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For a covering map with nonempty connected base, the degree of a covering map is the common cardinality of its fibres. Each evenly covered neighbourhood makes this number locally constant, so connectedness makes it constant. The total covering space need not be connected. On a path-connected base the path lifting theorem gives a fibre bijection by path lifting. A disconnected base may have different degrees on different connected components.

Choose a path $\gamma$ in $X$ from $x_0$ to $x_1$. For every $\hat{x}\in p^{-1}(x_0)$, the path lifting theorem gives a unique lift of $\gamma$ starting at $\hat{x}$. Send $\hat{x}$ to the endpoint of this lift. Lifting the reversed path gives the inverse map, so this is the fibre bijection by path liftingIn particular, all fibres have the same cardinality.

A connected covering is a normal covering map when its deck transformations act transitively on a fibre. Equivalently, for a choice of ${\hat{x}}_0$ above $x_0$,The relevant lifting criterion for a covering space says that a based map $f:(Y,y_0)\to (X,x_0)$ lifts through $p$ exactly whenFor a universal covering, $\hat{X}$ is simply connected, so the displayed covering subgroup is trivial and hence normal. Thus a universal covering map is normal.

Now consider connected finite covers of the closed orientable surface ${\Sigma }_g$. By the classification of connected covering spaces, degree-$n$ connected covers correspond to index-$n$ subgroups of the fundamental group of a closed orientable surface, and normal covers correspond to normal subgroups.

The cases in which normality is forced are:

It remains to show that these are the only forced cases. Let $g\ge 2$ and $n\ge 3$. WriteIn the symmetric group $S_n$, putDefine a homomorphism byand send all remaining generators to the identity. The relation is respected becauseThe cycle $\sigma$ and transposition $\tau$ generate $S_n$, so the homomorphism is surjective.

LetThe natural action of $S_n$ is transitive, so $H$ has index $n$. Its image is the point stabilizer $S_{n-1}$, which is not normal in $S_n$ for $n\ge 3$; hence $H$ is not normal. The connected covering corresponding to $H$ is therefore an explicit degree-$n$ nonnormal cover. This is the nonnormal finite cover of a higher-genus orientable surface.

Consequently, among connected covers that exist, normality is forced exactly whenwith the qualification that $g=0$ permits only $n=1$, as summarized by the forced normality of finite connected covers of orientable surfaces.

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.