Proof. Assume that \(X \in \operatorname{Gen}(E)\). Then there is a short exact sequence \[\begin{tikzcd} 0 & K & E & X& 0 \arrow[from=1-1, to=1-2] \arrow[ tail,from=1-2, to=1-3] \arrow["g",two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] By applying the \(\operatorname{Hom}(D,-)\) functor on this short exact sequence, we obtain the following long exact sequence \[\begin{tikzcd} 0 & \operatorname{Hom}(D,K) & \operatorname{Hom}(D,E) & \operatorname{Hom}(D,X) & \\ & \operatorname{Ext}^1(D,K) & \operatorname{Ext}^1(D,E) & \operatorname{Ext}^1(D,X) & 0, \arrow[from=1-1, to=1-2] \arrow[tail,from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=2-2] \arrow[from=2-2, to=2-3] \arrow[two heads,"g_{\bullet}",from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \end{tikzcd}\] where \(g_\bullet = \operatorname{Ext}^1(D,g)\). As \(\mathscr{A}\) is hereditary, we have that \(\operatorname{Ext}^2(D,K)=0\). So \(g_\bullet\) is surjective, and therefore, for any short exact sequence \(\xi \in \operatorname{Ext}^1(D,X)\), there exists \(\eta \in \operatorname{Ext}^1(D,E)\) such that \(g_\bullet(\eta) = \xi\). Since \(g_\bullet\) is the pushout along \(g\), we proved the assertion \((a)\).

By hypothesis, \(\operatorname{Ext}^1(D,E) \subseteq \mathcal{E}\). So we have \(\operatorname{Ext}^1(D,X) \subseteq \mathcal{E}\) as \(\mathcal{E}\) is an exact structure. If moreover \(X \in \operatorname{Gen}_\mathcal{E}(E)\), then \(g\) is obviously an \(\mathcal{E}\)-epimorphism. ◻

Proof. Suppose we have the following short exact sequence \[\begin{tikzcd} \xi: & 0 & F & E' & X& 0. \arrow[from=1-2, to=1-3] \arrow[tail, from=1-3, to=1-4] \arrow[two heads,from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \end{tikzcd}\] Since \(X \in \operatorname{Gen}_\mathcal{E}(E)\), there is a short exact sequence \[\begin{tikzcd} 0 & K & E & X& 0 & \in \mathcal{E}. \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,"\varphi",from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] We can construct the following diagram. \[\begin{tikzcd} & & & 0 & \\ 0 & F & E' & X & 0 \\ & & & E & \\ & & & K & \\ & & & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=5-4, to=4-4] \arrow[tail, from=4-4, to=3-4] \arrow[two heads,"\varphi",from=3-4, to=2-4] \arrow[from=2-4, to=1-4] \end{tikzcd}\] We complete the diagram by taking the pullback along \(\varphi\) and by using the kernel lifting property. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & F & E' & X & 0 \\ 0 & F & PB & E & 0 \\ 0 & 0 & K' & K & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=5-2, to=4-2] \arrow[tail, from=4-2, to=3-2] \arrow[equals,from=3-2, to=2-2] \arrow[from=2-2, to=1-2] \arrow[from=5-3, to=4-3] \arrow[tail, from=4-3, to=3-3] \arrow[two heads,from=3-3, to=2-3] \arrow[from=2-3, to=1-3] \arrow[from=5-4, to=4-4] \arrow[tail, from=4-4, to=3-4] \arrow[two heads,"\varphi",from=3-4, to=2-4] \arrow[from=2-4, to=1-4] \end{tikzcd}\]

As the short sequences given by the three columns and those given by the first two rows of the above diagram are exact, by 1, the short sequence given by the last row is exact too. So \(K' \cong K\).

Moreover, the short exact sequences given by the third column and the middle row are in \(\mathcal{E}\) by hypothesis. As \(\mathcal{E}\) is an exact structure, the short exact sequence given by the middle column is in \(\mathcal{E}\). As the short exact sequence given by the first column and the one provided by the last row are split, they are in \(\mathcal{E}\).

By 1, we have that \(\xi \in {\mathcal{E}}\). ◻

Proof. Let \(X\in \operatorname{Sub}(E)\). Then, there is a short exact sequence \[\begin{tikzcd} 0 & X & E & C& 0. \arrow[from=1-1, to=1-2] \arrow["f",tail, from=1-2, to=1-3] \arrow[two heads, from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] We obtain the following long exact sequence by applying the \(\operatorname{Hom}(-,F)\) functor on this short exact sequence. \[\begin{tikzcd} 0 & \operatorname{Hom}(C,F) & \operatorname{Hom}(E,F) & \operatorname{Hom}(X,F) & \\ & \operatorname{Ext}^1(C,F) & \operatorname{Ext}^1(E,F) & \operatorname{Ext}^1(X,F) & 0, \arrow[from=1-1, to=1-2] \arrow[from=1-2, to=1-3] \arrow[from=1-3, to=1-4] \arrow[from=1-4, to=2-2] \arrow[from=2-2, to=2-3] \arrow["\scriptscriptstyle{f^{\bullet}}",from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \end{tikzcd}\] where \(f^{\bullet} = \operatorname{Ext}^1(f,F)\). As \(\mathscr{A}\) is hereditary, we have \(\operatorname{Ext}^2(C,F)=0\). So \(f^{\bullet}\) is surjective: for any short exact sequence \(\xi \in \operatorname{Ext}^1(X,F)\), there exists \(\eta \in \operatorname{Ext}^1(E,F)\) such that \(f^{\bullet}(\eta) = \xi\). Since \(f^{\bullet}\) is the pullback along \(f\), it means that every short exact sequence in \(\operatorname{Ext}^1(X,F)\) is obtained by a pullback of a short exact sequence in \(\operatorname{Ext}^1(E,F)\).

By hypothesis, \(\operatorname{Ext}^1(E,F) \subseteq \mathcal{E}\). So we have \(\operatorname{Ext}^1(X,F) \subseteq \mathcal{E}\) as \(\mathcal{E}\) is an exact structure. If moreover \(X \in \operatorname{Sub}_\mathcal{E}(E)\), then \(f\) is an \(\mathcal{E}\)-monomorphism. ◻

Proof. We have the following short exact sequence \[\begin{tikzcd} \varsigma: & 0 & X & E'' & D& 0. \arrow[from=1-2, to=1-3] \arrow[tail, from=1-3, to=1-4] \arrow[two heads,from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \end{tikzcd}\] Since \(X \in \operatorname{Sub}_\mathcal{E}(E)\), there is a short exact sequence \[\begin{tikzcd} 0 & X & E & C & 0& \in \mathcal{E}. \arrow[from=1-1, to=1-2] \arrow[tail,"\psi", from=1-2, to=1-3] \arrow[,two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] We can construct the following diagram. \[\begin{tikzcd} & 0 & & & \\ 0 & X & E'' & D & 0 \\ & E & & & \\ & C & & & \\ & 0 & & & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=4-2, to=5-2] \arrow[two heads, from=3-2, to=4-2] \arrow[tail,"\psi",from=2-2, to=3-2] \arrow[from=1-2, to=2-2] \end{tikzcd}\] We complete the diagram by taking the pushout along \(\psi\) and by using the cokernel lifting property. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & X & E'' & F & 0 \\ 0 & E & PO & F & 0 \\ 0 & C & C'' & 0 & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=4-2, to=5-2] \arrow[two heads, from=3-2, to=4-2] \arrow[tail,"\psi",from=2-2, to=3-2] \arrow[from=1-2, to=2-2] \arrow[from=4-3, to=5-3] \arrow[two heads, from=3-3, to=4-3] \arrow[tail,from=2-3, to=3-3] \arrow[from=1-3, to=2-3] \arrow[from=4-4, to=5-4] \arrow[two heads, from=3-4, to=4-4] \arrow[equals,from=2-4, to=3-4] \arrow[from=1-4, to=2-4] \end{tikzcd}\]

As the short sequences given by the three columns and those given by the first two rows of the above diagram are exact, by 1, the short sequence given by the last row is exact too. So \(C'' \cong C\).

Moreover, the short exact sequences given by the first column and the middle row are in \(\mathcal{E}\) by hypothesis. As \(\mathcal{E}\) is an exact structure, the short exact sequence given by the middle column is in \(\mathcal{E}\). As the short exact sequence given by the first column and the one provided by the last row are split, they are in \(\mathcal{E}\).

By 1, we have that \(\varsigma \in {\mathcal{E}}\). ◻

Proof. Consider \(X\) an object of \(\operatorname{GS}^1_{\mathcal{E}}(\mathscr{C})\). Then \(X\) is in \(\operatorname{Gen}_{\mathcal{E}}(\mathscr{C})\) or in \(\operatorname{Sub}_{\mathcal{E}}(\mathscr{C})\).

Assume that \(X \in \operatorname{Gen}_{\mathcal{E}}(\mathscr{C})\). We can then suppose that \(X \in \operatorname{Gen}_{\mathcal{E}}(E) \subseteq \operatorname{Gen}(E)\) for \(E \in \mathscr{C}\). By hypothesis, we know for any \(D,F\in \mathscr{C}\), \(\operatorname{Ext}^1(D,E),\operatorname{Ext}^1(E,F) \subseteq \mathcal{E}\). By 4, \(\operatorname{Ext}^1(D,X) \subseteq \mathcal{E}\) for any \(D \in \operatorname{GS}^{1}_{\mathcal{E}}(\mathscr{C})\). By 5, \(\operatorname{Ext}^1(X,F) \subseteq \mathcal{E}\) for any \(F \in \mathscr{C}\).

Assume that \(X \in \operatorname{Sub}_{\mathcal{E}}(\mathscr{C})\). Similarly, we have that \(X \in \operatorname{Sub}_{\mathcal{E}}(E) \subseteq \operatorname{Sub}(E)\) for some \(E \in \mathscr{C}\). By hypothesis, we know for any \(D,F \in \mathscr{C}\), \(\operatorname{Ext}^1(D,E),\operatorname{Ext}^1(E,F) \subseteq \mathcal{E}\). By 6, \(\operatorname{Ext}^1(X,F) \subseteq \mathcal{E}\) for any \(F \in \mathscr{C}\). By 7, \(\operatorname{Ext}^1(D,X) \subseteq \mathcal{E}\) for any \(D \in \mathscr{C}\).

Therefore, we get to the conclusion that, for any \(X\in \operatorname{GS}^1_{\mathcal{E}}(\mathscr{C})\) and \(E' \in \mathscr{C}\), we have \(\operatorname{Ext}^1(X,E'),\operatorname{Ext}^1(E',X)\subseteq \mathcal{E}\). ◻

Proof. We prove by induction that \(\operatorname{GS}^{i}_{\mathcal{E}}(\mathscr{C})\) is \({\mathcal{E}}\)-adpated, for all \(i \geq 0\). For \(i=0\), the statement holds as \(\mathcal{C}\) is \(\mathcal{E}\)-adapted.

Fix \(k \geqslant 0\). Assume that \(\operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C})\) is \({\mathcal{E}}\)-adapted. Suppose that \(X,Y \in \operatorname{GS}^{k+1}_{\mathcal{E}}(\mathscr{C})\). Then \(X \in \operatorname{Gen}_{\mathcal{E}}(\operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C}))\) or \(X \in \operatorname{Sub}_{\mathcal{E}}(\operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C}))\).

If \(X \in \operatorname{Gen}_{\mathcal{E}}(\operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C}))\). So \(X \in \operatorname{Gen}_{\mathcal{E}}(E) \subseteq \operatorname{Gen}(E)\) for some \(E \in \operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C})\). By induction hypothesis and 15, we get that \(\operatorname{Ext}^1(Y,E),\operatorname{Ext}^1(E,Y) \subseteq \mathcal{E}\). Therefore, by 4, \(\operatorname{Ext}^1(Y,X) \subseteq \mathcal{E}\). Using 5, we also have \(\operatorname{Ext}^1(X,Y) \subseteq \mathcal{E}\).

If \(X \in \operatorname{Sub}_{\mathcal{E}}(\operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C}))\). Let \(E \in \operatorname{GS}^{k}_{\mathcal{E}}(\mathscr{C})\) such that \(X \in \operatorname{Sub}_{\mathcal{E}}(E) \subseteq \operatorname{Sub}(E)\). By induction hypothesis and 15, we get that both \(\operatorname{Ext}^1(Y,E) \subseteq \mathcal{E}\) and \(\operatorname{Ext}^1(E,Y) \subseteq \mathcal{E}\). Therefore, by 7, we also have that \(\operatorname{Ext}^1(Y,X) \subseteq \mathcal{E}\).

In either case, we prove that \(\operatorname{Ext}^1(Y,X) \subseteq \mathcal{E}\). So \(\operatorname{GS}^{k+1}_{\mathcal{E}}(\mathscr{C})\) is \(\mathcal{E}\)-adapted. We get the desired result on \(\operatorname{GS}^{i}_{\mathcal{E}}(\mathscr{C})\) for all \(i \geq 0\). So, \(\operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) is \(\mathcal{E}\)-adapted. ◻

Proof. It is enough to show that, for any \(m \in \mathbb{N}\), if we have a short exact sequence \[\begin{tikzcd} 0 & X & E & U & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(X\in \operatorname{GS}^m_{\mathcal{E}}(\mathscr{C})\) and \(U \in \mathscr{C}\), then \(E \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\). We proceed by induction on \(m \in \mathbb{N}\).

If \(m=0\), then \(X \in \operatorname{GS}^0_{\mathcal{E}}(T) = \mathscr{C}\). As \(\mathscr{C}\) is closed under extensions, the result follows. Fix \(m \in \mathbb{N}\). Assume that if there is a short exact sequence \[\begin{tikzcd} 0 & X & E & U & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(X \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\) and \(U \in \mathscr{C}\), then \(E \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\).

Consider a short exact sequence \[\begin{tikzcd} 0 & Y & E' & U' & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(Y \in \operatorname{GS}^{m+1}_{\mathcal{E}}(\mathscr{C})\) and \(U' \in \mathscr{C}\). Then we have that \(Y \in \operatorname{Sub}_\mathcal{E}(\operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C}))\) or \(Y \in \operatorname{Gen}_\mathcal{E}(\operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})).\)

Suppose \(Y \in \operatorname{Gen}_\mathcal{E}(\operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C}))\). Let \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\) such that \(Y \in \operatorname{Gen}_{\mathcal{E}}(F)\). Note that \(\operatorname{Gen}_{\mathcal{E}}(F) \subseteq \operatorname{Gen}(F)\). By 16, we have that \(\operatorname{Ext}^1(U,F) \subseteq \mathcal{E}\). Using 4, we have:

1. any \(\xi \in \operatorname{Ext}^1(U',Y)\) is obtained by a pushout of some \(\eta \in \operatorname{Ext}^1(U',F)\) along an \(\mathcal{E}\)-epimorphism \(\begin{tikzcd} g: F & Y; \arrow[two heads,from=1-1, to=1-2] \end{tikzcd}\) and,

2. \(\operatorname{Ext}^1(U',Y) \subseteq \mathcal{E}\).

We obtain the following commutative diagram. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & K' & K'' & 0 & 0 \\ 0 & F & E'' & U' & 0 \\ 0 & Y & PO & U' & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=4-2, to=5-2] \arrow[two heads, from=3-2, to=4-2] \arrow[tail,from=2-2, to=3-2] \arrow[from=1-2, to=2-2] \arrow[from=4-3, to=5-3] \arrow[two heads, from=3-3, to=4-3] \arrow[tail,from=2-3, to=3-3] \arrow[from=1-3, to=2-3] \arrow[from=4-4, to=5-4] \arrow[equals, from=3-4, to=4-4] \arrow[tail,from=2-4, to=3-4] \arrow[from=1-4, to=2-4] \end{tikzcd}\]

The short sequences given by the three columns and those given by the last two rows are exact. By 1, the short sequence given by the first row is exact, too. Therefore \(K' \cong K''\).

As \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\), we have that \(E'' \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) by induction hypothesis. So, in the diagram above, the short sequences given by the three rows and those given by the first and third columns are \(\mathcal{E}\)-exact. By 1, the short sequence given by the middle column is \(\mathcal{E}\)-exact. Thus \(PO \in \operatorname{Gen}_\mathcal{E}(E'') \subseteq \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\).

As every short exact sequence in \(\operatorname{Ext}^1(U',Y)\) is obtained by a pushout of a short exact sequence in \(\operatorname{Ext}^1(U',F)\), we get that \(PO \cong E' \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\).

Assume that \(X \in \operatorname{Sub}_\mathcal{E}(\operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C}))\). There exists a short exact sequence \[\begin{tikzcd} 0 & Y & F & C& 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\). We obtain the following pushout diagram. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & C' & C'' & 0 & 0 \\ 0 & F & PO & U' & 0 \\ 0 & Y & E' & U' & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=5-2, to=4-2] \arrow[tail, from=4-2, to=3-2] \arrow[two heads,from=3-2, to=2-2] \arrow[from=2-2, to=1-2] \arrow[from=5-3, to=4-3] \arrow[tail, from=4-3, to=3-3] \arrow[two heads,from=3-3, to=2-3] \arrow[from=2-3, to=1-3] \arrow[from=5-4, to=4-4] \arrow[equals, from=4-4, to=3-4] \arrow[two heads,from=3-4, to=2-4] \arrow[from=2-4, to=1-4] \end{tikzcd}\]

The short sequences given by the three columns and those given by the last two rows are exact. By 1, the short sequence given by the first row is exact, too. So \(C' \cong C''\).

As \(F,Y,T \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\), the two last rows are in \(\mathcal{E}\) by 16. Moreover we know that \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\), which implies that \(PO \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) by induction hypothesis.

The short sequences given by the three rows and those given by the first and third columns are \(\mathcal{E}\)-exact. By 1, the short sequence given by the middle column is also \(\mathcal{E}\)-exact. Then, \(E' \in \operatorname{Sub}_\mathcal{E}(PO) \subseteq \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\).

In either case, we get \(E' \in \operatorname{GS}_\mathcal{E}(\mathscr{C})\). This completes the proof by induction. ◻

Proof. Let us prove by induction that, for any \(m \in \mathbb{N}\), if we have a short exact sequence \[\begin{tikzcd} 0 & X & E & Y & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(X\in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) and \(Y\in \operatorname{GS}^m_{\mathcal{E}} (\mathscr{C})\), then \(E \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\). The \(m=0\) case follows 8.

Fix \(m \in \mathbb{N}\). Assume that if we have the short exact sequence \[\begin{tikzcd} 0 & U & E' & V & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(U \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) and \(V \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\), then \(E' \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\). Consider a short exact sequence \[\begin{tikzcd} 0 & X & E & Y & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(X \in \operatorname{GS}_\mathcal{E}(\mathscr{C})\) and \(Y \in \operatorname{GS}_\mathcal{E}^{m+1}(\mathscr{C})\). As before, we must consider two cases.

Suppose \(Y \in \operatorname{Sub}_\mathcal{E}(F)\) for some \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\). By 16, we have that \(\operatorname{Ext}^1(F,X) \subseteq \mathcal{E}\). By 6, we have:

1. any \(\xi \in \operatorname{Ext}^1(Y,X)\) is obtained by a pushout of some \(\eta \in \operatorname{Ext}^1(F,X)\) along an \(\mathcal{E}\)-monomorphism \(\begin{tikzcd} f: F & X; \arrow[tail,from=1-1, to=1-2] \end{tikzcd}\) and,

2. \(\operatorname{Ext}^1(Y,X) \subseteq \mathcal{E}\).

We obtain the following commutative diagram. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & 0 & C'' & C' & 0 \\ 0 & X & E'' & F & 0 \\ 0 & X & PB & Y & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=5-2, to=4-2] \arrow[equals, from=4-2, to=3-2] \arrow[two heads,from=3-2, to=2-2] \arrow[from=2-2, to=1-2] \arrow[from=5-3, to=4-3] \arrow[tail, from=4-3, to=3-3] \arrow[two heads,from=3-3, to=2-3] \arrow[from=2-3, to=1-3] \arrow[from=5-4, to=4-4] \arrow[tail, from=4-4, to=3-4] \arrow[two heads,from=3-4, to=2-4] \arrow[from=2-4, to=1-4] \end{tikzcd}\]

The short sequences given by the three columns and those given by the last two rows are exact. By 1, the short sequence given by the first row is exact, too. Then \(C'' \cong C\). Moreover, as \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\), we have that \(E'' \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) by induction hypothesis.

The short exact sequences given by the three rows and those given by the first and third columns are in \({\mathcal{E}}\). By 1, the short sequence given by the middle column is also \(\mathcal{E}\)-exact. So \(PB \in \operatorname{Sub}_\mathcal{E}(E'') \subseteq \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\). By the fact that every short exact sequence in \(\operatorname{Ext}^1(Y,X)\) is obtained by a pullback of a short exact sequence in \(\operatorname{Ext}^1(E',X)\), we get that \(PB \cong E \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\).

Now, assume that \(Y \in \operatorname{Gen}_\mathcal{E}(F)\) for some \(F \in \operatorname{GS}_\mathcal{E}^m(\mathscr{C})\). Therefore, there exists a short exact sequence \[\begin{tikzcd} 0 & K' & F & Y & 0 & \in \mathcal{E}. \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] We obtain the following pullback diagram. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & X & E & Y & 0 \\ 0 & X & PB & F & 0 \\ 0 & 0 & K & K' & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=5-2, to=4-2] \arrow[tail, from=4-2, to=3-2] \arrow[equals,from=3-2, to=2-2] \arrow[from=2-2, to=1-2] \arrow[from=5-3, to=4-3] \arrow[tail, from=4-3, to=3-3] \arrow[two heads,from=3-3, to=2-3] \arrow[from=2-3, to=1-3] \arrow[from=5-4, to=4-4] \arrow[tail, from=4-4, to=3-4] \arrow[two heads,from=3-4, to=2-4] \arrow[from=2-4, to=1-4] \end{tikzcd}\] Once again, by 1, the short sequence given by the last row is exact, and so \(K \cong K'\). Furthermore, we have that \(F,X,Y \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\). So the first two rows are in \(\mathcal{E}\) by 16. Moreover \(F \in \operatorname{GS}^{m}_{\mathcal{E}}(\mathscr{C})\). So we have that \(PB \in \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\) by induction hypothesis. Using 1, the short sequence given by the middle column is \({\mathcal{E}}\)-exact. Therefore \(E \in \operatorname{Gen}_\mathcal{E}(PB) \subseteq \operatorname{GS}_{\mathcal{E}}(\mathscr{C})\).

This completes the induction proof, and allows us to get that \(\operatorname{GS}_\mathcal{E}(\mathscr{C})\) is closed under extensions. ◻

Proof. It follows from 17 and 9. ◻

Proof. Let \(\mathscr{C} \subseteq \mathscr{A}\) be an \(\mathcal{E}\)-adapted subcategory closed under extensions which contains \(T\). By 8, and by 3 \((iv)\), to prove that \(\mathscr{C} \subseteq \operatorname{GS}_\mathcal{E}(T)\), it is enough to prove that all the objects in \(\mathscr{C}\) are in \(\operatorname{GS}_\mathcal{E}(T)\).

Suppose \(V \in \operatorname{proj}(\mathscr{A}) \subseteq \operatorname{Sub}(T)\). By 20, there is a short exact sequence \[\begin{tikzcd} \xi: & 0 & V & \Theta_0 & \Theta_1 & 0, & \arrow[from=1-2, to=1-3] \arrow[tail, from=1-3, to=1-4] \arrow[two heads,from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \end{tikzcd}\] with \(\Theta_0,\Theta_1 \in \operatorname{add}(T)\). Since \(V,\Theta_1 \in \mathscr{C}\) which is \(\mathcal{E}\)-adapted, then \(\xi \in \mathcal{E}\). Thus \(V \in \operatorname{Sub}_\mathcal{E}(T_0) \subseteq \operatorname{GS}_\mathcal{E}(T)\).

Assume that \(V \notin \operatorname{proj}(\mathscr{A})\). Consider the projective resolution of \(V\) \[\begin{tikzcd} 0 & P_1 & P_0 & V & 0 & \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(P_0,P_1 \in \operatorname{proj}(\mathscr{A})\). Since \(T\) is tilting, there is a short exact sequence \[\begin{tikzcd} 0 & P_1 & \Theta_0 & \Theta_1 & 0, & \arrow[from=1-1, to=1-2] \arrow[tail,"\psi", from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(\Theta_0,\Theta_1 \in \operatorname{add}(T)\). We obtain the following diagram by taking the pushout along \(\psi\) and by using the cokernel lifting property. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & P_1 & P_0 & V & 0 \\ 0 & \Theta_0 & PO & V & 0 \\ 0 & \Theta_1 & C & 0 & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=4-2, to=5-2] \arrow[two heads, from=3-2, to=4-2] \arrow[tail,"\psi",from=2-2, to=3-2] \arrow[from=1-2, to=2-2] \arrow[from=4-3, to=5-3] \arrow[two heads, from=3-3, to=4-3] \arrow[tail,from=2-3, to=3-3] \arrow[from=1-3, to=2-3] \arrow[from=4-4, to=5-4] \arrow[two heads, from=3-4, to=4-4] \arrow[equals,from=2-4, to=3-4] \arrow[from=1-4, to=2-4] \end{tikzcd}\] As the short sequences given by the three columns and those given by the first two rows of the above diagram are exact, by 1, the short sequence given by the last row is exact too. So \(C \cong \Theta_1\). Since \(V,\Theta_0 \in \mathscr{C}\), by assumption on \(\mathscr{C}\), then the middle row lies in \(\mathcal{E}\), and \(W = PO \in \mathscr{C}\).

Since \(T\) is tilting, there is also a short exact sequence \[\begin{tikzcd} 0 & P_0 & \Theta_2 & \Theta_3 & 0, & \arrow[from=1-1, to=1-2] \arrow[tail,"\phi", from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(\Theta_2,\Theta_3 \in \operatorname{add}(T)\). Given the short exact sequence and the middle column of the diagram above, we obtain the following diagram by taking the pushout along \(\phi\) and applying the cokernel lifting property again. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & P_0 & \Theta_2 & \Theta_3 & 0 \\ 0 & W & PO' & C' & 0 \\ 0 & \Theta_1 & \Theta_1 & 0 & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail,"\phi", from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[equals, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=4-2, to=5-2] \arrow[two heads, from=3-2, to=4-2] \arrow[tail,from=2-2, to=3-2] \arrow[from=1-2, to=2-2] \arrow[from=4-3, to=5-3] \arrow[two heads, from=3-3, to=4-3] \arrow[tail,from=2-3, to=3-3] \arrow[from=1-3, to=2-3] \arrow[from=4-4, to=5-4] \arrow[two heads, from=3-4, to=4-4] \arrow[from=2-4, to=3-4] \arrow[from=1-4, to=2-4] \end{tikzcd}\] As the short sequences given by the three rows and those given by the first two columns of the above diagram are exact, by 1, the short sequence given by the last column is exact too. So \(C' \cong \Theta_3\). Since \(T\) is tilting, the middle column splits, hence \(PO' \cong \Theta_1 \oplus \Theta_2\).

We get the following short exact sequence. \[\begin{tikzcd} \varsigma: & 0 & W & \Theta_1 \oplus \Theta_2 & \Theta_3 & 0 & \arrow[from=1-2, to=1-3] \arrow[tail, from=1-3, to=1-4] \arrow[two heads,from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \end{tikzcd}\] As \(W,\Theta_3 \in \mathscr{C}\), as \(\mathscr{C}\) is \(\mathcal{E}\)-adapted, then \(\varsigma \in \mathcal{E}\) and \(W \in \operatorname{Sub}_\mathcal{E}(\Theta_1 \oplus \Theta_2) \subseteq \operatorname{GS}_\mathcal{E}(T)\).

We finally obtain the following short exact sequence from the middle row of the first diagram \[\begin{tikzcd} \mu: & 0 & T_0 & W & V & 0 & \arrow[from=1-2, to=1-3] \arrow[tail, from=1-3, to=1-4] \arrow[two heads,from=1-4, to=1-5] \arrow[from=1-5, to=1-6] \end{tikzcd}\] As \(\mu \in \mathcal{E}\) and \(W \in \operatorname{GS}_\mathcal{E}(T)\), we get that \(V \in \operatorname{Gen}_\mathcal{E}(W) \subseteq \operatorname{GS}_\mathcal{E}(T)\).

So all the objects in \(\mathscr{C}\) are in \(\operatorname{GS}_\mathcal{E}(T)\), and therefore, we get the desired result. ◻

Proof. The assertion \((a)\) follows directly from the proof of 22. The assertion \((b)\) is a direct consequence of both 22 and 18 \((d)\). ◻

Proof. To prove this proposition, it is enough to show that \(\operatorname{GS}_{\mathcal{E}}(T') = \operatorname{GS}_{\mathcal{E}}(T)\) whenever \(T' \cong \pmb{\mu}_{U,\mathcal{E}}(T)\) for some \(U \in \operatorname{ind}(Q)\). We obtain the desired result by induction on the length of the sequence of \(\mathcal{E}\)-mutations, which allows us to have \(T'\approx_{\mathcal{E}} T\).

We have three cases to treat.

1. If \(\pmb{\mu}_{U,\mathcal{E}}(T) \cong T\), then the result is obvious.

2. If \(\pmb{\mu}_{U,\mathcal{E}}(T) = \pmb{\mu}_{U,\mathcal{E}}^-(T)\), by writing \(T = \widetilde{T} \oplus U\), we have the following short \(\mathcal{E}\)-exact sequence: \[\begin{tikzcd} 0 & U & \Theta & \operatorname{Coker}(f)& 0, \arrow[from=1-1, to=1-2] \arrow["f",tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] where \(f\) is a left minimal \((\operatorname{add}(\widetilde{T}), \mathcal{E})\)-approximation of \(U\). As \(\Theta \in \operatorname{add}(T) \subset \operatorname{GS}_\mathcal{E}(T)\), and \(f\) is an \(\mathcal{E}\)-monomorphism, by 3, we have that \(\operatorname{Coker}(f) \in \operatorname{GS}_{\mathcal{E}}(T)\). Therefore \(T' \cong \widetilde{T} \oplus \operatorname{Coker}(f) \in \operatorname{GS}_{\mathcal{E}}(T)\).

3. If \(\pmb{\mu}_{U,\mathcal{E}}(T) = \pmb{\mu}_{U, \mathcal{E}}^{+}(X)\), by writing \(T = \widetilde{T} \oplus U\), we have the following short \(\mathcal{E}\)-exact sequence: \[\begin{tikzcd} 0 & \operatorname{Ker}(f) & \Theta' & U & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow["g",two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] where \(g\) is a right minimal \((\operatorname{add}(\widetilde{T}), \mathcal{E})\)-approximation of \(U\). As \(\Theta' \in \operatorname{add}(\widetilde{T})\) and \(g\) is an \(\mathcal{E}\)-epimorphism, by 3, we have that \(\operatorname{Ker}(f) \in \operatorname{GS}_{\mathcal{E}}(T)\). So \(T' \cong \widetilde{T} \oplus \operatorname{Ker}(f) \in \operatorname{GS}_{\mathcal{E}}(T)\).

In either case, we got that \(T' \sim_{\mathcal{E}} T\). ◻

Proof. By 33, the map \(g\) is an epimorphism and we get the short exact sequence \[\begin{tikzcd} \xi : 0 & K_M & E_M & X & 0, \arrow[from=1-1, to=1-2] \arrow["f",tail, from=1-2, to=1-3] \arrow["g",two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(E_M \in \operatorname{add}(M)\).

Consider the morphism \(\text{id}_{E_M} - fh \in \operatorname{End}(E_M)\) with \(h \in \operatorname{Hom}(E_M,K_M)\). We have \(g(\text{id}_{E_M} - fh) = g - gfh = g\). Since \(g\) is minimal, \(\text{id}_{E_M} - fh\) is invertible for any \(h \in \operatorname{Hom}(E_M,K_M)\). Hence, \(f \in \operatorname{Rad}_{\mathscr{A}}(K_M,E_M)\). ◻

Proof. Consider \(X \in \operatorname{Gen}_\mathcal{E}(T)\). By 32 and 11, there exists a right minimal \((\operatorname{add}(T),\mathcal{E})\)-approximation \(g\) of X : \[\begin{tikzcd} \xi : 0 \arrow[r] & K_T \arrow[r,tail,"f"] & E_T \arrow[r,two heads,"g"] & X \arrow[r] & 0 \end{tikzcd}\] with \(E_T\in\operatorname{add}(T)\). Given that \(X \notin \operatorname{add}(T)\), \(\xi\) does not split and \(K_T \neq 0\). Moreover, \(K_T \in \operatorname{add}(T)\) by 14 and \(f \in \operatorname{Rad}_{\mathscr{A}}(K_T,E_T)\) by 13.

We will show that \(\operatorname{add}(K_T) \cap \operatorname{add}(E_T) = 0\).

Consider any morphism \(\alpha \in \operatorname{Hom}(K_T,E_T)\). Given that \(X\) is indecomposable and \(Q\) is of \(ADE\) Dynkin type, then \(X\) is rigid. By applying the functor \(\operatorname{Hom}(-,X)\), we get the short exact sequence

\[\begin{tikzcd} \xi : 0 & \operatorname{Hom}(X,X) & \operatorname{Hom}(E_T,X) & \operatorname{Hom}(K_T,X) & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow["{\scriptstyle \operatorname{Hom}(f,X)}",two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] Therefore, there exist \(\beta \in \operatorname{Hom}(E_T,X)\) such that \(\beta f = g\alpha\).

We can construct the following commutative diagram:

\[\begin{tikzcd}[column sep=large] & 0 \arrow[r] & K_T \arrow[r, tail, "f"] \arrow["\alpha",d] & E_T \arrow[r, two heads, "g"] \arrow["\beta",d, dashed] & X \arrow[r] & 0 \\ 0 \arrow[r] & K_T \arrow[r, tail, "f"'] & E_T \arrow[r, two heads, "g"'] & X \arrow[r] & 0 & \end{tikzcd}\]

Since \(g\) is a right \(\operatorname{add}(T)\)-approximation of \(X\), there is a morphism \(\beta' \in \operatorname{End}(E_T)\) such that \(\beta = g\beta'\). We get that \(g(\alpha - \beta'f) = g\alpha - g\beta' f = \beta f - \beta f = 0\). By the kernel universal property, there is a morphism \(\alpha' \in \operatorname{End}(K_T)\) such that \((\alpha - \beta'f) = f\alpha'\). Therefore, \(\alpha = f\alpha' + \beta'f\). By 13, \(f \in \operatorname{Rad}_{\operatorname{rep}Q}(K_T,E_T)\), so is \(\alpha\) because \(\operatorname{Rad}_{\operatorname{rep}Q}\) is a two-sided ideal of \(\operatorname{rep}Q\).

Given that any morphism \(\alpha \in \operatorname{Hom}(K_T,E_T)\) is in \(\operatorname{Rad}_{\operatorname{rep}Q}(K_T,E_T)\), by 12, we get that \(\operatorname{add}(K_T) \cap \operatorname{add}(E_T) = 0\).

Now, consider an indecomposable \(U \in \operatorname{add}(K_T) \subseteq \operatorname{add}(T)\). Since \(U \notin \operatorname{add}(E_T)\), we have \(E_T \in \operatorname{add}(M)\) with \(T = U \oplus M\). Therefore, since \(\xi\) is an \(\mathcal{E}\)-admissible short exact sequence, then \(K_T \in \operatorname{Sub}_\mathcal{E}(M)\) and \(X \in \operatorname{Gen}_\mathcal{E}(M)\). Hence, \(U \in \operatorname{Sub}_\mathcal{E}(M)\) and \(X \in \operatorname{Gen}_\mathcal{E}(M)\). ◻

Proof. Suppose that \(\operatorname{add}(T) \subsetneq \operatorname{Gen}_\mathcal{E}(T)\). Given that, there exist \(X \in \operatorname{Gen}_\mathcal{E}(T)\) such that \(X \notin \operatorname{add}(T)\). By 15, there exist an indecomposable \(U \in \operatorname{add}(T)\) such that \(U \in \operatorname{Sub}_\mathcal{E}(\widetilde{T})\) with \(T = \widetilde{T} \oplus U\). By 32 and 11, \(U\) admits a left minimal \((\operatorname{add}(\widetilde{T}),\mathcal{E})\)-approximation \(f\). In the sense of 12, \(T\) admits a left \(\mathcal{E}\)-mutation at \(U\) which is a contradiction. Hence, \(\operatorname{Gen}_\mathcal{E}(T) =\operatorname{add}(T)\). ◻

Proof. The assertions \((i)\) and \((ii)\) are dual statements. It suffices for us to show \((i)\).

First, we will show that \(\operatorname{add}(T) \oplus \Bigl(\,\bigoplus_{T \ThisStyle{\mathrel{ \stackinset{r}{5.4pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\lessdot_\mathcal{E}}}}T'} \operatorname{Gen}_{\mathcal{E}}(T') \Bigr) \subseteq \operatorname{Gen}_{\mathcal{E}}(T)\).

Let \(X \in \operatorname{add}(T) \oplus \Bigl(\,\bigoplus_{T \ThisStyle{\mathrel{ \stackinset{r}{5.4pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\lessdot_\mathcal{E}}}}T'} \operatorname{Gen}_{\mathcal{E}}(T') \Bigr)\) be an indecomposable representation. If \(X \in \operatorname{add}(T)\), then obviously \(X \in \operatorname{Gen}_\mathcal{E}(T)\). Assume that \(X \in \operatorname{Gen}_\mathcal{E}(T')\) for some \(T \ThisStyle{\mathrel{ \stackinset{r}{6pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\lessdot_\mathcal{E}}}}T'\). Set \(U \in \operatorname{ind}(Q)\) such that \(T \ncong \pmb{\mu}_{U,\mathcal{E}}^{-}(T) = T'\). We have the following \(\mathcal{E}\)-admissible short exact sequence \[\begin{tikzcd} \xi : 0 \arrow[r] & U \arrow[r,tail,"f"] & \Theta \arrow[r,two heads,"g"] & \operatorname{Coker}(f) \arrow[r] & 0, \end{tikzcd}\] with \(f\) a left minimal \((\operatorname{add}(\widetilde{T}),\mathcal{E})\)-approximation of \(U\) with \(T=\widetilde{T}\oplus U\). Since \(\Theta \in \operatorname{add}(\widetilde{T}) \subset \operatorname{add}(T)\), then \(\operatorname{Coker}(f) \in \operatorname{Gen}_\mathcal{E}(T)\). Given that \(T' = \widetilde{T} \oplus \operatorname{Coker}(f)\), then \(\operatorname{Gen}_\mathcal{E}(T') \subseteq \operatorname{Gen}_\mathcal{E}(T)\), so \(X \in \operatorname{Gen}_\mathcal{E}(T)\).

Now, we will show that \(\operatorname{Gen}_{\mathcal{E}}(T) \subseteq \operatorname{add}(T) \oplus \Bigl(\,\bigoplus_{T \ThisStyle{\mathrel{ \stackinset{r}{5.4pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\lessdot_\mathcal{E}}}}T'} \operatorname{Gen}_{\mathcal{E}}(T') \Bigr)\).

Let \(X \in \operatorname{Gen}_\mathcal{E}(T)\) be an indecomposable representation. If \(X \in \operatorname{add}(T)\), we are done. Otherwise, by 15, there is an indecomposable representation \(U \in \operatorname{add}(T)\) such that \(X \in \operatorname{Gen}_\mathcal{E}(\widetilde{T})\) and \(U \in \operatorname{Sub}_\mathcal{E}(\widetilde{T})\) with \(T = \widetilde{T} \oplus U\). By 32 and 11, \(U\) admits a left minimal \((\operatorname{add}(\widetilde{T}),\mathcal{E})\)-approximation \(f\). In the sense of 12, \(T\) admits a left \(\mathcal{E}\)-mutation at \(U\), and we get \(\pmb{\mu}_{U,\mathcal{E}}^{-}(T) = T'\) for some tilting object \(T' \in T^-_\mathcal{E}\) with \(\operatorname{add}(\widetilde{T}) \subseteq \operatorname{add}(T')\). Therefore, \(X \in \operatorname{Gen}_\mathcal{E}(\widetilde{T}) \subseteq \operatorname{Gen}_\mathcal{E}(T')\) implies \(X \in \operatorname{Gen}_\mathcal{E}(T')\) for some tilting object \(T \ThisStyle{\mathrel{ \stackinset{r}{6pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\lessdot_\mathcal{E}}}}T'\). Hence \(\operatorname{Gen}_{\mathcal{E}}(T) \subseteq \operatorname{add}(T) \oplus \Bigl(\,\bigoplus_{T' \in T^-_\mathcal{E}} \operatorname{Gen}_{\mathcal{E}}(T') \Bigr)\). ◻

Proof. Since \(Q\) is a representation-finite, the result follows directly from Propositions 42 and 43 by induction. ◻

Proof. Since \(\operatorname{Gen}_\mathcal{E}(T_1) = \operatorname{Gen}_\mathcal{E}(T_2)\), we have the two \(\mathcal{E}\)-admissible short exact sequences: \[\begin{tikzcd} \xi_1 : 0 & K & \Theta_1 & T_2 & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(\Theta_1 \in \operatorname{add}(T_1)\), and \[\begin{tikzcd} \xi_2 : 0 & K' & \Theta_2 & T_1 & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(\Theta_2 \in \operatorname{add}(T_2)\). Applying the functors \(\operatorname{Hom}(T_1,-)\) to \(\xi_1\) and \(\operatorname{Hom}(T_2,-)\) to \(\xi_2\), because \(\operatorname{Ext}^1(T_1,\Theta_2)\) = \(\operatorname{Ext}^1(T_2,\Theta_2) = 0\) and because \(\operatorname{rep}Q\) is hereditary, we obtain \(\operatorname{Ext}^1(T_1,T_2)\) = \(\operatorname{Ext}^1(T_2,T_1)=0\). Since \(T_1,T_2\) are maximally rigid objects, then \(T_1 \cong T_2\). ◻

Proof. It is clear that \([T]_{\approx_{\mathcal{E}}}\) has minimal elements for \(\ThisStyle{\mathrel{ \stackinset{r}{6pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\leqslant_\mathcal{E}}}}\). We must only show that there is indeed a unique one. Let \(T' \in [T]_{\approx_{\mathcal{E}}}\) be a minimal element. By 39, we already have \(\operatorname{Gen}_\mathcal{E}(T') \subseteq \operatorname{GS}_{\mathcal{E}}(T')=\operatorname{GS}_{\mathcal{E}}(T)\).

Let us show that \(\operatorname{GS}_{\mathcal{E}}(T') \subseteq \operatorname{Gen}_\mathcal{E}(T')\) by induction. First, we have that \(\operatorname{GS}^0_{\mathcal{E}}(T') = \operatorname{add}(T') \subseteq \operatorname{Gen}_\mathcal{E}(T')\). Assume that, for a fixed \(n \in \mathbb{N}\), we have \(\operatorname{GS}^n_{\mathcal{E}}(T') \subseteq \operatorname{Gen}_\mathcal{E}(T')\). Consider \(X \in \operatorname{GS}^{n+1}_{\mathcal{E}}(T')\). We have two cases to treat.

If \(X \in \operatorname{Gen}_{\mathcal{E}}(F)\) for some \(F \in \operatorname{GS}^{n}_{\mathcal{E}}(T')\), by induction hypothesis, \(F \in \operatorname{Gen}_\mathcal{E}(T')\). As \(\mathcal{E}\)-epimorphisms are closed under composition, so is \(X\).

Now assume that \(X \in \operatorname{Sub}_{\mathcal{E}}(F)\) for some \(F \in \operatorname{GS}^{n}_{\mathcal{E}}(T')\). Then we have the following short \(\mathcal{E}\)-exact sequence \[\begin{tikzcd} 0 & X & F & C & 0. \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] By induction hypothesis, \(F \in \operatorname{Gen}_\mathcal{E}(T')\), and therefore, there exists a short \(\mathcal{E}\)-exact sequence \[\begin{tikzcd} 0 & K & \Theta & F & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] with \(\Theta \in \operatorname{add}(T')\). We obtain the following commutative pull-back diagram. \[\begin{tikzcd} & 0 & 0 & 0 & \\ 0 & 0 & C' & C & 0 \\ 0 & K & \Theta & F & 0 \\ 0 & K & PB & X & 0 \\ & 0 & 0 & 0 & \arrow[from=2-1, to=2-2] \arrow[tail, from=2-2, to=2-3] \arrow[two heads,from=2-3, to=2-4] \arrow[from=2-4, to=2-5] \arrow[from=3-1, to=3-2] \arrow[tail, from=3-2, to=3-3] \arrow[two heads,from=3-3, to=3-4] \arrow[from=3-4, to=3-5] \arrow[from=4-1, to=4-2] \arrow[tail, from=4-2, to=4-3] \arrow[two heads,from=4-3, to=4-4] \arrow[from=4-4, to=4-5] \arrow[from=5-2, to=4-2] \arrow[equals, from=4-2, to=3-2] \arrow[two heads,from=3-2, to=2-2] \arrow[from=2-2, to=1-2] \arrow[from=5-3, to=4-3] \arrow[tail, from=4-3, to=3-3] \arrow[two heads,from=3-3, to=2-3] \arrow[from=2-3, to=1-3] \arrow[from=5-4, to=4-4] \arrow[tail, from=4-4, to=3-4] \arrow[two heads,from=3-4, to=2-4] \arrow[from=2-4, to=1-4] \end{tikzcd}\] The short sequences given by the three columns and those given by the last two rows are exact. By 1, the short sequence given by the first row is also exact. Thus \(C' \cong C\). Moreover, the short exact sequences given by the three rows and those given by the first and third columns are in \({\mathcal{E}}\). By 1, the short sequence given by the middle column is also \(\mathcal{E}\)-exact. So \(PB \in \operatorname{Sub}_\mathcal{E}(\Theta)\).

Since \(\Theta \in \operatorname{add}(T')\) and \(T'\) admits no right \(\mathcal{E}\)-mutation (as \(T'\) is minimal for \(\ThisStyle{\mathrel{ \stackinset{r}{6pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\leqslant_\mathcal{E}}}}\)), then \(\operatorname{Sub}_\mathcal{E}(\Theta) \subseteq \operatorname{Sub}_\mathcal{E}(T') = \operatorname{add}(T')\) by 42. Hence, \(PB \in \operatorname{add}(T')\). Given that the last row is \(\mathcal{E}\)-exact, \(X \in \operatorname{Gen}_\mathcal{E}(PB) \subseteq \operatorname{Gen}_\mathcal{E}(T')\).

Thus, this completes the induction proof and we get \(\operatorname{GS}_{\mathcal{E}}(T') \subseteq \operatorname{Gen}_\mathcal{E}(T')\). Thus \(\operatorname{GS}_{\mathcal{E}}(T) = \operatorname{GS}_{\mathcal{E}}(T') = \operatorname{Gen}_\mathcal{E}(T')\). The uniqueness directly follows from 16. Recall that \(\operatorname{Gen}_\mathcal{E}(T') \subseteq \operatorname{GS}_{\mathcal{E}}(T)\) and \(\operatorname{Sub}_\mathcal{E}(T') = \operatorname{add}(T') \subseteq \operatorname{GS}_{\mathcal{E}}(T)\) from Propositions 42 and 39. So \(\operatorname{GS}^1_{\mathcal{E}}(T') \subseteq \operatorname{GS}_{\mathcal{E}}(T) = \operatorname{Gen}_\mathcal{E}(T') \subseteq \operatorname{GS}^1_{\mathcal{E}}(T')\). ◻

Proof. By 39, we have that if \(T' \approx_{\mathcal{E}} T\), then \(T' \sim_{\mathcal{E}} T\). Consider \(T,T' \in \operatorname{Tilt}(Q)\) such that \(T' \sim_{\mathcal{E}} T\). By definition, \(\operatorname{GS}_{\mathcal{E}}(T) = \operatorname{GS}_{\mathcal{E}}(T')\). Due to 45, there exist \(S \in [T]_{\approx_{\mathcal{E}}}\) and \(S' \in [T']_{\approx_{\mathcal{E}}}\) such that \(\operatorname{GS}_{\mathcal{E}}(T) = \operatorname{Gen}_\mathcal{E}(S)\) and \(\operatorname{GS}_{\mathcal{E}}(T') = \operatorname{Gen}_\mathcal{E}(S')\). We obtain \(\operatorname{Gen}_\mathcal{E}(S) = \operatorname{Gen}_\mathcal{E}(S')\) and, by 16, \(S \cong S'\). We conclude that \([T]_{\approx_{\mathcal{E}}} = [T']_{\approx_{\mathcal{E}}}\). ◻

Proof. First of all, \[\operatorname{add}\left(\bigoplus_{T' \in [T]_{\approx_{\mathcal{E}}}} T' \right) \subseteq \operatorname{GS}_{\mathcal{E}}(T)\] by 39. By 45, there exists (a unique) \(S \in [T]_{\approx_{\mathcal{E}}}\) such that \(\operatorname{GS}_{\mathcal{E}}(T) = \operatorname{Gen}_\mathcal{E}(S)\). By 44, we can write \[\operatorname{GS}_\mathcal{E}(T) = \operatorname{Gen}_{\mathcal{E}}(S) \;=\; \bigoplus_{\substack{S \ThisStyle{\mathrel{ \stackinset{r}{5.4pt+.15\LMpt}{t}{.1\LMpt}{\rule{.4pt}{1.1\LMex+.2ex}}{\SavedStyle\leqslant_\mathcal{E}}}}T'}} \operatorname{add}(T') \subseteq \operatorname{add}\left(\bigoplus_{T' \in [T]_{\approx_{\mathcal{E}}}} T' \right).\] ◻

Proof. Let \(\mathscr{C} \subseteq \operatorname{rep}(Q)\) be a maximal canonically Jordan recoverable subcategory. By 28, consider \(\mathbf{B},\mathbf{E}\subseteq \{1, \ldots,n\}\) such that:

1. the pair \((\mathbf{B},\mathbf{E}[1])\) is a set partition of \(\{1,\ldots,n+1\}\); and,

2. we have \(\mathscr{C} = \operatorname{\pmb{\mathscr{C}}}(\mathbf{B},\mathbf{E})\).

To check that \(\mathscr{C}\) is extension closed, by the fact that \(\operatorname{Ext}^1\) is a bilinear functor, it is enough to prove that for any pair \((D,F)\) of indecomposable representations in \(\mathscr{C}\), and for any nonsplit short exact sequence \[\begin{tikzcd} 0 & D & E & F & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] we have \(E \in \mathscr{C}\).

Given such a short exact sequence, by 7 29, we have that \(E \cong E_1 \oplus E_2\) and there exist \(K,L \in \mathcal{I}_n\) with \(K \cap L \neq \varnothing\) such that:

1. \(K,L,K \cap L\) and \(K \cup L\) are four different intervals in \(\mathcal{I}_n\); and,

2. either \(\{D,F\} = \{X_K, X_L\}\) and \(\{E_1,E_2\} = \{X_{K\cap L}, X_{K \cup L} \}\), or
   \(\{D,F\} = \{X_{K \cap L}, X_{K \cup L}\}\) and \(\{E_1,E_2\} = \{X_{K}, X_{L} \}\) .

Set \(K = \llbracket b_1, e_1 \rrbracket\) and \(L = \llbracket b_2,e_2 \rrbracket\). Up to exchanging the role of \(K\) and \(L\), assume that \(e_1 \leqslant e_2\). To make sure that \(K,L,K \cap L\) and \(K \cup L\) are four distinct intervals nonempty in \(\mathcal{I}_n\), we must have either

1. \(b_1 < b_2 \leqslant e_1 < e_2\); or,

2. \(b_2 < b_1 \leqslant e_1 < e_2\).

Assume that we are in the case \((1)\). If \(\{D,F\} =\{X_K, X_L\}\), then \(b_1,b_2 \in \mathbf{B}\) and \(e_1,e_2 \in \mathbf{E}\). Therefore, \(K \cap L = \llbracket b_2;e_1 \rrbracket \in \operatorname{Int}(\mathscr{C})\) and \(K \cup L = \llbracket b_1, e_2 \rrbracket \in \operatorname{Int}(\mathscr{C})\). So \(E \in \mathscr{C}\). The case where \(\{D,F\} = \{X_{K\cap L},X_{K \cup L}\}\) can be treated similarly. These arguments yield the same result if we start with case \((2)\).

In either case, we get \(E \in \mathscr{C}\), and so the desired result. ◻

Proof. It follows from 29. ◻

Proof. By 28, consider \(\mathbf{B},\mathbf{E}\subseteq \{1,\ldots,n\}\) such that:

1. the pair \((\mathbf{B},\mathbf{E}[1])\) is a set partition of \(\{1,\ldots,n+1\}\); and,

2. we have \(\mathscr{C} = \operatorname{\pmb{\mathscr{C}}}(\mathbf{B},\mathbf{E})\).

Let \(\pmb{\mathscr{T}}(\mathbf{B}, \mathbf{E}) \subseteq \operatorname{\pmb{\mathscr{J}}}(\mathbf{B},\mathbf{E}) = \operatorname{Int}(\mathscr{C}\) be the set such that:

1. for any \(e \in \mathbf{E}\), \(\llbracket 1,e \rrbracket \in \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})\); and,

2. for any \(b \in \mathbf{B}\setminus \{1\}\),

   1. if, for any \(e \in E\), the interval \(\llbracket b,e \rrbracket\) is either on the top or at the bottom of \(\llbracket 1,n \rrbracket\), then \(\llbracket b,n \rrbracket \in \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})\);

   2. otherwise, by setting \(e_{(b)} \in \mathbf{E}\) the minimal element in \(\mathbf{E}\) such that \(\llbracket b,e_{(b)} \rrbracket\) is neither on the top and at the bottom of \(\llbracket 1,n \rrbracket\), then \(\llbracket b,e_{(b)} \rrbracket \in \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})\).

Set \[T = T_{(\mathbf{B},\mathbf{E})} = \bigoplus_{K \in \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})} X_{K}.\] Let us show that \(T\) is tilting.

First \(\operatorname{rep}(Q)\) is a hereditary category. Therefore, to show that \(T\) is a tilting representation, it suffices to show that:

1. \(\# \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E}) = n\); and,

2. \(\operatorname{Ext}^1(X_K, X_L) = 0\) for all \(K,L \in \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})\).

The first point is obvious, as one can check that \[\#\pmb{\mathscr{T}}(\mathbf{B},\mathbf{E}) = \#\mathbf{E}+ \#\mathbf{B}- 1 = n.\]

By 7, the short exact sequences \[\begin{tikzcd} 0 & D & E & F & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] where \(D,F \in \operatorname{ind}(Q)\), are parametrized by only two kind of pair of intervals \((K,L) \in \mathcal{I}_n\), namely:

1. the case where \(K \cap L = \varnothing\) and \(K \cup L \in \mathcal{I}_n\); or,

2. the case where \(K \cap L \neq \varnothing\), and \(K,L,K\cap L\) and \(K \cup L\) are four distinct intervals.

The case \((1)\) is equivalent to say that \(K\) and \(L\) are adjacent, and as \(\pmb{\mathscr{T}}(\mathbf{B},\mathbf{E}) \subset \operatorname{\pmb{\mathscr{J}}}(\mathbf{B},\mathbf{E})\) which is adjacency-avoinding, this cannot happen, with \(D,F \in \operatorname{add}(T)\) two indecomposable representations.

Assume the case \((2)\) holds with \(D,F \in \operatorname{add}(T)\) two indecomposable representations. To exchange the role of \(K\) and \(L\), assume that \(e(K) < e(L)\). Then \(b(K) < b(L) \leqslant e(K) < e(L)\). By construction of \(\pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})\), both of the following assertions holds:

1. \(b(L) \neq 1\);

2. if \(e(L) \neq n\), then \(L\) is neither on the top nor at the bottom of \(\llbracket 1,n \rrbracket\); and,

3. \(K \cap L \notin \pmb{\mathscr{T}}(\mathbf{B},\mathbf{E})\); and,

4. \(K \cap L\) is either on the top or at the bottom of \(\llbracket 1,n \rrbracket\), and, therefore, of \(K \cup L\).

So, in such case, we have \(\{D,F\} = \{X_{K}, X_{L}\}\). However, by the last point, we cannot have a non-zero morphism between \(X_K\) and \(X_L\) which factors through \(X_{K \cap L} \oplus X_{K \cup L}\). We sought a contradiction.

Therefore, any short exact sequence \[\begin{tikzcd} 0 & D & E & F & 0, \arrow[from=1-1, to=1-2] \arrow[tail, from=1-2, to=1-3] \arrow[two heads,from=1-3, to=1-4] \arrow[from=1-4, to=1-5] \end{tikzcd}\] where \(D,F \in \operatorname{add}(T)\) are indecomposable, must split, and so \(\operatorname{Ext}^1(G, D) = 0\). We proved that \(T \in \mathscr{C}\) is tilting. ◻

Proof. It follows from 16 and 29. ◻

Proof. Let \(\mathscr{C}\) be a maximal canonically Jordan recoverable subcategory of \(\operatorname{rep}(Q)\). By 52, there exists a tilting representation \(T \in \operatorname{rep}(Q)\) such that \(T \in \mathscr{C}\). By 29, the category \(\mathscr{C}\) is \(\mathcal{E}_\diamond\)-adapted. Moreover, \(\mathscr{C}\) is closed under extensions by 49. Therefore, we have \(\operatorname{GS}_{\mathcal{E}_{\diamond}}(T) \subseteq \mathscr{C}\). By 22, we get \(\operatorname{GS}_{\mathcal{E}_\diamond}(T) = \mathscr{C}\). The converse is obvious by 22 and 29. ◻

Proof. It follows directly from 54 48. ◻

Proof. The proof of [aCKGS] follows as in 17, noting that the subcategory is no longer necessarily closed under arbitrary \(\mathcal{E}\)-subobjects and \(\mathcal{E}\)-quotients, but it remains closed under taking kernels of \(\mathcal{E}\)-epimorphisms and cokernels of \(\mathcal{E}\)-monomorphisms within itself.

The proof of 22 can be adapted to the setting of \(\mathrm{CK}_{\mathcal{E}}(T)\) for showing [bCKGS]. Together with the result from [aCKGS], this implies that both \(\mathrm{CK}_{\mathcal{E}}(T)\) and \(\operatorname{GS}_{\mathcal{E}}(T)\) are maximal \(\mathcal{E}\)-adapted extension-closed subcategories with \(\mathrm{CK}_{\mathcal{E}}(T) \subseteq \operatorname{GS}_{\mathcal{E}}(T)\), which yields the desired equality. ◻

Proof. By 28 54, those equivalence classes are parametrized by pairs \((\mathbf{B},\mathbf{E})\) such that \(\{\mathbf{B},\mathbf{E}[1]\}\) is a bipartition of \(\{1,\ldots,n+1\}\). Then there is a one-to-one correspondence \(\phi\) from subsets of \(\{2,\ldots,n\}\) to such pairs given by \[\forall \mathbf{A}\subseteq \{2,\ldots,n\},\;\phi(\mathbf{A}) = (\{1\} \cup \mathbf{A}, \mathbf{A}\setminus \{2,\ldots, n+1\}).\] We got the desired result. ◻