CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passaudited 2026-08-31

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A round annulus is connected but not simply connected

Statement refuted

Every connected plane domain is simply connected.

Facts & Assumptions

Given: The annulus $Ω = {z \in C : \frac{1}{2} < ∣ z ∣ < 2}$ and the unit circle $γ (t) = e^{i t}$ on $[0, 2 π]$ .

[L1]

In the grand theorem, connected spherical complement is equivalent to simple connectivity (For a plane domain, the complement, homology, primitive, logarithm, conjugate, conformal, homotopy, and contractibility conditions are equivalent).

[L2]

The circle $γ (t) = e^{i t}$ has winding number $1$ about $0$ (A circle traversed $k$ times has winding number $k$ inside and $0$ outside).

Counterexample

technique · direct

1.1

The annulus $Ω$ is open and connected, and the unit circle $γ$ lies in $Ω$ . The spherical complement of $Ω$ has two pieces: the closed inner disc ${∣ z ∣ \leq \frac{1}{2}}$ and the exterior region ${∣ z ∣ \geq 2} \cup {\infty}$ . So it is disconnected.

given

2.1

By [L1], step 1.1 already shows that $Ω$ is not simply connected. Fact [L2] gives the familiar loop witness: the unit circle still winds once around the omitted origin.

step 1.1L1L2

3.1

Therefore connectedness of the domain does not force simple connectivity.

step 1.1step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

42 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

L. V. Ahlfors, Complex Analysis, 3rd ed., Ch. 4, §4.3 (standard reference, not scraped)