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The exponential map has no finite proper-map degree

Statement refuted

Every nonconstant holomorphic map of Riemann surfaces that is a local biholomorphism at every point is proper, and therefore has finite fibres and a finite degree in the sense of Degree of a proper holomorphic map of Riemann surfaces.

Facts & Assumptions

Given: The complex exponential $\exp : C \to C^{\times} = C ∖ {0}$ .

[F1]

Both $C$ and $C^{\times}$ are plane domains, hence Riemann surfaces with their identity atlases (Atlases on the sphere, plane, disc and annulus); for these atlases a map is holomorphic exactly when it is holomorphic as a map of plane domains (Holomorphic maps and meromorphic functions on Riemann surfaces).

[F2]

The complex exponential is entire with $\exp^{'} = \exp$ , so $\exp^{'} (z) = \exp z \neq 0$ for every $z \in C$ (The complex exponential is entire and its complex derivative is itself).

[F3]

The exponential is surjective onto $C^{\times}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

[F4]

If $f$ is holomorphic near $a$ and $f^{'} (a) \neq 0$ , then $f$ is biholomorphic between a neighbourhood of $a$ and a neighbourhood of $f (a)$ (A nonzero complex derivative gives a local biholomorphism).

[F5]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[F6]

A subset is compact when every open cover of the subspace has a finite subcover; consequently a one-point subset ${y}$ of any space is compact, since an open cover of ${y}$ has a member containing $y$ that alone covers it (Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right).

[F7]

With $f : X \to Y$ proper, meaning $f^{- 1} (K)$ is compact for every compact $K \subseteq Y$ , a nonconstant holomorphic map of Riemann surfaces is onto, has finite fibres, and has a degree $d$ (Degree of a proper holomorphic map of Riemann surfaces).

Counterexample

technique · direct

1.1F1F2F3F4

( $\exp$ is a holomorphic local biholomorphism onto $C^{\times}$ .) Being entire, $\exp$ is holomorphic as a map of Riemann surfaces in the identity atlases of [F1]; by [F2] its derivative is nowhere zero, so [F4] makes it a biholomorphism between a neighbourhood of each $z$ and a neighbourhood of $\exp z$ , and by [F3] it maps $C$ onto $C^{\times}$ .

1.2F6

({1} is compact.) The singleton ${1} \subseteq C^{\times}$ is a one-point space, so by [F6] every open cover of it has a one-member subcover; hence ${1}$ is compact.

1.3F5

(The fibre of $1$ is infinite.) By [F5], $\exp^{- 1} ({1}) = \ker (\exp) = 2 π i Z = {2 π i k : k \in Z}$ ; the assignment $k \mapsto 2 π i k$ is injective on $Z$ and $Z$ is infinite, so the fibre is infinite.

2.1F5F6step 1.3

(The fibre of $1$ is not compact.) Suppose $2 π i Z$ were compact. For each $k \in Z$ let $U_{k}$ be the open disc of radius $2 π$ about $2 π i k$ ; the sets $U_{k} \cap 2 π i Z$ form an open cover of the subspace $2 π i Z$ in the sense of [F6]. By compactness finitely many of them cover $2 π i Z$ , say for $k$ in a finite set $F$ . But distinct points $2 π i m, 2 π i k$ of the fibre satisfy $∣ 2 π i m - 2 π i k ∣ = 2 π ∣ m - k ∣ \geq 2 π$ , so the disc $U_{k}$ contains exactly one point of the fibre, namely $2 π i k$ ; the finitely many discs with $k \in F$ therefore cover at most the finitely many points $2 π i k$ , $k \in F$ , contradicting the infinitude of the fibre from step 1.3. Hence $\exp^{- 1} ({1})$ is not compact.

3.1F3F5F7step 1.2step 2.1∎

( $\exp$ is not proper, and the degree theorem does not apply.) Since ${1}$ is compact by step 1.2 while $\exp^{- 1} ({1})$ is not compact by step 2.1, the exponential is not proper; consequently the hypothesis of [F7] fails, no finite degree exists, and every fibre $\exp^{- 1} (w)$ , $w \in C^{\times}$ , is infinite: by [F3] write $w = \exp z_{0}$ and by [F5] the fibre is $z_{0} + 2 π i Z$ . Thus a holomorphic local biholomorphism of Riemann surfaces need not be proper and need not have finite fibres, so the properness hypothesis in Degree of a proper holomorphic map of Riemann surfaces cannot be dropped.

Remarks

The failure is exactly the failure of finiteness of fibres: the degree theorem Degree of a proper holomorphic map of Riemann surfaces would give the fibre of $1$ finite if it applied, but properness fails because the fibre $2 π i Z$ escapes to infinity inside $C$ . On the source side the exponential is as regular as possible — entire, nowhere-vanishing derivative, local biholomorphism at every point — so the example isolates properness as the hypothesis doing the work, and it contrasts with the compact-source case $X = \hat{C}$ of Riemann–Hurwitz for the sphere power map, where the same local model $z \mapsto z^{n}$ does give a finite degree.

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Sources

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