The projective module in the quiver representation I study the quiver just some weeks and I cannot understand the projective module in the quiver representation well.Here are some questions:
Suppose $Q$ is a quiver, $a\in Q_{0}$.
1)Show that the projective $KQ$-module $P(a)$ is simple iff $a$ is a sink.
2)If $Q$ is acyclic, $P(a)=(P(a)_{b},\phi_{\beta})$ be the indecomposable projective. Show that for each arrow $\beta$, the map $\phi_{\beta}$ is injective.
 A: I'll assume for now that your quiver is unbound, i.e. you aren't taking a quotient by an admissible ideal.
1) I'm assuming from the phrasing of question 2) that you know what the indecomposable projectives are and the simples are. Then if $a$ is a sink, there is only the constant path $a\to a$, so $P(a)_a=K$, and there are no paths $a\to b$ for any $b\ne a$, so $P(a)_b=0$ when $b\ne a$. But this is exactly the description of the simple module $S(a)$.
Conversely, if $a$ is not a sink, then $\dim{P(a)_a}\geq1$, as there is at least the constant path at $a$. Assume first that $Q$ is acyclic. Then as $a$ is not a sink, there must be a vertex $b\ne a$ and an arrow $a\to b$, so $\dim P(a)_b\geq1$, and again $P(a)$ cannot be simple, as simples of an acyclic quiver have non-zero vector spaces at only a single vertex.
Now if $Q$ is cyclic, but there are no paths from $a$ to a vertex involved in an cycle, then all of the vector spaces in $P(a)$ associated to vertices on the cycle are zero (as there are no paths). So we induce a representation of a quiver obtained by deleting edges in the cycles (which have associated map the zero map between zero vector spaces) to obtain an acyclic subquiver $\widetilde{Q}$. The representation $P(a)$ on $Q$ restricts to the representation $P(a)$ on $\widetilde{Q}$, so we can apply the argument from the acyclic case to find a proper submodule. Adding back the deleted edges, and their associated zero maps, we obtain a proper submodule of $P(a)$ on $Q$.
Finally, if $Q$ is cyclic and there is a path from $a$ to a vertex on a cycle $C$, then the dimension of $P(a)_b$ is infinite for any $b$ such that there is a path to $b$ from a vertex of $C$. Take $m$ to be the length of the shortest path from $a$ to $c$, and $n$ to be the number of edges in $C$. We get a subrepresentation $R$ by choosing $R_v=0$ if there is no path to $v$ from a vertex of $C$, and otherwise taking $R_v$ to be the subspace of the infinite dimensional space $P(a)_v$ spanned by paths $a\to v$ of length at least $m+n$ (this is to ensure the subspace is proper - we have excluded the direct path from $a\to v$ that does not include a full lap of the cycle). The restriction of the maps in $P(a)$ to $R$ are either zero, or increase the length of paths, so we have a proper subrepresentation and therefore $P(a)$ is not simple.
(Note: The acyclic case is more complicated than I was expecting, and you should check carefully for any errors. If anybody can think of a simpler argument I'd be interested to see it.)
2) Note that $P(a)_b$ is the $K$-vector space spanned by paths $w$ from $a\to b$. For each arrow $\beta:a\to c$, the map $\phi_\beta$ takes $w$ to $w\beta$, a path from $a$ to $c$. So let $v=(a|v_1,v_2,\ldots,v_k|b)$ and $w=(a|w_1,w_2,\ldots,w_l|b)$ be paths $a$ to $b$. Then if $v\beta=w\beta$, the sequence $(w_1,\ldots,w_l,\beta)$ is equal to the sequence $(v_1,\ldots,v_k,\beta)$, and so $(w_1,\ldots,w_l)=(v_1,\ldots,v_k)$ and $v=w$. So the map is injective.
I think the reason that acyclic-ness is introduced here is that if the quiver has a directed cycle, the zero ideal is not admissible, so you're forced to take a non-trivial quotient of the path algebra by an admissible ideal $\mathcal{I}$, which means the vector space $P(a)_b$ is no longer generated by paths $a\to b$, but by equivalence classes of paths modulo this ideal. In this case the result in 2) may no longer apply, as although $w\beta$ and $v\beta$ are distinct paths, it is possible that $v\beta-w\beta\in\mathcal{I}$.
