 Solution

For a covering map over a nonempty connected space, the degree of a covering map isthe cardinality of any fibre of a function. To show independence of $x$, choose an evenly covered neighbourhood $U$. Each covering sheet contains exactly one point over every $y\in U$, so $|p^{-1}(y)|$ equals the number of covering sheets over $U$. The fibre of a function cardinality is therefore locally constant. For a fixed cardinal number $\kappa$, the points with that cardinality form an open subset, and their complement is also open. Connectedness forces the value to be constant everywhere.

For a path-connected base there is also an explicit fibre bijection by path lifting: lift a path from $x$ to $y$ starting at each point over $x$ and take its endpoint. The path lifting theorem gives existence and uniqueness; path reversal gives the inverse bijection. The bijection may depend on the path, but the resulting equality of cardinalities does not.

A connected base is needed for one global degree. Without it, use a degree on each connected component on which the cardinality is constant. For example, a covering map of a discrete two-point topological space can have one point over the first point and two over the second. Thus the unqualified printed definition needs the usual connected-base convention, or the explicit assumption that every fibre of a function has the same cardinality.