We say that two finite dimensional algebras $A$ and $B$ are stably equivalent if there is an equivalence $F:\underline{mod} A\longrightarrow \underline{mod} B$ between the associated module categories modulo projective modules, where $mod A$ and $mod B$ are finitely generated modules categories over $A$ and $B$, respectively. Question: For any exact sequence $$0 \longrightarrow X_{1}\longrightarrow X_{2}\longrightarrow X_{3}\longrightarrow 0$$ in $mod A$, can we obtain an exact sequence $$0 \longrightarrow F(X_{1})\oplus P_{1}\longrightarrow F(X_{2})\oplus P_{2}\longrightarrow F(X_{3})\oplus P_{3}\longrightarrow 0$$ in $mod B$, where $P_{i}$ is projective module in $mod B$ for $i=1,2,3$ ?
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Let $A=kQ$ be the path algebra of Dynkin type $\mathcal{A_2}$ and $B=K[x]/(x^2)$. Both algebras have a unique simple non-projective module and all other indecomposable modules are projective. Thus their stable module categories consists of just one indecomposable objects and are isomorphic. Let $S$ be the simple non-projective $A$-module and M be the simple $B$-module. Then $S$ has projective dimension one and there is a short exact sequence: $0 \rightarrow P_1 \rightarrow P_0 \rightarrow S \rightarrow 0$. This sequence is necessarily mapped to a sequence of the form: $0 \rightarrow Q_1 \rightarrow Q_0 \rightarrow M \oplus Q_2 \rightarrow 0$, with $Q_1$, $Q_2$ and $Q_0$ projective. But this sequence can not be exact or else $M$ would have finite projective dimension. But the algebra $B$ is selfinjective and not semi-simple and thus every non-projetive $B$-module has infinite projective dimension.
