= Solution
{c}
For a <covering map> over a nonempty <connected space>, the <degree of a covering map> is
$$
\boxed{\deg p=|p^{-1}(x)|,}
$$
the <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.
\b[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.
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