Projective modules of the path algebra 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"?
 A: 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 $j$:


*

*$f \mapsto fj$

*$g \mapsto gj$

A: There is some wordings in your question which is not quite correct, and I don't really understand what you are trying to ask.  Please tell me if I am not explaining the thing you want to know.
Let $P(a)$ be the projective indecomposable $kQ$-module corresponding to vertex $a$.  Suppose you work with right modules and composing arrows from left to right, then $P(a) = e_a kQ$ with $e_a$ the trivial path at $a$.  In fact, $P(a) = \bigoplus_{b \in Q_0} P(a)_b$, where $P(a)_b$ is the vectorspace with basis containing all paths from $a$ to $b$.  Using Schur's Lemma, then $Hom_{kQ}(P(b),P(a)) = e_a kQ e_b = P(a)_b$.  (I think this is what you are asking?)
Example 1) Consider the Kronecker quiver $1 \rightrightarrows 2$, then $kQ = P(1) \oplus P(2)$ with $P(1) = k \rightrightarrows k^2$ and $P(2) = 0 \rightrightarrows k$.  Now $P(1)_2 = k^2 = k-span\{ \alpha_1,\alpha_2\}$, where $\alpha_1$ and $\alpha_2$ are the two arrows of $Q$, these also correspond to the basis of $Hom_{kQ}(P(2),P(1))$, given by $\alpha_i: P(2)\to P(1)$ which sends $e_2\alpha$ to $e_1\alpha_i\alpha$ with $i=1,2$.
Example 2) Let $Q$ be $1\to 2\to 3$ with an extra arrow $1\to 3$.  Then $P(1) = (k,k,k^2)$,  $P(2)=(0,k,k)$, $P(3)=(0,0,k)$.  (I use $i$-th entry in the triple denote the vectorspace on vertex $i$.)  $P(2)$ and $P(3)$ are probably easy to see.  For $P(1)$, we have two dimensional vectorspace assigned to vertex 3 (i.e. $P(1)_3$), because there are two paths from 1 to 3, one via 2, the other is the arrow $1\to 3$.  Now you can do the exercise like previous example to see how these corresponds to maps between projective indecomposables.
EDIT: Perhaps a clearer exposition using your example:
The projective indecomposable $P(3)$ is given by $e_3 kQ = k$-span$\{$ all paths starting from 3 $\} = k$-span$\{ e_3, f, g, fh, gh, fj, gj\}$.  (I will use $\langle x \rangle$ instead of $k$-span$\{x\}$ now.)  This is in fact a representation of the quiver $Q$ by assigining (1) vectorspace $P(3)_b$ to the vertices $b$, and (2) right multiplication maps to paths (I think this is what you mean by "the induced $k$-linear map"?), as follows
$$
\langle e_3\rangle 
\underset{(-)g}{\overset{(-)f}{\rightrightarrows}} \langle f,g\rangle \underset{(-)j}{\overset{(-)h}{\rightrightarrows}} \langle fh,gh,fj,gj\rangle
$$
Hope this clarifies things a bit.
