Note: All arcs and curves are assumed to be smooth in this entry.

Let f be a complex function defined in a disk D about a point z₀ ∈ ℂ. In this entry, we shall show how to construct a Riemann surface such that f may be analytically continued to a function on this surface by considering paths in the complex plane.

Let 𝒫 denote the class of paths on the complex plane having z₀ as an endpoint along which f may be analytically continued. We may define an equivalence relation ∼ on this set — C₁ ∼ C₂ if C₁ and C₂ have the same endpoint and there exists a one-parameter family of paths along which f can be analytically continued which includes C₁ and C₂.

Define 𝒮 as the quotient of 𝒫 modulo ∼. It is possible to extend f to a function on 𝒮. If C ∈𝒫, let f(C) be the value of the analytic continuation of f at the endpoint of C (not z₀, of course, but the other endpoint). By the monodromy theorem, if C₁ ∼ C₂, then f(C₁) = f(C₂). Hence, f is well defined on the quotient 𝒮.

Also, note that there is a natural projection map π: 𝒮→ ℂ. If C is an equivalence class of paths in 𝒮, define π(C) to be the common endpoint of those paths (not z₀, of course, but the other endpoint).

Next, we shall define a class of subsets of 𝒮. If f can be analytically continued from along a path C from z₀ to z₁ then there must exist an open disk D′ centered about z₁ in which the continuation of f is analytic. Given any z ∈ D′, let C(z) be the concatenation of the path C from z₀ to z₁ and the straight line segment from z₁ to z (which lies inside D′). Let N(C,D′) ⊂𝒮 be the set of all such paths.

We will define a topology of 𝒮 by taking all these sets N(C,D′) as a basis. For this to be legitimate, it must be the case that, if C₃ lies in the intersection of two such sets, N(C₁,D₁) and N(C₂,D₂) there exists a basis element N(C₃,D₃) contained in the intersection of N(C₁,D₁) and N(C₂,D₂). Since the endpoint of C₃ lies in the intersection of D₁ and D₂, there must exist a disk D₃ centered about this point which lies in the intersection of D₁ and D₂. It is easy to see that N(C₃,D₃) ⊂ N(C₁,D₁) ∩ N(C₂,D₂).

Note that this topology has the Hausdorff property. Suppose that C₁ and C₂ are distinct elements of 𝒮. On the one hand, if π(C₁)≠π(C₂), then one can find disjoint open disks D₁ and D₂ centered about C₁ and C₂. Then N(C₁,D₁) ∩ N(C₂,D₂) = ∅ because π(N(C₁,D₁)) ∩π(N(C₂,D₂)) = D₁ ∩D₂ = ∅. On the other hand, if π(C₁) = π(C₂), then let D₃ be the smaller of the disks D₁ and D₂. Then N(C₁,D₃) ∩ N(C₂,D₃) = ∅.

To complete the proof that 𝒮 is a Riemann surface, we must exhibit coordinate neighborhoods and homomorphisms. As coordinate neighborhoods, we shall take the neighborhoods N(C,D) introduced above and as homomorphisms we shall take the restrictions of π to these neighborhoods. By the way that these neighborhoods have been defined, every element of 𝒮 lies in at least one such neighborhood. When the domains of two of these homomorphisms overlap, the composition of one homomorphism with the inverse of the restriction of the other homomorphism to the overlap region is simply the identity map in the overlap region, which is analytic. Hence, 𝒮 is a Riemann surface.