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 $I= < \alpha \beta \gamma >$
I calculated the following injective resolution for the representation $\begin{matrix}
  2 &\\
  3 
 \end{matrix}$, 
$0 \longrightarrow \begin{matrix}
  2 &\\
  3 
 \end{matrix} \longrightarrow I(3)=\begin{matrix}
  1 &\\
  2 &\\
  3 
 \end{matrix} \longrightarrow I(1)= 1 \longrightarrow 0$
applying the Inverse Nakayama functor $\nu^{-1}$ I get
$0 \longrightarrow \nu^{-1} \begin{matrix}
  2 &\\
  3 
 \end{matrix} =0 \longrightarrow P(3) = \begin{matrix}
  3 &\\
  4 
 \end{matrix} \longrightarrow P(1) =\begin{matrix}
  1 &\\
  2 &\\
  3 
 \end{matrix} \longrightarrow \tau^{-1} \begin{matrix}
  2 &\\
  3 
 \end{matrix} \longrightarrow 0$
BUT the sequence is not exact  from $P(3) \longrightarrow P(1)$ since is not injective.
What is wrong here? I think that is because $\nu^{-1} \begin{matrix}
  2 &\\
  3 
 \end{matrix} \neq 0$ but I thought the inverse nakayama functor is zero on non- injectives?
 A: The inverse Nakayama functor is not zero on non-injective representations.  Think about what this would imply: it would mean that for every non-injective representation $M$, the projective presentation of $\tau^{-1} M$ computed in your post would show that it has projective dimension at most $1$... which cannot be true for all representations of the form $\tau^{-1}M$.
So let's redo the computation.  What is $\nu^{-1}\begin{matrix}
  2 &\\
  3 
 \end{matrix}\!\!\!\!\!$? By definition, it is
$$
\nu^{-1}\begin{matrix}
  2 &\\
  3 
 \end{matrix} = Hom_A(DA,\begin{matrix}
  2 &\\
  3 
 \end{matrix}\!\!\!\!\!) = Hom_A(I(1)\oplus I(2)\oplus I(3) \oplus I(4), \begin{matrix}
  2 &\\
  3 
 \end{matrix}\!\!\!\!\!).
$$
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 $\nu^{-1}\begin{matrix}  2 &\\  3  \end{matrix}\!\!\!\!\! = 4$.
Thus the exact sequence is
$$
0 \longrightarrow \nu^{-1} \begin{matrix}
  2 &\\
  3 
 \end{matrix}\!\!\!\!\! =4 \longrightarrow P(3) = \begin{matrix}
  3 &\\
  4 
 \end{matrix} \longrightarrow P(1) =\begin{matrix}
  1 &\\
  2 &\\
  3 
 \end{matrix} \longrightarrow \tau^{-1} \begin{matrix}
  2 &\\
  3 
 \end{matrix} \longrightarrow 0,
$$
which is more what we would expect.  Thus $\tau^{-1} \begin{matrix}
  2 &\\
  3 
 \end{matrix}\!\!\!\!\! = \begin{matrix}
  1 &\\
  2 
 \end{matrix}\!\!\!\!\!.$
