= Solution
For a finite <acyclic quiver>, the <arrow ideal of a path algebra> $R$ is nilpotent, and $A/R\cong\prod_{i\in Q_0}k$. If $S$ is a <simple module>, its <submodule> $RS$ is either zero or $S$. The latter would imply $R^dS=S$ for every $d$, contradicting nilpotence. Thus $RS=0$, and a simple module over the product of fields is supported at one coordinate. Therefore \b[the simples are exactly $S(i)$], with $k$ at $i$, zero elsewhere, and zero arrows; the vertex $i$ is unique.
A finite-dimensional <semisimple module> is consequently $\bigoplus_iS(i)^{\oplus n_i}$, where $n_i=\dim X_i$. Its <dimension vector of a quiver representation> determines its isomorphism class.
For an arbitrary finite quiver, cycles allowed, the <vertex projective module of a path algebra> is $P(i)=Ae_i$. Its space at vertex $j$ has <basis> all paths from $i$ to $j$, and an arrow acts by adjoining that arrow at the end of the path. Its <endomorphism ring> is
$$
\boxed{\operatorname{End}_A(P(i))\cong(e_iAe_i)^{\mathrm{op}}},
$$
where $e_iAe_i$ is spanned by the closed paths based at $i$. The opposite multiplication appears because endomorphisms act by right multiplication.
The <evaluation isomorphism for a vertex projective> is
$$
\boxed{\operatorname{Hom}_Q(P(i),X)\longrightarrow X_i,\qquad h\longmapsto h(e_i)}.
$$
For $x\in X_i$, its inverse sends a path $p$ starting at $i$ to $px$. This proves both injectivity and surjectivity, and is natural in $X$. Vertex evaluation is exact, so $P(i)$ is a <projective module>; alternatively it is a direct summand of the free module $A$.
The <closed-path corner of a path algebra is a domain>: in a product of two nonzero linear combinations, choose their longest path lengths. Concatenation in that top degree has a unique cut at those lengths, so a product of two nonzero top coefficients cannot cancel. Hence its only <idempotents> are zero and one. The same is true of the opposite ring, proving $P(i)$ is an <indecomposable module>, even when cycles make it infinite-dimensional.
For $1\to2\leftarrow3$, the paths starting at vertex $1$ are $e_1$ and the arrow $1\to2$. Thus the displayed $k\xrightarrow{1}k\leftarrow0$ is \b[$P(1)$ and is projective].
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