How to compute Nakayama functor explicitly? I am reading the book Elements of representation theory of associative algebras, volume 1. I have some questions related to the Nakayama functor. On page 113-114 of the book, the Auslander-Reiten translation $\tau M$ of a module $M$ of the path algebra of the Kronecker quiver $Q=1\underset{\beta}{\overset{\alpha}{\Leftarrow}} 2$ is computed. The module $M$ is defined by $K \underset{0}{\overset{1}{\Leftarrow}} K$. 
It is said that a minimal projective presentation of $M$ is given by
$$
0 \to P(1) \overset{p_1}{\to} P(2) \overset{p_2}{\to} M \to 0,
$$
where $$P(1)=S(1)= (K \underset{0}{\overset{0}{\Leftarrow}} 0), \quad P(2)= (K^2 \underset{\left( \begin{matrix}0 \\ 1\end{matrix} \right)}{\overset{\left( \begin{matrix}1 \\ 0\end{matrix} \right)}{\Leftarrow}} K).$$
We know that a basis of $K^2$ in $P(2)$ is $K\alpha\oplus K\beta$. The basis of $K$ in $P(1)$ is $K \varepsilon_1$. The map $p_1$ from $P(1)$ to $P(2)$ is
$$
\begin{matrix} K & \underset{0}{\overset{0}{\Leftarrow}} & 0 \\ 
f \downarrow & & \downarrow g \\
K^2 & \underset{\left( \begin{matrix} 0 \\ 1\end{matrix} \right)}{\overset{\left( \begin{matrix}1 \\ 0\end{matrix} \right)}{\Leftarrow}} & K
\end{matrix}
$$
Since the domain of $g$ is $0$, we must have $g=0$. How to compute the map $f$? Does $f$ send $\varepsilon_1$ to $\alpha$ or $\beta$ or $0$? 
If we know the map $f$, then we know the map $p_1$. How to compute $\nu p_1$? Where $\nu=DHom_A(\cdot, A)$ is the Nakayama functor.
We know $I(1)$ and $I(2)$. If we know the map $\nu p_1: I(1) \to I(2)$, how to compute the kernel of $\nu p_1$? I know that the kernel of $\nu p_1$ must be of the form $K \underset{h_2}{\overset{h_1}{\Leftarrow}} K$. But I don't know how to show that $h_1=1$ and $h_2=0$. Thank you very much.

 A: For your first question, $f$ is given by $\epsilon_1\mapsto \beta$, or equivalently $p_1:\epsilon_1 x \mapsto \beta x$ for any $x\in kQ$.  You just need to stare at your diagram long enough to realise this...Note that the squares have to commute.  I usually prefer computing in (sort of) Loewy diagram form, as the algebra is basic and the field (so all simples are of 1 dimensional) is algebraically closed.  This technique which can be demonstrate as follows.  Turning your exact sequence into Loewy diagram:
$$
0 \to 1 \to
\left(\begin{array}{ccc} & 2 & \\  & \swarrow  \searrow & \\ 1 &  & 1 \end{array}\right) \to 
\left(\begin{array}{c} 2  \\ {\scriptsize 1}\Downarrow {\scriptsize 0} \\ 1 \end{array}\right) \to 0 
$$
Each number here denote the composition factor (simple module), so for example the middle $P(2)$, has composition factor $S(2)$ on top, and two copies of $S(1)$ in the bottom; the left hand side arrow correspond to multiplication by $\alpha$ on the right, the right hand side arrow correspond to multiplication by $\beta$.  So you see the top $S(2)$ of $P(2)$ is mapped to the top $S(2)$ of $M$, and the left hand side $S(1)$ in socle of $P(2)$ is mapped to $S(1)$ in the socle of $M$, because we need to arrows in the middle to match up (corresponding to the commuting square).  So now the thing in $P(2)$ that has not been mapped (i.e. the kernel) is the right hand side $S(1)$ in the socle.  So the map you want is to map $S(1)$ to the right hand side $S(1)$, i.e. given by $\epsilon_1 x\mapsto \beta x$.
Normally, you would need to compute $DHom_A(X,A)$ explicitly to see what the maps resulting maps look like.  Sometimes, we can just "clever guess", because it maps (ind) projective to (ind) injective "in a compatible way".  By the description of injective modules (this is in earlier proposition in the book), $\nu(S(1)\to P(2))= I(1) \to I(2) = D( \epsilon_1 A \to \epsilon_2 A) $ has Loewy diagram
$$
 \left(\begin{array}{ccc} 2 &  & 2\\  & \searrow  \swarrow & \\ & 1 & \end{array}\right) \to 2 
$$
where left hand arrows corresponding to "multiplying $\alpha$ on the left, before taking $K$-linear dual $D$". "The compatibility" means that the map $\nu p_1$ is given by $D(x \beta \to x \epsilon_2)$, i.e. to map the right hand side composition factor $S(2)$ in the top of $I(1)$ to $I(2)=S(2)$.  Therefore the kernel, can be seen from the combinatorics easily, is given by "deleting the right hand side of the diagram of $P(2)$":
$$
\begin{array}{c} 2 \\ {\scriptsize 1}\Downarrow{\scriptsize 0} \\ 1 \end{array}
$$
so is just $M$ itself.
