TITLE: Projective modules of the path algebra
QUESTION [5 upvotes]: Let $Q$ be a finite and acyclic quiver. Let $a,b$ be vertices then $P(a)_{b}$, the projective indecomposable $kQ$-module is equal to the vector space having as a basis the set of all paths from $a$ to $b$. 
What I don't understand is how to obtain the induced $k$-linear maps between each vertex? Can you please explain with an example?
Here's an example of what I am asking. Suppose we have the quiver
$$1 \underset{j}{\overset{h}{\leftleftarrows}} 2 \underset{g}{\overset{f}{\leftleftarrows}} 3$$
and we look at $P(3)$ then we get:
$$K^{4} \leftleftarrows K^{2} \leftleftarrows K.$$
I want to describe explicitly the induced $K$-maps between the representations.
From what I understand the maps from $K \rightarrow K^{2}$ consist of $x \rightarrow (x,0)$ and $x \rightarrow (0,x)$. How do we find the maps from $K^{2} \rightarrow K^{4}$. Can you please ellaborate more on "right multiplication"?

REPLY [3 votes]: If $\alpha$ is an arrow from $b$ to $c$ then the vector space map $P(a)_b \to P(a)_c$ assigned to $\alpha$ in the representation $P(a)$ is defined as follows: A basis element of $P(a)_b$ is a path from $a$ to $b$.  Append $\alpha$ to this path to get a path from $a$ to $c$, hence a basis element of $P(a)_c$.
Put another way, if you are working with right modules over the path algebra $kQ$ (and hence reading paths from left to right) then $P(a) = e_akQ$, $P(a)_b = e_akQe_b$, and $P(a)_c = e_akQe_c$.  The map $P(a)_b \to P(a)_c$ corresponding to $\alpha$ is just right multiplication by $\alpha$.
Edit: In your example $P(3)_2$ has basis $\{f, g\}$, the two maps $3 \to 2$, and $P(3)_1$ has basis $\{fh, fj, gh, gj\}$, the paths $3 \to 1$.  Note that I'm assuming that you're doing right modules which means you read paths left to right, hence $fh$ instead of $hf$).  Now the map assigned to $h$ is right multiplication by $h$:

$f \mapsto fh$
$g \mapsto gh$

and the map assigned to $j$ is right multiplication by $j$:

$f \mapsto fj$
$g \mapsto gj$