Exercise 6, § 3

Let $G$ and $G'$ be graphs and let $\varphi : G'\rightarrow G$. We say that $(G', \varphi )$ is a covering of G if, for every vertex $a\in$ Som(G$'$) and every arrow $f\in$ Fl(G) with origin Som($\varphi$ )$(a)$, there exists a unique arrow $f'\in$ Fl(G$'$) with origin $a$ such that Fl($\varphi$ )$(f') =f$.

Suppose that $(G', \varphi )$ is a covering of G.

a) Prove that there exists a unique operation of Grp(G) into Som(G$'$), relative to the mapping Som($\varphi$ ), such that $x\cdot$ Fl($\varphi$ )$(f) =t(f)$, for every vertex $x\in$ Som(G$'$) and every arrow $f\in$ Fl(G$'$) with origin $x$.

b) Deduce that if $x$ and $y$ are vertices of G belonging to the same connected component of G, one has Card(Som($\varphi$ )$^{-1}(x)$) $=$ Card(Som($\varphi$ )$^{-1}(y)$).