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.
I find it confusing when I have to think of a Hom-space as itself being a module. In this case, though, $Hom(DA, {2 \atop 3})$ is one-dimensional by the given information, so it must be a simple module over some vertex, and the only question is which one. We can answer that question by analysing which $e_i$ doesn't annihilate it.
I am going to follow the conventions of Assem-Simson-Skrowroński, so I am assuming we are using right modules over $A$ which is the quotient of the path algebra. We get a right module structure on $Hom(DA,{2 \atop 3})$ by considering the left $A$-module structure on $DA$, and, defining $fa$ by $(fa)(x)=f(ax)$. The left $A$-module structure on $DA$ comes from the right $A$-module structure on $A$. $$Hom(DA,{2\atop 3})e_i=Hom(e_i(DA),{2\atop 3})= Hom(D(Ae_i),{2\atop 3})=Hom(I_i,{2\atop 3}).$$ It therefore follows that it is only $e_4$ which doesn't annihilate $Hom(DA,{2\atop 3})$, and thus $Hom(DA,{2\atop 3})$ as an $A$-module is isomorphic to the simple $S_4$.