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?