TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24

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Open mapping theorem for holomorphic functions

Statement

Every nonconstant holomorphic function on a complex domain is an open map.

Thus, if $f : Ω \to C$ is nonconstant and holomorphic on a complex domain $Ω$ , then $f [O]$ is open in $C$ for every open subset $O \subseteq Ω$ (Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological).

Facts & Assumptions

Given: A nonconstant holomorphic function $f$ on a complex domain $Ω$ (A complex domain is a nonempty connected open subset of $C$ ) and an open subset $O \subseteq Ω$ .

[L1]

If two holomorphic functions on a complex domain agree on a set with an accumulation point in the domain, then they agree everywhere on the domain (Identity theorem for holomorphic functions).

[L2]

If $Ω$ is a complex domain, $f : Ω \to C$ is nonconstant and holomorphic, $a \in Ω$ , and $m = \deg_{a} f$ , then near $a$ there is a biholomorphic coordinate $ϕ$ with $ϕ (a) = 0$ and $f (z) - f (a) = ϕ (z)^{m}$ (Local normal form of a nonconstant holomorphic map).

[L3]

If $Ω$ is a complex domain, $f : Ω \to C$ is nonconstant and holomorphic, $a \in Ω$ , and $m = \deg_{a} f$ , then every neighbourhood $N$ of $a$ contains an open neighbourhood $V$ for which some $ρ > 0$ gives exactly $m$ preimages in $V$ for $0 < ∣ w - f (a) ∣ < ρ^{m}$ , while $f (a)$ has only the preimage $a$ , counted with multiplicity $m$ (A local degree-m holomorphic map has m nearby sheets).

Proof

technique · direct

1.1

The function $f$ is not constant on any neighbourhood of any $a \in Ω$ : if it were constant on one, [L1] would make it constant on the connected domain $Ω$ .

L1given

2.1

Fix $a \in O$ . Shrink the neighbourhood in [L2] so that it lies in $O$ . The local multiplicity conclusion [L3], including its central value, gives a disc about $f (a)$ contained in the image of that neighbourhood and therefore in $f [O]$ .

step 1.1L2L3

3.1

Every point of $f [O]$ is therefore interior. Hence $f [O]$ is open; when $O = \emptyset$ , its image is empty and the same conclusion holds.

step 2.1∎

Depends on

Used by

Dependency tree · two levels

23 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Guide to Cultivating Complex Analysis, Theorem 5.5.1 (standard reference, not scraped)
B. V. Shabat, Introduction to Complex Analysis, Theorem 1.8 (standard reference, not scraped)