Proof. (i) $\Rightarrow$ (ii) Let $r\,:\, B \to C$ be a retraction with corresponding section $s\colon C \to B$ (so that $rs= \mathrm{id}_{C}$ ) and let $k\colon K \to B$ be a kernel of $r$ , which exists by assumption. It follows that $(k \ s)\,\colon K\oplus C \to B$ is an isomorphism (see [Reference Bühler16, Rem. 7.4]). Then we have a commutative diagram

which is an equivalence of composable morphisms (see (2.2)). Our claim follows since the upper sequence is always an extriangle [Reference Nakaoka and Palu45, Rem. 2.11].

(ii) $\Rightarrow$ (i) Let $r\colon B \to C$ be a retraction and suppose that there is an extriangle in $({\mathcal{A}}, {\mathbb{E}}, {\mathfrak{s}})$ . Note that this already implies $x$ is a weak kernel of $r$ by [Reference Nakaoka and Palu45, Prop. 3.3(2)]. Let $s\colon C \to B$ be a morphism in $\mathcal{A}$ with $rs=\mathrm{id}_{C}$ . Then we apply (ET3) to obtain the morphism

of extriangles. By the commutativity of the right hand square, we have that $ (0 \ c) = r (x \ s) = (0 \, \,\mathrm{id}_{C})$ , whence $c = \mathrm{id}_{C}$ . It follows from [Reference Nakaoka and Palu45, Cor. 3.6] that $(x \ s) \colon A \oplus C \to B$ is an isomorphism, and we deduce that $x$ is a monomorphism as the composite of two monomorphisms. Thus, $x$ is a kernel of $r$ .

The equivalence of (i) and (iii) is similar. The equivalence of (i) and (iv) is [Reference Klapproth37, Prop. 2.7].

Proof. Since $(0 \ b_{2}) = b_{2}\circ (0 \ \mathrm{id}_{B_{2}})$ is a deflation, we know that $b_{2}$ is a deflation by (WIC) as $({\mathcal{A}}, {\mathbb{E}}, {\mathfrak{s}})$ is weakly idempotent complete (see Proposition 2.5). Thus, there is an extriangle of the form Taking the direct sum with and applying (ET $3)^{\mathrm{op}}$ , there is a morphism of extriangles

where $(x,s)$ is an isomorphism by [Reference Nakaoka and Palu45, Cor. 3.6]. By commutativity of the diagram above, we see $a = a_{2}s$ and we are done.

Proof. Since $\mathfrak{s}$ is an additive realisation, we know that $ (0 \ \mathrm{id}_{C_{2}}) \colon C_{1}\oplus C_{2} \to C_{2}$ is a deflation. Since $(0 \ g_{2}) = (0 \ \mathrm{id}_{C_{2}}) \circ \begin{pmatrix}g_{1} \quad 0\\ 0 \quad g_{2}\end{pmatrix}$ is the composition of deflations, it is a deflation itself by [Reference Nakaoka and Palu45, Rem. 2.16]. Like in the proof of Lemma 2.7, it then follows that $g_{2}$ is a deflation and we have an extriangle of the form By symmetry, we also have an extriangle . Using (ET $3)^{\mathrm{op}}$ and [Reference Nakaoka and Palu45, Cor. 3.6], we obtain an isomorphism of extriangles

which completes the proof.

Proof. Assume that $Z \,:\!= Z_{1} \oplus Z_{2} \in {\mathcal{X}} \ast \mathcal{Y}$ . Since $Z\in {\mathcal{X}}\ast \mathcal{Y}$ , there is an extriangle

(2.4)

where $X\in {\mathcal{X}}$ and $Y\in \mathcal{Y}$ . Note that $f$ is a right $\mathcal{X}$ -approximation of $Z$ since ${\mathcal{A}}({\mathcal{X}},\mathcal{Y}) = 0$ .

If $f=0$ , then by the dual of Lemma 2.7, we have that (2.4) is isomorphic to

where $Z_{1} \oplus Y' \cong Y$ . We see that $Z_{1}\in \mathcal{Y}$ because $\mathcal{Y}$ is closed under direct summands. By symmetry, we also have $Z_{2}\in \mathcal{Y}$ . Noting that $\mathcal{Y}\subseteq {\mathcal{X}}\ast \mathcal{Y}$ concludes the case when $f=0$ .

Thus, now assume $f\neq 0$ . By [Reference Krause and Saorín41, Cor. 1.4], we may assume that $f$ is of the form $ (f' \ f'') \colon X'\oplus X'' \to Z$ , where $f'\neq 0$ is right minimal and $f''=0$ . So, as $f = (f' \ 0 )$ and $\begin{pmatrix}\mathrm{id}_{X'}\\ 0 \end{pmatrix}\colon X' \to X'\oplus X''$ are inflations, we see that $f' = f\circ \begin{pmatrix}\mathrm{id}_{X'} \\ 0\end{pmatrix}$ is an inflation by [Reference Nakaoka and Palu45, Rem. 2.16]. Furthermore, since $\mathcal{X}$ is closed under direct summands, we have $X'\in {\mathcal{X}}$ and $f'\colon X'\to Z$ is a minimal right $\mathcal{X}$ -approximation of $Z$ . Moreover, we have that there is an extriangle

(2.5)

where $Y'$ is a direct summand of $Y$ and hence $Y'\in \mathcal{Y}$ .

Note that the compositions $(\mathrm{id}_{Z_{1}} \ 0) \circ f' \colon X'\to Z_{1}$ and $(0 \ \mathrm{id}_{Z_{2}}) \circ f' \colon X'\to Z_{2}$ are both right $\mathcal{X}$ -approximations. First, assume $(\mathrm{id}_{Z_{1}}\,\,0) \, f' = 0$ . Then we may apply the dual of Lemma 2.7 to (2.5) to obtain an extriangle where $Z_{1} \oplus Y'' \cong Y'$ . In this case, we see that $Z_{1}\in \mathcal{Y}$ and $Z_{2}\in {\mathcal{X}}\ast \mathcal{Y}$ , and we are done. By symmetry, we are also done if $(0 \ \mathrm{id}_{Z_{2}}) f' = 0$ .

Therefore, assume now that both $(\mathrm{id}_{Z_{1}} \ 0) f'$ and $(0 \ \mathrm{id}_{Z_{2}}) f'$ are non-zero. In particular, we may take minimal right $\mathcal{X}$ -approximations $f_{1}\colon X_{1} \to Z_{1}$ and $f_{2}\colon X_{2} \to Z_{2}$ of $Z_{1}$ and $Z_{2}$ , respectively. This yields a minimal right $\mathcal{X}$ -approximation $f_{1}\oplus f_{2} \colon X_{1} \oplus X_{2} \to Z_{1}\oplus Z_{2}$ of $Z = Z_{1} \oplus Z_{2}$ . (Indeed, since $\mathcal{A}$ is Krull-Schmidt, by [Reference Krause and Saorín41, Cor. 1.4] there is a direct sum decomposition of $X_1 \oplus X_2 \cong A \oplus B$ such that $\left .(f_1 \oplus f_2) \right |_{A}$ is right minimal and ${\left .(f_1 \oplus f_2) \right |_{B}} = 0$ , so if $B \neq 0$ then this would contradicts the minimality of $f_1$ or $f_2$ .) This implies $f_{1}\oplus f_{2}$ is isomorphic to $f'$ by uniqueness of minimal approximations (see e.g. [Reference Auslander and Reiten4, Prop. 2.2]), and hence, $f_{1}\oplus f_{2}$ is an inflation. Using Lemma 2.8, we see that there are extriangles

where $Y_{1} \oplus Y_{2} \cong Y'$ . As $\mathcal{Y}$ is closed under direct summands, this implies $Z_{i}\in {\mathcal{X}}\ast \mathcal{Y}$ for $i=1,2$ .

Proof. Assume there is an extriangle in $\mathcal{A}$ . Since is also an extriangle, we obtain the direct sum

As $S$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple, the morphism $(0 \;\;\; \mathrm{id}_{S})$ is either zero or an isomorphism. It cannot be zero because $S$ is non-zero, and hence we have an isomorphism $(0 \;\; \mathrm{id}_{S}) \colon A\oplus S \to S$ . This implies there is a morphism $\begin{pmatrix}x\\y\end{pmatrix} \,\colon S \to A\oplus S$ , such that $\begin{pmatrix}\mathrm{id}_{A} \quad 0 \\ 0 \quad \mathrm{id}_{S}\end{pmatrix} = \begin{pmatrix}x \\ y \end{pmatrix} (0 \;\; \mathrm{id}_{S}) = \begin{pmatrix}0 \;\;\; x \\0 \;\;\; y\end{pmatrix}$ . Thus, $\mathrm{id}_{A} = 0$ and hence $A= 0$ . Therefore, $0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple-like.

Proof. It is clear that (ii) implies (i).

(i) $\Rightarrow$ (ii): Suppose $S\neq 0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple and let

(3.1)

be an extriangle in $({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ . If $a$ is an isomorphism, then we are done, so suppose that $a=0$ . From the dual of Lemma 2.7, we deduce that (3.1) is isomorphic to the direct sum of two extriangles, with one of the form and the other of the form . Since $0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple-like by Proposition 3.6, we must have that $A=0$ .

(ii) $\Rightarrow$ (iii) : If there were an extriangle with both $A$ and $C$ non-zero, then $a$ would be an isomorphism by (ii) and $C$ would be zero, a contradiction.

(iii) $\Rightarrow$ (ii): Suppose is an extriangle. By (iii), either $A=0$ or $C=0$ . In the former, we are done immediately. The latter implies $a$ is an isomorphism, and we are also done.

Proof. If (2.6) is an $\mathcal{X}$ -filtration of $M\in {\mathcal{A}}$ , then we have $t\,{\geqslant }\,1$ and that there is an extriangle in ${\mathcal{F}}({\mathcal{X}})$ with $X_{t}\in {\mathcal{X}}$ . We deduce that $M_{t-1} = 0$ or that $a$ is an isomorphism by Lemma 3.7. The latter implies that $X_{t}$ is a zero object, which is contrary to our assumption on $\mathcal{X}$ . Thus, we see that $M_{t-1} = 0$ and hence $M\cong X_{t}$ .

Proof. (i) $\Rightarrow$ (ii): Suppose $\xi = (\xi _{0},\xi _{1},\ldots ,\xi _{t})$ from (2.6) is an $({\mathbb{E}},{\mathfrak{s}})$ -composition series of $M$ . If $t=0$ , then $M=0$ which is a contradiction. If $t=1$ , then we are done. Now assume for contradiction that $t\,{\geqslant }\, 2$ . From the $({\mathbb{E}},{\mathfrak{s}})$ -composition series $\xi$ , we see that there is an extriangle with $X_{t}$ an $({\mathbb{E}},{\mathfrak{s}})$ -simple. Since we are supposing $M$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple, we have that $M_{t-1}$ is zero or $a$ is an isomorphism by Lemma 3.7. The latter would imply $X_{t} = 0$ , which is impossible, so we must have $M_{t-1}=0$ . In particular, we see that we have an extriangle , so $M_{t-2}$ is forced to be the zero object as $0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple-like by Proposition 3.6. This implies that $X_{t-1} = 0$ , which is a contradiction so $t \ngeqslant\, 2$ .

(ii) $\Rightarrow$ (i): By assumption, $M$ has an $({\mathbb{E}},{\mathfrak{s}})$ -composition series of length $1$ . That is, there is an ${\operatorname {\mathsf{Sim}}\nolimits }({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ -filtration of the form

in which $X_{1}$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple. In particular, $M \cong X_{1}$ and $M$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple.

Proof. (i) If $M\cong 0$ , then $[M]=[0]$ in $\operatorname {Iso}\nolimits {\mathcal{A}}$ and hence also in ${\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ . Conversely, assume $[M]=[0]$ in ${\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ . By Proposition 3.14 there is a sequence of objects $ 0 = A_{0}, A_{1}, \ldots , A_{n} = M$ in $\mathcal{A}$ such that $A_{i-1}$ and $A_{i}$ have isomorphic $({\mathbb{E}},{\mathfrak{s}})$ -inflation series for each $1\,{\leqslant }\, i\,{\leqslant }\,n$ . Since $0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple-like, any $({\mathbb{E}},{\mathfrak{s}})$ -inflation series of $0$ has $M_{i}/M_{i-1} = 0$ for all $1\,{\leqslant }\, i\,{\leqslant }\, t$ . Since $A_{1}$ has an $({\mathbb{E}},{\mathfrak{s}})$ -inflation series isomorphic to one of $0$ , it has an $({\mathbb{E}},{\mathfrak{s}})$ -inflation series in which $M^{1}_{i}/M^{1}_{i-1} = 0$ for all $1\,{\leqslant } \,i\,{\leqslant }\, t_{1}$ . This implies, $0 = M^{1}_{0}\cong M^{1}_{1}\cong \cdots \cong M^{1}_{t_{1}} = A_{1}$ . Continuing in this way, we see that $M = A_{n} \cong 0$ .

(ii) Assume $[X] + [Y] = [0]$ for some $X,Y\in {\mathcal{A}}$ . Thus, $[X\oplus Y] = [0]$ , so by part (i) we have $X\oplus Y\cong 0$ . Thus, $X\cong 0\cong Y$ , whence $[X]=[0]$ and $[Y]=[0]$ by (i).

(iii) In light of Proposition 3.14, Lemma 3.7 and part (i), one may carefully recycle the argument of [Reference Enomoto25, Prop. 3.6] to our setting. See also [Reference Berenstein and Greenstein9, Lem. 5.3].

Proof. In light of Proposition 3.16, the equivalence of (ii) and (iii) follows from [Reference Enomoto25, Thm. A.20]. Thus, it suffices to prove (i) $\Leftrightarrow$ (ii). Furthermore, if $\mathcal{A}$ has no $({\mathbb{E}},{\mathfrak{s}})$ -simple objects, then we observe that (i) and (ii) are trivially equivalent because it forces $\mathcal{A}$ to be zero in these equivalent cases. Thus, we also assume $\mathcal{A}$ has at least one $({\mathbb{E}},{\mathfrak{s}})$ -simple object in the following. In particular, by Proposition 3.6, we know that $0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple-like.

(i) $\Rightarrow$ (ii): By Proposition 3.16 (iii), we have ${\operatorname {\mathsf{Atom}}\nolimits }( {\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}}) ) = \left \{[S] \,|\, S\in {\operatorname {\mathsf{Sim}}\nolimits }({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}}) \right \}$ . The remainder of the argument is as in the proof of $(1)\Rightarrow (2)$ of [Reference Enomoto25, Thm. 4.9], using Lemma 3.17 in place of [Reference Enomoto25, Prop. 2.5(2)].

(ii) $\Rightarrow$ (i): By Proposition 3.16, we have that ${\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ is reduced and again the equality ${\operatorname {\mathsf{Atom}}\nolimits }( {\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}}) ) = \left \{[S] \,|\, S\in {\operatorname {\mathsf{Sim}}\nolimits }({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}}) \right \}$ . Since ${\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ is free, and hence factorial [Reference Enomoto25, Prop. A.9], we have that there is a length-like function $\nu \colon {\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}}) \to \mathbb{N}$ satisfying $\nu ([S]) = 1$ for each $[S]\in {\operatorname {\mathsf{Atom}}\nolimits }( {\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}}) )$ by [Reference Enomoto25, Lem. A.18].

To show each $M\in {\mathcal{A}}$ admits an $({\mathbb{E}},{\mathfrak{s}})$ -composition series, we induct on $\nu ([M])\in \mathbb{N}$ . Since $\nu$ is length-like and $0$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple-like, we have that $\nu ([M]) = 0$ if and only if $[M] = 0$ in ${\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ . In this case, $M$ has an $({\mathbb{E}},{\mathfrak{s}})$ -composition series of length $0$ . Assume $\nu ([M]) = 1$ . We claim that this implies $M$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple. To this end, suppose there is an extriangle . Then $1 = \nu ([M]) = \nu ([A]) + \nu ([C])$ since $\nu$ is a monoid homomorphism. Therefore, $\nu ([A]) = 0$ or $\nu ([C]) = 0$ , which implies $A\cong 0$ or $C\cong 0$ , and hence that $M$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple. Now suppose $\nu ([M]) \gt 1$ and that the claim holds for objects $Y$ with $\nu ([Y])\lt \nu ([M])$ . If $M$ were $({\mathbb{E}},{\mathfrak{s}})$ -simple, then we would have that $[M]$ is an atom and $\nu ([M]) = 1$ , which is not possible. So there exists an extriangle with $A$ and $C$ both non-zero and, in particular, $\nu ([A]),\nu ([C])\,{\geqslant }\, 1$ . Since $\nu ([M]) = \nu ([A]) + \nu ([C])$ and $\nu$ takes values in $\mathbb{N}$ , we see that $\nu ([A]),\nu ([C]) \lt \nu ([M])$ . Hence, $A$ and $C$ both admit an $({\mathbb{E}},{\mathfrak{s}})$ -composition series. Then we may apply Lemma 3.17 (with ${\mathcal{X}} = {\operatorname {\mathsf{Sim}}\nolimits }({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ ) to obtain an $({\mathbb{E}},{\mathfrak{s}})$ -composition series of $M$ .

Now we show $({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ is a length extriangulated category. So suppose (2.6) is an $({\mathbb{E}},{\mathfrak{s}})$ -composition series of $M$ . Then $\nu ([M]) = \nu ([X_{1}]) + \cdots + \nu ([X_{t}]) = t$ as each $X_{i}$ is $({\mathbb{E}},{\mathfrak{s}})$ -simple. In particular, each $({\mathbb{E}},{\mathfrak{s}})$ -composition series of $M$ is of finite length $\nu ([M])\in \mathbb{N}$ .

To see that $({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ is Jordan-Hölder, let $(\xi _{1},\ldots ,\xi _{t})$ and $(\xi '_{1},\ldots ,\xi '_{s})$ be two $({\mathbb{E}},{\mathfrak{s}})$ -composition series of $M\in {\mathcal{A}}$ . Since ${\mathcal{M}}({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ is a free monoid, we must have that $s=t$ and there is a permutation $\sigma \in \operatorname {Sym}\nolimits ({n})$ such that $[X_{i}] = [X'_{\sigma (i)}]$ . It follows that $X_{i} \cong X'_{\sigma (i)}$ by Proposition 3.16 (iii).

Proof. (i): It follows from (S2) that ${\mathcal{F}}(\Phi _{i}) = \operatorname {\mathsf{add}}\nolimits \Phi _{i}$ for each $1\,{\leqslant }\, i\, {\leqslant }\, n$ , and $\operatorname {\mathsf{add}}\nolimits \Phi _{i}$ is clearly closed under direct summands.

(ii): For $t=0,1$ , there is nothing to show. Thus, suppose $t\,{\geqslant }\, 2$ . If $j_{1} \,{\geqslant }\, j_{2} \,{\geqslant }\, \ldots \,{\geqslant }\, j_{t}$ , then we are done. Otherwise, let $s\in \{1,\ldots ,t-1\}$ be the largest integer such that $j_{s} \lt j_{s+1}$ . Applying (ET $4$ ) to the extriangles $\xi _{s}$ and $\xi _{s+1}$ yields

Since $\Phi$ satisfies (S2), we have ${\mathbb{E}}(\Phi _{j_{s+1}}, \Phi _{j_{s}})=0$ , and thus $\delta = {}_{\Phi _{j_{s}}}0_{\Phi _{j_{s+1}}}\!$ . Without loss of generality, we may assume $E = \Phi _{j_{s+1}} \oplus \Phi _{j_{s}}$ . We now apply (ET $4)^{\mathrm{op}}$ to $\eta$ and ${}_{\Phi _{j_{s}}}0_{\Phi _{j_{s+1}}}$ to obtain

Now replace $\xi _{s}$ and $\xi _{s+1}$ by $\xi '_s$ and $\xi '_{s+1}$ , respectively. Note that this yields a $\Phi$ -filtration $(\xi _{0}, \ldots , \xi _{s-1}, \xi '_{s}, \xi '_{s+1}, \xi _{s+2}, \ldots , \xi _{t})$ , which has exactly the same $\Phi$ -factors as $(\xi _{0}, \xi _{1}, \ldots , \xi _{t})$ just with $\Phi _{j_{s}}$ and $\Phi _{j_{s}+1}$ switched. By repeating this process we obtain the desired $\Phi$ -filtration of $X$ .

For the last assertion, note that the $\Phi$ -filtration $(\xi '_0, \xi '_1, \ldots , \xi '_t)$ implies $X\in \operatorname {\mathsf{add}}\nolimits \Phi _{j'_1} \ast \cdots \ast \operatorname {\mathsf{add}}\nolimits \Phi _{j'_t}$ with $j'_{1} \,{\geqslant }\, j'_{2} \,{\geqslant }\, \ldots \,{\geqslant }\, j'_{t}$ . Therefore, for any $X\in {\mathcal{F}}(\Phi )$ , we have $X\in \operatorname {\mathsf{add}}\nolimits \Phi _{n} \ast \ldots \ast \operatorname {\mathsf{add}}\nolimits \Phi _1$ since each $\operatorname {\mathsf{add}}\nolimits \Phi _{i}$ is extension-closed by (S2). Equation (4.2) then follows from (i).

Proof. By Proposition 4.4, we have ${\mathcal{F}}(\Phi ) = \operatorname {\mathsf{add}}\nolimits \Phi _{n} \ast \cdots \ast \operatorname {\mathsf{add}}\nolimits \Phi _{1}$ and that each $\operatorname {\mathsf{add}}\nolimits (\Phi _{i})$ is closed under direct summands. By (S1) we have ${\mathcal{A}}(\operatorname {\mathsf{add}}\nolimits \Phi _{i}, \operatorname {\mathsf{add}}\nolimits \Phi _{j}) = 0$ for all $i \gt j$ . Thus, ${\mathcal{F}}(\Phi )$ is closed under direct summands by Lemma 2.9.

Notation 4.7. If $\Phi = \{\Phi _{i}\}_{i=1}^{n}$ is an $\mathbb{E}$ -stratifying system and $i_{0}\in \{1,\ldots ,n\}$ , then we denote the subset $\left \{\Phi _{j} \,|\, j \gt i_{0} \right \}$ by $\Phi _{j\gt i_{0}}$ . We define other subsets (e.g. $ \Phi _{j{\leqslant } i_{0}} \,:\!= \left \{ \Phi _{j} \,|\, j \,{\leqslant } \,i_{0} \right \}$ ) of $\Phi$ in a similar way.

Proof. We will construct extriangles with $0\, {\leqslant }\, j\, {\leqslant } \,r$ for some $r\in \{0,\ldots ,n\}$ dependent on $X$ , such that $m_{j} \,{\geqslant }\, 0$ , and:

1. (1) $1\,{\leqslant }\, a_{j} \,{\leqslant }\, n$ is minimal such that ${\mathbb{E}}(X_{j},\Phi _{a_{j}}^{\oplus m_{j}}) \neq 0$ and ${\mathbb{E}}(X_{j},\Phi _{s}) = 0$ for all $s\lt a_{j}$ ;

2. (2) $a_{j+1}\gt a_{j}$ for $0\,{\leqslant }\, j\, {\leqslant }\, r-1$ ; and

3. (3) ${\mathbb{E}}(X_{r+1}, \Phi ) = 0$ .

Set $X_{0}\,:\!= X$ . If ${\mathbb{E}}(X_{0},\Phi ) = 0$ , then take $\delta _{0}$ to be the extriangle and stop with $X_{1} \,:\!= X_{0} = X$ . Else, take $a_{0} \in \{1,\ldots , n\}$ to be minimal such that ${\mathbb{E}}(X_{0},\Phi _{a_{0}}) \neq 0$ . Apply Lemma 4.11 to obtain a non-split extriangle such that $(\delta _{0}^{\sharp })_{\Phi _{a_{0}}} \colon {\mathcal{A}}(\Phi _{a_{0}}^{\oplus m_{0}}, \Phi _{a_{0}}) \to {\mathbb{E}}(X, \Phi _{a_{0}})$ is surjective.

If ${\mathbb{E}}(X_{1},\Phi ) = 0$ , then we are done with $r=0$ . Else, set

\begin{equation*} a_{1} \,:\!= \text{min}\left \{ 1\, {\leqslant }\, s \,{\leqslant }\, n \,|\, {\mathbb{E}}(X_{1},\Phi _{s}) \neq 0\text{ and } {\mathbb{E}}(X_{1}, \Phi _{s'}) =0 \;\; \forall s' \lt s \right \}. \end{equation*}

There is an extriangle such that $(\delta _{1}^{\sharp })_{\Phi _{a_{1}}} \colon {\mathcal{A}}(\Phi _{a_{1}}^{\oplus m_{1}}, \Phi _{a_{1}}) \to {\mathbb{E}}(X_{1}, \Phi _{a_{1}})$ is surjective by Lemma 4.11.

Assume for contradiction that $a_{1} \,{\leqslant }\, a_{0}$ . The extriangle $\delta _{0}$ induces an exact sequence

If $a_{1} \lt a_{0}$ , then ${\mathbb{E}}(X_{0}, \Phi _{a_{1}}) = 0$ by the minimality of $a_{0}$ and ${\mathbb{E}}(\Phi _{a_{0}}^{\oplus m_{0}}, \Phi _{a_{1}}) = 0$ by (S2). If $a_{1} =a_{0}$ , then we would still have ${\mathbb{E}}(\Phi _{a_{0}}^{\oplus m_{0}}, \Phi _{a_{1}}) = 0$ by (S2), and now also that ${\mathbb{E}}(y_{1}, \Phi _{a_{1}}) = 0$ since $(\delta _{0}^{\sharp })_{\Phi _{a_{1}}} = (\delta _{0}^{\sharp })_{\Phi _{a_{0}}}$ is surjective. In either case, we deduce that ${\mathbb{E}}(X_{1}, \Phi _{a_{1}}) = 0$ , which is a contradiction, so we must have $a_{1} \gt a_{0}$ .

Then we can repeatedly apply the above argument to construct the desired extriangles, noting that this process will terminate since $\{1,\ldots ,n\}$ is finite. Lastly, by repeatedly applying (ET4 $)^{\mathrm{op}}$ to the deflations $y_{j}$ , we obtain an extriangle with $y\,:\!= y_{1}y_{2}\ldots y_{r+1}$ , $A \in {\mathcal{F}}(\Phi _{j\,{\geqslant }\, a_{0}}) \subseteq {\mathcal{F}}(\Phi )$ and ${\mathbb{E}}(X_{r+1},\Phi ) = 0$ .

The last assertion follows from ${\mathcal{F}}(\Phi )$ being extension-closed in $({\mathcal{A}},{\mathbb{E}},{\mathfrak{s}})$ , noting that if $X\in {\mathcal{F}}(\Phi )$ then $X_{r+1}$ must also lie in ${\mathcal{F}}(\Phi )$ .

Proof. We first show that each $\mathbb{E}$ -stratifying system $\Phi = \{\Phi _{i}\}_{i=1}^{n}$ is part of a projective one. Then we will show that we may refine this projective $\mathbb{E}$ -stratifying system to a minimal one. Apply Proposition 4.12 to $\Phi _{1},\Phi _{2},\ldots ,\Phi _{n}$ in that order to obtain a sequence of objects $Q_{1},Q_{2},\ldots ,Q_{n}$ . Thus, for each $1\,{\leqslant }\, i \,{\leqslant }\, n$ , we have an extriangle

(4.4)

with $Q_{i}$ projective in $({\mathcal{F}}(\Phi ),{\left .{\mathbb{E}} \right |_{{\mathcal{F}}(\Phi )}},{\left .{\mathfrak{s}} \right |_{{\mathcal{F}}(\Phi )}})$ . Furthermore, the fact that $K_{i}$ lies in ${\mathcal{F}}(\Phi _{j\gt i})$ is a consequence of the construction of $a_{0}$ in the proof of Proposition 4.12, noting that (S2) implies $i \lt a_{0}$ . Moreover, with $Q \,:\!= \{Q_{i}\}_{i=1}^{n}$ , the pair $(\Phi , Q)$ is a projective $\mathbb{E}$ -stratifying system.

We now show that, for each $i \in \{1, \ldots , n\}$ , we may take the morphism $q_{i}$ in (4.4) to be right minimal. If $q_{i} = 0$ is zero, then by Lemma 2.7 we have that (4.4) is isomorphic to

where $K'_{i} \oplus Q_{i}\cong K_{i}$ . Note that the morphism $q'_{i} \colon 0\to \Phi _{i}$ is a monomorphism and hence right minimal. Thus, in this case, we can replace $\eta _i$ by $\eta _i'$ , noting that $K_{i}' \in {\mathcal{F}}(\Phi _{j\gt i})$ as ${\mathcal{F}}(\Phi _{j\gt i})$ is closed under direct summands by Corollary 4.5.

Now suppose $q_{i}$ is non-zero. Since $\mathcal{A}$ is a Krull-Schmidt category and $q_{i}$ is a right $\mathcal{P}(\Phi )$ -approximation of $\Phi _{i}$ (see Remark 4.9(iv)), there is a direct sum decomposition $Q_{i} \cong Q'_{i} \oplus P$ such that $q'_{i}\,:\!= {\left . q_{i} \right |_{Q'_{i}}} \neq 0$ is right minimal and ${\left . q_{i} \right |_{P}} = 0$ by [Reference Krause and Saorín41, Cor. 1.4]. Again by Lemma 2.7, we have that (4.4) is isomorphic to