I'm trying to prove proposition 2.3 of Quiver Representations by Ralf Schiffler. To any vertex $i$ finite acyclic quiver $Q$ we can associate the indecomposable projective $P(i)$, this proposition asks us to show that $P(i)$ is indeed projective.
So let $g:M\rightarrow N$ be a surjective morphism of quiver representations and let $f:P(i)\rightarrow N$ be a morphism. We have to show that there exists a morphism $h:P(i)\rightarrow M$ such that $gh=f$.
I'm going to outline my proof, but I would be grateful if someone could tell me whether this is the correct way of doing it.
First of all, $P(i)_i\cong k$ as it is generated by the idempotent $e_i$. Since $g_i:M_i\rightarrow N_i$ is surjective, there exists an $m\in M_i$ such that $g_i(m)=f_i(e_i)$. I define $h_i:P(i)_i\rightarrow M_i$ by mapping $e_i$ to $m$ and I claim that this choice completely determines a desired map $h$.
At this point I need some more notation: The quiver representation $M=(\left\{M_i\right\}_{i\in Q_0},\left\{\lambda_{\alpha}\right\}_{\alpha\in Q_1})$. Now let $\alpha:i\rightarrow j$ be a fixed arrow (that is a path of length $1$). Then $\lambda_{\alpha}:M_i\rightarrow M_j$ and $g_j(\lambda_{\alpha}(m))=f_j(\alpha)$ (here I used that $P(i)=(P(i)_j,\phi_{\beta})$ and $\phi_{\beta}$ is given by the multiplication by $\beta$ and I used the fact that $f$ and $g$ are module morphisms/morphisms of quiver representations). Hence for $\alpha\in P(i)_j$, we should define $h_j(\alpha)=\lambda_{\alpha}(m)$.
Thus we now know how to define $h$ on arrows. Now suppose that $\mathfrak{p}=(i\mid \alpha_1\dots \alpha_n\mid j)$ is a path of length $l$ from $i$ to $j$, then we define $h_j(\mathfrak{p}):=\lambda_{\alpha_n}\dots\lambda_{\alpha_2}\lambda_{\alpha_1}(m)$.
The remainder of the proof is devoted to checking that $h$ is indeed a morphism of quiver representations and checking that $gh=f$.
Thank you in advance.