Branched covering

Remarks

Every branched covering $f$ is surjective, and its ramification points form a closed discrete set $R \subseteq X$. There is an integer $d \ge 1$, the degree, such that $f$ restricts to a $d$-sheeted covering map from $X \setminus f^{-1}(f(R))$ onto $Y \setminus f(R)$. Every fibre has exactly $d$ points when counted with multiplicity, where the multiplicity at $x$ is the $k$ such that $f$ looks like $z \mapsto z^k$ near $x$ in suitable charts. If $X$ is compact, every non-constant holomorphic map $X \to Y$ is a branched covering. In topology, a branched covering of surfaces is defined in the same way with continuous maps that look locally like $z \mapsto z^k$.

Example: $f \colon \C \to \C$, $f(z) = z^2$. It is holomorphic and non-constant, and $f^{-1}(K)$ is closed and bounded for compact $K$, since $\abs{z}^2 = \abs{f(z)}$. Its degree is $2$, its only ramification point is $0$, and the only branch point is $0$.

Non-example: $\exp \colon \C \to \C$ is holomorphic and non-constant but not proper. The preimage of the compact set $\set{1}$ is $2\pi i \Z$, which is not compact; clause 2 fails.