I am confused at the answer of the question: Why is the Auslander Reiten theory not working in this example?
Consider the quiver
$ 1 \xrightarrow{\alpha} 2 \xrightarrow{\beta} 3 \xrightarrow{\gamma} 4$
with relations $I= < \alpha \beta \gamma >$.
The answer talks about how to calculate $\nu^{-1} \begin{matrix} 2 \\ 3\end{matrix}$:
$$ \nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix} = Hom_A(DA,\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = Hom_A(I(1)\oplus I(2)\oplus I(3) \oplus I(4), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!). $$
Since $Hom_A(I(1), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = Hom_A(I(2), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = Hom_A(I(3), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!) = 0$ and $Hom_A(I(4), \begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\!)$ is one-dimensional, we get that $\nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! = 4$.
Who can tell me how to get $\nu^{-1}\begin{matrix} 2 &\\ 3 \end{matrix}\!\!\!\!\! = 4$ just by these things? Thank you.