 Solution

Assume $p$ has a common finite degree of a covering map $d$. Declare the lifted vertices to be $p^{-1}(V(X))$, so there are exactly $dv(X)$ of them. Each open edge is homeomorphic to $(0,1)$, a simply connected interval. The restriction of the covering map above it is a disjoint union of $d$ copies of that open edge.

These copies acquire endpoints as follows. Lift the parametrizing path $[0,1]\to X$ of each oriented edge, starting at each of the $d$ points over its initial vertex. The path lifting theorem gives a unique lift all the way to $1$. Its endpoint lies over the terminal vertex, and the interiors of these lifted paths are exactly the $d$ open edges already described. This also handles a graph loop, whose two endpoints may coincide downstairs or upstairs. Local covering sheets above a small star at a vertex show that the lifted incidence relations have precisely the original topology. Thus the graph structure lifted through a covering map is a finite topological graph withSubtracting gives the Euler characteristic under a finite covering formulaThe formula holds even when $\tilde{X}$ is disconnected. If a disconnected base is allowed to have degrees $d_C$ that vary with its connected components $C$, the precise replacement isThe printed formula with a single $d$ uses the constant-degree convention explained in part (a)(ii).