# Prove that a covering projection of graphs is a local isomorphism

This question comes from section 1.3 of Topological Graph Theory by Tucker and Gross.

Let $$B$$ be a group acting freely on a graph $$G$$ meaning that the only automorphism defined from the group action with a fixed point (on both vertices and edges) is the identity automorphism. The quotient $$G/B$$ is the graph with vertex orbits as vertices and edge orbits as edges. The endpoints of an edge orbit $$[e]$$ are the vertex orbits containing vertices incident to any edges in $$[e]$$. We define the quotient map $$q_B \colon G \rightarrow G/B$$ to be the map that maps vertices and edges to their orbits.

A covering projection is defined to be a map between graphs $$p \colon \tilde{G} \rightarrow G$$ such that there exists an isomorphism $$i \colon \tilde{G}/B \colon \rightarrow G$$ with $$p = i \circ q_B$$.

The star of a vertex $$v$$, $$\text{star}(v)$$, is the subgraph consisting of $$v$$, the neighbors of $$v$$, and each edge connecting $$v$$ to a neighbor. A map between graphs $$f$$ is a local isomorphism if for each vertex $$v$$ the image $$f(\text{star}(v))$$ is isomorphic to $$\text{star}(v)$$.

The exercise asks to show that if $$p \colon \tilde{G} \rightarrow G$$ is a covering projection and $$G$$ is a simple graph then $$p$$ is a local isomorphism.

Here's my approach so far. It should be sufficient to show that $$q_B$$ is a local isomorphism. This means that no two vertices or edges in $$\text{star}(v)$$ should lie in the same orbit under the action of $$B$$. I have been trying to show that two vertices from $$\text{star}(v)$$ lie in the same orbit, then the action defined by $$B$$ must have a fixed point coming from an automorphism other than the identity. If anyone could help me fill in the details for this argument, or has a completely different approach, it would be much appreciated.