In the book quiver representation and quiver varieties he lefts Lemma 6.13 to the reader, but I'm having trouble with it.
If $Q$ is a tree, any orientation of $Q$ can be obtained from a slice in $\mathbb{Z} Q$. Two slices $T, T'$ give the same orientation iff $T'=\tau^k(T)$ for some $k\in \mathbb{Z}$.
If I have an orientation of $Q$ and fix a vertice $a\in I$ and choose some $n\in \mathbb{Z}$ such that $(a,n)\in \mathbb{Z}Q$ then we can define for every edge conecting $a$ and $b$ in the graph we put in the slice like this: $$(b,n-1)\rightarrow (a,n) \text{ if } a\rightarrow b$$ and $$(a,n)\rightarrow (b.n+1) \text{ if } b\rightarrow a$$
With this I have defined the slice for all edges connected to $a$, how do I finish this?
And for the second part, is there some ideia of how to do it easily ass the author said? I'm thinking about doing induction on it.