= Solution
{c}
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> with
$$
v(\widetilde X)=dv(X),\qquad e(\widetilde X)=de(X).
$$
Subtracting gives the <Euler characteristic under a finite covering> formula
$$
\boxed{\chi(\widetilde X)=d\,\chi(X).}
$$
The formula holds even when $\widetilde X$ is disconnected. If a disconnected base is allowed to have degrees $d_C$ that vary with its <connected components> $C$, the precise replacement is
$$
\chi(\widetilde X)=\sum_C d_C\chi(C).
$$
The printed formula with a single $d$ uses the constant-degree convention explained in part (a)(ii).
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