Let $Q=(Q\_{0},Q\_{1},h,t)$ be a finite quiver where $Q\_{0}$ are the vertices, $Q\_{1}$ the arrows and we have two maps $h: Q\_{1} \rightarrow Q\_{0}$ (head) and $t: Q\_{1} \rightarrow Q\_{0}$ (tail). Fix a field $K$ and associative to $Q$ two vector spaces $R=K^{Q\_{0}}$ and $A=K^{Q\_{1}}$ i.e vector spaces consisting of $K$-valued functions on $Q\_{0}$ and $Q\_{1}$ respectively.


View $A$ as an $R$-bimodule as follows: $(e \cdot f)(a)=e(ha)f(a)$ and $(f \cdot e)(a)=f(a)e(ta))$ for all $a \in Q\_{1}$, $e \in R$, $f \in A$


For each $\tau \in Q\_{1}$ consider the map $\gamma\_{\tau}: Q\_{1} \rightarrow K$ by $\gamma\_{\tau}(\beta)=0$ if $\beta \neq \tau$
and $1$ if $\beta = \tau$, i.e the characteristic function.


It is clear then that $A = \bigoplus\_{\tau \in Q\_{1}} K\tau$.


Question: 


Consider now the tensor product $A \otimes\_{R} A$, let $\alpha$,$\tau$ in $Q\_{1}$ and suppose we consider an element of this tensor product, say $\gamma\_{\alpha} \otimes\_{R} \gamma\_{\tau}$. Why can this tensor be "identified" with a path of length $2$? Which quiver are we considering, are we constructing a quiver on the tensor product?



Quiver on tensor product
There is a basis of the tensor product $A^{\otimes n}$ given by $\gamma\_1\otimes \cdots \otimes \gamma\_n$ where $\gamma\_1,\dots, \gamma\_n$ are a list of $n$ elements of $Q\_1$ that are composable. That is, where concatenating $\gamma\_1\cdots \gamma\_n$ gives a path of length $n$. So, just as $A$ has a basis given by paths of length 1, $A^{\otimes n}$ has a basis given by paths of length $n$ in the original quiver. 


I'm not sure what you mean by "a quiver on the tensor product" but you never use any quiver other than the original one.


