Let $Q$ be a finite quiver without oriented cycles and let $k$ be a field. Then $A = kQ$ is a finite-dimensional path algebra. Let $e_1, \dots, e_n$ be the idempotents corresponding to the vertices $1, \dots, n$ of $Q$. Then the modules $I(i) = D(e_iA)$ are precisely the indecomposable injective modules.
Question: Assume $I(i)$ is not projective. Then there is an Auslander-Reiten sequence $0 \to \tau I(i) \to E(i) \to I(i) \to 0$. How can we describe the module $E(i)$?
In the article "Representations of Wild quivers" by Otto Kerner, the author seems to implicitly say in the proof of theorem 3.8 that $E(i) = \bigoplus I(j) \oplus \bigoplus \tau I(j')$, where $j$ and $j'$ together run through all vertices in the quiver which are adjacent to $i$. Is this correct? Do you know a reference for this?
Yes, that's correct.
For any finite dimensional algebra, the left minimal almost split morphism from an indecomposable injective $I$ is the natural epimorphism $I\to I/\text{soc }I$. This is Prop. IV.3.5 in Assem, Simson and Skowronski's Elements of the Representation Theory of Associative Algebras, vol. I.
In the case of the path algebra of a quiver, $I(i)/\text{soc }I(i)$ is the direct sum of the $I(j')$, as $j'$ runs over the tails of all arrows into $i$. So the arrows in the Auslander-Reiten quiver starting from $I(i)$ are $I(i)\to I(j')$. None of the $I(j')$ are projective, or they would be proper direct summands of $I(i)$, so there are corresponding arrows $\tau I(j')\to I(i)$. Then there are also the irreducible maps from indecomposable injectives to $I(i)$, which as shown above are maps $I(j)\to I(i)$ where $j$ runs over the heads of all arrows from $i$.