LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)audited 2026-10-08

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A degree-one holomorphic map of compact Riemann surfaces is an isomorphism

Statement

Let $f : X \to Y$ be a nonconstant holomorphic map between compact connected Riemann surfaces (Holomorphic maps and meromorphic functions on Riemann surfaces, Riemann surfaces and holomorphic atlases). Then $f$ is proper and has a degree $d$ as in Degree of a proper holomorphic map of Riemann surfaces. If $d = 1$ , then $f$ is bijective and its set-theoretic inverse $g : Y \to X$ is holomorphic. Thus $g \circ f = {i d}_{X}$ and $f \circ g = {i d}_{Y}$ , so $f$ is an isomorphism of Riemann surfaces.

Facts & Assumptions

Given: A nonconstant holomorphic map $f : X \to Y$ between compact connected Riemann surfaces.

[F1]

The source $X$ is compact, and the target $Y$ is Hausdorff because it is a Riemann surface (Riemann surfaces and holomorphic atlases).

[F2]

A holomorphic map of Riemann surfaces is continuous (Holomorphic maps and meromorphic functions on Riemann surfaces).

[F3]

Every compact subset of a Hausdorff space is closed (In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones).

[F4]

For a proper nonconstant holomorphic map between connected Riemann surfaces, every fibre is nonempty and finite, and the degree is the constant weighted count $\sum_{x \in f^{- 1} (y)} e_{x} (f)$ (Degree of a proper holomorphic map of Riemann surfaces).

[F5]

Each ramification index $e_{x} (f)$ is a positive integer, and $e_{x} (f) = 1$ exactly when $f$ is a local biholomorphism at $x$ (Ramification index, ramification order and branch value).

[F6]

In suitable centred charts, $f$ is locally $z \mapsto z^{e_{x} (f)}$ ; in particular, if $e_{x} (f) = 1$ , the local inverse is holomorphic (Local power-map normal form on Riemann surfaces, Biholomorphic maps between complex domains).

Proof

technique · direct

1.1F1F2F3F4

Let $K \subseteq Y$ be compact. By [F3], $K$ is closed in $Y$ , and by [F2] its preimage $f^{- 1} (K)$ is closed in $X$ . Since $X$ is compact by [F1], $f^{- 1} (K)$ is compact. This holds for every compact $K$ , so $f$ is proper and the degree in [F4] is defined.

1.2F4F5

Suppose $d = 1$ . For every $y \in Y$ , [F4] gives a nonempty finite fibre with $\sum_{x \in f^{- 1} (y)} e_{x} (f) = 1$ . Each summand is a positive integer by [F5], so the fibre has exactly one point, with ramification index $1$ . Thus $f$ is bijective and is unramified at every point.

2.1F6step 1.2∎

For each $x \in X$ , [F6] gives charts in which $f$ is $z \mapsto z$ near $x$ , so it has a holomorphic local inverse near $f (x)$ . The set-theoretic inverse $g$ from step 1.2 agrees with each such local inverse on its domain, hence is holomorphic on all of $Y$ . Its defining identities $g \circ f = {i d}_{X}$ and $f \circ g = {i d}_{Y}$ show that $f$ is an isomorphism.

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Sources

Curtis T. McMullen, Riemann Surfaces, Harvard Math 213b course notes (2026) (standard reference, not scraped)
Eduard Looijenga, Riemann Surfaces (2007 author lecture notes) (standard reference, not scraped)