Let $Q$ be a quiver without oriented cycles. Suppose in addition that $Q$ is non-Dynkin. I am looking for the reference or proof of the following fact:
Let $I$ be an injective $kQ$-module and $M$ be a preprojective $kQ$-module. Then $Hom_{kQ}(I, M)=0.$
For Dynkin quivers this shouldn't be true. For example, let
$$Q = 2 \stackrel{a}\longrightarrow 1$$ and denote $A:=kQ$.
Then we have a nontrivial $A$-module map $f:A^*\to A$ such that $f(e_1^*)=a, f(a^*)=e_2$ and $f(e_2^*)=0.$
It suffices to show that $Hom_{KQ}(M,P)=0$ for any non-projective indecomposable module $M$ and any projective module $P$.
Let $f:M\to P$ be a non-zero map. Then we have a short exact sequence $$ 0\to \ker(f)\to M\to \mathrm{im}(f)\to 0.$$
$\mathrm{im}(f)$ is a submodule of $P$, so the fact that $KQ$ is hereditary means that $\mathrm{im}(f)$ is projective. Since $\mathrm{Ext}_A^1( X, Y)=0$ for any algebra $A$, any projective $A$-module $X$, and any $A$-module $Y$. The short exact sequence above splits. Since $M$ is indecomposable, splitness implies that $M\cong \mathrm{im}(f)$, so $M$ is projective.