Given a torsion pair $(\T,\F)$ in an abelian category $\A$, there is a t-structure $(\U_\T,\V_\T)$ determined by $\T$ on the derived category $\db\A$. The existence of derived equivalence between heart $\B$ of the t-structure and $\A$ which naturally extends the embedding $\B\to \db\A$ is determined by the completeness of the torsion pair [7]. When $\A$ is the module category of a finite-dimensional hereditary algebra and $\U_\T$ is closed under Serre functor, then there exists a triangle equivalence  $\db \B\to\db\A$ [22]. In this case, we give a straightforward proof of the fact torsion pair $(\T,\F)$ is complete if and only if $\U_\T$ is closed under the Serre functor.

§ INTRODUCTION

Given a t-structure on a triangulated category $\D$, the Serre functor plays an important role in the relation between the derived category of the heart and $\D$ [22]. On the other hand, given an abelian category $\A$ with a torsion pair $(\T,\F)$, the heart $\B$ of the t-structure $(\U_\T,\V_\T)$ on $\db{\A}$ defined by the torsion pair is an abelian category[12, 3]. We call the category $\B$ Happel-Reiten-Smalø tilt (HRS tilt) of $\A$ with respect to $(\T,\F)$. By [3], there is a realization functor $F\colon \db\A\to \db \B$ which extends the embedding $\B\to \db\A$. In general, the realization functor $F$ is not an equivalence. There is a criterion on $(\T,\F)$ for the functor $F$ being an equivalence [7].
In this paper, we consider how the aisle $\U_\T$ closed under the Serre functor relates to the completeness property of $(\T,\F)$.

We call a torsion pair $(\mathcal T,\mathcal F)$ of $\mathcal A$ is complete if for each object $X$ in $\mathcal A$, there is an exact sequence
$0\to F_1\to F_2\to X\to T_1\to T_2\to 0$ with $T_i\in \T$ and $F_i\in \F$ such that the corresponding class in $\Ext^3_{\A}(T_2,F_1)$ vanishes.  According to <cit.>, the realization functor $F$ is an equivalence if and only if $(\T,\F)$ is a complete torsion pair. In [22], the authors consider the sufficient and necessary condition for $F$ being equivalence from the point of view of bounded t-structure. Assume that the derived category $\db\A$ admits a Serre functor $\mathbb S$. The functor $F$ being an equivalence is closely related to the aisle $\U$ satisfying $\S\U\subseteq \U$. However, it is sufficient and necessary for the module category $\A$ of a finite-dimensional hereditary algebra.

From [7] and [22], there does not exist a straightforward approach to prove the equivalence between $\S\U_\T\subseteq \U_\T$ and completeness of $(\T,\F)$. The completeness of $(\T,\F)$ is sufficient to imply $\S\U_\T\subseteq \U_\T$. Conversely, there is no explicit construction of the 5-term exact sequence in (<ref>). The reason is that the completeness of $(\T,\F)$ depends on the calculation of the preimage of objects in $\A$ under the realization functor $F$. The main result of the paper gives a straightforward proof of the equivalence which also provides a construction of preimages of the realization functor.

Let $\Lambda$ be a finite-dimensional hereditary algebra and $(\T,\F)$ be a torsion pair in $\A=\mod\Lambda$.  Then the following statements are equivalent.

* The aisle $\U$ that determined by $\T$ is closed under the Serre functor.

* The torsion pair $(\mathcal T,\F)$ is complete.

Note that one could not generalize the result to any hereditary abelian category $\A$, see <cit.> and <cit.>. From the proof of Theorem <ref>, we actually construct the preimage of objects in $\A$ under the realization functor $F\colon \db \B\to\db \A$.

In Section 2, we recall basic facts about tilting theory and perpendicular categories defined by exceptional objects. The twist functors associated with exceptional objects play an essential role in proving Theorem <ref>. In Section 3, we prove the completeness of torsion pair $(\T,\F)$ implies the aisle $\U_\T$ closed under the Serre functor. Furthermore, we prove the converse holds either $\T$ is finitely generated or without any Ext-projective objects. In Section 4 and Section 5, we apply the indecomposable exceptional objects to reduce the general case of $\T$ to the special case $\T$ without any Ext-projective objects.

§ PRELIMINARY

For any category $\D$, a subcategory $\C$ of $\D$ is called reflective if the embedding $\iota:\C\hookrightarrow\D$ has a left adjoint $F$, i.e., there is a bifunctorial isomorphism

\[\Hom_\D(X,\iota Y)\cong\Hom_\C(FX,Y), \forall X\in\D, \forall Y\in\C.\]

The coreflective subcategory is defined dually.

For any subcategory $\C$ of a triangulated category $\D$, let $\add\C$ be the additive closure of $\C$. For any pair of subcategories $(\C_1,\C_2)$, we write $\C_1*\C_2$ the full subcategory of $\D$ consisting of those objects $Z$ such that there is a triangle $C_1\to Z\to C_2\to C_1[1]$ in $\D$, where $C_1\in\C_1$ and $C_2\in\C_2$.

We denote by $\mod\Lambda$ the category of finite-dimensional right $\Lambda$-modules over a finite-dimensional algebra $\Lambda$.

§.§ Tilting theory and wide subcategory

A full subcategory $\T$ of an abelian category $\A$ is called a torsion class if it is reflective and closed under quotients and extensions. A torsion class $\T$ is called tilting if for all $X\in \A$, there is an monomorphism $X\to T_X$ for some $T_X\in \T$.

A pair $(\T,\F)$ of full subcategories of $\A$ is called a torsion pair if $\T$ is a torsion class and $\Hom_\A(T,F)=0$ for any $T\in\T$ and $F\in\F$. The subcategory $\F$ is called a torsion free class.

For an abelian category $\A$ and a subcategory $\mathcal C$, an object $M\in\mathcal C$ is called an Ext-projective in $\mathcal C$ if $\Ext^i_{\mathcal C}(M,C)=0$ for each $C\in\mathcal C$. A torsion class $\T$ is called finitely generated if each object is a quotient of an object in $\add E$ for an Ext-projective object $E$ in $\T$.
We call a torsion pair $(\mathcal T,\mathcal F)$ of $\mathcal A$ is complete if for each object $X$ in $\mathcal A$, there is an exact sequence

\begin{equation}\label{eq:seq}
0\to F_1\to F_2\to X\to T_1\to T_2\to 0
\end{equation}

with $T_i\in \T$ and $F_i\in \F, i=1,2$ such that the corresponding class in $\Ext^3_{\A}(T_2,F_1)$ vanishes.

In [16], Krause and Šťovíček defined the complete Ext-orthogonal pair. The definition of complete torsion pair in their setting is not equivalent to the notation here.

Wide subcategories of a hereditary abelian category have been extensively studied in the representation theory of algebras. There are close relations between the wide subcategory and combinatorics of the Coxeter groups [10, 14]. A full additive subcategory $\mathcal W\subseteq\A$ is called a wide if it is closed under kernels, cokernels, and extensions. For any full subcategory $\C$ of $\A$, the wide closure of $\C$ in $\A$ is the intersection of all wide subcategories of $\A$ containing $\C$. We denote $\a_{\A}(\C)$ the wide closure of $\C$ in $\A$.

In general, there is no canonical way to obtain the wide closure $\C$ for a subcategory$\C$. The following result gives a construction of $\a_{\A}(\C)$ in the case $\A$ is hereditary.

If $\A$ is a hereditary abelian category and $\C$ is a full subcategory closed under extensions and cokernels. Then each object in $\a_{\A}(\C)$ is the kernel of a map in $\C$.

§.§ t-structure and co-t-structure

Denote by $\db{\A}$ the bounded derived category of $\A$. If $\A$ is hereditary, we have

\[\D^b(A)=\bigcup_{n\in\mathbb{Z}}\A[n].\]

A pair $(\U,\V)$ of full subcategories of $\db\A$ is called a t-structure (resp. co-t-structure) if

* $\U[1]\subseteq\U$ and $\V[-1]\subseteq\V$ (resp. $\U[-1]\subseteq\U$ and $\V[1]\subseteq\V$);

* $\Hom_{\db{\A}}(\U,\V)=0$, and

* for any $X\in\db{\A}$ there is a triangle

\[X_\U\to X\to X_\V\to X_\U[1]\]

with $X_\U\in\U,X_\V\in\V$.

We call subcategories $\U$ and $\V$ an aisle and coaisle, respectively. A t-structure $(\U,\V)$ is said to be bounded if $\bigcup_{n\in\mathbb{Z}}\U[n]=\db{\A}=\bigcup_{n\in\mathbb{Z}}\V[n]$.

An object $E\in\db{\A}$ is called presilting if $\Hom_{\db{\A}}(E,E[1])=0$. A bounded t-structure $(\U,\V)$ is called finitely generated $\U=\bigcup_{n\in\mathbb{Z}}\add E[n]\ast\cdots\add E[1]\ast\add E$ for some presilting object $E$.

The heart of a t-structure $(\U,\V)$ is defined to be the full subcategory $\U\cap\V[1]$ of $\db{\A}$, which is an abelian category by <cit.>.

For each torsion pair $(\T,\F)$ in $\A$, let

\[\U_{\T}=\{X\in\db\A\mid H^0(X)\in\T,H^n(X)=0\text{ for any } n>0\}\]

\[\V_{\T}=\{Y\in\db\A\mid H^0(Y)\in\F, H^n(Y)=0 \text{ for any } n<0\}.\]

It is direct to check that $(\U_{\T},\V_{\T})$ is a bounded t-structure in $\db{\A}$ <cit.>. The heart $\B=\F[1]\ast\T$ of $(\U,\V)$ is called the HRS-tilt of $\A$.

Conversely, given a bounded t-structure $(\mathcal{D}^{\leq 0}, \mathcal{D}^{\geq 0})$ on a triangulate category $\D$ with heart $\mathcal H$, each t-structure $(\U,\V)$ with $\D^{\leq0}[1]\subset\U\subset\D^{\leq 0}$ determines a torsion class $\T:=H^0_{\mathcal H}(\U)$ in $\mathcal H$, where $H^0_{\mathcal H}$ denotes the cohomological functor corresponding to $\mathcal H$. The following lemma shows that such a correspondence is a bijection.
Let $\H$ be the heart of a bounded t-structure $(𝒟^≤0, 𝒟^≥0)$ on a triangulate category $$. Then there is a bijection between the set of all torsion classes $$ of $$̋ and the set of t-structures $(\U,\V)$ of $\D$ satisfying

\[\D^{\leq 0}[1]\subseteq \U\subseteq \D^{\leq 0},\]

given by $\T\mapsto \U_{\T}$

Moreover, if $\T$ is finitely generated with a generator $E$, one may check that $E$ is exactly the presilting object that generates $\U$.

§.§ Serre functor
Let $\C$ be a Hom-finite $\k$-linear category, where $$̨ is an algebraically closed field. A Serre functor is a $\k$-linear autoequivalence $$ of $$ such that for any objects $A,B∈$, there exists an isomorphism
\[\eta_{A,B}\colon \Hom(A,B)\to \Hom(B,\S A)^*,\]
where $(-)^*=_(̨-,)̨$ is the standard duality.

Recall that an abelian category $$ is called \emph{Ext-finite} if $_^i_(X,Y)<∞$ for any $X,Y ∈$ and all $i∈ℤ$. An Ext-finite abelian category $𝒜$ has \emph{Serre duality} if $D^b(𝒜)$ admits a Serre functor $$.
A t-structure $(𝒰,𝒱)$ or an aisle $$ is called closed under the Serre functor $$ if $⊆$.

\begin{lemma}[{[22]}]\label{lem:trans}
Let $\Lambda$ be a finite-dimensional hereditary algebra and $(\T,\F)$ be a torsion pair in $\mod\Lambda$. The aisle $\U_{\mathcal T}$ determined by $\T$ is closed under the Serre functor if and only if for each projective object $P\in\mathcal T$, the corresponding injective object $P\otimes_{\Lambda}\Lambda^*$ is in $\mathcal T$.
\end{lemma}
%The converse of Lemma~\ref{lem:trans} holds if $\Lambda$ is hereditary, but fails in general, see, e.g., \cite[Example 4.8]{CHZ}.

\subsection{Twist functors and perpendicular categories} \label{sec:exc-perp}
Let $𝒟$ be a triangulated category and $𝒮$ a full triangulated subcategory, the \emph{right orthogonal} $𝒮 ^⊥$ of $𝒮$ is the full subcategory consisting of those objects
\[\{X\in \mathcal D\mid \Hom(Y,X)=0, \forall Y\in \mathcal S\}.\]
The \emph{left orthogonal} $^⊥𝒮$ is defined similarly. A full triangulated subcategory of $$ is called \emph{thick} if it is closed under taking direct summands. For any object $X$ in $𝒟$, we write $⟨X⟩$ for the thick subcategory generated by $X$. Furthermore, we write $X^⊥$ and $^⊥X$ for $⟨X⟩^⊥$ and $^⊥⟨X⟩$ respectively.

For a $$̨-linear, Hom-finite, algebraic triangulated category $\D$, an object $E$ in $\mathcal D$ is called exceptional if $\Hom(E,E)\cong \k$ and $(E,E[n])=0$ for all $n≠0$. An object $E$ in an abelian category $𝒜$ is exceptional if it is exceptional in $𝒜$.

%For an exceptional object $E\in \D$, the thick subcategory $\langle E\rangle$ is exactly $\add (E[i]\mid i\in\mathbb Z)$. By \cite[Theorem~3.2]{B89}, the embedding $\langle E\rangle \to \D$ has a left and right adjoint.

%For any object $X\in \mathcal D$, We define two functors
%\[\mu_X:=\Hom(-,X)^*\otimes X\colon \mathcal D\to \mathcal D \] and
%\[\lambda_X:=\Hom(X,-)\otimes X \colon \mathcal D\to \mathcal D.\]
%We shall define the \emph{twist functors} associated with the functors $\mu_S$ and $\lambda_S$.

Let $𝒜$ be an Ext-finite abelian category of finite global dimension and $S∈$ an exceptional object. Then the thick subcategory $⟨S⟩$ generated by $S$ in $$ is
equivalent to $$ and the embedding
$⟨S⟩→$ has a left and right adjoint, given on objects by $λ_S(X)=R(S,X)⊗_^̨LS$ and $μ_S(X)=R(X,S)^*⊗^L_S̨$, see \cite[Theorem~3.2]{B89} and \cite[Section 2.4]{SR16}.

The embedding $^⊥ S→$ has a right adjoint $T^*_S→^⊥S $ and left adjoint $T_S →^⊥S$, which are defined by the following triangles
\[T^*_S(X)\to X\xrightarrow{\alpha_X} \mu_S(X) \to T^*_S(X)[1]\]

\[T_S(X)[-1]\to\lambda_S(X)\xrightarrow{\beta_X} X\to T_S(X),\]
see \cite[Lemma~3.1]{B89}. By the assumption of $S$, we have that \[\lambda_S(X)\cong \oplus_{i\in\mathbb Z}\Hom(S,X[i])\otimes_\k S[-i], \quad \mu_S(X)=\oplus_{i\in\mathbb Z} \Hom(X,S[i])^*\otimes_\k S[i].\]

The functors $T_S$ and $T^*_S$ could be viewed as endofunctors of $$
%$T_S\colon \db{\mathcal A}\to \db{\mathcal A}$ and $T^*_S\colon \db{\mathcal A}\to \db{\mathcal A}$
by composing the corresponding embedding. We call the functors $T^*_S$ and $T_S$ the \emph{twist functors} associated with $S$. The functor $T_S$ is the right adjoint of $T^*_S$ \cite[Section~3.2]{HKP}.

\begin{lemma}[{\cite[Lemma~1]{K}}]\label{lem:se}
Keep the notation as above, the subcategory $\ps$ has the Serre functor $\S'=T^*_S\circ \S\circ T_S$.
\end{lemma}

%\hp{The functor $T_S$ (resp.  $T^*_S$) is the left (resp. right) adjoint to the inclusion functor $\ps\to \db{\A}$ (excessive?).} %By \cite[Lemma~1.9]{BK}, the functors $T_S$ and $T^*_S$ induce equivalences $T_S\colon \ps\to S^{\perp}$ and  $T_S^*\colon S^{\perp}\to \ps$. They are quasi-inverse to each other. The following result is obvious and useful.
\begin{lemma}[{\cite[Lemma~1.9]{BK}}]\label{lem:iso} Keep the notation as above, the twist functors on $\db\A$ induces equivalent functors $T_S\colon \ps\to S^{\perp}$ and $T_S^*\colon S^{\perp}\to \ps$ which are quasi-inverse to each other.
\end{lemma}

For an abelian category $𝒜$ and a class $𝒞$ of objects in $𝒜$, there is a full subcategory $𝒞^⊥_𝒜$ consisting of objects ${X∈𝒜|∀n∈ℤ, ^n_(𝒞,X)=0}$. The subcategory $^⊥𝒞_𝒜$ is defined similarly. In general, the categories $𝒞^⊥_𝒜$ and $^⊥𝒞_𝒜$ are neither wide subcategories of $𝒜$ nor abelian categories.

If $=Λ$ for some finite-dimensional algebra $Λ$ and $𝒞$ consisting of $M∈Λ$ with projective dimension $M≤1$, then $𝒞^⊥_𝒜$ is an abelian subcategory of $𝒜$. Note that the condition $(M) ≤1$ is necessary for $M^⊥$ to be abelian. If $𝒜$ is hereditary, then both $𝒞^⊥_𝒜$ and $^⊥𝒞_𝒜$ are wide subcategories of $𝒜$ \cite[Proposition~1.1]{GL}.

\begin{lemma}[{\cite[Proposition~3]{HRS2}}]\label{lem:red}
Let $\mathcal A=\mod \Lambda$ for some finite-dimensional hereditary algebra $\Lambda$ and $S$ be an indecomposable exceptional object in $\mathcal A$. Then $^\perp S\subseteq \db{\Lambda}$ is equivalent to $\db{\Lambda'}$ for a finite-dimensional hereditary algebra with one fewer distinct simple objects of $\Lambda$ and $^{\perp}S_{\mathcal A}$ is equivalent to $\mod \Lambda'$.
\end{lemma}

\section{The equivalence for special torsion classes}

For a finite-dimensional algebra $Λ$, we denote the category $Λ$ by $$. In this section, we prove that for a hereditary algebra $Λ$, the torsion pair $(,)$ in a module category $$ of a hereditary algebra $Λ$ is complete implies that the aisle $_$ in $Λ$ is closed under the Serre functor. We also prove the converse holds for the torsion class $$ either is finitely generated or has no non-zero Ext-projective objects.
\subsection{Necessity}
For a finite-dimensional algebra $Λ$ with a finite global dimension, the category $=Λ$ has Serre duality. The Serre functor $$ is given by $≅-⊗_^LΛ^*→$. The category $$ has a Serre functor $$ if and only if $$ has Auslander-Reiten triangles. For every indecomposable object $X∈$, the Auslander-Reiten triangle is
$[-1]X→Y→X→(X)$, see \cite[Proposition~I.2.3]{RV1} and \cite[Proposition~2.8]{C}.

If $Λ$ is a hereditary algebra, then the Auslander-Reiten translation functor on $$ and the Auslander-Reiten translation on $$ coincide on non-projective indecomposable objects in $$. We write the Auslander-Reiten translation on $$ as $τ$.
\begin{lemma}\label{lem:inj-proj}
Let $\Lambda$ be a finite-dimensional algebra of finite global dimension. $(\mathcal T,\mathcal F)$ a torsion pair on $\mod \Lambda$. Assume that $P\in \mathcal T$ is projective, and the corresponding injective object $I$ fits in a 5-term exact sequence of the form (\ref{eq:seq}). Then $I$ is also in $\mathcal T$.
\end{lemma}
\begin{proof}
Considering the 5-term exact sequence \ref{eq:seq} with $X=I$, we have that $F_2\to I$ is zero. Otherwise, there exists a non-trivial map $P\to F_2$ by Serre duality, which contradicts the fact that $P\in\T$. So the sequence \ref{eq:seq} for $I$ is a short exact sequence of the form
\begin{equation}\label{eq:3 to 2}
0\to I\to T_1\to T_2\to 0.
\end{equation}
Since $I$ is injective, the sequence~\eqref{eq:3 to 2} splits. This implies that $I$ is a direct summand of $T_1$ and $I\in \T$.
\end{proof}
Now assume that $Λ$ is a hereditary algebra. Then $$ has a Serre functor $=τ[1]$. So $X⊆[≥1]$ for any $X∈[≥0]∖$, and $P=P⊗_ΛΛ^*$ for any $P∈$. Combining Lemma~\ref{lem:trans} and Lemma~\ref{lem:inj-proj}, we have that $_$ is closed under the Serre functor. By Lemma~2.4, we have the following result.
\begin{corollary}\label{cor:suff}
Let $\Lambda$ be a finite-dimensional hereditary algebra. If a torsion pair $(\mathcal T,\mathcal F)$ on $\mod \Lambda$ is complete. Then the aisle $\U_\T$ is closed under the Serre functor.
\end{corollary}
\begin{remark}
The corollary holds for any finite-dimensional algebra with finite global dimension by \cite[Theorem~A]{CHZ} and \cite[Corollary~4.13]{SR16}. We do not have a straightforward proof of this argument.
\end{remark}
\subsection{ Aisle determined by special torsion classes}
\begin{lemma}\label{lem:enough}
Let $\Lambda$ be a finite-dimensional algebra of finite global dimension. Assume $\T$ is a finitely generated torsion class in $\mod \Lambda$ and the aisle $\U_\T$ is closed under the Serre functor. Then the torsion pair $(\T,\F)$ is complete.
\end{lemma}
\begin{proof}
Recall that for each t-structure $(\U,\V)$ with $\U$ finitely generated, we have that $(\V,\S\U)$ is a co-t-structure, see \cite[Proposition~2.23]{AI}.

If $\mathcal T$ is finitely generated, then so is $\U_{\T}$. Now we take the the co-t-structure $(\V_{\T},\S\U_{\T})$. For any $X\in\A$, let
\[N\to X\to M\to N[1]\]
be the triangle of $X$ with respect to $(\V_\T,\S\U_\T)$. Taking cohomology with respect to canonical t-structure, we have the following long exact sequence
\[H^{-1}(X)=0\to H^{-1}(M)\to H^0(N)\to X\to H^0(M)\to H^1(N)\to H^1(X)=0\]
where the first term $H^{-1}(M)$ lies in $\F$ because $N\in\V_{\T}$ and $H^0(N)\in\F$. Similarly, $H^1(N)\in\T$ implies that $H^0(M)\in\T$. Thus $(\T,\F)$ is complete.
\end{proof}

If a torsion class $$ is not finitely generated, then we do not have the co-t-structure $(_,_)$. In this case, there is no direct approach to prove $(,)$ is complete under the assumption $_$ is closed under the Serre functor. For the category $$ with $Λ$ being a hereditary algebra, we could show that $_⊆_$ implies $(,)$ is complete under the assumption the torsion class $$ without Ext-projectives.

\begin{lemma}\label{lem:torp}
Assume that $\mathcal A$ is a hereditary category and $\T$ is a torsion class without Ext-projective objects, then all projective objects of $\A$ are in the torsion-free class.
\end{lemma}
\begin{proof}
For any projective object $P\in \A$, there is a short exact sequence
\[0\to P_{\T}\to P\to P_{\F}\to 0\]
with $P_{\T}\in \T$, and $P_{\F}\in \F$. Since $\A$ is hereditary, we have that $P_{\F}$ is either zero or projective. Since $\T$ does not contain any nonzero projective objects, we have that $P_{\T}=0$ and hence $P\in\F$.
\end{proof}

\begin{lemma}\label{lem:no}
Assume that $\A=\mod \Lambda$ for a hereditary algebra $\Lambda$. Let $(\T,\F)$ be a torsion pair in $\A$ with $\T$ having no nonzero Ext-projectives. If $\U_{\T}$ is closed under the Serre functor then $(\T,\F)$ is complete.
\end{lemma}
\begin{proof}
If $\T$ is a torsion class of $\A$, then the wide closure $\a_{\A}(\T)$ is coreflective and reflective wide subcategory of $\A$ by \cite[Proposition~7.3]{SR19}. It implies that there is a right approximation $X\to W$ with $W\in\a_{\A}(\T)$ for any object $X\in\A$. On the other hand,  $\T$ is a tilting torsion class in $\a_{\A}(\T)$ by \cite[Corollary~3.7]{SR19}. So there exists an object $T_1$ in $\T$ such that there exists an embedding $W\hookrightarrow T_1$. We denote the composition $X\to W\to T_1$ by $g\colon X\to T_1$.

Denote $T_2:=\coker g$, then $T_2\in\T$. Thus we get a sequence from the morphism $g\colon X\to T_1$
\[0\to \ker g\to X\xrightarrow{g} T_1\to T_2\to 0 \]
with $T_1, T_2\in \T$.
By Lemma~\ref{lem:torp}, all projective objects are contained in $\F$. We take the projective cover $P_2\to \ker g$ of $\ker g$. Denote $P_1:=\ker(P_2\to \ker g)$, we have the desired exact sequence
\[0\to P_1\to P_2\to X\to T_1\to T_2\to 0.\]
Note that $P_1\in \F$ since $P_2\in \F$ and $\A$ is hereditary. By definition, $(\T,\F)$ is complete.
\end{proof}

\section{The Reduction via exceptional objects}
Throughout this section, we assume that category $𝒜$ is an abelian category with Serre duality $$. Let $(,)$ be a torsion pair of $𝒜$ such that the corresponding aisle $_$ is closed under the Serre functor, i.e. $_⊂_$. The HRS-tilt of $$ with respect to $(,)$ is denoted by $$.

We will consider an Ext-projective object $E$ in $_$. It is actually a projective object in $$, and we fix its simple top $S$ in $$. The object $S$ is an exceptional object in $$. If $Λ$ is hereditary, we could reduce the category $$ to a new category $$ and the category $$ to the left orthogonal $$.

\subsection{The aisle on \texorpdfstring{$\ps$}{}}
Like the Ext-projective objects in an abelian category, we consider the Ext-projective object in a subcategory of a triangulated category. Given a triangulated category $$ and a full additive category $$, an object $E$ in $$ is called Ext-projective if $_(E,C[i])=0$ for all $i>0$. The following result shows that the indecomposable Ext-projective object $E∈_$ is projective in $$ and has a simple top $S$.

\begin{proposition}[{\cite[Proposition~6.4]{SR16}}]\label{prop:SE}
Let $\A$ be an abelian category with Serre duality and $(\U,\V)$ the bounded t-structure in $\db\A$ closed under the Serre functor. Then for the Ext-projective $E$ of $\U$, and the Auslander-Reiten triangle $\tau E\to M\to E\to \S E$ given by $E$, we have that
\begin{enumerate}
\item $E$ is projective in $\B$ and $\S E$ is injective in $\B$;
\item  The object $E$ has a simple top $S$ in $\B$;
\item there is a short exact sequence $0\to S\to \S E\to M_\V[1]\to 0$ in $\B$;
\item $\Hom_{\db{\A}}(S,S[n])=0$ for $n\neq 0$, and $\End S=\k$.
\end{enumerate}
\end{proposition}
Recall that a torsion pair $(\T,\F)$ of $\A$ is called \emph{splitting }if $\Ext^1_{\A}(F,T)=0$  for all $F\in F$ and $T\in \T$. In this case, any object in $\A$ is isomorphic to a direct sum $F\oplus T$ for some $F\in \F$ and $T\in \T$. The following result is obvious.
\begin{lemma}\label{lem:split}
If $\mathcal A$ is a hereditary category with a torsion pair $(\mathcal T,\mathcal F)$, then HRS-tilt $\B$ has a splitting torsion pair $(\mathcal F[1],\mathcal T)$.
\end{lemma}

In the following, we assume that $\A=\mod \Lambda$ for a finite-dimensional hereditary algebra $\Lambda$. By Lemma~\ref{lem:split}, we have that either $S\in \F[1]$ or $S\in \T$. Let $R$ be $S$ if $S\in\T$ and $S[-1]$ if $S\in\F[1]$. Since $S$ is exceptional, so is $R$. Hence, by \cite[Proposition~1.1]{GL} and Lemma~\ref{lem:red}, the subcategory
\[\W=\{X\in\A|\Hom(X,R)=0,\Ext^1_\A(X,R)=0\}\]
is a wide subcategory of $\A$ that equivalent to the module category of some hereditary algebra. The following lemma is dual to \cite[Lemma~3.6]{RV}.
\begin{lemma}\label{lem:perp}
There exists equivalences $\db{\W}\cong D^b_\W(\A)\cong\ps$, where $D^b_\W(\A)$ the full subcategory of $\db\A$ consisting of objects $\{X\in\db\A|H^i(X)\in\W, \forall i\in\mathbb{Z}\}$.
\end{lemma}
\begin{proof}
Denote by $\Sigma$ the suspension functor in $\db\W$. Since $\W$ is hereditary, each object of $\db\W$ is isomorphic to a direct sum of shifts of objects in $\W$.
%$\{\Sigma^nX|X\in\W \text{ indecomposable, }n\in\mathbb{Z}\}$.
Let $\Phi\colon \db\W\to D^b_\W(\A)$ be the functor which is induced by $ \Sigma^nX\mapsto X[n]$ for $X\in \W$. It is obvious that $\Phi$ is dense.
For any two objects $X,Y\in \W$, $\Hom_{\db\W}(\Sigma^iX,\Sigma^j)\neq 0$ if and only if $i=j$ or $j-i=1$.
%For any two indecomposable objects $X,Y\in\db\W$, $\Hom_{\db\W}(X,Y)\neq0$ if and only if $X=\Sigma^n\tilde{X},Y=\Sigma^n\tilde{Y}$, or $X=\Sigma^n\tilde{X},\Y=\Sigma^{n+1}Y$ for some $\tilde{X},\tilde{Y}\in\W, n\in\mathbb{Z}$.
In the former case, we have
\[\Hom_{\db\W}(\Sigma^iX,\Sigma^jY)\cong\Hom_{\db\A}(X[i],Y[j]).\]
In the  case $j=i+1$, we have
\[\Hom_{\db\W}(\Sigma^iX,\Sigma^jY)\cong\Ext^1_\W(X,Y)\cong\Ext^1_\A(X,Y)\cong\Hom_{\db\A}(X[i],Y[j]),\]
where the second isomorphism follows because $\W$ is closed under extensions. So $\Phi$ is fully faithful and hence $\db\W\cong D^b_\W(\A)$.

Since $\A$ is hereditary, we have $X=\oplus_{i\in\mathbb{Z}}H^i(X)[-i]$ for any $X\in\db\A$. Thus, $X\in\ps$ if and only if $\Hom(H^i(X),S[n])=0$, $\forall i,n\in\mathbb{Z}$ if and only if $H^i(X)\in\W,\forall i\in\mathbb{Z}$. So $\ps\cong D^b_\W(\A)$.
\end{proof}

Recall that $\ps$ admits a Serre functor
%the embedding $^{\perp}S\subseteq \db{\A}$ gives rise a Serre functor
$\S'=T^{*}_S\circ \S\circ T_S$ by Lemma~\ref{lem:se}. Let $\U'=\U\cap\ps$. By \cite[Lemma 9.1]{SR16}, $T^*_S(X)\in \U'$ for any $X\in \U$.
\begin{lemma}[{\cite[Proposition~9.2]{SR16}}]\label{lem:induction step}
For any bounded aisle $\U\subseteq \db{\A}$ closed under the Serre functor $\S$, then  $\U'$ is an aisle in $\ps$ and  $\S'\U'\subseteq \U'$.
\end{lemma}
Note that the coaisle $\V'$ is not $\V\cap\ps$, see \cite[Lemma~9.3]{SR16}. The following result controls the $\add R$-approximation in the category $\A$.
\begin{lemma}\label{lem:ker}
\begin{enumerate}
\item If $S\in\T$ then for any short exact sequence $0\to M\to S^n\xrightarrow{f} T\to 0, n\in\mathbb{Z}_{>0}$ in $\A$, we have $M=M'\oplus S^l$ with $l\leq n$ and $M'\in\F$. In particular, $T\in\add S$ if $M'=0$.

\item If $S\in\F[1]$ then for any $0\to F\xrightarrow{g} S[-1]^n\to M\to 0, n\in\mathbb{Z}_{>0}$ in $\A$, we have $M=M''\oplus S[-1]^l$ with $l\leq n$ and $M''\in\T$. In particular, $F\in\add S[-1]$ if $M''=0$.
\end{enumerate}
\end{lemma}
\begin{proof}
We only prove the first assertion since the proof of the second statement is similar. Since $S\in\T$ and $\T$ is closed under images, we have $T\in\T$. Write $f=(f_1,\cdots,f_{n-l},0,\cdots,0)$ with $f_i\in\Hom_{\A}(S, T),1\leq i\leq n-l$ nonzero. Since $S$ is simple in $\B$, $f'=(f_1,\cdots, f_{n-1}):S^{n-l}\to T$ is injective in $\B$. There is a short exact sequence $0\to S^{n-1}\xrightarrow{f'}T\to\coker f'\to0$ in $\B$ which induces a triangle
\[T[-1]\to \coker f'[-1]\oplus S^l\to S^{n-l}\oplus S^l\xrightarrow{f} T\]
in $\db{\A}$.

The short exact sequence $0\to M\to S^l\xrightarrow{f} T\to 0$ in $\A$ gives another triangle
\[T[-1]\to M\to S^n\xrightarrow{f} T\]
in $\db\A$. It implies that $S^l\oplus\coker f'[-1]\cong M\in\A$. So $\coker f'[-1]\in\F$ and the assertion follows by taking $M'=\coker f'[-1]$.
\end{proof}

\subsection{The torsion pairs on \texorpdfstring{$\W$}{}}
By \cite[Proposition~2.2]{SR16}, $\W$ is a reflective and coreflective wide subcategory of $\A$. The torsion class $(\T,\F)$ induces a torsion pair $(\T',\F')$ on $\W$ with the torsion class $\T'=\T\cap \W$. %In general, the torsion-free class $\F'\neq \F\cap \W$.
By Lemma~\ref{lem:perp} and Lemma~\ref{lem:induction step}, the aisle $\U_\T\subseteq \db \A$ induces the aisle  $\U'=\bigcup_{n\in\mathbb{Z}}\W[n]\ast\cdots\ast\W[1]\ast\T'$ in $\ps$. Moreover, $\T'$ and $\T$ are related by the twist functor $T^*_S$ as the following.
\begin{lemma}\label{lem:T*_S(T)}
Suppose $S\in\T$ as above. Then $T^*_S(T)\in\T'$ for any $T\in\T$.
\end{lemma}
\begin{proof}
By the definition of the functor $T^*_S$, we have the triangle $\mu_S(T)[-1]\xrightarrow{g} T^*_S(T)\to T\xrightarrow{\alpha_T} \mu_S(T)$ for any $T\in \T$. Since $\A$ is hereditary, we have $\mu_S(T)=\Hom(T,S)^*\otimes S\oplus\Hom(T,S[1])^*\otimes S[1]$. Taking cohomology, we have $H^i(T^*_S(T))=0$ for $i\neq 0,1$, and a long exact sequence
\[0\to H^{-1}(\mu_S(T))\to H^0(T^*_S(T))\to T\xrightarrow{h} H^0(\mu_S(T))\to H^1(T^*_S(T))\to 0.\]
So we get short exact sequences
\begin{equation}\label{eq:1}
0\to\ker h\to T\xrightarrow{p}\im h\to 0,
\end{equation}
and thus $\im h\in\T$. By Lemma~\ref{lem:ker} and the short exact sequence
\[0\to\im h\xrightarrow{i} H^0(\mu_S(T))\to H^1(T^*_S(T))\to 0\]
we have that $\im h\in\add S$ and hence $H^1(T^*_S(T))\in\add S$. Since $T^*_S(T)\in \ps$, we have $H^1(T^*_S(T))=0$. So $\im h\cong H^0(\mu_S(T))\in\add S$. Since $S$ is simple in $\B$, we have that $p$ in ~\eqref{eq:1} is also an epimorphism in $\B$. So ~\eqref{eq:1} is an exact sequence in $\B$, which implies $\ker h\in\T$ by Lemma~\ref{lem:split}. So our claim follows from the short exact sequence
\[0\to H^{-1}(\mu_S(T))\to H^0(T^*_S(T))\to\ker h\to 0.\]
\end{proof}
The torsion class $\T'$ in $\W$ is given by $\T\cap\W$ but the corresponding torsion free class $\F'\neq \F\cap \W$. Indeed, $\F'$ is determined by the exceptional object $S$ as the following lemmas.

\begin{lemma}\label{lem:F'}
Suppose $S\in\T$. Then $X\in\F'$ if and only if $X\cong T^*_S(F)$ for some $F\in \F$ with $\Hom(F,S)=0$.
\end{lemma}
\begin{proof}
For $F\in\F$ such that $\Hom(F,S)=0$, we have a short exact sequence in $\A$
\[\zeta: 0\to\Hom(F,S[1])^*\otimes S \to T^*_S(F)\to F\to0.\]
Applying functor $\Hom(T',-)$ for any $T\in\T'=\T\cap\W$ to the short exact sequence, we have $\Hom(T',T^*_S(F))=0$. Thus $T^*_S(F)\in \F'$.

Conversely, for any $X\in\F'$, we consider the following diagram, where the second row is given by the canonical triangle of $X$ with respect to $(\T,\F)$.
T^*_S(X_{\T})\ar[r]^-{\beta}\ar[d]_-{h} & X_{\T}\ar[r]^-{\alpha}\ar@{=}[d] &\mu_S(X_{\T})\ar[d]^-{f_1}\ar[r]\ar[dl]_{\alpha'}& T^*_S(X_{\T})[1]\ar[d]\\
X_{\F}[-1]\ar[r]_-{\beta'} & X_{\T}\ar[r]_-{f} & X\ar[r]_-{\delta} & X_{\F} \\
By Lemma~\ref{lem:T*_S(T)}, we have $T^*_S(X_{\T})\in\T'$.

Since $X\in\F'$, we have $f\beta=0$. So there exists a morphism $f_1\in\Hom(\mu_S(X_{\T}),X)$ such that $f=f_1\alpha$. Since $\mu_S(T)\in\add(S\oplus S[1])$, $S\in\T$ and $X_{\F}\in\F$, we have $\delta f_1=0$. It implies there exists $\alpha'\in\Hom(\mu_S(X_{\T}),X_{\T})$ such that $f_1=f\alpha'$. Thus $f(\id_{X_{\T}}-\alpha'\alpha)=0$.

Since $\alpha'\alpha$ is a morphism in $\A$ and $f$ is monomorphism in $\A$, we have $\alpha'\alpha=\id_{X_{\T}}$ and $X_{\T}\in\add S$. Since $X\in{}^{\perp}S$, applying $\Hom(-,S)$ to the second row of the above commutative diagram, we have that $\Hom (X_{\F},S)=0$ and $\Hom(X_{\T},S)\cong\Hom(F,S[1])$. This implies $X_{\T}\cong\Hom(F,S[1])^*\otimes S$ and   $X\cong T^*_S(X_{\F})$.
\end{proof}

If the simple top $S$ belongs to $ \F[1]$, then for some $F\in \F$, the object $T^*_S(F)$ may be not in $\A$.  The torsion free class $\F'$  has one more constraint as follows.
\begin{lemma}
Suppose $S\in\F[1]$ as above. Then $X\in \F'$ if and only if there exists some  $F\in \F$ with $\Hom(F,S)=0$ such that $X$ is a direct summand of $T^*_S(F)$ in $\mathcal A$.
\end{lemma}
\begin{proof}
Let $F\in\F$ such that $\Hom(F,S)=0$. By the definition of $\mu_S$, we have $\mu_S(F)=\Hom(F,S[-1])^*\otimes S[-1]$. Since $S$ is simple in $\B$, the canonical morphism  $\alpha_F[1]\colon F[1]\to \mu_S(F)[1]$ is an epimorphism in $\B$. Hence it gives a short exact sequence
\[0\to T^*_S(F)[1]\to F[1]\to\mu_S(F)[1]\to 0\]
in $\B$. Thus we could write $T^*_S(F)=K_1\oplus K_2$ with $K_1\in\F$ and $K_2\in\T[-1]$ by Lemma~\ref{lem:split}. It follows that $K_1\in\ps\cap\F=\F'$.

Conversely, for any $X\in\F'$, we have the following commutative diagram
\[\xymatrix{
\mu_S(F)[-1]\ar[r]\ar@{=}[d]&T^*_S(F)\ar[d]\ar[r]^-{g}&F\ar[r]^-{\alpha}\ar[d]&\mu_S(F)\ar@{=}[d]\\
\mu_S(F)[-1]\ar[r]& M\ar[r]\ar[d] & X_{\T}[1]\ar[d]\ar[r]_-{h}&\mu_S(F)\\
where the third column is given by the canonical sequence of $X$ with respect to $(\T,\F)$ on $\A$.
Since $X\in\F'$, we have $\alpha f=0$ and  a morphism $f'\in\Hom(X,T^*_S(F))$ such that $gf'=f$. The commutative diagram is followed by the octahedral axiom.

Since $\mu_S(F)\in\add(S[-1]\oplus S)\subseteq\F\cup\F[1]$ and $X_\T\in\T$, we have that $h=0$ and thus $\alpha=0$. It implies that $\mu_S(F)=0$. Applying $\Hom(-,S)$ to the third column of the commutative diagram, we have that $\Hom(X_{\T},S)=0$ and $X_{\T}\in \ps$. It follows that $X_{\T}\in \T'$ and hence $X_{\T}=0$.  Hence $X\cong F$, which is exactly $T^*_S(F)$.
\end{proof}

\begin{corollary}\label{cor:F' in F}
If $S\in\F[1]$ then the torsion class $\F'\subseteq\F$.
\end{corollary}

\begin{proposition}\label{prop:SF}
Let $X$ be an object in $\A$ such that $T^*_S(X)\in\F'\ast\T'[-1]$ for a simple object  $S$ of $\B$ which lies in $\T$,  then $X\in \add S*\F$.

\end{proposition}
\begin{proof}
Assume $S\in\T$. Let $0\to X_\T\to X\to X_\F\to 0$ be the canonical sequence of $X$ with respect to $(\T,\F)$. Applying the twist functor $T^*_S$ to $X$ and $X_\T$ respectively, we have the following diagram
T^*_S(X_\T)\ar[r]^-{\beta}\ar[d]_-{h} & X_\T\ar[r]^-{\alpha}\ar[d]_{f} & \mu_S(X_\T)\ar[d]^-{p}\ar[r]& T^*_S(X_\T)[1]\ar[d]\\
T^*_S(X)\ar[r]\ar[d] & X\ar[r]_-{g}\ar[d] &  \mu_S(X) \ar[d]\ar[r] & T^*_S(X)[1]\ar[d]\\
M\ar[r]_{(\delta_1,\delta_2)}\ar[d] & X_\F\ar[r]\ar[d] & N\ar[r]\ar[d] & M[1]\ar[d]\\
T^*_S(X_\T)[1]\ar[r]& X_\T\ar[r] & \mu_S(X_\T)\ar[r] & T^*_S(X_\T)[1]
By Lemma~\ref{lem:T*_S(T)}, $T^*_S(X_\T)\in \T'$. Since $\mu_S(X)\in\langle S\rangle$, we have $gf\beta=0$, and hence there exists a morphism $h\colon T^*_S(X_\T)\to T^*_S(X)$ such that the top-left square commutes. So we get the whole commutative diagram by $3\times 3$ lemma.

The assumption $T^*_S(X)\in \F'*\T'[-1]$ implies $h=0$. Thus $M=T^*_S(X)\oplus T^*_S(X_\T)[1]$. Since $\delta_2\in\Hom(T^*_S(X_\T)[1], X_\F)=0$, we get that $T^*_S(X_\T)[1]$ is a direct summand of $N\in\langle S\rangle$. Since  $T^*_S(X_\T)\in \ps$, we have $T^*_S(X_\T)=0$. So the first row shows that $X_\T\cong\mu_S(X_\T)\in\add S$ and the assertion follows.

%Suppose $S\in\F[1]$ then consider the following diagram obtained from the octahedral axiom
%    &S[-1]^l\ar@{=}[r]\ar[d]^-{(0,1)'}&S[-1]^l\ar[d]&\\
%    X\ar[r]^-{(f,g)}\ar@{=}[d]&S[-1]^l\oplus S^k\ar[r]\ar[d]^-{(1,0)}&T^*_S(X)[1]\ar[d]\ar[r]&X[1]\ar@{=}[d]\\
%    X\ar[r]_-{f}& S^k\ar[r]\ar[d] &M[1]\ar[d]\ar[r] &X[1]\\
%    &S^l\ar@{=}[r]&S^l&
%where $l=\dim\Hom(X,S[-1])$ and $k=\dim\Hom(X,S)$. Note that in this case, $T^*_S(X)\in\F'\subseteq\F$. So $M\in\T^*_S(X)\ast S[-1]\subseteq\F$. So $X\in M\ast S\subseteq\F\ast\add S$ and the assertion follows.
\end{proof}

\section{Proof of Theorem \ref{thm:main}}
Keep the notation as in the previous subsection. Throughout this section, we assume that $\Lambda$ is a finite-dimensional hereditary algebra. Let $\B'=\F'[1]\ast\T'$ the HRS tilt of $\W$ with respect to $(\T',\F')$.  In this section, we will prove the following proposition and Theorem \ref{thm:main}. We actually give a construction of the objects $B$ and $C$ in the following results. This construction closely relates to the image of the realization functor, see Section \ref{sec:rel}.

\begin{proposition}\label{prop:main}
If $(\T',\F')$ is complete in $\W$, then $(\T,\F)$ is complete in $\A$
%Suppose for each $A'\in\W$, there is a triangle $A'\rightarrow B'\rightarrow C'\rightarrow A'[1]$ in $\db{\W}$ such that $B',C'\in\B'$. Then for each $A\in\A$, there is a triangle $A\rightarrow B\rightarrow C\rightarrow A[1]$ in $\db{\A}$ such that $B,C\in\B$.
\end{proposition}
\begin{remark}
For each $A\in \A$, the torsion pair $(\T,\F)$ is complete in $\A$ if and only if there is a triangle $A\to B\to C\to A[1]$ in $\db{\A}$ such that $B,C\in \B$. Such triangles and the 5-term exact sequences determined by $(\T,\F)$ are the same thing in different forms, see \cite[Proposition~3.2]{CHZ} for more details. We will use this equivalence in this section without mentioning that.
\end{remark}

\subsection{Proof of Proposition~\ref{prop:main}}
\subsubsection{The case \texorpdfstring{$S\in\F[1]$}{}}
%Throughout this subsection, we assume the simple top $S$ is in $\F[1]$.
\begin{lemma}\label{lem:red-t}
Assume $S\in\F[1]$. If $(\T',\F')$ is complete in $\W$ then for each $X\in\A$ with $\Hom(X,S[-1])=0$, there is a triangle $X\rightarrow M\rightarrow N\rightarrow X[1]$ in $\db{\A}$ such that $M,N\in\B$.
\end{lemma}

\begin{proof}
Let $X\in\A$ such that $\Hom(X,S[-1])=0$. Then $\mu_S(X)=\Hom(X,S)^*\otimes S$. By the triangle $\mu_S(X)[-1]\to T^*_S(X)\to X\to \mu_S(X)$, we have $T^*_S(X)\in\ps\cap\A=\W$.  By the octahedral axiom, we have the following diagram, where the second column is from the assumption, i.e. $B',C'\in \B'$.
\mu_S(X)[-1]\ar@{=}[d]\ar[r]^-{g}&T^*_S(X)\ar[r]\ar[d]^-{h} &X\ar[d]\ar[r]^-{\alpha}&\mu_S(X)\ar@{=}[d]\\
\mu_S(X)[-1]\ar[r]_-{hg}&B'\ar[r]\ar[d]&M\ar[r]\ar[d]&\mu_S(X)\\
Note that $\B'=\F'[1]\ast\T'\subseteq\F[1]\ast\T=\B$ by Corollary~\ref{cor:F' in F}. Taking the cohomology of the third row, we have a long exact sequence in $\A$
\[0\to B'_1\to M_1\to \mu_S(X)[-1]\xrightarrow{f} B'_0\to M_0\to 0.\]
Since $B'_0\in\T'\subseteq\T$, we have $M_0\in\T$. Taking the cohomology of the third column, we have a long exact sequence
\[0\to M_1\xrightarrow{g} C'_1\to X\xrightarrow{f} M_0\to C'_0\to 0.\]
By Corollary~\ref{cor:F' in F}, $C'_1\in\F'\subseteq\F$. Let $N=C'$ then the triangle in the third column is the desired triangle.
\end{proof}

\begin{lemma}\label{lem:finf}
Assume that $S\in\F[1]$. If $(\T',\F')$ is complete in $\W$ then for each $A\in\A$, there is a triangle $A\rightarrow B\rightarrow C\rightarrow A[1]$ in $\db{\A}$ such that $B,C\in\B$.
\end{lemma}
\begin{proof}
For any indecomposable $A\in\A$, consider the triangle
\[X\to A\xrightarrow{\alpha} G\xrightarrow{g}X[1]\]
where $G=\Hom(A,S[-1])^*\otimes S[-1]$ and $\alpha$ is the canonical map. Applying $\Hom(-,S[-1])$ to the triangle, we get $\Hom(X,S[-1])=0$. Taking cohomology, we have a long exact sequence
\[0\to H^0(X)\to A\to G\xrightarrow{f} H^1(X)\to 0\]
in $\A$. By Lemma~\ref{lem:ker}, we have $H^1(X)=S[-1]^l\oplus T$ with $l\in\mathbb{Z}_{>0}$ and $T\in\T$. Since $\alpha$ is minimal, $g$ belongs to radical. So does the component of $g$ between $G$ and $S[-1]^l$. This implies $l=0$ because $S$ is exceptional. Thus we have $H^1(X)\in\T$.

Since $H^0(X)\in\A$ and $\Hom(H^0(X),S[-1])=0$, by Lemma~\ref{lem:red-t}, there is a triangle $H^0(X)\to M\to N\to H^0(X)[1]$ with $M,N\in\B$. Now we replace the second column and second row of the commutative diagram in Lemma~5.3 by $N[-1]\oplus H^1(X)[-1]\to X\to M\to N\oplus H^1(X)$ and $G[-1]\to X\to A\to G$. Then we obtained a triangle $N[-1]\oplus H^1(X)[-1]\to A\to B\to N\oplus H^1(X)$ in the third column by the octahedral axiom. By calculation of the cohomology, we have $B\in\B$ and $C=N\oplus H^1(X)\in\B$.
%So we have the following commutative diagram via the octahedral axiom
%	&N[-1]\oplus H^1(X)[-1]\ar[d]\ar@{=}[r]&N[-1]\oplus H^1(X)[-1]\ar[d]&\\
%	G[-1]\ar@{=}[d]\ar[r]&X\ar[r]\ar[d] &A\ar[d]\ar[r]^-{\alpha}&G\ar@{=}[d]\\
%	G[-1]\ar[r]&M\ar[r]\ar[d]&B\ar[r]\ar[d]&G\\
%	&N\oplus H^1(X)\ar@{=}[r]&N\oplus H^1(X)&   }$$
%    Let $C=N\oplus H^1(X)$. Taking cohomology, a similar discussion as in Lemma~\ref{lem:red-t} shows that $B, C\in\B$.
\end{proof}

\subsubsection{The case \texorpdfstring{$S\in\T$}{}}
\begin{lemma}\label{lem:2-term}
For any nonzero $X\in\F'$, we have $T_S(X)\in\A$. In particular, $T_S(B')\in\A\ast\A[1]$ for any $B'\in\B'$.
\end{lemma}
\begin{proof}
Without loss of generality, suppose that $X$ is indecomposable. Taking cohomology of the triangle $\lambda_S(X)\to X\to T_S(X)\to \lambda_S(X)[1]$, we have a long exact sequence,
\[0\to H^{-1}(T_S(X))\to H^0(\lambda_S(X))\to X\to H^0(T_S(X))\to H^1(\lambda_S(X))\to0.\]
By Lemma~\ref{lem:iso}, we have $T_S(X)$ is indecomposable. So either $H^0(T_S(X))=0$ or $H^{-1}(T_S(X))=0$. In the former case, since $H^0(\lambda_S(X))\in\add S\subseteq\T$, we have $X\in\T\cap\W=\T'$, a contradiction. So $H^{-1}(T_S(X))=0$ and the first assertion follows.

For any $B'\in\B'$, note that $B'=B'_1[1]\oplus B'_0$ with $B'_0\in\T'$ and $B'_1\in\F'$, and $T_S(B'_0)\in\A\ast\A[1]$ by definition. So the last assertion follows from the first assertion.
\end{proof}

%The HRS tilt of $\W$ with respect to $(\T',\F')$ is $\B'\cong\F'[1]\ast\T'$. For any $B'\in \B'$, we have $B'=B'_1[1]\oplus B’_0$, where $B'_i=H^{-i}_\A(B')$. Note that $T_S(B')=T_S(B'_0)\oplus T_S(B'_1)[1]$. Moreover, since $B'_1=T^*_ST_S(B'_1)$ by Lemma~\ref{lem:iso}, we have $T_S(B'_1)\in\A$ by Lemma~\ref{lem:2-term} and Lemma~\ref{lem:split}. Thus $T_S(B')\in\A\ast\A[1]$.

Now we could prove the Proposition \ref{prop:main} for $S\in \T$ in the following steps.
\begin{lemma}\label{lem:step 1}
If $(\T',\F')$ is complete in $\W$ then for each $A\in\A$ such that $\Hom(A,S)=0$ and $\Hom(S,A)=0$, there is a triangle $A\rightarrow B\rightarrow C\rightarrow X[1]$ in $\db{\A}$ such that $B,C\in\B$.
\end{lemma}

\begin{proof}
Let $A\in\A$ such that $\Hom(A,S)=0$. Then $\mu_S(A)=\Hom(A,S[1])^*\otimes S[1]$, which implies $T^*_S(A)\in \W$. By assumption, there is a triangle $T^*_S(A)\rightarrow B'\rightarrow C'\rightarrow T^*_S(A)[1]$ in $\db{\W}$ such that $B',C'\in\B'$.

Consider the following diagram. Since $\beta'h\alpha'\in\Hom(\mu_S(A)[-1],T_S(B'))=0$, we have a morphism $f\in\Hom(\mu_S(A),\lambda_{S}(B'))$ such that the top-left square commutes. By $3\times 3$ Lemma, we have the following commutative diagram
\mu_S(A)[-1]\ar[r]^{\alpha'}\ar@{->}[d]_-{ f[-1]}&T^*_S(A)\ar[d]^{h}\ar[r]&A\ar[r]^-{\alpha}\ar[d]^-{g}&\mu_S(A)\ar[d]^{f}\\
\lambda_{S}(B')[-1]\ar@{->}[r]\ar[d]&B'\ar[r]^{\beta'}\ar[d]&T_S(B')\ar[r]^-{\beta}\ar[d]&\lambda_{S}(B')\ar[d]\\		V[-1]\ar[d]\ar[r]&C'\ar[r]\ar[d]&N\ar[r]^{\delta}\ar[d]&V\ar[d]\\
\mu_S(A)\ar[r]&T^*_S(A)[1]\ar[r]&A[1]\ar[r]&\mu_S(A)[1]
Note that $V$ is in the category $\langle S\rangle$. We could assume that $V\cong \oplus_{k\in \mathbb Z}V_k\otimes S[k]$.

We will fit $A$ into a 5-term exact sequence which is equivalent to the triangle $A\to B\to C\to A[1]$. We construct the exact sequence by computing the cohomologies of $T_S(B')$ and $N$.

Firstly, we claim that $C'\cong T^*_S(N)$. Since $T_S(B')\in \A*\A[1]$ by Lemma~\ref{lem:2-term}, we have that $N\in \A*\A[1]$. With $\Hom(-,S)^*$ acting on the last column, we get a long exact sequence of vector spaces, where $f_i=\Hom(f, S[i])^*, i=0,1$.
\[\begin{array}{rl}
0& \to \Hom(\lambda_S(B'),S[2])^*\to V_2\to\Hom(\mu_S(A),S[1])^*\xrightarrow{f_1}\Hom(\lambda_S(B'),S[1])^*\\
&\to  V_1 \to \Hom(\mu_S(A),S)^*\xrightarrow{f_0}\Hom(\lambda_S(B'),S)^*\to V_0\to 0\end{array}\]
On the other hand, with $\Hom(-,S)^*$ acts on the third column, we have a long exact sequence, where $g_i=\Hom(g,S[i])^*$
\[\begin{array}{rl}
0&\to\Hom(T_S(B'),S[2])^*\to\Hom(N,S[2])^*\to \Hom(A,S[1])^*\\
&\xrightarrow{g_1}\Hom(T_S(B'),S[1])^*\to\Hom(N,S[1])^*\to \Hom(A,S)^* \\  &\xrightarrow{g_0}\Hom(T_S(B'),S)^*\to \Hom(N,S)^*\to 0
\end{array}\]
Since $\alpha_i=\Hom(\alpha,S[i])^*$ and $\beta_i=\Hom(\beta,S[i])^*$ are isomorphisms, and $\beta g=f\alpha$, we have $g_i=\beta^{-1}_if_i\alpha_i$. Thus, comparing these two sequences, we have $V_i\cong\Hom(N,S[i])^*$. Note that components of $\delta$ are non-zero, otherwise $C'\in\ps$ has common summands with $V[-1]\in\langle S\rangle$, a contradiction. So $\delta\colon N\to V$ is the canonical map and the claim follows.

Secondly, we construct $B$ and $C$ via $N$. Note that $N=H^{-1}(N)[1]\oplus H^0(N)$. Taking the cohomology of the second row and noticing $S\in\T$, we have $H^0(T_S(B'))\in\T$. Taking the cohomology of the third column, we have a long exact sequence
\[0\to H^{-1}(T_S(B'))\to H^{-1}(N)\xrightarrow{f} A\to H^0(T_S(B'))\to H^0(N)\to 0\]
in $\A$. Since $H^0(T_S(B'))\in\T$, we have $H^0(N)\in\T$.

On the other hand, since $T^*_S(H^{-1}(N))[1]$ is a direct summand of $C'\in\F'[1]\ast\T'$, by Proposition~\ref{prop:SF}, we have a short exact sequence $0\to S_N\xrightarrow{i} H^{-1}(N)\xrightarrow{p}F_N\to 0$ with $S_N\in\add S$ and $F_N\in\F$.
Since $\Hom(S,A)=0$ by assumption, we have $fi=0$ and there exists $f'\in\Hom(F_N,A)$ such that $f=f'p$. So we get the 5-term exact sequence
\[0\to \ker f' \to F_N\xrightarrow{f'} A\to H^0(T_S(B'))\to H^0(T_S(C'))\to 0.\]
Thus $B,C\in \B$ are determined by this exact sequence. \end{proof}

Now we could remove the constraint that $\Hom(S,A)=0$ in Lemma~\ref{lem:step 1}.
\begin{lemma}\label{lem:step 2}
If $(\T',\F')$ is complete in $\W$ then for each $A\in\A$ such that $\Hom(A,S)=0$, there is a triangle $A\rightarrow N\rightarrow D\rightarrow X[1]$ in $\db{\A}$ such that $D,N\in\B$.
\end{lemma}

\begin{proof}
Suppose without loss of generality that $A$ is indecomposable. Let $S_A\xrightarrow{\alpha}A\to A'\to S_A[1]$ be the triangle with $S_A=\Hom(S,A)\otimes S$ and $\alpha$ the canonical map. Then $\Hom(S,A')=0$ and $A'\in\A[1]\ast\A$. In particular, $H^0(A')\in\A$, and $\Hom_{\A}(S,H^0(A'))=0$. Applying $\Hom(-,S)$ to the triangle, we have also $\Hom_{\A}(H^0(A'),S)=0$. Thus, by Lemma~\ref{lem:step 1}, there is a triangle $H^0(A')\to B\to C\to H^0(A')[1]$ with $B,C\in\B$.

By Proposition~\ref{prop:SE}, there is a short exact sequence $0\to S\to\S E\to M[1]\to0$ in $\B$ with $E$ the projective cover of $S$ in $\B$. Consider the following diagram, where $M_A=\Hom(S,A)\otimes M$ and $\S E_A=\Hom(S,A)\otimes\S E$.
M_A\ar[d]\ar[r]&D[-1]\ar[r]\ar[d] & C[-1]\ar[d]_-{f}\ar[r]^-{u}& M_A[1]\ar[d]^-{i}\\
S_A\ar[r]^-{\alpha}\ar[d] & A\ar[r]\ar[d] & H^0(A')\oplus H^{-1}(A')[1]\ar[r]^-{g}\ar[d] & S_A[1]\ar[d]^-{h}\\
\S E_A\ar[d]\ar[r]&N\ar[r]\ar[d] & B\oplus H^{-1}(A')[1]\ar[r]\ar[d] & \S E_A[1]\ar[d]\\
M_A[1]\ar[r]&D\ar[r] & C\ar[r]& M_A[2]
Since $\Hom(C[-1],\S E[1])=0$, we have $hgf=0$ and hence there exists a morphism $u\in\Hom(C[-1],M_A[1])$ such that the top-right square commutes. So we get the whole commutative diagram by $3\times 3$ Lemma.

Since $M[1]\in\B$, we have $H^{-1}(M[1])\in\F$. Taking the cohomology of the first row, we get $H^{-1}(D)\in\F$ by $H^{-1}(C)\in\F$. Similarly, since $\S E\in\B$, we have $H^0(\S E_A)\in\T$. Taking the cohomology of the third row, we have $H^0(N)\in\T$ by $H^0(B)\in\T$. Thus we have that $D,N\in \B$.
\end{proof}

\begin{lemma}\label{lem:fint}
If $(\T',\F')$ is complete in $\W$ then for each $A\in\A$, there is a triangle $A\rightarrow B\rightarrow C\rightarrow X[1]$ in $\db{\A}$ such that $B,C\in\B$.
\end{lemma}
\begin{proof}
Consider the triangle $S^A[-1]\to X\to A\xrightarrow{\alpha} S^A$ with $S^A=\Hom(A,S)^*\otimes S$ and $\alpha$ the canonical map. Taking the cohomology of this triangle, we have $H^{1}(X)\in\T$. Note that $H^0(X)\in\A$ and $\Hom(H^0(X),S)=0$, there is a triangle $H^0(X)\to N\to D\to H_0(X)[1]$ with $N,D\in\B$ by Lemma~\ref{lem:step 2}.

Combining the above triangles, we have the following commutative diagram by the octahedral axiom, where $C=D\oplus H^{1}(X)$
S^A[-1]\ar[r] &N\ar[r]\ar[d]&B\ar[r]\ar[d]&S^A\\
Taking the cohomology of the third row and noticing that $H^0(N)\in\T$, we have $H^0(B)\in\T$. Taking cohomology of the third column and noticing that $H^{-1}(C)=H^{-1}(D)\in\F$, we have $\H^{-1}(B)\in\F$. So the third column is the desired triangle.
\end{proof}

Proposition~\ref{prop:main} follows from  Lemma~\ref{lem:finf} and Lemma~\ref{lem:fint}.

\subsection{Proof of Theorem~\ref{thm:main}}
If a torsion pair $(\T,\F)$ is complete in $\A$, then Corollary \ref{cor:suff} shows that the aisle $\U_\T$ is closed under the Serre functor.

Conversely, we assume that $\U_{\T}$ is closed under the Serre functor. If $\T$ either is finitely generated  or has no
Ext-projective objects, then $(\T,\F)$ is complete by Lemma~\ref{lem:enough} and Lemma~\ref{lem:no}.

In general, we suppose $\T$ is not finitely generated and has non-zero Ext-projective object. Then $\U_{\T}$ has Ext-projective object $E$ with its simple top $S$ in $\B$. Then for $\db{\W}$, $\U_\T$ induces a bounded t-structure $(\U',\V')$ (and hence a torsion pair $(\T',\F')$) such that $\U'$ has one fewer Ext-projective generators than $\U_{\T}$.

Repeating this procedure finitely many times, we could get a aisle $\U_{0}$ without non-zero Ext-projective and thus its corresponding torsion class $\T_0$. Applying finitely many times Proposition \ref{prop:main} on $\T_0$, we have that the torsion pair $(\T,\F)$ is complete in $\A$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%     The main structure of the paper

\subsection{Relation with realization functor}\label{sec:rel}
Given a triangulated category $\mathcal D$ with a bounded t-structure $(\U,\V)$ and its heart $\B$, the \emph{realization functor} $F\colon \db\B\to \mathcal D$ is a triangle functor which  naturally extends the embedding $\B\to \mathcal D$ \cite[Section 3.1]{BBD}. In [3], the existence of the realization functor depends on filtered structures over $\mathcal D$. Note that a realization functor exists if $\mathcal D$ is algebraic [8, 17].
It follows that the realization functor $F\colon \db{\B}\to \db \A$ exists for the t-structure $(\U_\T,\V_\T)$ given by a torsion pair $(\T,\F)$ on $\A$.  For more details about tilting theory and realization functor, we refer to [19]. We remark that the uniqueness of the functor is not known [7, 22].

We fix a realization functor $F\colon\db\B\rightarrow \db\A$ with respect to the heart $\mathcal{B}$. Let $A'$ be the HRS titling of $\B$ with respect to $(\F[1],\T)$. The functor $F$ is $t$-exact, where $\db\B$ is endowed with the $t$-structure given by the heart $\mathcal{A}'$ and $\db\A$ has the canonical $t$-structure. Moreover, the restriction $F|_{\mathcal{A}'}\colon \mathcal{A}'\rightarrow \mathcal{A}$ is exact \cite[Section 3]{CHZ}.

We have the following description of the essential image of the realization functor  $F$, i.e. the full subcategory of $\db{\A}$ consisting of those objects $X$, which are isomorphic to $F(Y)$ for some object $Y$ in $\db{\B}$.

\begin{proposition}[{\cite[Proposition~3.2]{CHZ}}]\label{prop:ff2}
Keep the notation as above. Assume that $A\in \mathcal{A}$. Then the following statements are equivalent:
\begin{enumerate}
\item The object $A$ belongs to ${\rm Im}\;  F$.
\item There is an exact triangle $A\rightarrow B^0\rightarrow B^1\rightarrow \Sigma(A)$ in $\db\A$ with $B^i\in \mathcal{B}$;

\end{enumerate}
\end{proposition}

In general, it is difficult to construct the essential image or preimage of a realization functor. In the simplest case of realization functor, i.e., the $t$-structure $(\U,\V)$ is the standard $t$-structure, we have that the realization functor is the identity functor. Except for this trivial case, we do not know how to construct the preimage of the realization functor for the $t$-structure $(\U_{\T},\V_{\T})$ given by a torsion pair $(\T,\F)$.

If the aisle $\U_\T$ is closed under the Serre functor $\S$ or $(\T,\F)$ is complete, then realization functor $F$ is an equivalence and each object $X$ in $\A$ has preimage in $\db{\B}$. By Proposition \ref{prop:main} and Proposition \ref{prop:ff2}, we could construct the preimage of the realization functor for any object $X$ in $\A$. Unfortunately, we do not know how to construct the preimage of realization functor in general.

\begin{remark}In [18], Neeman considered the problem when a long exact sequence is obtained from the cohomology of a triangle in a derived category.
Given a 5-term exact sequence
\[0\to X_0\to X_1\to X_2\to X_3\to X_4\to 0  \] in derived category $\db{\A}$, this sequence defines a class in $\Ext^3(X_4,X_0)$.  Neeman showed that the long exact sequence is obtained from taking the cohomology of a triangle if and only if the above class in $\Ext^3(X_4,X_0)$ under realization functor vanishes.
Note that Neeman's argument is in a more general setting. He also claimed the class in  $\Ext^3(X_4,X_0)$ vanishing is a necessary condition but not known whether it suffices. It looks interesting to find the relationship between the Serre functor and the completeness of the torsion class in a more general case.
\end{remark}

\vskip 5pt

\noindent {\bf Acknowledgements}\quad  The work is supported by the National Natural Science Foundation of China (No.12001164). The authors would like to thank Xiao-Wu Chen, Yifei Cheng, and Yiyu Cheng for their discussions and comments. We special thank Yu Zhou for his inspiration discussions and suggestions.
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\end{document}

\begin{comment}
\begin{proof}
\hp{Suppose $S\in\T$ then $(2)\Rightarrow(1)$.} We prove the proposition $R=R^*_S(A')$. By Lemma~\ref{lem:F'} we have $\F'=\add S\ast\F$ the following diagram.
&A'[-1]\ar[d]\ar[r] & R[-1]\ar[d]\ar[r] &  \\
<EMAIL_ADDRESS><EMAIL_ADDRESS>&<EMAIL_ADDRESS>& \\
\Hom(S,C'_1)\otimes S\ar@{.>}[d]_{h'}\ar[r]_-{(\alpha_C,0)'} & C'[-1]\ar[r]\ar[d] & C[-1]\ar[r]\ar[d]^-{h} &\\
\oplus_{i\in\mathbb{Z}}\Hom(S[i],A')\otimes S[i]\ar[r]_-{\alpha}&A'\ar[r]&R\ar[r]& \oplus_{i\in\mathbb{Z}}\Hom(S[i],A')\otimes S[i+1]
where $\alpha$ is the $S[\mathbb{Z}]$-approximation of $A'$, we have $h'$. Since $S$ is rigid and brick, we have $V=S[-2]^m\oplus S[-1]^n\oplus S^l$ for some $m,n,l\in\mathbb{Z}$. Taking homology of the second column we see that $B\in\A[1]\ast\A$. Taking homology of the second row, we have
$B'_0\to B_0\to S^m\to 0$ which implies $B_0\in\T$.

By Lemma~\ref{lem:F'}, we have that $\F'=\add S\ast\F$. By octahedral axiom, we have the following diagram.
&A'[-1]\ar[d]\ar@{=}[r] & A'[-1]\ar[d] &  \\
\Hom(S,C'_1)\otimes S\ar<EMAIL_ADDRESS><EMAIL_ADDRESS>&<EMAIL_ADDRESS>& \\
\Hom(S,C'_1)\otimes S\ar[r]_-{(\alpha_C,0)'} & C'[-1]\ar[r]\ar[d] & C[-1]\ar[r]\ar[d] &\\
where $\alpha_C$ is the $S$-approximation of $C_1\in\F$. By Lemma~\ref{lem:F'}, we have $C[-1]=C_1\oplus C_0[-1]$ with $C_1\in\F$ and $C_0\in\T$. Taking homology of the second row, we have
\[0\to \Hom(S,C'_1)\otimes S\to B'_1\to H^{-1}(B)\to 0\to B'_0\to H^0(B)\to 0\]
So we have $B[-1]=H^{-1}(B)\oplus H^0(B)[-1]$ with $B_0\in\T$.
%Note that, by admissible, we have $f=\begin{pmatrix}f_{11}&\ast\\0&\ast\end{pmatrix}$ with $f_{11}:B'_1\to C'_1$. Since $\alpha_C$ is nonzero, we have $h\neq0$ and hence be the $S$-approximation of $B'_1$. Thus we have $B[-1]=B_1\oplus B_0[-1]$ with $B_1\in\F$ and $B_0\in\T$.

Consider the following diagram
&  C[-1]\ar@{=}[r]\ar[d] &<EMAIL_ADDRESS>&   \\
\oplus_{i\in\mathbb{Z}}\Hom(A,S[i])^*\otimes S[i-1]\ar[r]\ar@{=}[d]&A'\ar[d]\ar[r]&A\ar[r]^-{\alpha}\ar@{.>}[d]&\oplus_{i\in\mathbb{Z}}\Hom(A,S[i])^*\otimes S[i]\\
\oplus_{i\in\mathbb{Z}}\Hom(A,S[i])^*\otimes S[k-1]\ar[r]&B\ar[r]\ar@{.>}[d]&X\ar[r]\ar@{.>}[d]&\oplus_{i\in\mathbb{Z}}\Hom(A,S[i])^*\otimes S[k]\\
where $\oplus_{i\in\mathbb{Z}}\Hom(A,S[i])^*\otimes S[i]=\Hom(A,S)^*\otimes S\oplus\Hom(A,S[1])^*\otimes S[1]$. Taking homology of the third column, we have that $H^i(X)=0$ for $i\neq 0,-1$ and
\[0\to H^{-1}(X)\to C_1\to A\to H^0(X)\to C_0\to 0.\]
Since $C_1\in\F$, we have $H^{-1}(X)\in\F$. On the other hand, taking homology of the third row, we have
\[0\to H^{-1}(B)=B_1\to H^{-1}(X)\to \Hom(A,S[1])^*\otimes S\xrightarrow{f} B_0\to H^0(X)\to \Hom(A,S)^*\otimes S\to 0\]
Since $B_0\in\T$ which is closed under image and extension, we have that $H^0(X)\in\T$. Hence the assertion follows.

%Note that $C=C_1[1]\oplus C_0$ with $C_0=C'_0\in\T'\subseteq\T$ and $C_1\in\F$. Since $S$ is simple in $\B$ and rigid, we have that $V_i=0$ except for $i=0,1,2$. Taking homologies, we have $B\in\A[1]\ast\A$ with $B_1=B'_1\ast S$ and $B_0=B'_0\ast S\in\T$Taking homologies, we have
%\[0\to B_1\to C_1\to A\to B_0\to C_0\to 0\]
%gives the desired sequence.
\end{proof}

\begin{corollary}[Induction Lemma]\label{lem:idu}
With the above notation, the following are equivalent.
\begin{enumerate}
\item the aisle $\U$ (or torsion class $\T$) induces an exact sequence
\begin{equation}
0\to F_1\to F_2\to X\to T_1\to T_2\to 0
\end{equation}
with $T_i\in \T$ and $F_i\in \F$.
\item  the aisle $\U'$ (or torsion class $\T'$) induces an exact sequence
\begin{equation}
0\to F'_1\to F'_2\to X'\to T'_1\to T'_2\to 0
\end{equation}
with $T'_i\in \T'$ and $F'_i\in \F'$.
\end{enumerate}
\end{corollary}

\begin{proof}
If there is an sequence
\[\begin{tikzcd}
A \arrow{d} \arrow{r}[near start]{\phi}[near end]{\psi}
& B \arrow[red]{d}{\xi} \\
C \arrow[red]{r}[blue]{\eta}
& D
\end{tikzcd}\]
\end{proof}

\end{comment}

\begin{comment}
\newpage

For any $B'\in\B'=\F'[1]\ast\T'$, we have $B'=B'_1[1]\oplus B_0$, where $B_i=H^{-i}_\A(B')$. Note that $R^*_S$ is additive, we have $R^*_S(B')=R^*_S(B'_0)\oplus R^*_S(B'_1)[1]$. Moreover, since $B'_1=T^*_SR^*_S(B'_1)$ by Lemma~\ref{lem:iso}, we have $B'_1\in\A$ by Lemma~\ref{lem:F'}.

For any $A\in\A$, such that $\Hom(A,S)=0$, consider the following diagram, where $A'=T^*_S(A)$.
Then, since $\A$ is hereditary, we have $A_S=\Hom(A,S[1])^*\otimes S[1]$. So $A'\in A_S[-1]\ast A\in\A$ and hence belongs to $\W$.

Since $\beta$ is the $\thick S$-approximation of $B'$, there exists $f\in\Hom(A_S[-1],{}_{S}B)$ such the up-left square commutes. So we get the whole commutative diagram by $4\times 4$-lemma.

Since $\Ext^n(S,S[n])=0$ and $\End S=\k$, we have $V=\oplus_{k\in\mathbb{Z}}V_{k-1}\otimes S[k-1]$. Since $N\in R^*_S(B')\ast A[1]\subset\A\ast\A[1]$, we may write $N=N_0\oplus N_1[1]$ with $N_i=H^{-i}(N),i=0,1$. Taking the homology of the third column, we have a long exact sequence in $\A$
\[0=H^{-1}(A)\to B_1\to H^{-1}(N)\to A\to B_0\to H^0(N)\to H^1(A)=0,\]
where $B_i=H^{-i}(R^*_S(B')), i=0,1$. Since $B_1\in B'_1\ast S\subseteq \F$, we have $N_0\in\T$.

By Lemma~\ref{lem:iso}, we can replace the second row by a new triangle as the following.
\oplus_{k\in\mathbb{Z}}V_{k-1}\otimes S[k-1]\ar@{.>}[d]\ar@{.>}[r]&C'\ar[r]\ar<EMAIL_ADDRESS>S[k]\\
where $R^*_S(X)_S=\oplus_{i=0,1,2}\Hom(R^*_S(X),S[i])\otimes S[i]$. With $\Hom(-,S)^*$ acts on the last column, we have a long exact sequence of vector space.
\[\begin{array}{rl}
&  0\to \Hom(B,S[2])^*\to V_1\to\Hom(A,S[1])^*\to \Hom(R^*_S(B'),S[1])^*\\
\to & V_0 \to \Hom(A,S)^*\to\Hom(R^*_S(B'),S)^*\to V_{-1}\to 0
\end{array}\]
On the other hand, with $\Hom(-,S)^*$ acts on the third column, we have a long exact sequence
\[\begin{array}{rl}
& 0\to\Hom(B[-2],S)^*\to\Hom(N[-2],S)^*\to \Hom(A[-1],S)^*\to \Hom(B[-1],S)^* \\
\to & \Hom(N[-1],S)^*\to \Hom(A,S)^*\to \Hom(B,S)^*\to \Hom(N,S)^*\to 0
\end{array}\]
\hp{Comparing these two sequence we have $V_i\cong\Hom(N,S[i+1])^*$, which implies $C'=T^\ast_S(N)$.}

Let $N_1=\oplus^n_{i=1} N^i_1$ with $N^i_1$ indecomposable. Then $T^*_S(N_1)=\oplus^n_{i=1} T^*_S(N^i_1)$. Since $T^*_S(N^i_1)\cong N^i_1$ in $\ps$, we have $T^*_S(N^i_1), 1\leq i\leq n$ are indecomposable. Since $T^*_S(N^i_1)[1]$ is a summand of $T^*_S(N)\cong C'\in\A\ast\A[1]$, we have $T^*_S(N^i_1)\in\A[-1]$ or $\A$. Let
\[I=\{1\leq i\leq n|T^*_S(N^i_1)\in\A[-1]\}\text{ and }J=\{1\leq j\leq n|T^*_S(N^j_1)\in\A\}.\]
By Corollary~\ref{cor:SF}, we have $N^j_1\in\add S\ast F$ for any $j\in J$. For any $i\in I$, by admissibility, we have that the corresponding triangle is
\[\Hom(N^i_1,S)\otimes S[-1]\xrightarrow{f} T^*_S(N^i_1)\to N^i_1\to \Hom(N^*_1,S)\otimes S.\]
In particular, $\Hom(N^i_1,S[1])=0$. But then, since $T^*_S(N^i_1)[1]\in\A$ is a summand of $T^*_S(N)=C'$, we have $T^*_S(N^i_1)[1]$ be a summand of $C'_0\in\T'\subset\T$. Since $S$ is simple in $\B$, we have $N^i_1\in\F$.

Suppose $\Hom(S,A)=0$. Then take the homology of the third column, we have a long exact sequence in $\A$
\[0\to B_1\to N^I_1\oplus N^J_1\xrightarrow{(f,g)} A\to B_0\to N_0\to 0\]
with $N^I_1=\oplus_{i\in I}N^i_1$ and $N^J_1=\oplus_{j\in J}N^j_1$. Since $N^J_1\in\add S\ast\F$, there exists a short exact sequence $0\to S_J\to N^J_1\to F_J\to 0$. Consider the following diagram in $\A$
0\ar[r]&B_1\ar[r]&N^I_1\oplus N^J_1\ar[r]^-{\overline{(f,g)}}\ar[d]_{(p,1)}&\im(f,g)\ar@{=}[d]\ar@{^{(}->}[r]&A\\
0\ar[r]&\ker f'\ar[r]&N^I_1\oplus F_J\ar[r]^-{f'}&\im(f,g)&
Since $\Hom(S,A)=0$, we have $\Hom(S_N,\im(f,g))=0$. So there exists $f'\in\Hom(N^I_1\oplus F_J,\im(f,g))$ such that $\overline{(f,g)}=f'(p,1)$.
% and hence the following commutative diagram.
Note that $f'$ is a surjection, and $\ker f'\in\F$ because $F_N\oplus N^J_1$ does. So we get the sequence as desired
\[0\to \ker f'\to N^I_1\oplus F_J\xrightarrow{\iota f'} A\to B_0\to N_0\to 0.\]

By Proposition~\ref{prop:SE}, we have a short exact sequence $0\to S\to\S E\to M_\V[1]\to0$ in $\B$.  For any indecomposable $A\in\A$ such that $\Hom(A,S)=0$, consider the following diagram, where $l=\dim\Hom(S,A)$ and $\alpha$ is the right $\add S$-approximation of $A$.
M_\V^l\ar[d]\ar[r]&D[-1]\ar[r]\ar[d] &<EMAIL_ADDRESS>M_\V[1]^l\ar[d]^-{i}\\
S^l\ar[r]^-{\alpha}\ar[d] & A\ar[r]\ar[d] & A'\ar[r]^-{g}\ar[d] & S[1]^l\ar[d]^-{h}\\
\S E^l\ar[r]&M\ar[r] & B\ar[r] & \S E[1]^l
Since $\S E\in\B$ and $C_0\in\T$, we have $\Hom(C_0[-1],\S E[1])=0$. Since $E\in\B$ and $C_1\in\F$, we have $\Hom(C_1,\S E[1])=\Hom(E[1],\F)=0$. Thus $hgf\in\Hom(C[-1],\S E)=0$. So there exists a map $u$ such that the up-right square commutes. Hence the whole diagram commutes by $4\times4$-lemma.

Since $M_\V[1]\in\B$, we have $H^0(M_\V)\in\F$ and $H^1(M_\V)\in\T$. Taking the homology of the first row, we have a long exact sequence in $\A$
\[0\to H^0(M^l_\V)\to D_1\to C_1\to H^1(M^l_\V)\to D_0\to C_0\to 0.\]
Since $C_1\in\F$, we have $H^{-1}(D)\in\F$. %Since $C_0\in\T$ which is closed under images, we have $H^0{D}\in\T$.  %$D\in\F[1]\ast\T$, where $D_i=H^{-i}(D),i=0,1$.
Taking the homology of the third row, we have the long exact sequence in $\A$
\[0\to M_1\to B_1\to \S E\to M_0\to B_0\to 0\]
if $\S E\in\T$, and
\[0\to \S E[-1]\to M_1\to B_1\to 0\to M_0\to B_0\to 0\]
if $\S E\in\T$. In either case, we have $M_0\in\T$. Thus, taking the homology of the second column, we have the sequence
\[0\to M_1\to D_1\to A\to M_0\to D_0\to0\]
as desired.

Now for arbitrary $X\in\A$, consider the following diagram.
S[-1]^d\ar[r] &B\ar[r]\ar[d]&Y\ar[r]\ar[d]&S^d\\
where $\alpha$ is the left $\add S$-approximation of $X$. With $\Hom(-,S)$ acts on the second row, we have $\Hom(A,S)=0$. So by previous discussion, we have the second row with $B,C\in\B$. Taking the homology of the third row, we have a long exact sequence in $\A$
\[0\to B_1\to Y_1\to 0\to B_0\to Y_0\to S\to 0.\]
Since $S\in\T$, we have $Y_0\in\T$. Taking the homology of the third column, we get the sequence as desired
\[0\to Y_1\to C_1\to X\to Y_0\to C_0\to 0.\]

\subsection{The proof when $S\in\F[1]$}
Denote by $T=S[-1]$. Suppose $A\in\A$ such that $\Hom(A,T)=0$. Then for $T^l\to A'\to A\xrightarrow{\alpha} T[1]^l$ where $l=\dim\Hom(A,T[1])$, we have $A'\in\W$. Consider the following diagram

\end{comment}

\begin{comment}
So $N_1=N'_1\oplus N''_1$ such that $C_1=T^*_S(N'_1)$ and
\[\Hom(N''_1,S)\otimes S[-1]\to T^*_S(N''_1)\to N''_1\to\Hom(N''_1,S)\otimes S,\]
with $N''_1\in\F$ and $T^*_S(N''_1)\in\T[-1]$.
%Hence $f$ split (because $S$ is simple in $\T$) and $T^*_S(N^i_1)\cong S$ (because it is indecomposable), a contradiction. Thus $\T^*_S(N_1)[1]\in\A[1]$ is exactly $C'_1[1]\in\F'[1]$.

To show our main theorem (that is, $A$ admits a $B=\F[1]\ast\T$-presentation), it is enough to show $N_1\in\F$. We may assume $N_1=N'_1$ since $N''_1$ is already in $\F$.
%Note also that $C_1=H^{-1}T^*_S(N)=H^0T^*_S(N_1)\in\F'$. To prove the thm above, we need the following.
\begin{lemma}
Let $X\in\A$. If $T^*_S(X)\in\F'$ then $X\in S\ast\F$.
\end{lemma}
\begin{proof}
We have the following commutative diagram, where the first column comes from Lemma~\ref{lem:F'}, the second column is the canonical decomposition of $X$ with respect to $(\T,\F)$ and the up-left square commutes because $S\in\T$.
\Hom(F,S[1])^*\otimes<EMAIL_ADDRESS>& T_X\ar[r]^-{g}\ar[d]_-{h'} & Y\ar[r]\ar[d]^-{h} &\\
T^*_S(X)\ar[r]\ar[d] & X\ar[r]_-{g'}\ar[d] & \oplus_{i=0,1}\Hom(X,S[i])\otimes S[i]\ar[r]\ar[d] & \\
F\ar[r]\ar[d]^{\alpha_F} & X_{\F}\ar[r]\ar[d] & M\ar[r]\ar[d] & \\

Note that $M\in X_{\F}\ast F[1]\subset \A\ast\A[1]$, so $M=H^{-1}(M)[1]\oplus H^0(M)$. Taking homology of the third row, we have a long exact sequence in $\A$.
\[0=H^{-1}(X_{\F})\to H^{-1}(M)\to F\to X_{\F}\to H^0(M)\to H^1(F)=0\]
In particular, we have $M_1:=H^{-1}(M)\in\F$.

On the other hand, $Y\in T_X\ast S[1]\subset\A\ast \A[1]$. So we have $Y\cong Y_1[1]\oplus Y_0$, where $Y_i=H^{-i}(Y),i=0,1$. Note that $Y_0$ is a factor of $T_X$ hence belongs to $\T$. Taking homology of the third column, we have a long exact sequence in $\A$.
\[0=H^{-2}(M)\to Y_1\to\Hom(X,S[1])\otimes S\xrightarrow{f} M_1\to Y_0\to\Hom(X,S)\otimes S\to M_0\to H^1(Y)=0\]
Since $S\in\T$, we have $f=0$ and hence $Y_1\cong\Hom(X,S[1])\otimes S\in\add S$. Note that we have also $M_0\in\T$. Since $S$ is simple in $\T\subseteq\B$, we have $M_0\cong\Hom(M,S)^*\otimes S$. So we get a ses $0\to M_1\to Y_0\to V\otimes S\to 0$ in $\A$ with $V$ a vector space of dimension $\dim\Hom(X,S)-\dim\Hom(M_0,S)$. Hence a ses
\[0\to Y_0\to V\otimes S\to M_1[1]\to0\in\B.\]
Since $S$ simple in $\B$, we have $M_1[1]\in\add S\cap\F[1]=0$. Hence $Y_0\in\add S$.

Taking homology of the first row, we have a sequence
\[0=H^{-1}(T_X)\to Y_1\to \Hom(F,S[1])^*\otimes S\to T_X\to Y_0\to 0.\]
Thus a ses $0\to W\otimes S\to T_X\to Y_0\to0$ with $W$ a vector space of dimension $\dim\Hom(F,S[1])^*-\dim\Hom(X,S[1])^*$. So we have $T_X\in\add S$ since $Y_0$ does. Hence $X\in S\ast\F$.

For any summand $S$ of $T_X$ not in the image of $f$, we have $hg(S)=S$. But $g'h'(S)=0$ because $S$ is a brick and factor through $X$. So we have $f$ is a retraction and $Y_0=0$, i.e., $Y\in\add S[1]$.
\end{proof}

\begin{corollary}
Let $A\in\A$ such that $\Hom(S,A)=0$. Then $A$ admits a $\B$-presentation.
\end{corollary}
\begin{proof}
Construct the presentation as above (the third column of the second diagram in this subsection). Taking homology, we have the following sequence

\end{proof}
\begin{remark}
Important that $\Hom(F_N,S)=0$.
\end{remark}\end{comment}

%, we have $0 \to S\to \S E\to M_\V[1]\to 0$.
%So we have either $\S E,M_\V\in\F[1]$, or
%    \item $\S E\in\T$, $S\in\T$ and $M_\V\in\T[-1]$;
%    \item $\S E\in\T$, $S\in\T$ and $M_\V\in\F$;
%    \item $\S E\in\F[1]$, $S\in\T$ and $M_\V\in\F$;
%    \item $\S E\in\F[1]$, $S\in\F[1]$ and $M_\V\in\F$;

%Firstly, note that since $S$ is simple in $\B=\F[1]\ast\T$, if $S\in\T$ then we have either $0\to F_S\xrightarrow{i} S\xrightarrow{p} T_S\to0$ for some $0\neq F_S\in\F, T_S\in\T$; or $0\to S\to T_S\to T_1\to0$ for $T_0,T_1\in\T$. \hp{Here, since $\T'$ closed under Serre functor, we take $T_S$ be the injective of $S$ in $\B$.}

%Recall that $C[-1]=C_0[-1]\oplus C_1\in\F\ast\T[-1]$, and $C_1=F_N\oplus N''_1$ such that $\Hom(F_N,S)=0$ and $N''_1\in\F$ with $\Hom(N''_1,S[1])=0$, and $\Hom(C_0[-1],T_S)=0$ (because $T_S$ is injective in $\B$). Note that $\Hom(F,\S E[1])=\Hom(E[1],F)=0$. So \[hgf\in\Hom(C[-1],T_S[1])=\Hom(C_0[-1],T_S[1])\oplus \Hom(S[1],T_S[1])\circ\Hom(N''_1,S[1])\]\[\oplus \Hom(F_N,T_S[1])=0\]
%So we have $\alpha$, and hence the following commutative diagram by $4\times 4$ lemma, where $l=\dim\Hom(S,A)$.
%    M_\V^l\ar[d]\ar[r]&D[-1]\ar[r]\ar[d] &<EMAIL_ADDRESS>M_\V[1]^l\ar[d]^-{i}\\
%    S^l\ar[r]\ar[d] & A\ar[r]\ar[d] & A'\ar[r]^-{g}\ar[d] & S[1]^l\ar[d]^-{h}\\
%    \S E^l\ar[r]&M\ar[r] & B\ar[r] & \S E[1]^l
%By hereditary, we have that $\Hom(C[-1],S[1])=\Hom(C_1,S[1])$. Since $C_1\in\F$ and $\Hom(F[-1],T_S)=0$, we have that $\Hom(C_1,S[1]^l)\to\Hom(C_1,F_S[1]^l)$ is surjective and hence the right-up diagram commutes. So we have the diagram by $4\times4$ lemma.

\begin{comment}
Since which implies $M\in\F[1]\ast\T$, where $M_i=H^{-i}(M)$. So we get the $\B$-presentation of $A$.

So we just need to show our claim.
\begin{lemma}
If there exists $0\to F_S\to S\xrightarrow{i} T_S\to 0$ with $F_S\neq0$ then $\Hom(F,T_S[1])=0$.
\end{lemma}
\begin{proof}
Suppose to the contrary we have $0\to T_S\to M\to F_N\to 0$ then we have a nonzero map $S\twoheadrightarrow T\hookrightarrow M$ in $\A$. Extend this map to be a triangle. Then by octahedral axiom, we have the following diagram
By heredity, we have $W=F_N\oplus F_S[1]$. So we have the following diagram
S\ar[r]&M\ar[d]\ar[r]^-{(f_1,f_2)'}&F_S[1]\oplus F_N\ar[d]_-{(0,1)}\ar[r]^-{(g_1,g_2)}&S[1]\ar@{=}[d]\\
We claim that $\dim\Hom(F_N,S[1])=1$. If $g_2$ is nonzero then it equals $\alpha$. So the down-right square and hence the whole diagram commutes. But then $N'_1=F_S[1]\oplus M$, a contradiction to $N'_1\in\A$. So we have $g_2=0$, hence $f_2$ is an isomorphism, which implies $\beta$ is an isomorphism and $M=T_S\oplus F_N$.

So just need to show $\dim\Hom(F_N,S[1])=1$. Cosider the following diagram
With $\Hom(-,S)$ acts on the third row, note that $N'_1\in\F'$ we have
\[0=\Hom(N'_1,S)\to\Hom(F_S,S)\to\Hom(Z[-1],S)\to\Hom(N'_1,S[1])=0\]
With $\Hom(-,S)$ acts on the second row, we have
\[0=\Hom(T_S,S)\to\Hom(S,S)\to\Hom(F_S,S)\to\Hom(T_S,S[1])\]
$\Hom(-,S)$ on the third column,
\[0=\Hom(T_S,S)\to\Hom(F_N[-1],S)\to\Hom(Z[-1],S)\to\Hom(T_S[-1],S)\]

Firstly, we show that $\Hom(T_S,S)=0$. Indeed, for any $f\in\Hom(T_S,S)$, we have the following diagram
S\ar[r]^-{i}\ar[d]_-{fi} &T_S\ar[r]\ar[d]^-{f} & F_S[1]\ar[r]&\\
S\ar@{=}[r] &S\ar[r] & 0\ar[r]&
Note that $fi\neq0$ since otherwise $f$ factors through $F_S[1]$ and hence be 0, a contradiction. But then, since $S$ is a brick, we have $fi$ is an isomorphism. Thus $F_S[1]=0$, a contradiction.

With $\Hom(F,-[1])$ acts, we have
\[\Hom(F,F_S[1])\to\Hom(F,S[1])\to\Hom(F,T_S[1])\to0.\]
Just need to show $gf=0$ for any $f\in\Hom(F,S[1])$, where $g=i[1]$. Consider the following diagram, where $M=\Cone(gf)[-1]$
M\ar[r]^-{h}\ar[d]_-{l} & F\ar[r]^-{gf}\ar[d]^-{f} & T_S[1]\ar@{=}[d]\\
F_S[1]\ar[r]_-{p} &S[1]\ar[r]_-{g}& T_S[1]
With $\Hom(-,S[1])$ acts on the first row, we have
\[0=\Hom(T_S[1],S[1])\to\Hom(F,S[1])\to\Hom(M,S[1]).\]
So $fh\neq0$ but $gfh=0$. So there exists $l:M\to F_S[1]$ such that $pl=fh$.
\end{proof}

Then for arbitrary $A\in\A$, we have the following diagram by $4\times4$ lemma, where $l=\dim\Hom(S,A)$, and $X=F^l$ or $T_1[-1]^l$ and $Y\in\T$.
X\ar[d]_-{f}\ar[r]^-{(1,0)'}&X\oplus C[-1]\ar[r]^-{(0,1)}\ar[d]^{gf} & C[-1]\ar[d]\ar[r]& X[1]^l\\
S^l\ar[r]^-{g}\ar[d] & A\ar[r]\ar[d] & A'\ar[r]\ar[d] & S[1]^l\\
Y\ar[r]&M\ar[r] & B\ar[r] &
Note that, in either case, we have $X\oplus C[-1]\in T[-1]\ast F$. Taking homology of the third row, we have
\[0=H^{-1}(Y)\to M_1\to B_1\to Y\to M_0\to B_0\to 0\]
where $M_i=H^{-i}(M)$ and $B_i\in H^{-i}(B)$.

On the other hand, consider the triangle
\[S[-1]^l\xrightarrow{\alpha} A\to A'\xrightarrow{f} S^l\]
with $l=\dim\Hom(S[-1],A)\neq0$ and $\alpha$ minimal $S[-1]$-approximation. We claim that $\Hom(S,A')=0$. Indeed, since $l\neq0$ and $A$ is indecomposable, we have $f\neq 0$. If $i\in\Hom(S,A')$ nonzero then $fi\in\End S=\k$ is invertible, which implies $S$ is a summand of $A'$, contradicts to $\alpha$ being minimal. So $A'$ admits a $\B$-approximation.

Consider the following diagram given by the octahedral axiom, we have $G\in\A$.
With $\Hom(S[1],-)$ acts on the third row, we have
\[0=\Hom(S[1],T_S)\to\Hom(S[1],G[1])\to\Hom(S[1],A'[1])=0\]
So $G$ admits a $\B$-presentation $G\to D\to C\to G[1]$. Hence we have the following diagram
Tanking homology of the second row, we have
\[0\to X_1\to C_1\to F_S\to X_0\to C_0\to 0.\]
So $X_1\in\F$. Taking homology of the second column, we have
\[0\to D_1\to X_1\to A\to D_0\to X_0\to0,\]
So we have $X_0\in\T$. So $A\to D\to X\to A[1]$ is the triangle we desire.
we have up-right square commutes. Hence we get the whole diagram commutes by $4\times 4$ lemma.
X\ar[d]_-{f}\ar[r]&D[-1]\ar[r]\ar[d] &<EMAIL_ADDRESS>X[1]^l\ar[d]\\
S^l\ar[r]^-{g}\ar[d] & A\ar[r]\ar[d] & A'\ar[r]\ar[d] & S[1]^l\ar[d]\\
T_S^l\ar[r]&M\ar[r] & B\ar[r] & T_S[1]^l

Taking the homology of the first row, we have
\[0\to D_1\to C_1\to F\to D_0\to C_0\to0\]
if $X=F$. So $D_1\in\F$. Taking the homology of the third row, we have
\[0\to M_1\to B_1\to 0\to M_0\to B_0\to T_S\to 0\]
So $M_0\in\T$ (because $B_0\to T_S\in\B$, whose kernel must be in $\T$. Remark that $\T$ closed under kernel?)
Taking the homology of the second column, we have
\[0\to M_1\to D_1\to A\to M_0\to D_0\to0\]
we have $D_0\in\T$ since it is the factor of $M_0$.

%Consider the following diagram
%    &M^{-1}\ar[d]\ar@{=}[r]&M^{-1}\ar[d]&\\
%    T_X\ar[r]\ar@{=}[d]&A^0\ar[r]\ar[d]&W\otimes S[1]\ar[r]\ar[d]&T_X[1]\ar@{=}[d]\\
%    T_X\ar[r]&V^*\otimes S\ar[d]\ar[r]&N[1]\ar[r]\ar[d]&T_X[1]\\
%    &M^{-1}[1]\ar@{=}[r]&M^{-1}[1]&
%}$$Note that

\end{comment}

%    $R^\ast_S(Y)\in\A$ for any $Y\in\T'$ and $R^\ast_S(Y)\in\A[1]$ if $Y\in\F'[1]$.
%Taking homology, we have long exact sequence
%\[0=H^{-1}(Y)\to H^{-1}(R^\ast_S(Y))\to \Hom(S,Y)\otimes S\to Y\to\]
%\[H^0(R^\ast_S(Y))\to\Hom(S[-1],Y)\otimes S\to H^1(Y)=0 \]
%Since $\A$ is hereditary, we have $H^i(R^\ast_S(Y))=0$ for $i\neq0,-1$. If $Y\in\T$ then the sequence above shows that $H^{-1}(R^\ast_S(Y))=0$ and the assertion follows.

%Similarly, if $Y[-1]\in\F'$, taking homology we have that
%\[0=H^{-2}(Y)\to H^{-2}(R^\ast_S(X))\to \Hom(S[1],Y)\otimes S\xrightarrow{f} Y[-1]\to H^{-1}(R^\ast_S(Y))\to \Hom(S,Y)\otimes S\to 0\]
%From the $\zeta$ and the last sequence in Lemma~\ref{lem:F'}, we see that $f$ is injective \hp{if $S\in\T$. If $S\in\F$ then $Y[-1]\in\F'=\F\cap\ps$. Since $S[1]$ simple in $F[1]\ast T$ and hence $F[1]$ we have also that $f$ is injective}. So the assertion also follows.

%where $f=0$ if $S\in\T$ (because $Y[-1]\in F$) and $f$ is injective if $S\in\F$ (because $S[1]$ is simple in $F[1]\ast\T$ and $Y\in F[1]$).

\begin{comment}
Indeed, since $S\in\T$ we have $(X_{\T})_S=\Hom(X_{\T},S)^*\otimes S\oplus\Hom(X_{\T},S[1])^*\otimes S[1]$. Taking homology of the first row, we have $H^i(T^*_S(X_{\T}))=0$ when $i\neq 0,1$, and a long exact sequence
\[0\to H^{-1}((X_{\T})_S)\to H^0(T^*_S(X_{\T}))\to X_{\T}\xrightarrow{h} H^0((X_{\T})_S)\to H^1(T^*_S(X_{\T}))\to 0.\]
So we get short exact sequences
\begin{equation}\label{eq:1}
0\to\ker h\to X_{\T}\xrightarrow{p}\im h\to 0
\end{equation}
\[0\to\im h\xrightarrow{i} H^0((X_{\T})_S)\to H^1(T^*_S(X_{\T}))\to 0.\]
in $\A$. Since $H^0((X_{\T})_S)\in\add S$, by Lemma~\ref{lem:ker}, we have $\im h=S^n\oplus W$ with $W\in\F$. On the other hand, since $X_{\T}\in\T$, we have $\im h\in\T$. So $W$ belongs to $\T\cap\F=0$. Again by Lemma~\ref{lem:ker}, we have $H^1(T^*_S(X_{\T}))$ is a summand of $H^0((X_{\T})_S)\subseteq \add S$. But then we have $g=(g_1,g_2)'$ with $g_2\in\Hom((X_{\T})_S[-1],H^1(T^*_S(X_{\T}))[-1])$. If $H^1(T^*_S(X_{\T}))\neq 0$ then, since $\End S=\k$, components of $g_2$ are either zeros or isomorphisms, a contradiction to $\alpha$ is the minimal left $\thick S$-approximation. So we have $H^1(T^*_S(X_{\T}))=0$ and hence $T^*_S(X_{\T})\in\A$. Since $\im h\in\add S$ which is simple in $\B$, we have ~\eqref{eq:1} is also an exact sequence in $\B$, which implies $\ker h\in\T$. So our claim follows via the short exact sequence
\[0\to H^{-1}((X_{\T})_S)\to H^0(T^*_S(X_{\T}))\to\ker h\to 0.\]
\end{comment}