58.1. PERIODIC FUNCTIONS 1849

Proof: Let a ∈V and choose r0 small enough that f (B(a,r0)) contains neither 0 nor 1.You need only let B(a,r0)⊆V . Now there exists a unique point in Q,τ0 such that λ (τ0) =f (a). By Corollary 58.1.24, λ

′ (τ0) ̸= 0 and so by the open mapping theorem, Theorem52.6.3 on Page 1665, There exists B(τ0,R0) ⊆ P+ such that λ is one to one on B(τ0,R0)and has a continuous inverse. Then picking r0 still smaller, it can be assumed f (B(a,r0))⊆λ (B(τ0,R0)). Thus there exists a local inverse for λ , λ

−10 defined on f (B(a,r0)) having

values in B(τ0,R0)∩ λ−1 ( f (B(a,r0))). Then defining g0 ≡ λ

−10 ◦ f , (g0,B(a,r0)) is a

function element. I need to show this can be continued along every curve starting at a insuch a way that each function in each function element has values in P+.

Let γ : [α,β ]→ V be a continuous curve starting at a,(γ (α) = a) and suppose that ift < T there exists a nonnegative integer m and a function element (gm,B(γ (t) ,rm)) whichis an analytic continuation of (g0,B(a,r0)) along γ where gm (γ (t))∈ P+ and each functionin every function element for j ≤ m has values in P+. Thus for some small T > 0 this hasbeen achieved.

Then consider f (γ (T )) ∈ C\{0,1} . As in the first part of the argument, there existsa unique τT ∈ Q such that λ (τT ) = f (γ (T )) and for r small enough there is an analyticlocal inverse, λ

−1T between f (B(γ (T ) ,r)) and λ

−1 ( f (B(γ (T ) ,r)))∩B(τT ,RT )⊆ P+ forsome RT > 0. By the assumption that the analytic continuation can be carried out for t < T,there exists {t0, · · · , tm = t} and function elements (g j,B(γ (t j) ,r j)) , j = 0, · · · ,m as justdescribed with g j (γ (t j)) ∈ P+,λ ◦ g j = f on B(γ (t j) ,r j) such that for t ∈ [tm,T ] ,γ (t) ∈B(γ (T ) ,r). Let

I = B(γ (tm) ,rm)∩B(γ (T ) ,r) .

Then since λ−1T is a local inverse, it follows for all z ∈ I

λ (gm (z)) = f (z) = λ

(λ−1T ◦ f (z)

)Pick z0 ∈ I . Then by Lemma 58.1.18 on Page 1838 there exists a unimodular mapping ofthe form

φ (z) =az+bcz+d

where (a bc d

)∼(

1 00 1

)mod2

such thatgm (z0) = φ

(λ−1T ◦ f (z0)

).

Since both gm (z0) and φ

(λ−1T ◦ f (z0)

)are in the upper half plane, it follows ad− cb = 1

and φ maps the upper half plane to the upper half plane. Note the pole of φ is real andall the sets being considered are contained in the upper half plane so φ is analytic where itneeds to be.

Claim: For all z ∈ I,gm (z) = φ ◦λ

−1T ◦ f (z) . (58.1.40)