Omitting strictly semistable objects breaks Hall factorization at the A₂ wall

Let $E=({\mathbb{F}}_q\stackrel{\mathrm{i}\mathrm{d}}{\to }{\mathbb{F}}_q)$, $S_1=({\mathbb{F}}_q\to 0)$, and $S_2=(0\to {\mathbb{F}}_q)$. A subrepresentation of $E$ is a pair $U_1\subseteq U_2\subseteq {\mathbb{F}}_q$, so its only proper nonzero possibility is $S_2$. Since $z_1=z_2$, $E$ and $S_2$ have the same phase. Thus $E$ is semistable but not stable. The sequence $0\to S_2\to E\to S_1\to 0$ is nonsplit because $E$ has no subrepresentation isomorphic to $S_1$.

Every representation of dimension vector $(1,1)$ is a scalar arrow map. There are exactly two isomorphism classes: the zero map gives $S_1\oplus S_2$, and every nonzero scalar is isomorphic by vertex base changes to $E$. Both classes are semistable at phase $\varphi$; neither is stable, because $S_1\oplus S_2$ has a proper subobject of the same phase and step 1.1 handles $E$. Hence the coefficient of ${\delta }_E$ in $A_{\mathrm{a}\mathrm{l}\mathrm{l}}$ is $1$.

All nonzero dimension vectors at this wall have phase $\varphi$, since $Z(d_1,d_2)=(d_1+d_2)z_1$. Therefore the ordered product over distinct phases has only the single factor at $\varphi$. If that factor is replaced by $S_Z^{\mathrm{s}\mathrm{t}}(\varphi )$, its coefficient at ${\delta }_E$ is $0$: step 2.1 lists every isomorphism class of dimension $(1,1)$ and neither is stable. This contradicts the coefficient $1$ from step 2.1. The omitted term is the strictly semistable nontrivial extension $E$, so the stable-only replacement cannot satisfy the HN factorization.