Proof Consider the exact triangle

$$ \begin{align*}\tau^{\geqslant\ell+1}(X)[-1]\rightarrow\tau^{\leqslant\ell}(X)\rightarrow X\rightarrow\tau^{\geqslant\ell+1}(X).\end{align*} $$

As $\mathrm {Hom}_{\mathcal {T}}(P[-\ell ],\tau ^{\geqslant \ell +1}(X)[-1])=0=\mathrm {Hom}_{\mathcal {T}}(P[-\ell ],\tau ^{\geqslant \ell +1}(X))$ , $\mathrm {Hom}_{\mathcal {T}}(P[-\ell ],$ $\tau ^{\leqslant \ell }(X))\stackrel {\cong }\rightarrow \mathrm {Hom}_{\mathcal {T}}(P[-\ell ],X)$ . Consider the exact triangle

$$ \begin{align*}\tau^{\leqslant\ell-1}(X)\rightarrow\tau^{\leqslant\ell}(X)\rightarrow \mathcal{H}^\ell_M(X)[-\ell]\rightarrow\tau^{\leqslant\ell-1}(X)[1].\end{align*} $$

As $\mathrm {Hom}_{\mathcal {T}}(P[-\ell ],\tau ^{\leqslant \ell -1}(X))=0=\mathrm {Hom}_{\mathcal {T}}(P[-\ell ],\tau ^{\leqslant \ell -1}(X)[1])$ , $\mathrm {Hom}_{\mathcal {T}}(P[-\ell ],$ $\tau ^{\leqslant \ell }(X))\stackrel {\cong }\rightarrow \mathrm {Hom}_{\mathcal {T}}(P[-\ell ],\mathcal {H}^\ell _M(X)[-\ell ])$ . Consider the exact triangle

$$ \begin{align*}\tau^{\leqslant-1}(P)\rightarrow P\rightarrow\mathcal{H}^0_M(P)\rightarrow\tau^{\leqslant-1}(P)[1].\end{align*} $$

As $\mathrm {Hom}_{\mathcal {T}}(\tau ^{\leqslant -1}(P),\mathcal {H}^\ell _M(X))=0=\mathrm {Hom}_{\mathcal {T}}(\tau ^{\leqslant -1}(P)[1],\mathcal {H}^\ell _M(X))$ , it implies that

$$ \begin{align*}\mathrm{Hom}_{\mathcal{H}_M}(\mathcal{H}^0_M(P),\mathcal{H}^\ell_M(X))\stackrel{\cong}\rightarrow\mathrm{Hom}_{\mathcal{T}}(P,\mathcal{H}^\ell_M(X)).\end{align*} $$

Since $\mathrm {Hom}_{\mathcal {T}}(M[\ell ],M)\cong \mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^{-\ell }_M(M))=0$ for $\ell \leqslant 0$ , one has $\mathrm {Hom}_{\mathcal {T}}(M[-\ell ],M^\ast )\cong \mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^{\ell }_M(M^\ast ))=0$ for $\ell \leqslant 0$ . Also $\mathrm {Hom}_{\mathcal {T}}(M,M)\neq 0$ , $\mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^{0}_M(M^\ast ))\neq 0$ . So $M^\ast \in \mathcal {T}^{\geqslant 0}$ and $\mathrm {inf}M^\ast =0$ by Lemma 2.4. The last two isomorphisms follow by dual arguments.

Proof By Lemmas 2.4 and 2.5, $\mathcal {H}^{0}_M(M)$ is a projective generator for $\mathcal {H}_M$ and $\mathcal {H}^{0}_M(M^\ast )$ is an injective cogenerator for $\mathcal {H}_M$ , as claimed.

Proof By Lemma 2.5, one has

$$ \begin{align*}\begin{aligned}\mathrm{Hom}_{\mathcal{T}}(M,\mathcal{H}^\ell_M(\coprod_{\lambda\in\Lambda}X_\lambda)) &\cong\mathrm{Hom}_{\mathcal{T}}(M[-\ell],\coprod_{\lambda\in\Lambda}X_\lambda)\\ &\cong\coprod_{\lambda\in\Lambda}\mathrm{Hom}_{\mathcal{T}}(M[-\ell],X_\lambda)\\ &\cong\coprod_{\lambda\in\Lambda}\mathrm{Hom}_{\mathcal{T}}(M,\mathcal{H}^\ell_M(X_\lambda))\\ &\cong\mathrm{Hom}_{\mathcal{T}}(M,\coprod_{\lambda\in\Lambda}\mathcal{H}^\ell_M(X_\lambda)).\end{aligned}\end{align*} $$

Similarly, one has $\mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^\ell _M(\prod _{\lambda \in \Lambda }X_\lambda )) \cong \mathrm {Hom}_{\mathcal {T}}(M,\prod _{\lambda \in \Lambda }\mathcal {H}^\ell _M(X_\lambda ))$ . The desired isomorphisms follow by Lemma 2.4.

Proof (1) By Lemma 2.5, $\mathcal {H}^{\ell }_M(X)=0$ for all $\ell \in \mathbb {Z}$ if and only if $\mathrm {Hom}_{\mathcal {T}}(M[-\ell ],X)=0$ for all $\ell \in \mathbb {Z}$ if and only if $X\cong 0$ .

(2) For $\ell>n$ , if $\mathcal {H}^{\ell }_M(X)=0$ , then $\mathrm {Hom}_{\mathcal {T}}(M,X[\ell ])=0$ by Lemma 2.5, so $X\in \mathcal {T}^{\leqslant n}$ . Conversely, let $X\in \mathcal {T}^{\leqslant n}$ . For $\ell>n$ , consider the exact triangle

$$ \begin{align*}\tau^{\leqslant \ell-1}(X)\rightarrow \tau^{\leqslant \ell}(X)\rightarrow\mathcal{H}^{\ell}_M(X)[-\ell]\rightarrow\tau^{\leqslant \ell-1}(X)[1].\end{align*} $$

As $\tau ^{\leqslant \ell -1}(X)\cong \tau ^{\leqslant \ell }(X)\cong X$ by [Reference Kashiwara and Schapira16, Proposition 10.1.6], it implies that $\mathcal {H}^{\ell }_M(X)=0$ . Similarly, one can show that $X\in \mathcal {T}^{\geqslant n}$ if and only if $\mathcal {H}^{\ell }_M(X)=0$ for all $\ell <n$ .

Proof (1) $\Rightarrow $ (2). For any sppj resolution of X,

$$ \begin{align*}\cdots\longrightarrow P_n\stackrel{d_n}\longrightarrow P_{n-1}\longrightarrow\cdots\longrightarrow P_1\stackrel{d_1}\longrightarrow P_0\longrightarrow X,\end{align*} $$

let $e\geqslant 0$ be such that $X_e\in \mathcal {P}[-\mathrm {sup}X_e]$ and $e\leqslant d-\mathrm {sup}P_0+\mathrm {sup}P_e\leqslant d$ . Then

$$ \begin{align*}X_e\stackrel{g_e}\longrightarrow P_{e-1}\stackrel{d_{e-1}}\longrightarrow\cdots\stackrel{d_1}\longrightarrow P_0\stackrel{f_0}\longrightarrow X\end{align*} $$

is an sppj resolution of length e. For $U\in \mathcal {T}^{-}$ , as $\mathrm {Hom}_{\mathcal {T}}(M,U[i])\cong \mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^i_M(U))=0$ for $i>\mathrm {sup}U$ by Lemma 2.5, it follows by the corresponding triangle $(\dagger _n)$ that (2) holds true.

(2) $\Rightarrow $ (4). Set $s=\mathrm {sup}X$ . We proceed by induction on d. For $d=0$ , consider the exact triangle $Y\rightarrow P\stackrel {f}\rightarrow X[s]\rightarrow Y[1]$ with f an sppj morphism. Then $\mathrm {sup}Y\leqslant 0$ , and so $\mathrm {Hom}_{\mathcal {T}}(X[s],Y[1])=0$ . Thus, this triangle splits and $X\in \mathcal {P}[-s]$ . Assume that $d\geqslant 1$ and $Y\rightarrow Q\stackrel {g}\rightarrow X\rightarrow Y[1]$ is an exact triangle such that g is an sppj morphism. For any $U\in \mathcal {T}^{-}$ and $i\in \mathbb {Z}$ , one has an exact sequence

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(Q,U[i])\rightarrow\mathrm{Hom}_{\mathcal{T}}(Y,U[i])\rightarrow\mathrm{Hom}_{\mathcal{T}}(X[-1],U[i]),\end{align*} $$

it implies that $\mathrm {Hom}_{\mathcal {T}}(Y,U[i])=0$ for $i>-s+d+\mathrm {sup}U-1$ . By induction, $Y\in \mathcal {P}[-s]\ast \cdots \ast \mathcal {P}[-s+d-1]$ , and hence $\mathrm {pd}_{\mathcal {T}}X\leqslant d$ .

(4) $\Rightarrow $ (3). It follows by $\mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}_M[\neq 0])=0$ and $X\in \mathcal {P}[-s]\ast \cdots \ast \mathcal {P}[-s+d]$ .

(3) $\Rightarrow $ (2). For $L\in \mathcal {T}^{\mathrm {b}}$ , by induction on $\mathrm {amp}L$ , one can show that $\mathrm {Hom}_{\mathcal {T}}(X,L[i])=0$ for $i\not \in [-s+\mathrm {inf}L,-s+d+\mathrm {sup}L]$ . Let $U\in \mathcal {T}^{-}$ . There is a sequence in $\mathcal {T}$

$$ \begin{align*}\cdots\stackrel{\nu_2}\longrightarrow\tau^{\geqslant\mathrm{sup}U-2}(U)\stackrel{\nu_1}\longrightarrow\tau^{\geqslant\mathrm{sup}U-1}(U)\stackrel{\nu_0}\longrightarrow\tau^{\geqslant\mathrm{sup}U}(U),\end{align*} $$

where $\tau ^{\geqslant \mathrm {sup}U-j}(U)\in \mathcal {T}^{\mathrm {b}}$ and $\mathrm {sup}\tau ^{\geqslant \mathrm {sup}U-j}(U)=\mathrm {sup}U$ for $j\geqslant 0$ . Put

$$ \begin{align*}1-\nu=\left[\begin{smallmatrix}\ddots &\vdots&\vdots&\vdots&\vdots\\ \dots &1&0&0&0\\ \dots&-\nu_2&1&0&0\\ \dots &0&-\nu_1&1&0\\0 &0&0&-\nu_0&1\end{smallmatrix}\right]:\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U)\rightarrow\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U),\end{align*} $$

it yields the following exact triangle:

$$ \begin{align*}\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U)[-1]\longrightarrow \bar{U}\longrightarrow\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U)\xrightarrow{1-\nu}\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U).\end{align*} $$

For $\ell \in \mathbb {Z}$ , the above sequence induces an inverse system of epimorphisms in $\mathcal {H}_M$

$$ \begin{align*}\cdots\rightarrow\mathcal{H}^\ell_M(\tau^{\geqslant\mathrm{sup}U-2}(U))\rightarrow\mathcal{H}^\ell_M(\tau^{\geqslant\mathrm{sup}U-1}(U))\rightarrow\mathcal{H}^\ell_M(\tau^{\geqslant\mathrm{sup}U}(U)).\end{align*} $$

By Lemma 2.7, $0\rightarrow \mathcal {H}^\ell _M(\bar {U})\rightarrow \prod _{j\geqslant 0}\mathcal {H}^\ell _M(\tau ^{\geqslant \mathrm {sup}U-j}(U))\rightarrow \prod _{j\geqslant 0}\mathcal {H}^\ell _M (\tau ^{\geqslant \mathrm {sup}U-j}(U)) \rightarrow 0$ is exact. Also $0\rightarrow \underleftarrow {\textrm {lim}}\mathcal {H}^\ell _M(\tau ^{\geqslant \mathrm {sup}U-j}(U))\rightarrow \prod _{j\geqslant 0}\mathcal {H}^\ell _M(\tau ^{\geqslant \mathrm {sup}U-j}(U))\rightarrow \prod _{j\geqslant 0}\mathcal {H}^\ell _M(\tau ^{\geqslant \mathrm {sup}U-j}(U))\rightarrow 0$ is exact, and $\mathcal {H}^\ell _M(\tau ^{\geqslant \mathrm {sup}U-j}(U))\cong \mathcal {H}^\ell _M(U)$ for $j\geqslant \mathrm {max}\{0,\mathrm {sup}U-\ell \}$ , it implies that $\mathcal {H}^\ell _M(\bar {U})\cong \underleftarrow {\textrm {lim}}\mathcal {H}^\ell _M(\tau ^{\geqslant \mathrm {sup}U-j}(U))=\mathcal {H}^\ell _M(U)$ , so $U\cong \bar {U}$ . Thus, the exact sequence

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}\Big(X,\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U)[i-1]\Big) &\rightarrow\mathrm{Hom}_{\mathcal{T}}(X,U[i])\\ &\rightarrow\mathrm{Hom}_{\mathcal{T}}\Big(X,\prod_{j\geqslant0}\tau^{\geqslant\mathrm{sup}U-j}(U)[i]\Big)\end{align*} $$

implies that $\mathrm {Hom}_{\mathcal {T}}(X,U[i])=0$ for $i>-s+d+\mathrm {sup}U+1$ . Consider the exact triangle $\tau ^{\leqslant \mathrm {sup}U-1}(U)\rightarrow U\rightarrow \mathcal {H}^{\mathrm {sup}U}_M(U)[-\mathrm {sup}U]\rightarrow \tau ^{\leqslant \mathrm {sup}U-1}(U)[1]$ . For $i'=-s+d+\mathrm {sup} U+1$ , it follows by the exact sequence

$$ \begin{align*}0=\mathrm{Hom}_{\mathcal{T}}(X,\tau^{\leqslant\mathrm{sup}U-1}(U)[i']) &\rightarrow\mathrm{Hom}_{\mathcal{T}}(X,U[i']) \\ &\rightarrow \mathrm{Hom}_{\mathcal{T}}(X,\mathcal{H}^{\mathrm{sup}U}_M(U)[-\mathrm{sup}U+i'])=0\end{align*} $$

that $\mathrm {Hom}_{\mathcal {T}}(X,U[i])=0$ for $i>-s+d+\mathrm {sup}U$ .

(2) $\Rightarrow $ (1). We proceed by induction on d. If $d=0$ , then $e=0$ , the statement holds. Assume that $d\geqslant 1$ . For any sppj resolution $P_\bullet $ of X, consider the exact triangle

$$ \begin{align*}X_1\rightarrow P_0\rightarrow X\rightarrow X_1[1].\end{align*} $$

For $U\in \mathcal {T}^{-}$ , $\mathrm {Hom}_{\mathcal {T}}(X_1,U[i])\cong \mathrm {Hom}_{\mathcal {T}}(X,U[i+1])=0$ for $i>-\mathrm {sup}X+d-1+\mathrm {sup}U=-\mathrm {sup}X_1+(d-1-\mathrm {sup}P_0+\mathrm {sup}P_1)+\mathrm {sup}U$ . As $\cdots \rightarrow P_n\rightarrow \cdots \rightarrow P_1\rightarrow X_1$ is an sppj resolution of $X_1$ and $\mathrm {sup}P_1\leqslant \mathrm {sup}P_0$ , it follows by the induction that there is $e\geqslant 0$ such that $X_e\in \mathcal {P}[-\mathrm {sup}X_e]$ and $e-1+\mathrm {sup}P_1-\mathrm {sup}P_e\leqslant d-1-\mathrm {sup}P_0+\mathrm {sup}P_1$ , as desired.

Proof We just prove (1) since (2) is dual. Set $s=\mathrm {sup}X$ .

“Only if” part. If $\mathrm {Hom}_{\mathcal {T}}(X,\mathcal {H}_M[-s+d])=0$ , then $\mathrm {Hom}_{\mathcal {T}}(X,\mathcal {H}_M[i])=0$ for $i\not \in [-s,-s+d-1]$ , so $\mathrm {pd}_{\mathcal {T}}X\leqslant d-1$ by Theorem 3.3, a contradiction.

“If” part. If $X\in \mathcal {P}[-s]\ast \cdots \ast \mathcal {P}[-s+d-1]$ , then $\mathrm {Hom}_{\mathcal {T}}(X,\mathcal {H}_M[-s+d])=0$ by Theorem 3.3, a contradiction. Thus, $\mathrm {pd}_{\mathcal {T}}X=d$ .

Proof Again we will just prove (1) since (2) is dual. Set $\mathrm {pd}_{\mathcal {T}}X=d$ and $s=\mathrm {sup}X$ .

If $d\geqslant 1$ , then $\mathrm {Hom}_{\mathcal {T}}(X',U[i])=0$ for $U\in \mathcal {T}^{-}$ and $i>-s+d+\mathrm {sup}U-1$ by the proof (2) $\Rightarrow $ (4) of Theorem 3.3, it implies that $\mathrm {pd}_{\mathcal {T}}X'\leqslant d-s+\mathrm {sup}X'-1$ . By Corollary 3.5, there is $H\in \mathcal {H}_M$ , such that $\mathrm {Hom}_{\mathcal {T}}(X,H[-s+d])\neq 0$ . Then the exact sequence

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(X',H[-s+d-1]) &\rightarrow\mathrm{Hom}_{\mathcal{T}}(X[-1],H[-s+d-1]) \\ &\rightarrow\mathrm{Hom}_{\mathcal{T}}(P[-1],H[-s+d-1])=0\end{align*} $$

yields that $\mathrm {Hom}_{\mathcal {T}}(X',H[-s+d-1])\neq 0$ , so $\mathrm {pd}_{\mathcal {T}}X'=d-s+\mathrm {sup}X'-1$ by Corollary 3.5.

If $d=0$ , then $\mathrm {Hom}_{\mathcal {T}}(X,X'[1])=0$ as $\mathrm {sup}X'\leqslant s$ , so the triangle splits. Thus, $X'$ is a retract of P, and so $X'\cong 0$ or $\mathrm {pd}_{\mathcal {T}}X'=0$ .

Proof (1) $\Rightarrow $ (2). The corresponding exact triangles

$$ \begin{align*}X_{n+1}\rightarrow F_n\rightarrow X_n\rightarrow X_{n+1}[1]\end{align*} $$

imply that $X\in \mathcal {F}[-\mathrm {sup}X]\ast \mathcal {F}[-\mathrm {sup}X+(\mathrm {sup}X-\mathrm {sup}X_1)+1]\ast \cdots \ast \mathcal {F}[-\mathrm {sup}X+(\mathrm {sup}X-\mathrm {sup}X_e)+e]$ , so $\mathrm {fd}_{\mathcal {T}}X\leqslant d$ .

(2) $\Rightarrow $ (1). Set $s=\mathrm {sup}X$ . We proceed by induction on d. If $d=0$ , then $e=0$ , the statement holds. Assume that $d\geqslant 1$ . There is an exact triangle $Y\rightarrow F\stackrel {f}\rightarrow X\rightarrow Y[1]$ , such that $Y\in \mathcal {F}[-s]\ast \cdots \ast \mathcal {F}[-s+d-1]$ and $F\in \mathcal {F}[-s]$ . Then Y has an spft resolution $F_\bullet $ such that $Y_{e'}\in \mathcal {F}[-\mathrm {sup}Y_{e'}]$ , $e'+s-\mathrm {sup}Y_{e'}\leqslant d-1$ by induction. As $\mathrm {sup}Y\leqslant s$ , f is an spft morphism by Lemma 2.3, it implies that

$$ \begin{align*}Y_{e'}\rightarrow F_{e'-1}\rightarrow\cdots\rightarrow F_0\rightarrow F\rightarrow X\end{align*} $$

is the desired spft resolution with $e=e'+1$ .

Proof Set $s=\mathrm {sup}X$ . (1) $\Rightarrow $ (2) is trivial.

(2) $\Rightarrow $ (3). Let $0\neq F\in \mathcal {F}$ . For any $T\in \mathcal {T}^{c}$ , if $\mathrm {Hom}_{\mathcal {T}}(F,T^\ast [i])=0$ for $i\geqslant \mathrm {sup}T^\ast $ , then $\mathrm {Hom}_{\mathcal {T}}(F[1],T^\ast [i])=0$ for $i>\mathrm {sup}T^\ast $ , so $F[1]\in \mathcal {F}$ and $\mathrm {sup}F=-1$ as $\mathcal {F}=\mathcal {F}'$ , a contradiction. Thus, there exists $T'\in \mathcal {T}^c$ such that $\mathrm {Hom}_{\mathcal {T}}(F,T^{\prime \ast }[\mathrm {sup}T^{\prime \ast }])\neq 0$ . For $i>d-s+\mathrm {sup}T^\ast $ , the corresponding exact triangle implies that $\mathrm {Hom}_{\mathcal {T}}(X,T^\ast [i])\cong \mathrm {Hom}_{\mathcal {T}}(X_{e-1},T^\ast [i-e+1]) \cong \mathrm {Hom}_{\mathcal {T}}(X_e,T^\ast [i-e])=0$ . Consider the exact triangle $X_e\rightarrow F_{e-1}\rightarrow X_{e-1}\rightarrow X_e[1]$ , it yields an exact sequence $\mathrm {Hom}_{\mathcal {T}}(F_{e-1},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast ])\rightarrow \mathrm {Hom}_{\mathcal {T}}(X_{e},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast ]\rightarrow \mathrm {Hom}_{\mathcal {T}}(X_{e-1},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast +1]\rightarrow 0$ . Assume that $\mathrm {Hom}_{\mathcal {T}}(X,T^\ast [d-s+\mathrm {sup}T^\ast ])=0$ for all $T\in \mathcal {T}^c$ . If $\mathrm {sup}X_{e-1}-\mathrm {sup}X_e>0$ , then $\mathrm {Hom}_{\mathcal {T}}(F_{e-1},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast ])=0$ , so $\mathrm {Hom}_{\mathcal {T}}(X_{e},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast ])=0$ for all $T\in \mathcal {T}^c$ , a contradiction. Thus, $\mathrm {sup}X_{e-1}=\mathrm {sup}X_e$ , and $\mathrm {Hom}_{\mathcal {T}}(X_{e-1},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast +1]) \cong \mathrm {Hom}_{\mathcal {T}}(X,T^\ast [d-s+\mathrm {sup}T^\ast ])=0$ . So $X_{e-1}\in \mathcal {F}[-\mathrm {sup}X_{e-1}]$ and $g_e$ is a pure monomorphism, a contradiction. Consequently, $\mathrm {Hom}_{\mathcal {T}}(X,T^{\prime \ast }[d-\mathrm {sup}X+\mathrm {sup}T^{\prime \ast }])\neq 0$ for some $T'\in \mathcal {T}^c$ .

(3) $\Rightarrow $ (4). We proceed by induction on d. The case of $d=0$ is trivial. Assume that $d\geqslant 1$ and $X_1\rightarrow F\stackrel {f}\rightarrow X\rightarrow X_1[1]$ is an exact triangle with f an spft morphism. For any $T\in \mathcal {T}^{c}$ , one has an exact sequence

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(F,T^\ast[i])\rightarrow\mathrm{Hom}_{\mathcal{T}}(X_1,T^\ast[i]) \rightarrow\mathrm{Hom}_{\mathcal{T}}(X[-1],T^\ast[i]),\end{align*} $$

it implies that $\mathrm {Hom}_{\mathcal {T}}(X_1,T^\ast [i])=0$ for $i>d-1-s+\mathrm {sup}T^\ast $ . Also there is $T'\in \mathcal {T}^c$ , such that $\mathrm {Hom}_{\mathcal {T}}(X,T^{\prime \ast }[d-s+\mathrm {sup}T^{\prime \ast }])\neq 0$ . As $\mathrm {Hom}_{\mathcal {T}}(F[-1],T^{\prime \ast }[d-1-s+\mathrm {sup}T^{\prime \ast }])=0$ , the following exact sequence

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(X_1,T^{\prime\ast}[d-1-s+\mathrm{sup}T^{\prime\ast}])\rightarrow\mathrm{Hom}_{\mathcal{T}}(X[-1],T^{\prime\ast}[d-1-s+\mathrm{sup}T^{\prime\ast}]) \rightarrow0\end{align*} $$

yields that $\mathrm {Hom}_{\mathcal {T}}(X_1,T^{\prime \ast }[d-1-s+\mathrm {sup}T^{\prime \ast }])\neq 0$ , so $X_1\in \mathcal {F}[-s]\ast \cdots \ast \mathcal {F}[-s+d-1]$ by the induction, and hence $\mathrm {fd}_{\mathcal {T}}X\leqslant d$ . If $\mathrm {fd}_{\mathcal {T}}X<d$ , then $\mathrm {Hom}_{\mathcal {T}}(X,T^\ast [d-s+\mathrm {sup}T^\ast ])=0$ by $\mathrm {Hom}_{\mathcal {T}}(\mathcal {F}[-s],T^\ast [i])=0$ for $i>-s+\mathrm {sup}T^\ast $ , and so $\mathrm {Hom}_{\mathcal {T}}(X_1,T^\ast [d-1-s+\mathrm {sup}T^\ast ])=0$ for all $T\in \mathcal {T}^{c}$ by the preceding proof, a contradiction. Thus, $\mathrm {fd}_{\mathcal {T}}X=d$ .

(4) $\Rightarrow $ (2). We proceed by induction on d. The case of $d=0$ is trivial. Assume that $d\geqslant 1$ . There is an exact triangle $Y\rightarrow F\stackrel {f}\rightarrow X\rightarrow Y[1]$ , such that $d-1$ is the smallest number that satisfies $Y\in \mathcal {F}[-s]\ast \cdots \ast \mathcal {F}[-s+d-1]$ and $F\in \mathcal {F}[-s]$ . Then Y has an spft resolution $F_\bullet $ such that $Y_{e'}\in \mathcal {F}[-\mathrm {sup}Y_{e'}]$ , $e'+s-\mathrm {sup}Y_{e'}=d-1$ and $Y_{e'-1}\not \in \mathcal {F}[-\mathrm {sup}Y_{e'-1}]$ by the induction. As $\mathrm {sup}Y\leqslant s$ , f is an spft morphism by Lemma 2.3, it yields that

$$ \begin{align*}Y_{e'}\longrightarrow F_{e'-1}\longrightarrow\cdots\longrightarrow F_0\longrightarrow F\longrightarrow X\end{align*} $$

is the desired spft resolution.

(3) $\Rightarrow $ (1). Let $F_\bullet $ be spft resolution of X. For any $T\in \mathcal {T}^{c}$ , the corresponding exact triangle implies that $\mathrm {Hom}_{\mathcal {T}}(X_e,T^\ast [i-e])\cong \mathrm {Hom}_{\mathcal {T}}(X,T^\ast [i])=0\, i>d-s+\mathrm {sup}T^\ast $ and $\mathrm {Hom}_{\mathcal {T}}(X_{e-1},T^\ast [-\mathrm {sup}X_e+\mathrm {sup}T^\ast +1]) \cong \mathrm {Hom}_{\mathcal {T}}(X,T^\ast [d-s+\mathrm {sup}T^\ast ])$ , it implies that $X_{e-1}\not \in \mathcal {F}[-\mathrm {sup}X_{e-1}]$ and $g_e$ is not a pure monomorphism.

Proof It suffices to prove that $\mathrm {pd}_{\mathcal {T}}X\leqslant \mathrm {fd}_{\mathcal {T}}X=d$ . As $X\in \mathcal {T}^{c}$ , one can choose an sppj morphism $P\rightarrow X$ such that $P\in \mathrm {add}(M)[-\mathrm {sup}X]$ , so $X'\rightarrow P\rightarrow X\rightarrow X'[1]$ is an exact triangle in $\mathcal {T}^{c}$ . If $d=0$ , then the pure exact triangle is split as $X\in \mathcal {T}^{c}$ . If $d\geqslant 1$ , then $\mathrm {fd}_{\mathcal {T}}X'=\mathrm {fd}_{\mathcal {T}}X-\mathrm {sup}X+\mathrm {sup}X'-1\leqslant d-1$ by Corollary 3.13. By the induction, $\mathrm {pd}_{\mathcal {T}}X'=\mathrm {fd}_{\mathcal {T}}X'$ , so $\mathrm {pd}_{\mathcal {T}}X=\mathrm {pd}_{\mathcal {T}}X'+\mathrm {sup}X-\mathrm {sup}X'+1 =\mathrm {fd}_{\mathcal {T}}X$ by Proposition 3.6.

Proof “ $d_3\leqslant d_1$ .” Let $X\in \mathcal {T}^{-}$ . We show $\mathrm {pd}_{\mathcal {T}}X\leqslant d_1+\mathrm {amp}X$ . In the case $\mathrm {amp}X=\infty $ , there is nothing to prove. We deal with the case $\mathrm {amp}X<\infty $ by the induction on $\mathrm {amp}X$ . The case of $\mathrm {amp}X=0$ is trivial as $\mathrm {pd}_{\mathcal {T}}H[n]=\mathrm {pd}_{\mathcal {T}}H$ for all $H\in \mathcal {H}_M$ and $n\in \mathbb {Z}$ . Assume that $\mathrm {amp}X>0$ and set $s=\mathrm {sup}X$ . Consider the exact triangle

$$ \begin{align*}\tau^{\leqslant s-1}(X)\rightarrow X\rightarrow \mathcal{H}^s_M(X)[-s]\rightarrow\tau^{\leqslant s-1}(X)[1].\end{align*} $$

As $\mathrm {amp}\tau ^{\leqslant s-1}(X)\leqslant \mathrm {amp}X-1$ , it follows by the induction that $\mathrm {pd}_{\mathcal {T}}\tau ^{\leqslant s-1}(X)\leqslant d_1+\mathrm {amp}X-1$ . So $\mathrm {pd}_{\mathcal {T}}X-\mathrm {sup}X\leqslant d_1-\mathrm {inf}X$ by Corollary 3.8, as required.

It is clear $d_1\leqslant d_3$ . Thus, $d_1=d_3$ . We can prove $d_2=d_4$ in the same way.

“ $d_1= d_2$ .” Let $H\in \mathcal {H}_M$ . Then $\mathrm {Hom}_{\mathcal {T}}(H,H'[i])\cong \mathrm {Hom}_{\mathcal {T}}(H[-i],H')=0$ for $-i\not \in [-d_2,0]$ by Theorem 3.4 for all $H'\in \mathcal {H}_M$ as $\mathrm {id}_{\mathcal {T}}H'\leqslant d_2$ , so $\mathrm {pd}_{\mathcal {T}}H\leqslant d_2$ by Theorem 3.3. Hence, $d_1\leqslant d_2$ . Similarly, one can show that $d_2\leqslant d_1$ .

Proof (1) $\Leftrightarrow $ (2) For $\ell \in \mathbb {Z}$ , Lemma 2.5 yields the following isomorphism:

If $X\rightarrow Y\rightarrow Z\rightarrow X[1]$ is M-ghost, then the lower row is exact, so is the upper row. Thus, $0\rightarrow \mathcal {H}^\ell _M(X)\rightarrow \mathcal {H}^\ell _M(Y)\rightarrow \mathcal {H}^\ell _M(Z)\rightarrow 0$ is exact in $\mathcal {H}_M$ by Lemma 2.4. If $0\rightarrow \mathcal {H}^\ell _M(X)\rightarrow \mathcal {H}^\ell _M(Y)\rightarrow \mathcal {H}^\ell _M(Z)\rightarrow 0$ is exact, then the upper row is exact by Lemma 2.4, so is the lower row. Thus, $X\rightarrow Y\rightarrow Z\rightarrow X[1]$ is M-ghost.

(2) $\Rightarrow $ (3) For $\ell \in \mathbb {Z}$ , extend $\tau ^{\leqslant \ell }(X)\rightarrow \tau ^{\leqslant \ell }(Y)$ and $\tau ^{\leqslant \ell }(Y)\rightarrow \tau ^{\leqslant \ell }(Z)$ to exact triangles, one has a commutative diagram:

By the third row and Lemma 2.3, $U\in \mathcal {T}^{\leqslant \ell }$ . As $\mathcal {H}^i_M(-)\cong \mathcal {H}^i_M(\tau ^{\leqslant \ell }(-))$ for $i\leqslant \ell $ and $-=X,Y,Z$ , it follows that $\mathcal {H}^i_M(U)\cong \mathcal {H}^i_M(\tau ^{\leqslant \ell }(Z))$ for $i\in \mathbb {Z}$ , so $\mathcal {H}^i_M(W)=0$ for $i\in \mathbb {Z}$ by the third column. Thus, $U\cong \tau ^{\leqslant \ell }(Z)$ by Lemma 2.8, and $\tau ^{\leqslant \ell }(X)\rightarrow \tau ^{\leqslant \ell }(Y)\rightarrow \tau ^{\leqslant \ell }(Z)\rightarrow \tau ^{\leqslant \ell }(X)[1]$ is an exact triangle.

(3) $\Rightarrow $ (4) For $\ell \in \mathbb {Z}$ , by $4\times 4$ Lemma, one has the following commutative diagram:

such that the square at the bottom right is anti-commutative, while the others are commutative, it implies that the third row is an exact triangle.

(4) $\Rightarrow $ (2) For $\ell \in \mathbb {Z}$ , by $4\times 4$ Lemma, one has the following commutative diagram:

such that the square at the bottom right is anti-commutative, while the others are commutative, it implies that the third row is an exact triangle. Hence, $0\rightarrow \mathcal {H}^\ell _M(X)\rightarrow \mathcal {H}^\ell _M(Y)\rightarrow \mathcal {H}^\ell _M(Z)\rightarrow 0$ is exact in $\mathcal {H}_M$ by Lemma 2.3.

Proof We just prove the statement (1) since (2) is dual.

If $Z\in \langle \mathcal {P}\rangle _{1}$ , then $Z\cong \bigoplus _{i\in \mathbb {Z}}P_i[-i]$ with $P_i\in \mathcal {P}$ . Since $0\rightarrow \mathrm {Hom}_{\mathcal {T}}(P_i[-i],X)\rightarrow \mathrm {Hom}_{\mathcal {T}}(P_i[-i],Y)\rightarrow \mathrm {Hom}_{\mathcal {T}}(P_i[-i],Z)\rightarrow 0$ is exact, one has $0\rightarrow \mathrm {Hom}_{\mathcal {T}}(\bigoplus _{i\in \mathbb {Z}}P_i[-i],X)\rightarrow \mathrm {Hom}_{\mathcal {T}}(\bigoplus _{i\in \mathbb {Z}}P_i[-i],Y)\rightarrow \mathrm {Hom}_{\mathcal {T}}(\bigoplus _{i\in \mathbb {Z}}P_i[-i],Z)\rightarrow 0$ is exact, so $h=0$ . Conversely, there is a morphism $a_i:M_i\rightarrow \mathcal {H}^i_M(Z)$ with $M_i\in \mathcal {P}$ for $i\in \mathbb {Z}$ such that $\mathcal {H}^0_M(a_i)$ is surjective by Lemma 2.6, it follows by Lemma 2.5 that

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(\bigoplus_{i\in\mathbb{Z}}M_i[-i],Z)\cong\prod_{i\in\mathbb{Z}}\mathrm{Hom}_{\mathcal{T}}(M_i[-i],Z)\cong \prod_{i\in\mathbb{Z}}\mathrm{Hom}_{\mathcal{T}}(M_i,\mathcal{H}^i_M(Z)).\end{align*} $$

As $\mathcal {H}^\ell _M(\bigoplus _{i\in \mathbb {Z}}M_i[-i])\rightarrow \mathcal {H}^0_M(M_\ell )$ is surjective for all $\ell \in \mathbb {Z}$ , it follows by the above isomorphism and Lemma 4.3 that

$$ \begin{align*}X\rightarrow\bigoplus_{i\in\mathbb{Z}}M_i[-i]\rightarrow Z\stackrel{h}\rightarrow X[1]\end{align*} $$

is an M-ghost triangle. Thus, $h=0$ and $Z\in \langle \mathcal {P}\rangle _{1}$ .

Proof (i) “Only if” part. The case of $n=0$ follows by Lemma 4.4. Assume that $X\in \langle \mathcal {P}\rangle _{n+1}$ , $P\stackrel {\alpha }\rightarrow \widetilde {X}\stackrel {\beta }\rightarrow T\rightarrow P[1]$ is an exact triangle with $P\in \langle \mathcal {P}\rangle _1$ and $T\in \langle \mathcal {P}\rangle _n$ , and $X\stackrel {a}\rightarrow \widetilde {X}\stackrel {b}\rightarrow X$ are morphisms such that $ba=1_X$ . Let $X\stackrel {f}\rightarrow Y\stackrel {g}\rightarrow Z$ be a composite of $f\in \mathcal {G}_M$ and $g\in \mathcal {G}^n_M$ . Then $fb\alpha \in \mathcal {G}_M$ and so $fb\alpha =0$ . Thus, there is $\delta :T\rightarrow Y$ such that $\delta \beta =fb$ . Note that $g\delta \in \mathcal {G}^n_M$ , so $gfb=g\delta \beta =0$ by the induction, and hence $gf=gfba=0$ .

“If” part. By Lemma 4.4 and [Reference Christensen12, Theorem 3.5], there is an exact triangle $T\rightarrow X\stackrel {h}\rightarrow Y\rightarrow T[1]$ with $T\in \langle \mathcal {P}\rangle _{n+1}$ and $h\in \mathcal {G}^{n+1}_M$ . By assumption, $gf=0$ . Thus, X is a retract of T and so $X\in \langle \mathcal {P}\rangle _{n+1}$ .

(ii) This follows by the dual of (i).

Proof Let $\mathcal {P}\textrm {-}\mathrm {tri.dim}\mathcal {T}^{\mathrm {b}}=n$ , and let $X\in \mathcal {T}^{\mathrm {b}}$ . By Lemma 2.6, there is a morphism $\mathcal {H}^\ell _M(X)\rightarrow I^\ell $ with $I^\ell \in \mathcal {I}$ for $\ell \in \mathbb {Z}$ . By Lemma 2.5,

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(X,\bigoplus_{\ell=\mathrm{inf}X}^{\mathrm{sup}X}I^{\ell}[-\ell])\cong \bigoplus_{\ell=\mathrm{inf}X}^{\mathrm{sup}X}\mathrm{Hom}_{\mathcal{T}}(\mathcal{H}^\ell_M(X),I^{\ell}),\end{align*} $$

it yields an exact triangle $Y\stackrel {f}\rightarrow X\rightarrow \bigoplus _{\ell =\mathrm {inf}X}^{\mathrm {sup}X}I^{\ell }[-\ell ]\rightarrow Y[1]$ in $\mathcal {T}^{\mathrm {b}}$ , such that $f\in \mathcal {G}_M$ . By the induction, one can obtain an exact triangle $Z\stackrel {g}\rightarrow X\rightarrow T\rightarrow Z[1]$ in $\mathcal {T}^{\mathrm {b}}$ , such that $g\in \mathcal {G}^{n+1}_M$ and $T\in \langle \mathcal {I}\rangle _{n+1}$ . As $Z\in \langle \mathcal {P}\rangle _{n+1}$ , it follows by Lemma 4.5 that $g=0$ , so X is a retract of T. Hence, $\mathcal {I}\textrm {-}\mathrm {tri.dim}\mathcal {T}^{\mathrm {b}}\leqslant n$ . Similarly, one can get the desired equality.

“ $n\leqslant \mathrm {gldim}\mathcal {T}=d$ .” Let $X\in \mathcal {T}^{\mathrm {b}}$ . We may assume that $\mathrm {amp}X>0$ . For $\ell \in \mathbb {Z}$ , $\mathrm {pd}_{\mathcal {T}}\mathcal {H}^{\ell }_M(X)\leqslant d$ . Set $K_{\ell ,0}=\mathcal {H}^{\ell }_M(X)$ and fix an sppj resolution

$$ \begin{align*}\cdots\rightarrow P_{\ell,n}\rightarrow\cdots\rightarrow P_{\ell,0}\rightarrow\mathcal{H}^{\ell}_M(X)\end{align*} $$

with the corresponding exact triangle $K_{\ell ,n+1}\rightarrow P_{\ell ,n}\stackrel {f_{\ell ,n}}\rightarrow K_{\ell ,n}\rightarrow K_{\ell ,n+1}[1]$ for $n\geqslant 0$ , such that $K_{\ell ,d+1}=0$ . Set $\mathrm {inf}X=i$ . Consider the exact triangle

$$ \begin{align*}\mathcal{H}^{i}_M(X)[-i]\rightarrow \tau^{\leqslant i+1}(X)\rightarrow \mathcal{H}^{i+1}_M(X)[-i-1]\rightarrow\mathcal{H}^{i}_M(X)[-i+1].\end{align*} $$

As $\mathrm {Hom}_{\mathcal {T}}(P_{i+1,t}[-i-1],K_{i,t}[-i+1])=0$ for $t\geqslant 0$ , by $4\times 4$ Lemma, one has a diagram of exact triangles

such that all squares are commutative, except the middle square above is anti-commutative, where $L_{i+1,0}=\tau ^{\leqslant i+1}(X)$ . As $K_{i,d+1}=0=K_{i+1,d+1}$ , it implies that $L_{i+1,d+1}=0$ and then $\tau ^{\leqslant i+1}(X)\in \mathrm {thick}^{d+1}(\mathcal {P})$ . As $\mathrm {Hom}_{\mathcal {T}}(P_{i+2,t}[-i-2],\tau ^{\leqslant i+1}(X)[1])=0$ for $t\geqslant 0$ , by repeating this process on the triangle $\tau ^{\leqslant i+1}(X)\rightarrow \tau ^{\leqslant i+2}(X)\rightarrow \mathcal {H}^{i+2}_M(X)[-i-2]\rightarrow \tau ^{\leqslant i+1}(X)[1]$ , one has $\tau ^{\leqslant i+2}(X)\in \mathrm {thick}^{d+1}(\mathcal {P})$ . Consequently, $X\in \mathrm {thick}^{d+1}(\mathcal {P})$ , as required.

Proof Let $H=H_0$ and fix an sppj resolution

$$ \begin{align*}\cdots\rightarrow P_n\rightarrow P_{n-1}\rightarrow\cdots\rightarrow P_1\rightarrow P_0\rightarrow H_0\end{align*} $$

with the corresponding exact triangle $H_{i+1}\rightarrow P_i\rightarrow H_i\stackrel {f_{i+1}}\rightarrow H_{i+1}[1]$ . As $P_i\in \mathcal {H}_M$ , $f_{i+1}\in \mathcal {G}_M$ and $H_i\in \mathcal {H}_M$ for $i\geqslant 1$ by Lemma 4.3. One has a commutative diagram of exact triangles

where $T_1=P_0$ , $g_1=f_1$ and $g_i=f_i[i-1]f_{i-1}[i-2]\cdots f_2[1]f_1\in \mathcal {G}^i_M$ for $i\geqslant 2$ . As

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(H_{n-1}[-1],\mathcal{H}_M)\cong\mathrm{Hom}_{\mathcal{T}}(H[-n],\mathcal{H}_M)\neq0,\end{align*} $$

it follows that $H_{n-1}\not \in \mathcal {P}$ , and then $H_0\not \in \mathcal {P}\ast \cdots \ast \mathcal {P}[n-1]$ by Proposition 3.6. Thus, $g_n\neq 0$ . As $\alpha _i:\mathrm {Hom}_{\mathcal {T}}(-,H_{i-1}[i-1])\rightarrow \mathrm {Hom}_{\mathcal {T}}(-,H_{i}[i])$ vanishes on $\mathrm {thick}^{1}(\mathcal {P})$ for $i\geqslant 1$ and $g_n\neq 0$ , it follows by [Reference Rouquier24, Lemma 4.11] that $H\not \in \mathrm {thick}^{n}(\mathcal {P})$ .

Proof If $\mathrm {pd}_{\mathcal {T}}H=t$ , then $H\in \mathrm {thick}^{t+1}(\mathcal {P})$ by Theorem 3.3. Consider the sppj resolution

$$ \begin{align*}\cdots\rightarrow P_n\rightarrow P_{n-1}\rightarrow\cdots\rightarrow P_1\rightarrow P_0\rightarrow H\end{align*} $$

with the corresponding exact triangle $H_{i+1}\rightarrow P_i\rightarrow H_i\stackrel {f_{i+1}}\rightarrow H_{i+1}[1]$ for $i\geqslant 0$ , where $H_0=H$ . Then $\mathrm {Hom}_{\mathcal {T}}(H_{t},\mathcal {H}_M[1])\cong \mathrm {Hom}_{\mathcal {T}}(H,\mathcal {H}_M[t+1])=0$ by Lemma 4.7. Note that $f_{i+1}\in \mathcal {G}_M$ , so $H_i\in \mathcal {H}_M$ for all $i\geqslant 0$ . Thus, the triangle $H_{t+1}\rightarrow P_t\rightarrow H_t\rightarrow H_{t+1}[1]$ splits, and then $H_{t}\in \mathcal {P}$ . Also by Lemma 4.3,

$$ \begin{align*}0\rightarrow H_t\rightarrow P_{t-1}\rightarrow\cdots\rightarrow P_1\rightarrow P_0\rightarrow H\rightarrow0\end{align*} $$

is exact in $\mathcal {H}_M$ , it implies that $\mathrm {pd}_{\mathcal {H}_M}H\leqslant t$ . Let $\mathrm {pd}_{\mathcal {H}_M}H=r$ and $0\rightarrow Q_r\rightarrow Q_{r-1}\rightarrow \cdots \rightarrow Q_0\rightarrow H\rightarrow 0$ be exact in $\mathcal {H}_M$ with each $Q_{i}\in \mathcal {P}$ . Set $K_i=\mathrm {coker}(Q_i\rightarrow Q_{i-1})$ . Then $Q_i\rightarrow K_i$ is an sppj morphism and $K_{i+1}\rightarrow Q_i\rightarrow K_i\rightarrow K_{i+1}[1]$ is an exact triangle for $0\leqslant i\leqslant r-1$ with $K_0=H$ and $K_r=Q_r$ , it implies that $\mathrm {pd}_{\mathcal {T}}H\leqslant r$ .

Proof The first equality follows by Lemma 4.8. By Theorem 4.6, it suffices to prove that $\mathrm {gldim}\mathcal {T}\leqslant \mathcal {P}\textrm {-}\mathrm {tri.dim}\mathcal {T}^{\mathrm {b}}=n$ . Let $X\in \mathcal {T}^{\mathrm {b}}$ and set $s=\mathrm {sup}X$ . If $\mathrm {amp}X=0$ , then $X[s]\in \mathcal {H}_M$ . Thus, $\mathrm {Hom}_{\mathcal {T}}(X,\mathcal {H}_M[-s+n+i])=0$ for $i\geqslant 1$ by Lemma 4.7, so $\mathrm {pd}_{\mathcal {T}}X\leqslant n$ by Theorem 3.3. Assume $\mathrm {amp}X>0$ , and consider the exact triangle

$$ \begin{align*}\tau^{\leqslant s-1}(X)\rightarrow X\rightarrow \mathcal{H}^s_M(X)[-s]\rightarrow\tau^{\leqslant s-1}(X)[1].\end{align*} $$

As $\mathrm {amp}\tau ^{\leqslant s-1}(X)\leqslant \mathrm {amp}X-1$ , it follows by the induction that $\mathrm {pd}_{\mathcal {T}}\tau ^{\leqslant s-1}(X)\leqslant n+\mathrm {amp}X-1$ . So $\mathrm {pd}_{\mathcal {T}}X\leqslant n+\mathrm {amp}X$ by Corollary 3.8, as desired.

Proof Let $\mathrm {gldim}\mathcal {T}\leqslant 1$ . Consider the exact triangle

$$ \begin{align*}\tau^{\leqslant-1}(M)\stackrel{\iota}\rightarrow M\rightarrow \mathcal{H}^0_M(M)\rightarrow\tau^{\leqslant-1}(M)[1].\end{align*} $$

As $\mathrm {pd}_{\mathcal {T}}\mathcal {H}^0_M(M)\leqslant 1$ and $\mathrm {sup}\tau ^{\leqslant -1}(M)\leqslant -1$ , $\mathrm {Hom}_{\mathcal {T}}(\mathcal {H}^0_M(M),\tau ^{\leqslant -1}(M)[1])=0$ by Theorem 3.3, so the triangle is split and $\tau ^{\leqslant -1}(M)\in \mathcal {P}$ . Thus, by Lemma 2.5,

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(\tau^{\leqslant-1}(M),M)\cong \mathrm{Hom}_{\mathcal{H}_M}(\mathcal{H}^0_M(\tau^{\leqslant-1}(M)),\mathcal{H}^0_M(M))=0,\end{align*} $$

it implies that $\iota =0$ and then $\tau ^{\leqslant -1}(M)=0$ . So M is tilting and $\mathrm {gldim}\mathcal {H}_M\leqslant 1$ by Theorem 4.9. The converse follows by Theorem 4.9.

Proof If $M\in \mathcal {T}^{\mathrm {b}}$ , then $\mathcal {T}^{\mathrm {b}}_p=\mathrm {thick}(\mathcal {P})\subseteq \mathcal {T}^{\mathrm {b}}$ and $\mathcal {T}^{\mathrm {b}}_i=\mathrm {thick}(\mathcal {I})\subseteq \mathcal {T}^{\mathrm {b}}$ by [Reference Kostas, Psaroudakis and Vitória17, Theorem 3.13]. Let $X\in \mathcal {T}^{\mathrm {b}}_p$ . Then $X\in \mathrm {thick}^{n+1}(\mathcal {P})$ for some $n\geqslant 0$ . If $n=0$ , then $X=\bigoplus _{\ell =\mathrm {inf}X}^{\mathrm {sup}X}M_\ell [-\ell ]$ with $M_\ell \in \mathcal {P}$ , it follows by Theorem 3.3 that $\mathrm {pd}_{\mathcal {T}}X\leqslant \mathrm {amp}X<\infty $ . Assume that $n\geqslant 1$ . Consider the exact triangle $Y\rightarrow X\rightarrow P\rightarrow Y[1]$ in $\mathcal {T}^{\mathrm {b}}$ with $P\in \mathrm {thick}^{1}(\mathcal {P})$ and $Y\in \mathrm {thick}^{n}(\mathcal {P})$ . Then $\mathrm {pd}_{\mathcal {T}}Y\leqslant n-1+\mathrm {amp}Y<\infty $ by the induction, it implies that $\mathrm {pd}_{\mathcal {T}}X<\infty $ by Corollary 3.8. Thus, $\mathcal {T}^{\mathrm {b}}_p=\{X\in \mathcal {T}^{\mathrm {b}}\hspace {0.02cm}|\hspace {0.02cm}\mathrm {pd}_{\mathcal {T}}X<\infty \}$ by Theorem 3.3. Similarly, one has $\mathcal {T}^{\mathrm {b}}_i=\{X\in \mathcal {T}^{\mathrm {b}}\hspace {0.02cm}|\hspace {0.02cm}\mathrm {id}_{\mathcal {T}}X<\infty \}$ .

Proof (1) $\Rightarrow $ (2) This follows by Lemma 4.8 and its dual.

(2) $\Rightarrow $ (1) For all $H\in \mathcal {I}$ , as $\mathrm {pd}_{\mathcal {H}_M}H<\infty $ , it follows by Lemma 4.8 that $\mathrm {pd}_{\mathcal {T}}H<\infty $ , so $\mathcal {T}^{\mathrm {b}}_i\subseteq \mathcal {T}^{\mathrm {b}}_p$ . For all $H'\in \mathcal {P}$ , as $\mathrm {id}_{\mathcal {H}_M}H'<\infty $ , it follows by the dual of Lemma 4.8 that $\mathrm {id}_{\mathcal {T}}H'<\infty $ , so $\mathcal {T}^{\mathrm {b}}_p\subseteq \mathcal {T}^{\mathrm {b}}_i$ , as required.

Proof As $\mathcal {T}$ is Noetherian, it follows that $M\in \mathcal {T}^{\mathrm {b}}$ and $\mathrm {End}_{\mathcal {T}}(M)$ is a Noetherian ring by [Reference Anderson and Fuller3, p. 131, Exercise 7]. Let $X\in \mathcal {T}^{\mathrm {b}}_c$ . Then $\bigoplus _{\ell \in \mathbb {Z}}\mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^\ell _M(X))\cong \bigoplus _{\ell \in \mathbb {Z}}\mathrm {Hom}_{\mathcal {T}}(M,X[\ell ])\in \mathrm {mod}k$ by Lemma 2.5, so $X\in \mathcal {T}^{\mathrm {b}}$ and $\mathrm {Hom}_{\mathcal {T}}(M,\mathcal {H}^\ell _M(X))$ is a finitely generated $\mathrm {End}_{\mathcal {T}}(M)$ -module for $\ell \in \mathbb {Z}$ , that is, there is an epimorphism

$$ \begin{align*}\mathrm{Hom}_{\mathcal{T}}(M,\mathcal{H}^0_M(M^{n_\ell}))\cong\mathrm{End}_{\mathcal{T}}(M)^{n_\ell}\rightarrow\mathrm{Hom}_{\mathcal{T}}(M,\mathcal{H}^\ell_M(X))\end{align*} $$

in $\mathrm {End}_{\mathcal {T}}(M)$ for some $n_\ell \geqslant 1$ . Thus, $\mathcal {H}^\ell _M(X)$ is finitely generated in $\mathcal {H}_M$ for all $\ell \in \mathbb {Z}$ . Conversely, it follows by the next epimorphism

$$ \begin{align*}\bigoplus_{\ell=\mathrm{inf}X}^{\mathrm{sup}X}\mathrm{Hom}_{\mathcal{T}}(M,\mathcal{H}^0_M(M^{n_\ell}))\twoheadrightarrow \bigoplus_{\ell=\mathrm{inf}X}^{\mathrm{sup}X}\mathrm{Hom}_{\mathcal{T}}(M,\mathcal{H}^\ell_M(X))\cong \bigoplus_{\ell=\mathrm{inf}X}^{\mathrm{sup}X}\mathrm{Hom}_{\mathcal{T}}(M,X[\ell])\end{align*} $$

and $\mathrm {Hom}_{\mathcal {T}}(M,M)\in \mathrm {mod}k$ that $\bigoplus _{\ell \in \mathbb {Z}}\mathrm {Hom}_{\mathcal {T}}(M,X[\ell ])\in \mathrm {mod}k$ , as desired.

Proof (1) $\Rightarrow $ (2). Let I be an ideal of $\mathrm {End}_{\mathcal {T}}(M)$ . Then there exists $H\in \mathcal {H}_M$ , such that $\mathrm {End}_{\mathcal {T}}(M)/I\cong \mathrm {Hom}_{\mathcal {T}}(M,H)$ by Lemma 2.4. As $\mathrm {End}_{\mathcal {T}}(M)$ is a Noetherian ring, $H\in \mathcal {T}^{\mathrm {b}}_c$ , so $\mathrm {pd}_{\mathcal {H}_M}H\leqslant d$ by Lemma 4.8. Thus, $\mathrm {gldim}\mathcal {H}_M=\mathrm {gldim}\mathrm {End}_{\mathcal {T}}(M)=d$ by Lemma 2.4.

(2) $\Rightarrow $ (1). Let $X\in \mathcal {T}^{\mathrm {b}}_c$ . By Lemma 5.7, one can choose a projective resolution

$$ \begin{align*}\cdots\rightarrow P_{\ell,n}\rightarrow\cdots\rightarrow P_{\ell,0}\rightarrow\mathcal{H}^{\ell}_M(X)\rightarrow0\end{align*} $$

in $\mathcal {H}_M$ , such that $P_{\ell ,d+1}=0$ and $P_{\ell ,n}\in \mathrm {add}\mathcal {H}^0_M(M)$ for all $n\geqslant 0$ and $\ell \in \mathbb {Z}$ . Thus, $X\in \mathrm {thick}^{d+1}(\mathrm {add}(M))$ by the proof of Theorem 4.6.

(2) $\Leftrightarrow $ (3) $\Leftrightarrow $ (4). These follow by Theorem 4.9.