0}}$${M}$${0.}$

  2. (2)

Any $M\in\mathrm{mod}A$ has a left $\mathcal{C}$-resolution

${0}$${M}$${C_{0}}$${C_{1}}$${\cdots}$${C_{d-1}}$${0.}$

Recall the stable module category $\underline{\mathrm{mod}}A$ is full
subcategory of $\mathrm{mod}A$ obtained by factoring out all morphisms that
factor through a projective module.

######  Theorem 2.3.

[16, Theorem 1.5] Let $A$ be a finite-dimensional algebra. Then:

  * •

If $\mathrm{Ext}^{i}_{A}(M,A)=0$ for all $0<i<d$, then
$\mathrm{Ext}^{i}_{A}(M,N)\cong D\mathrm{Ext}^{d-i}_{A}(N,\tau_{d}M)$ and
$\underline{\mathrm{Hom}}_{A}(M,N)\cong D\mathrm{Ext}^{d}_{A}(N,\tau_{d}M)$
for all $M\in\mathrm{mod}A$ and all $0<i<d$.

  * •

If $\mathrm{Ext}^{i}_{A}(DA,N)=0$ for all $0<i<d$, then
$\mathrm{Ext}^{i}_{A}(M,N)\cong D\mathrm{Ext}^{d-i}_{A}(\tau_{d}^{-}N,M)$ and
$\overline{\mathrm{Hom}}_{A}(M,N)\cong D\mathrm{Ext}^{d}_{A}(\tau_{d}^{-}N,M)$
for all $N\in\mathrm{mod}A$ for all $0<i<d$.

We may now generalise a result of Auslander–Smalø, the proof of which is
unchanged apart from indices.

######  Proposition 2.4 (c.f. Proposition 5.8 of [6]).

Let $A$ be a finite-dimensional algebra. Then for $X,Y\in\mathrm{mod}A$ the
following are equivalent:

  1. (i)

$\mathrm{Hom}_{A}(\tau_{2}^{-}Y,X)=0$.

  2. (ii)

$\underline{\mathrm{Hom}}_{A}(\tau_{2}^{-}Y,\mathrm{Sub}X)=0$ .

  3. (iii)

$\mathrm{Ext}^{2}_{A}(\mathrm{Sub}X,Y)=0$.

###### Proof.

Firstly, statements $(ii)$ and $(iii)$ are equivalent by Theorem 2.3 and
statement $(i)$ trivially implies $(ii)$. It remains to show $(ii)$ implies
$(i)$; we prove by contradiction. So suppose there is a morphism
$f:\tau_{2}^{-}Y\rightarrow X$ is a non-zero morphism with image
$\mathrm{Im}f\in\mathrm{Sub}X$, and such that the induced morphism
$f^{\prime}:\tau_{2}^{-}Y\twoheadrightarrow\mathrm{Im}f$ factors through a
projective module $P$. Since $f^{\prime}$ is surjective, we may assume $P$ is
the projective cover of $\mathrm{Im}f$. Let $g:\tau_{2}^{-}M\rightarrow P$ be
any morphism; since $\tau_{2}^{-}Y$ has no non-trivial projective summands, we
have $g(\tau_{2}^{-}M)\subset\mathrm{rad}P$, hence any composition
$\tau_{2}^{-}Y\rightarrow P\rightarrow\mathrm{Im}f$ is not an epimorphism.
Therefore the image of $f^{\prime}$ in
$\underline{\mathrm{Hom}}_{A}(\tau_{2}^{-}Y,\mathrm{Im}f)=0$ is not zero.
Hence $\mathrm{Hom}_{A}(\tau_{2}^{-}Y,X)=0$ if and only if
$\underline{\mathrm{Hom}}_{A}(\tau_{2}^{-}Y,\mathrm{Sub}X)=0$. ∎

Key homological tools for higher cluster-tilting categories are $d$-pushouts
and $d$-pullbacks, constructed as follows.

######  Proposition 2.5.

[21, Proposition 3.8] Let $A$ be a finite-dimensional algebra and
$\mathcal{C}\subseteq\mathrm{mod}A$. For any $d$-exact sequence in
$\mathcal{C}$

$0\rightarrow Y_{0}\rightarrow Y_{1}\rightarrow\cdots\rightarrow
Y_{d+1}\rightarrow 0$

and any morphism $f:X_{d+1}\rightarrow Y_{d+1}$ there exists a commutative
diagram in $\mathcal{C}$:

${0}$${Y_{0}}$${X_{1}}$${\cdots}$${X_{d}}$${X_{d+1}}$${0}$${0}$${Y_{0}}$${Y_{1}}$${\cdots}$${Y_{d}}$${Y_{d+1}}$${0}$$\scriptstyle{f}$

such that there is an induced $d$-exact sequence

$0\rightarrow X_{1}\rightarrow X_{2}\oplus Y_{1}\rightarrow\cdots\rightarrow
X_{d+1}\oplus Y_{d}\rightarrow Y_{d+1}\rightarrow 0$

The construction of this commutative diagram is obtained inductively as
follows: $X_{d}$ is defined to be the right $\mathcal{C}$-approximation of the
pullback of $((X_{d+1}\oplus Y_{d})\rightarrow Y_{d+1})$. Subsequently
$X_{d-1}$ is defined to be the right $\mathcal{C}$-approximation of the
pullback of $((X_{d}\oplus Y_{d-1})\rightarrow Y_{d})$. This continues until
$X_{0}=Y_{0}$ is reached.

######  Remark 2.6.

In this setting, it is actually immaterial whether $Y_{d+1}$ is in
$\mathcal{C}$ or not: Theorem 2.2 ensures that the so-called $d$-kernel of the
morphism $(X_{d+1}\oplus Y_{d})\rightarrow Y_{d+1})$ exists. Equally, it is
not important whether $Y_{0}\in\mathcal{C}$.

A technical result is the following:

######  Lemma 2.7.

Consider a module $X\in\mathcal{C}$ and two $d$-exact sequences:

$0\rightarrow L\rightarrow M\rightarrow N\rightarrow X\rightarrow 0,$
$0\rightarrow L^{\prime}\rightarrow M^{\prime}\rightarrow
N^{\prime}\rightarrow X\rightarrow 0.$

Then there exist $P,Q\in\mathcal{C}$ with no common non-zero summands and a
commutative diagram with exact rows and columns:

${0}$${0}$${L}$${L}$${Q}$${Q\oplus M}$${M}$${0}$${0}$${L^{\prime}}$${Q\oplus
M^{\prime}}$${P}$${N}$${0}$${0}$${L^{\prime}}$${M^{\prime}}$${N^{\prime}}$${X}$${0}$${0}$${0}$${0}$

###### Proof.

Just as in the above method for $d$-pullbacks, we may form the following
commutative diagram with exact rows and columns by taking the appropriate
right $\mathcal{C}$-approximations of pullbacks: $P,Q,R,S$

${0}$${0}$${L}$${L}$${Q}$${S}$${M}$${0}$${0}$${L^{\prime}}$${R}$${P}$${N}$${0}$${0}$${L^{\prime}}$${M^{\prime}}$${N^{\prime}}$${X}$${0}$${0}$${0}$${0}$

It suffices to show the surjection $R\twoheadrightarrow M^{\prime}$ is split
(by symmetry, also $S\twoheadrightarrow M$ splits). By the pullback property
of $P$, we find $M^{\prime}\rightarrow N^{\prime}$ factors through $P$. But
the pullback property of $R$ then implies both $M^{\prime}\rightarrow P$
factors through $R$ and that $M^{\prime}\rightarrow R\rightarrow
M^{\prime}\cong\mathrm{id}$. Hence $R\cong M^{\prime}\oplus Q$.

Finally, suppose $P$ and $Q$ have a common summand $Y$. Then any morphism
$Y\rightarrow N^{\prime}$ must factor through $M^{\prime}$. But as above,
$M^{\prime}\rightarrow N^{\prime}$ factors through $P$. Hence $Y$ must be a
summand of $M^{\prime}$. This is a contradiction, since $M^{\prime}$ is
already included a summand of $Q\oplus M^{\prime}$, and hence cannot be a
summand of $Q$. This finishes the proof. ∎

A second technical result will prove similarly useful.

######  Lemma 2.8.

In the setting of Lemma 2.7, in case there exists $P^{\prime}\in\mathcal{C}$
such that $P\cong P^{\prime}\oplus M^{\prime}$, then there is a commutative
diagram with exact rows and columns

${0}$${0}$${0}$${L}$${M}$${0}$${L^{\prime}}$${Q}$${P^{\prime}}$${N}$${0}$${0}$${L^{\prime}}$${M^{\prime}}$${N^{\prime}}$${X}$${0}$${0}$

###### Proof.

In this case we know that the morphism $L^{\prime}\rightarrow M^{\prime}$
factors through $Q$. Since we are dealing with a $2$-pullback diagram, the
$2$-exact sequence:

$0\rightarrow L\rightarrow M\oplus Q\rightarrow P^{\prime}\oplus
M^{\prime}\rightarrow N^{\prime}\rightarrow 0$

may simply be transcribed into a commutative diagram as stated. ∎

##  3\. Main results

In this section let $\mathcal{C}\subset\mathrm{mod}A$ be a fixed $2$-cluster-
tilting subcategory.

######  Definition 3.1.

Define a subcategory $\mathcal{X}\subseteq\mathcal{C}$ to be _$2$
-contravariantly finite in $\mathcal{C}$_ if for any $M\in\mathcal{C}$ there
exists a right $\mathcal{X}$-approximation $X_{1}\rightarrow M$ and object
$X_{2}\in\mathcal{X}$ and a sequence $X_{2}\rightarrow X_{1}\rightarrow M$ on
which $\mathrm{Hom}(C,-)$ is exact for all $C\in\mathcal{C}$.

Dually, define a subcategory $\mathcal{X}\subseteq\mathcal{C}$ to be _$2$
-covariantly finite in $\mathcal{C}$_ if for any $M\in\mathcal{C}$ there
exists a left $\mathcal{X}$-approximation $M\rightarrow X_{1}$ and object
$X_{2}\in\mathcal{X}$ and a sequence $M\rightarrow X_{1}\rightarrow X_{2}$ on
which $\mathrm{Hom}(-,C)$ is exact for all $C\in\mathcal{C}$. A subcategory
$\mathcal{X}\subseteq\mathcal{C}$ will be said to be _$2$ -functorially finite
in $\mathcal{C}$_ if it is both $2$-contravariantly finite in $\mathcal{C}$
and $2$-covariantly finite in $\mathcal{C}$.

######  Lemma 3.2.

Let $\mathcal{X}\subseteq\mathcal{C}$ be functorially finite. Then
$\mathcal{X}$ is 2-covariantly finite in $\mathcal{C}$ if and only if it is
2-contravariantly finite in $\mathcal{C}$.

###### Proof.

It suffices to show any subcategory $\mathcal{X}$ 2-contravariantly finite in
$\mathcal{C}$ is also $2$-covariantly finite in $\mathcal{C}$. So suppose
$\mathcal{X}$ is 2-contravariantly finite in $\mathcal{C}$. Let
$f:M\rightarrow X_{0}$ be a left $\mathcal{X}$ approximation and let
$X_{1}^{\prime}\rightarrow X_{0}^{\prime}\rightarrow\mathrm{coker}f$ be a
right $\mathcal{X}$-approximation. We have that
$X_{0}\rightarrow\mathrm{coker}f$ factors through $X_{1}^{\prime}$, and since
$X_{0}^{\prime}\not\cong\mathrm{coker}f$, this induces an additional non-zero
morphism $M\rightarrow X_{0}^{\prime}$. The sequence $M\rightarrow
X_{0}^{\prime}\rightarrow\mathrm{coker}f$ is now zero, and hence induces a
morphism $g:M\rightarrow X_{1}^{\prime}$. Since $f:M\rightarrow X_{0}$ is a
left $\mathcal{X}$-approximation, we have that $g$ factors through $X_{0}$.
But this means $X_{0}\rightarrow\mathrm{coker}f$ factors through
$X_{1}^{\prime}\rightarrow X_{0}^{\prime}$ and is hence zero, a contradiction.
This is illustrated below:

${M}$${X_{0}}$${\mathrm{coker}f}$${X_{1}^{\prime}}$${X_{0}^{\prime}}$${\mathrm{coker}f}$$\scriptstyle{f}$$\scriptstyle{g}$

∎

Recall that for a finite-dimensional algebra $A$, then a _torsion pair in
$\mathrm{mod}A$_ consists of two subcategories $(\mathcal{T},\mathcal{F})$
such that

  * •

$\mathrm{Hom}_{A}(T,F)=0$ for all $T\in\mathcal{T}$, $F\in\mathcal{F}$.

  * •

$\mathcal{T}=\\{X\in\mathrm{mod}A|\mathrm{Hom}_{A}(X,F)=0\ \forall\
F\in\mathcal{F}\\}$.

  * •

$\mathcal{F}=\\{Y\in\mathrm{mod}A|\mathrm{Hom}_{A}(T,Y)=0\ \forall\
T\in\mathcal{T}\\}$.

We may now define one of our primary objects of study.

######  Definition 3.3.

A _2-functorially-finite torsion pair in $\mathcal{C}$_ consists of two
subcategories $(\mathcal{T},\mathcal{F})$ each $2$-functorially finite in
$\mathcal{C}$ and such that

  * •

$\mathrm{Hom}_{A}(T,F)=0$ for all $T\in\mathcal{T}$, $F\in\mathcal{F}$.

  * •

$\mathcal{T}=\\{X\in\mathcal{C}|\mathrm{Hom}_{A}(X,F)=0\ \forall\
F\in\mathcal{F}\\}$.

  * •

$\mathcal{F}=\\{Y\in\mathcal{C}|\mathrm{Hom}_{A}(T,Y)=0\ \forall\
T\in\mathcal{T}\\}$.

For a $2$-functorially-finite torsion pair $(\mathcal{T},\mathcal{F})$ in
$\mathcal{C}$, we say $\mathcal{T}$ is a _torsion class_ and $\mathcal{F}$ a
_torsion-free class_.

######  Lemma 3.4.

For any 2-functorially-finite torsion pair $(\mathcal{T},\mathcal{F})$ in
$\mathcal{C}$ then $(\mathrm{Fac}\mathcal{T},\mathrm{Sub}\mathcal{F})$ is a
torsion pair in $\mathrm{mod}A$.

###### Proof.

Suppose $(\mathcal{T},\mathcal{F})$ is a 2-functorially-finite torsion pair in
$\mathcal{C}$. Clearly any $X\in\mathrm{mod}A$ satisfies that
$\mathrm{Hom}_{A}(\mathrm{Fac}\mathcal{T},X)=0$ if and only if
$\mathrm{Hom}_{A}(\mathcal{T},X)=0$. Let $f:X\rightarrow C_{X}$ be a left
$\mathcal{C}$-approximation of $X$, moreover $f$ must be injective. So
$\mathrm{Hom}_{A}(\mathcal{T},X)=0$ if and only if
$\mathrm{Hom}_{A}(\mathcal{T},C_{X})=0$ and hence whenever
$C_{X}\in\mathcal{F}$ and $X\in\mathrm{Sub}\mathcal{F}$. The result follows. ∎

######  Proposition 3.5.

Let $A$ be a finite-dimensional algebra and
$\mathcal{C}\subseteq\mathrm{mod}A$ a 2-cluster-tilting subcategory. Let
$\mathcal{T},\mathcal{F}\subseteq\mathcal{C}$ be 2-functorially-finite
subcategories in $\mathcal{C}$. The following are equivalent.

  1. (i)

$(\mathcal{T},\mathcal{F})$ is a 2-functorially-finite torsion pair in
$\mathcal{C}$.

  2. (ii)

For every $M\in\mathcal{C}$ there is an exact sequence

$T_{M}\rightarrow M\rightarrow F_{M}$

such that $M\rightarrow F_{M}$ is a left $\mathcal{F}$-approximation and
$T_{M}\rightarrow M$ is a right $\mathcal{T}$-approximation.

###### Proof.

(i)$\implies$ (ii): Given $M\in\mathcal{C}$, and a 2-functorially-finite
torsion pair $(\mathcal{T},\mathcal{F})$ in $\mathcal{C}$, if follows from
Lemma 3.4 that $(\mathrm{Fac}\mathcal{T},\mathrm{Sub}\mathcal{F})$ is a
torsion pair in $\mathrm{mod}A$. A classical property of torsion pairs (see
for example [5, Proposition V1.1.5]) there is an exact sequence in
$\mathrm{mod}A$: $0\rightarrow T\rightarrow M\rightarrow F\rightarrow 0$ for
some $T\in\mathrm{Fac}\mathcal{T}$ and $F\in\mathrm{Sub}\mathcal{F}$. By
taking appropriate approximations we obtain the result.

(ii)$\implies$ (i) This is trivial. ∎

We may characterise torsion classes in $\mathcal{C}$.

######  Proposition 3.6.

For a 2-functorially-finite subcategory $\mathcal{T}\subseteq\mathcal{C}$, the
following are equivalent:

  1. (i)

There is an inclusion
$(\mathrm{Fac}\mathcal{T}\cap\mathcal{C})\subseteq\mathcal{T}$ and for every
$d$-exact sequence in $\mathcal{C}$ with $T_{0},T_{3}\in\mathcal{T}$:

$0\rightarrow T_{0}\rightarrow X\rightarrow Y\rightarrow T_{3}\rightarrow 0$

there exists a $d$-pushout diagram with exact rows:

${0}$${T_{0}^{\prime}}$${T_{1}^{\prime}}$${T_{2}^{\prime}}$${T_{3}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$

such that $T_{0}^{\prime},T_{1}^{\prime},T_{2}^{\prime}\in\mathcal{T}$.

  2. (ii)

$\mathcal{T}$ is the torsion part of a 2-functorially-finite torsion pair in
$\mathcal{C}$.

###### Proof.

(i)$\implies$ (ii) For an arbitrary $M\in\mathcal{C}$, let $g:T_{M}\rightarrow
M$ be a right $\mathcal{T}$-approximation. By Proposition 3.5 it suffices to
show that $\mathrm{Hom}_{A}(\mathcal{T},\mathrm{coker}g)=0$. So assume there
exists a morphism $T_{3}\rightarrow\mathrm{coker}g$ for some
$T_{3}\in\mathcal{T}$. Then there exist $X,Y\in\mathcal{C}$ and a pullback
diagram by Remark 2.6

${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$${0}$${T_{0}}$${T_{M}}$${M}$${\mathrm{coker}g}$${0}$

By assumption there exist some
$T_{0}^{\prime},T_{1}^{\prime},T_{2}^{\prime}\in\mathcal{T}$ making the
following diagram commute:

${0}$${T_{0}^{\prime}}$${T_{1}^{\prime}}$${T_{2}^{\prime}}$${T_{3}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$${0}$${T_{0}}$${T_{M}}$${M}$${\mathrm{coker}g}$${0}$

Since the composition $T_{2}^{\prime}\twoheadrightarrow
T_{3}\rightarrow\mathrm{coker}g$ is non-zero there exists a non-zero morphism
$T_{2}^{\prime}\rightarrow M$, which by assumption factors through $T_{M}$.
But this implies the above non-zero morphism
$T_{2}^{\prime}\rightarrow\mathrm{coker}g$ factors through $T_{M}\rightarrow
M\rightarrow\mathrm{coker}g$, a contradiction.

(ii)$\implies$ (i) Clearly $\mathcal{T}$ is closed under factor modules in
$\mathcal{C}$. Now let

$(*):\ 0\rightarrow T_{0}\rightarrow X\rightarrow Y\rightarrow
T_{3}\rightarrow 0$

be $d$-exact with $T_{0},T_{3}\in\mathcal{T}$. By Proposition 3.5 there exist
$T_{X},T_{Y}\in\mathcal{T}$ and $F_{X},F_{Y}\in\mathcal{F}$ and exact
sequences

$T_{X}\rightarrow X\rightarrow F_{X},$ $T_{Y}\rightarrow Y\rightarrow F_{Y}.$

Applying $\mathrm{Hom}_{A}(-,F)$ to $(*)$ implies an isomorphism
$\mathrm{Hom}_{A}(Y,F)\cong\mathrm{Hom}_{A}(X,F)$ for any
$F\in\mathrm{Sub}\mathcal{F}$; this implies $\mathrm{Im}(X\rightarrow
F_{X})\cong\mathrm{Im}(Y\rightarrow F_{Y})$. We are in the situation of Lemma
2.8, so there exist $T_{X}^{\prime},T_{Y}^{\prime},T^{\prime}\in\mathcal{T}$
such that there there is a pullback diagram

${0}$${0}$${0}$${T^{\prime}_{X}}$${T^{\prime}_{Y}}$${0}$${T_{0}}$${T_{X}}$${T_{Y}}$${T^{\prime}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$${0}$

This induces the required $d$-pushout diagram

${0}$${T_{0}\oplus T_{X}^{\prime}}$${T_{X}\oplus
T_{Y}^{\prime}}$${T_{Y}}$${T_{3}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$

since we know $0\rightarrow T_{X}^{\prime}\rightarrow T_{X}\oplus
T_{Y}^{\prime}\rightarrow X\oplus T_{Y}\rightarrow Y\rightarrow 0$ is exact. ∎

Recall an object $T\in\mathcal{C}$ is $\tau$-tilting [1] if

  * •

$\mathrm{Hom}_{A}(T,\tau T)=0$

  * •

$|T|=|A|$, where $|\cdot|$ denotes the number of indecomposable summands.

If $T$ is $\tau$-tilting as an $A/\langle e\rangle$-module for some idempotent
$e$, then $T$ is _support $\tau$-tilting_. This can be generalised as follows,
using the generalised tilting theory of [13], [26]: recall that an $A$-module
$T$ is a _$d$ -tilting module_ if:

  1. (1)

$\mathrm{proj.dim}(T)\leq d$.

  2. (2)

$\mathrm{Ext}^{i}_{A}(T,T)=0$ for all $0<i\leq d$.

  3. (3)

there exists an exact sequence

$0\rightarrow A\rightarrow T_{0}\rightarrow T_{1}\rightarrow\cdots\rightarrow
T_{d}\rightarrow 0$

where $T_{0},\ldots,T_{d}\in\mathrm{add}T$.

We may now define the latter of our primary objects of study.

######  Definition 3.7.

An object $T\in\mathcal{C}$ is _$\tau_{2}$ -tilting_ if

  * •

$\mathrm{Hom}_{A}(T,\tau_{2}T)=0$

  * •

There exists an exact sequence

$0\rightarrow A\rightarrow T_{0}\rightarrow T_{1}\rightarrow T_{2}\rightarrow
0$

such that $T_{0},T_{1},T_{2}\in\mathrm{add}T$.

If $T$ is $\tau_{2}$-tilting as an $A/\langle e\rangle$-module for some
idempotent $e$, then we say that $T$ is _support $\tau_{2}$-tilting_.

We now show that support $\tau_{2}$-tilting modules are indeed $2$-tilting.
Recall that an $A$-module $T$ is _faithful_ if its right annihilator
$\mathrm{ann}T$ vanishes.

######  Lemma 3.8 (c.f. Lemma IV.2.7 of [5]; c.f Lemma VII.5.1 of [5]; c.f.
Proposition 2.2(b) of [1]).

Let $A$ be a finite-dimensional algebra with 2-cluster-tilting subcategory
$\mathcal{C}\subseteq\mathrm{mod}A$. Then

  1. (i)

For any $T\in\mathcal{C}$, $\mathrm{proj.dim}T\leq 2$ if and only if
$\mathrm{Hom}_{A}(DA,\tau_{2}T)=0$.

  2. (ii)

Let $T\in\mathcal{C}$ be a faithful $A$-module. If
$\mathrm{Hom}_{A}(T,\tau_{2}T)=0$, then $\mathrm{proj.dim}T\leq 2$.

  3. (iii)

Any $\tau_{2}$-tilting $A$-module $T$ is a tilting $(A/\mathrm{ann}T)$-module.

###### Proof.

$(i)$: Apply the left exact functor $\nu^{-1}=\mathrm{Hom}_{A}(DA,-)$ to the
exact sequence

$0\rightarrow\tau_{2}M\rightarrow\nu P_{2}\rightarrow\nu P_{1}\rightarrow\nu
P_{0}\rightarrow\nu M\rightarrow 0$

to obtain an exact sequence

$0\rightarrow\nu^{-1}\tau_{2}M\rightarrow\nu^{-1}\nu
P_{2}\rightarrow\nu^{-1}\nu P_{1}\rightarrow\nu^{-1}\nu
P_{0}\rightarrow\nu^{-1}\nu M\rightarrow 0$

It follows that $\mathrm{Hom}_{A}(DA,\tau_{2}M)=\nu^{-1}\tau_{2}M$ vanishes if
and only if $\mathrm{proj.dim}M\leq 2$.

$(ii)$: It is known (see [5, V1.2.2]) that an $A$-module $T$ is faithful if
and only if $DA$ is generated by $T$. So let $T^{i}\twoheadrightarrow DA$ be a
surjection. Applying the functor $\mathrm{Hom}_{A}(-,\tau_{2}T)$ results in a
monomorphism
$\mathrm{Hom}_{A}(DA,\tau_{2}T)\hookrightarrow\mathrm{Hom}_{A}(T,\tau_{2}T)$.
The result now follows from part (i) above.

$(iii)$ Note that for any idempotent ideal $\langle e\rangle$ of $A$, and any
$M,N\in\mathrm{mod}A$ there is a natural inclusion
$\mathrm{Ext}^{2}_{A/\langle
e\rangle}(M,N)\hookrightarrow\mathrm{Ext}^{2}_{A}(M,N)$. In this case let
$\langle e\rangle:=\mathrm{ann}T$. Since $\mathrm{Hom}_{A}(T,\tau_{2}T)=0$,
Proposition 2.4 implies $\mathrm{Ext}^{2}_{A}(T,\mathrm{Fac}T)=0$, and it
follows that $\mathrm{Ext}^{2}_{A/\langle e\rangle}(T,\mathrm{Fac}T)=0$. By
Proposition 2.4 once more, we have $\mathrm{Hom}_{A/\langle
e\rangle}(T,\tau_{2}T)=0$. Since $T$ is faithful as an
$(A/\mathrm{ann}T)$-module, it follows from part $(ii)$ that
$\mathrm{proj.dim}T\leq d$ and that $T$ is $2$-tilting as an
$(A/\mathrm{ann}T)$-module. ∎

######  Lemma 3.9 (Happel [13]).

Let $A$ be a finite-dimensional algebra and $T$ be a $d$-tilting $A$-module.
Assume that $M\in\mathrm{mod}(A)$ satisfies $\mathrm{Ext}_{A}^{i}(T,M)=0$ for
all $i>0$. Then there exists an exact sequence

$0\rightarrow T_{m}\rightarrow\cdots\rightarrow T_{1}\rightarrow
T_{0}\rightarrow M\rightarrow 0$

such that $T_{j}\in\mathrm{add}(T)$ for all $0\leq j\leq m$ and
$m\leq\mathrm{gl.dim}A$.

We may now prove our main result.

######  Theorem 1.1.

An object $T\in\mathcal{C}$ is a support $\tau_{2}$-tilting module if and only
if $\mathrm{Fac}\mathcal{T}\cap\mathcal{C}$ is the torsion part of a
2-functorially-finite torsion pair in $\mathcal{C}$.

###### Proof.

Let $\mathcal{T}$ be a torsion class in $\mathcal{C}$, then by 2-functorial
finiteness and closure under factor modules, there exists an exact sequence
$A\rightarrow T_{0}\rightarrow T_{1}\rightarrow T_{2}\rightarrow 0$ such that
$f:A\rightarrow T_{0}$ is a left $\mathcal{T}$-approximation and
$T_{1},T_{2}\in\mathcal{T}$. Clearly $\mathrm{Im}f\cong
A/\mathrm{Ann}\mathcal{T}$.

Now we need to prove that $T_{i}$ are $\mathrm{Ext}$-projective in
$\mathcal{T}$ for all $0\leq i\leq 2$, i.e. that
$\mathrm{Ext}^{2}_{A}(T_{i},T)=0$ for all $T\in\mathcal{T}$. First suppose
$\mathrm{Ext}_{A}^{2}(T_{0},T)\neq 0$. Then by Proposition 3.6 there exists
some $T_{0}^{\prime}\in\mathcal{T}$ and a non-split surjection
$T_{0}^{\prime}\twoheadrightarrow T_{0}$. But then $f$ must factor through
$T_{0}^{\prime}$, a contradiction.

We next claim that there is a surjection
$\mathrm{Ext}^{2}_{A}(T_{2},T)\twoheadrightarrow\mathrm{Ext}_{A}^{2}(T_{1},T)$
for any $T\in\mathcal{T}$. Let $X:=\mathrm{Im}(T_{0}\rightarrow T_{1})$. This
induces exact sequences

$\mathrm{Ext}^{2}_{A}(T_{2},T)\rightarrow\mathrm{Ext}^{2}_{A}(T_{1},T)\rightarrow\mathrm{Ext}^{2}_{A}(X,T)$
$\mathrm{Ext}^{1}_{A}(A/\mathrm{ann}\mathcal{T},T)\rightarrow\mathrm{Ext}^{2}_{A}(X,T)\rightarrow\mathrm{Ext}^{2}_{A}(T_{0},T)=0.$

Suppose $\mathrm{Ext}^{1}_{A}(A/\mathrm{ann}\mathcal{T},T)\neq 0$ and there
exists a short exact sequence

$0\rightarrow T\rightarrow E\rightarrow A/\mathrm{ann}\mathcal{T}\rightarrow
0.$

But then $E$ must be an $A/\mathrm{ann}\mathcal{T}$-module, since both $T$ and
$A/\mathrm{ann}\mathcal{T}$ are, meaning that this sequence splits, a
contradiction. So
$\mathrm{Ext}^{1}_{A}(A/\mathrm{ann}\mathcal{T},T)=0=\mathrm{Ext}^{2}_{A}(X,T)$,
and there is a surjection
$\mathrm{Ext}^{2}_{A}(T_{2},T)\twoheadrightarrow\mathrm{Ext}_{A}^{2}(T_{1},T)$
as claimed.

Now suppose $\mathrm{Ext}_{A}^{2}(T_{2},T^{\prime})\neq 0$ for some
$T^{\prime}\in\mathcal{T}$. By Proposition 3.6 there exists
$T_{0}^{\prime},T_{1}^{\prime},T_{2}^{\prime}\in\mathcal{T}$ and a $d$-exact
sequence $0\rightarrow T_{0}^{\prime}\rightarrow T_{1}^{\prime}\rightarrow
T_{2}^{\prime}\rightarrow T_{2}\rightarrow 0$. By Lemma 2.7 there exist
$P,Q\in\mathcal{C}$ and a commutative diagram with exact rows and columns:

${0}$${0}$${A/\mathrm{ann}\mathcal{T}}$${A/\mathrm{ann}\mathcal{T}}$${Q}$${Q\oplus
T_{0}}$${T_{0}}$${0}$${T_{0}^{\prime}}$${Q\oplus
T_{1}^{\prime}}$${P}$${T_{1}}$${0}$${0}$${T_{0}^{\prime}}$${T_{1}^{\prime}}$${T_{2}^{\prime}}$${T_{2}}$${0}$${0}$${0}$

Moreover $P,Q\in\mathcal{T}$ by assumption, since they have no common non-zero
summands and we may replace $T_{0}^{\prime}$ if necessary. This implies a
morphism $A\rightarrow Q$, which by assumption factors through $T_{0}$. But
this implies the morphism $T_{0}\rightarrow T_{1}$ also factors through $Q$,
and is hence the zero composition $Q\rightarrow P\rightarrow T_{1}$, a
contradiction.

On the other hand, let $T$ be a support $\tau_{2}$-tilting module. We show
that $\mathrm{Fac}T\cap\mathcal{C}$ satisfies the conditions of Proposition
3.6. Closure under factor modules is trivial, and 2-functorial finiteness in
$\mathcal{C}$ follows from Lemma 3.9. So let

$0\rightarrow T_{0}\rightarrow X\rightarrow Y\rightarrow T_{3}\rightarrow 0$

be a $d$-exact sequence in $\mathcal{C}$ with $T_{0},T_{3}\in\mathrm{Fac}T$.
By Lemma 2.4, we have that
$\mathrm{Hom}_{A}(T,\tau_{2}T)=0\iff\mathrm{Ext}^{2}_{A}(T,\mathrm{Fac}T)=0$,
so $T_{3}$ is not in $\mathrm{add}T$. Lemma 3.9 implies there exists an exact
sequence in $\mathrm{Fac}T$: $0\rightarrow T_{0}^{\prime}\rightarrow
T_{1}^{\prime}\rightarrow T_{2}^{\prime}\rightarrow T_{3}\rightarrow 0.$ By
Lemma 2.7 there exist $P,Q\in\mathcal{C}$ to form a commutative diagram with
exact rows and columns:

${0}$${0}$${T_{0}^{\prime}}$${T_{0}^{\prime}}$${Q}$${Q\oplus
T_{1}^{\prime}}$${T_{1}^{\prime}}$${0}$${T_{0}}$${Q\oplus
X}$${P}$${T_{2}^{\prime}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$${0}$${0}$

Applying Lemma 2.4 once more, we find
$\mathrm{Ext}^{2}_{A}(T_{2}^{\prime},T_{0})=0$. This implies the diagram
splits, $Q\cong T_{0}$, $P\cong T_{2}^{\prime}\oplus X$ and that we are in the
situation of Lemma 2.8, with a commutative diagram with exact rows and
columns:

${0}$${0}$${0}$${T_{0}^{\prime}}$${T_{1}^{\prime}}$${0}$${T_{0}}$${T_{0}}$${T_{2}^{\prime}}$${T_{2}^{\prime}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$${0}$

Hence we find the required $d$-pushout diagram

${0}$${T_{0}\oplus T_{0}^{\prime}}$${T_{0}\oplus
T_{1}^{\prime}}$${T_{2}^{\prime}}$${T_{3}}$${0}$${0}$${T_{0}}$${X}$${Y}$${T_{3}}$${0}$

since we know $0\rightarrow T_{0}^{\prime}\rightarrow T_{0}\oplus
T_{1}^{\prime}\rightarrow X\oplus T_{2}^{\prime}\rightarrow Y\rightarrow 0$ is
exact. ∎

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