Uniformization of two-dimensional metric surfaces

We establish uniformization results for metric spaces that are homeomorphic to the Euclidean plane or sphere and have locally finite Hausdorff 2-measure. Applying the geometric definition of quasiconformality, we give a necessary and sufficient condition for such spaces to be QC equivalent to the Euclidean plane, disk, or sphere. Moreover, we show that if such a QC parametrization exists, then the dilatation can be bounded by 2. As an application, we show that the Euclidean upper bound for measures of balls is a sufficient condition for the existence of a 2-QC parametrization. This result gives a new approach to the Bonk–Kleiner theorem on parametrizations of Ahlfors 2-regular spheres by quasisymmetric maps.


Background
One of the main problems in Analysis in Metric Spaces is to find conditions under which a metric space can be mapped to a Euclidean space by homeomorphisms with good geometric and analytic properties. In particular, non-smooth versions of the classical uniformization theorem have found applications in several different areas of mathematics. This problem is very difficult in general, and many basic questions remain open.
Without the presence of smoothness, the parametrizations one looks for are usually quasiconformal (QC) or quasisymmetric (QS) homeomorphisms, which distort shapes in a controlled manner (see Sects. 1.2 and 16 for definitions), or bi-Lipschitz homeomorphisms, which also distort distances in a controlled manner. We will here concentrate on QC and QS maps. Concerning the existence of bi-Lipschitz parametrizations, we only briefly note that interesting sufficient conditions and counterexamples have been found both in the 2-dimensional [13,22,38,44,52,53] and higher-dimensional cases [4,28,30,32,50].
Uniformization problems concerning QC and QS maps have received considerable attention in recent years, and they have found significant applications in geometry, complex dynamics, geometric topology and geometric measure theory, among other areas. In particular, several problems in the theory of hyperbolic groups can be interpreted as uniformization problems concerning boundaries of the groups in question, cf. [9,10,12,14,26,37].
The theory of QC and QS maps f : Y → R n can be roughly divided into two parts depending on the metric space Y . If Y "has dimension n", meaning that Y shares some metric or geometric properties with R n and in particular is not a fractal, then analytic methods can be used to study such maps f . On the other hand, if Y has fractal-like behavior, then one mainly has to rely on weaker methods. Also, the infinitesimal or analytic definitions of quasiconformality do not give a good theory in this case, and one has to concentrate on QS maps. See [54] and [27] for the basic properties of QS maps in metric spaces, and [15,18,41,42] for results on QS parametrizations of fractal spaces.
In this paper we consider the first part described above. A theory of QC maps between metric spaces equipped with the Hausdorff Q-measure has been established by Heinonen and Koskela [29], based on two assumptions: Ahlfors regularity and the Loewner condition. The first assumption requires balls B(x, r ) in the space to have mass comparable r Q . The second assumption is a certain estimate concerning the Q-modulus of path families (see Sect. 1.2) resembling change of variables by polar coordinates. These assumptions lead to strong results and can often be verified assuming purely geometric conditions on the space. Heinonen and Koskela showed that the Loewner condition is equivalent to a suitable Poincaré inequality. They also proved that the infinitesimal QC condition and the QS condition are equivalent under the above assumptions, at least locally.
We now come back to the uniformization problem. After previous results by Semmes [48] and David and Semmes [17], Bonk and Kleiner [11] gave a satisfactory answer in the case Y Ahlfors 2-regular and homeomorphic to S 2 . Namely, they proved that under these assumptions Y is QS equivalent to S 2 if and only if Y is linearly locally contractible. This is a geometric condition which in particular rules out cusp-like behavior, see Sect. 16. This result has been extended in several consequent works cf. [12,40,59,60]. There the Ahlfors regularity condition is combined with varying geometric conditions on the space Y .
In higher dimensions, the uniformization problem does not have a satisfactory answer even for Ahlfors regular spaces. Examples by Semmes [49] show that the result of Bonk and Kleiner mentioned above does not generalize to dimension 3. Heinonen and Wu [33] and Pankka and Wu [45] gave further examples of geometrically nice spaces without QS parametrizations.
In this paper we take a slightly different approach to the uniformization problem in dimension two. We would like to find minimal hypotheses under which a result resembling the classical uniformization theorem as much as possible could be proved. This means giving up the geometric conditions such as Ahlfors regularity and linear local contractibility, and instead of QS maps seek for parametrizations by conformal or QC maps which do not in general have good global properties.
There are two main reasons for using such an approach. First, while the geometric conditions are good tools to work with, assuming them and Ahlfors regularity in particular is too restrictive in many situations. Secondly, if one knows the existence of QC parametrizations in general spaces, then one can try to upgrade their properties using QC invariants together with whatever conditions the underlying spaces satisfy.
We consider metric spaces X homeomorphic to R 2 . Also, we work with the Hausdorff 2-measure and assume that it is locally finite on X . This is natural since the Hausdorff measure is related to the metric in X , but also to plane topology via coarea estimates and separation properties. This guarantees that QC maps in X are closely related to the metric and topology. Under these minimal assumptions, we define conformal and QC maps f : X → R 2 using the geometric definition. This is a standard definition of quasiconformality involving conformal modulus of path families, see Sect. 1.2. QC maps between general metric spaces are often defined using the metric definition, see Sect. 16. The advantage of the geometric definition is that it automatically gives a QC invariant that can be used to prove estimates in the presence of geometric or other conditions. This is not true with the metric definition which in general implies very few properties by itself. In the case of Ahlfors regular Loewner spaces the two definitions of quasiconformality coincide ( [29,55]). See [6] and [61] for more general results concerning the equivalence between different definitions.
The uniformization problem now asks for conditions on X under which there exists a QC map f : X → R 2 . It follows from the results mentioned above that such a map exists if X is Ahlfors 2-regular and linearly locally contractible. However, as discussed above, it is of great interest to consider more general spaces that do not satisfy such strong conditions. One could hope that a QC map always exists under the minimal assumptions that X be homeomorphic to R 2 with locally finite Hausdorff 2-measure. This is not true, however, as shown in Example 2.1.
Our main result, Theorem 1.4, gives a necessary and sufficient condition called reciprocality: whenever Q ⊂ X is a topological square, let M 1 be the modulus of all paths joining two opposite sides in Q, and M 2 the modulus of all paths joining the other two sides in Q. Then we require that M 1 · M 2 is bounded from above by κ and below by κ −1 , with constant κ depending only on X . We also assume that the modulus of a point is always zero, in a suitable sense.
A basic exercise in classical QC theory shows that planar rectangles satisfy the reciprocality condition with constant 1. Applying the Riemann mapping theorem, or arguing directly, one sees that this holds for all Jordan domains in the plane. Then it is easy to deduce that reciprocality is necessary for the existence of a QC map f : X → R 2 . Theorem 1.4 shows that it is also sufficient. Reciprocality of a general space X implies that X cannot be too "squeezed" and concentrated too much around a small set of zero Hausdorff measure, cf. Example 2.1.
Methods applying reciprocality in connection with quasiconformality have previously been used in Euclidean spaces and also in more general situations, cf. [16,47], although they usually do not appear explicitly. Indeed, reciprocality is connected to the fact that conjugate functions can be defined for harmonic functions. Also related is the fact that capacities are dual to the moduli of separating hypersurfaces, cf. [19,20,23,62]. In this paper we show that the reciprocality condition can be isolated and applied to prove uniformization results in a very general setting.
The reciprocality condition is much weaker than Ahlfors regularity. Although it is sometimes difficult to determine whether the condition holds, it can be verified in several important cases. In Theorem 1.6, we show that reciprocality holds if the measures of balls B(x, r ) are bounded from above by a constant times r 2 , without assuming further geometric conditions on X . Consequently, such spaces admit QC parametrizations. In Theorem 1.5 we show that if a QC parametrization exists, it can be always chosen to have dilatation bounded from above by 2.
As an application of our results, we can reprove the QS uniformization theorem of Bonk and Kleiner discussed above. Indeed, it follows from the theory of Heinonen and Koskela that QC maps between Ahlfors regular, linearly locally contractible spaces are QS. Now Theorem 1.6 gives a QC map even without the connectivity condition, so under its presence the quasiconformality can be "upgraded" to quasisymmetry.

Definitions
Throughout this paper, X denotes a metric space homeomorphic to R 2 . It then follows that the Hausdorff 2-measure H 2 of every ball B ⊂ X is positive (see Remark 3.4). In this paper we always assume that H 2 (B) is also finite whenever B ⊂ X is compact. Notice that X is not assumed to be complete or proper. Definition 1.1 Let be a family of continuous paths in X . The (conformal) modulus of is where the infimum is taken over all admissible functions for , i.e., all nonnegative Borel functions ρ satisfying mod(ζ 1 , ζ 3 ; Q) · mod(ζ 2 , ζ 4 ; Q) κ, and (1) mod(ζ 1 , ζ 3 ; Q) · mod(ζ 2 , ζ 4 ; Q) κ −1 . ( If a ∈ X and X \B(a, R) = ∅, then lim r →0 mod(B(a, r ), X \B(a, R); B(a, R)) = 0.
We say that X is reciprocal if X is κ-reciprocal for some κ.
It follows from the Riemann mapping theorem, or can be proved directly, that simply connected domains in R 2 are 1-reciprocal, as well as smooth surfaces. R 2 equipped with a non-Euclidean norm is always κ-reciprocal for some κ, and 1-reciprocal if and only if the norm is induced by an inner product. We discuss further examples in the next sections.

Main results
The main result of this paper reads as follows. The constant 2 in Theorem 1.5 is not best possible. The best constant for the space X = (R 2 , || · || ∞ ) is π/2, see Example 2.2. This suggests that π/2 may also be the sharp constant in the theorem. See Sect. 14 for further discussion. It follows from Theorems 1.4 and 1.5 that if X is reciprocal then X is always 4-reciprocal and if moreover X ⊂ R N then X is 1-reciprocal. Theorems 1.4 and 1.5 can be applied to the class of spaces satisfying upper Euclidean mass bounds. Theorem 1. 6 Suppose there exists C U > 0 such that for every x ∈ X and r > 0. Then X is reciprocal.
The proofs of Theorems 1.4, 1.5 and 1.6 show that Theorem 1.6 remains true if (4) is assumed for balls inside compact subsets E of X , such that the constant C U is allowed to depend on E. Several examples of reciprocal spaces can be constructed that do not satisfy (4) even locally. Theorems 1.4, 1.5 and 1.6 hold also when X is homeomorphic to the Riemann sphere S 2 , with obvious modifications.
Bonk and Kleiner [11] gave an excellent characterization for quasispheres among the topological spheres satisfying (4 16 for the definitions of quasisymmetry and linear local contractibility. Again, the actual content of Corollary 1.7 is the existence of the required quasisymmetric map. Corollary 1.7 is quantitative: f can be chosen to be η-quasisymmetric with η depending only on C U and the linear local contractibility constant. In contrast to Theorem 1.4, it is clear that Corollary 1.7 does not hold with a universal quasisymmetry function η.

Organization of the paper
In Sect. 2 we give two examples illustrating Theorems 1.4 and 1.5. In the first example we construct a surface X that cannot be parametrized by a QC map. This is done by fixing a Cantor set of positive Lebesgue measure in R 2 , and choosing a continuous weight vanishing on the Cantor set. Taking the path metric with respect to this weight yields a non-reciprocal space X . In the second example we consider R 2 equipped with the L ∞ -norm, and find the best possible dilatation for QC maps between this space and the Euclidean plane. Theorem 1.4 is proved in Sects. 3-13. First, in Sect. 3 we apply coarea estimates to find paths with positive modulus in X . Both the results and the methods in this section are frequently applied in the following sections.
We construct the map in Theorem 1.4 in several steps. We first show the existence of a QC map in a given topological square Q ⊂ X . We start in Sect. 4 by defining the real part u of f . Applying Heinonen and Koskela's notion of upper gradients, we show that u can be defined as an energy minimizer among functions taking value 0 on a fixed edge of ∂ Q and value 1 on the opposite edge. In R 2 this would mean finding the harmonic function with minimal energy under such boundary conditions. We also prove a maximum principle for u that later allows us to develop its main properties. The results in this section hold in great generality, and at this point we do not assume any of the reciprocality conditions.
In Sect. 5 we apply the maximum principle, together with conditions (2) and (3), to prove continuity of the function u in Q. Moreover, in Sect. 6 we show that under these conditions almost every level set of u is a simple curve. This helps us define a conjugate function v for u. Indeed, our method for defining v in R 2 would simply be integrating |∇u| over the level sets of u. It turns out that a similar approach works also in our generality, although the actual definition is more involved. In Sect. 7 we carry out the construction of v and prove continuity.
Once we have constructed the map f , we need to show that it is QC. In particular, we need to establish some analytic properties for f . Section 8 is the first step in this direction. There we apply a dyadic decomposition of the image to show that f maps sets of measure zero to sets of measure zero, and that the change in area is what corresponds to |∇u| 2 in R 2 . This leads to a change of variables formula that by itself does not imply quasiconformality of f but plays a role in the proof. The first application of the formula appears in Sect. 9 where we prove that f is a homeomorphism.
To prove quasiconformality of f , we need to show the validity of the modulus inequalities mod( ) K mod( f ) and mod( f ) K mod( ). These depend on the analytic (Sobolev) regularity of f and f −1 , respectively. To prove the regularity of f , we introduce in Sect. 10 a modification of conformal modulus, called variational modulus. Although the variational modulus is not as easy to work with as the conformal modulus, it has the advantage of being exactly the dual of conformal modulus, in a suitable sense. In Sect. 11, we use this duality together with the reciprocality conditions to prove regularity of f , and, consequently, the first of the modulus inequalities. It is worth noticing that this is the only step in the proof where condition (1) is assumed.
We complete the proof of quasiconformality of f in Sect. 12, by showing regularity of the map f −1 and the second modulus inequality. To prove Theorem 1.4, we exhaust the space X with squares Q as above, and give normal family arguments to show the existence of a QC map from the whole space X as a limit of maps f constructed above; we do this in Sect. 13.
We prove Theorem 1.5 in Sect. 14. In contrast to other parts of this paper, which are mostly elementary and self-contained, here we rely on results from different areas. We apply the differentiability results of Kirchheim [36], the measurable Riemann mapping theorem, and John's theorem on convex bodies to find a QC map in X with small dilatation.
In Sect. 15 we prove Theorem 1.6 by checking that spaces satisfying (4) satisfy the reciprocality conditions. In Sect. 16 we consider quasisymmetric maps and apply Theorems 1.4 and 1.6 to prove Corollary 1.7. Finally, in Sect. 17 we briefly discuss the absolute continuity properties of QC maps in the current generality, as well as the reciprocality conditions.

Examples
We first introduce some basic notation and terminology.
where a 1 = 1 and a 2 = π/4. H 2 coincides with the Lebesgue measure | · | in R 2 . We always assume that X = (X, d) is homeomorphic to R 2 and that At the second step, we repeat the process with the unit square replaced by each of the squares Q remaining after the first step. Continuing this way, after n steps we have 4 n squares remaining, each of sidelength Taking the intersection of all the remaining squares gives the Cantor set C. We choose the sequence (a j ) such that C has positive Lebesgue measure. Now let 0 ω 1 be a continuous function in R 2 such that ω(x) = 0 if and only if x ∈ C and ω = 1 near infinity. Define d = d ω in R 2 by setting where the infimum is taken over all rectifiable paths in R 2 joining x and y. We check that d is a metric in R 2 . First, since 0 ω 1, d(x, y) is finite for every x and y. Also, the triangle inequality follows directly from the definition. It remains to show that d(x, y) > 0 if x = y.
If both x, y ∈ C, then there is some step n remaining square Q in the construction of C such that x ∈ Q but y / ∈ Q. It follows that there is a slightly larger squareQ with same center as Q, such that Since ω is continuous and positive onQ\Q, there exists > 0 such that for every path γ joining the two boundary components, γ ω ds . Consequently, We conclude that d is a metric on R 2 . Moreover, d(x, y) |x − y| for all x, y ∈ R 2 , so the identity map I : R 2 → (R 2 , d) is a homeomorphism and d is a length metric. Recalling that ω = 1 near infinity and applying the Hopf-Rinow theorem, we conclude that (R 2 , d) is a geodesic metric space.
We have shown that (R 2 , d) is a geodesic metric space homeomorphic to R 2 . The 1-Lipschitz continuity of the identity map I also shows that the Hausdorff We then integrate over t and apply Hölder's inequality to get Minimizing over ρ, we get mod( 1 ) M. Similarly, if 2 is the family of paths joining the horizontal edges, we get mod( 2 ) M. Therefore, Letting M → ∞, we conclude that (R 2 , d) is not reciprocal.
If H 2 denotes Hausdorff measure on (R 2 , || · || ∞ ) and | · | the Lebesgue 2-measure, then H 2 (A) = π |A|/4 for every Borel set A ⊂ R 2 , see [36,Lemma 6]. We now claim that the identity map f : (R 2 , || · || ∞ ) → R 2 is π/2-QC, where it is understood that the image is equipped with Euclidean norm || · ||. We have for every x ∈ R 2 , for the maximal stretching L f and volume derivative J f of f . A standard change of variables argument now shows that mod( ) whenever is a path family in (R 2 , || · || ∞ ). Indeed, if ρ is an admissible function for f , then the function L f (ρ • f ) is admissible for , and moreover Since this holds true for all admissible functions ρ, (5) follows. Similarly, we see that mod( f ) for every path family . We conclude that f is π/2-QC. We next show that there are no K -QC maps with K < π/2. Denote by φ the counterclockwise rotation of R 2 by π/4, and let Q = φ([0, 1] 2 ). We give a lower bound for the modulus mod( 1 ) in (R 2 , || · || ∞ ) of the family of paths Integrating over t and applying the co-area formula [1,Theorem 9.4] (with the co-area factor of L ∞ ), we see that Minimizing over ρ gives mod( 1 ) π/2. Similarly, if 2 is the family of paths joining the other two sides of Q, then mod( 2 ) π/2. Hence, if f : Q → R 2 is K -QC, then using the 1-reciprocality of R 2 we get i.e., K π/2.

Existence of rectifiable paths
Recall that we assume that X is homeomorphic to R 2 and has locally finite 2-measure. In this section we show that under these mild conditions one can find large families of rectifiable paths in X (see [51] for much deeper results along these lines). We will later prove qualitative estimates, such as continuity, using such families. We will frequently use the following results. These are [ We next show that the family of paths joining two continua always has positive modulus. We need the following topological lemma, cf. [ (S(x, r )) > 0 for almost every 0 < r < r 0 . Applying Proposition 3.2 with m = d(·, x), we see that H 2 (B) > 0 for every ball B ⊂ X . See [35] for further connections between topological dimension and Hausdorff measures. Proof We first assume that both α and β lie in the interior of Q, henceforth denoted by int Q. Fix points a ∈ α and b ∈ β, and a continuous path η : Then m is 1-Lipschitz. Moreover, there exists > 0 such that F t := m −1 (t) ⊂ Q and F t separates ∂ Q and |η| for every 0 < t < . Applying Proposition 3.2 to m and g = 1, we see that H 1 (F t ) is finite for almost every t. Therefore, since Q is homeomorphic to a planar square, Lemma 3.3 shows that F t contains a continuum G t which also separates. Since α and β are nontrivial continua, there exists 0 < < such that for every 0 < t < there are points a t ∈ α ∩ G t and b t ∈ β ∩ G t . Applying Proposition 3.1, we find for almost every 0 < t < a rectifiable, injective path γ t joining a t and b t in G t . Denote by the family of all such γ t . Then ⊂ (α, β; Q). Now let g : Q → [0, ∞] be admissible for . Then, applying Proposition 3.2 and Hölder's inequality, we have Since the estimate holds for all admissible functions g, we conclude that mod(α, β; Q) mod( ) π 4H 2 (Q) 1/2 If α touches the boundary of Q but β does not, then we modify the proof as follows: if α contains a point in int Q, then we can find a subcontinuum in the interior and the proof above applies. Otherwise, α contains a topological line segment I ⊂ ∂ Q. Now, we can choose the point a to be the center of I , and we can choose η : [0, 1] → int Q ∪ {a} such that the -neighborhood of |η| does not intersect ∂ Q\I when is small enough. Now the proof above applies. We proceed similarly if both α and β touch the boundary of Q.

Energy minimizer u on a topological square
In this section we define a suitable energy minimizing, "harmonic" function u in our general setting. We also develop some basic properties for u. Later, we define a "conjugate function" v of u, and show that, under our reciprocality assumption, the resulting map f = (u, v) is QC.
Let ⊂ X . Recall that a Borel function g 0 is an upper gradient of a function u in , if for every x and y ∈ and every locally rectifiable path γ joining x and y in . Here by joining we mean that both x and y ∈ |γ |. Also, we agree that the left term in (6) equals ∞ if |u(x)| = ∞ or |u(y)| = ∞. We say that g is a weak upper gradient of u, if there exists a path family 0 with modulus zero such that (6) holds for every x and y and every γ / ∈ 0 . Similarly, we say that a property holds for almost every path in a path family , if there exists 0 ⊂ of modulus zero such that the property holds for all γ ∈ \ 0 . Furthermore, we say that a Borel function ρ is weakly admissible for , if the integral of ρ over γ is at least 1 for almost every γ ∈ .
We now construct the function u. Let Q ⊂ X be homeomorphic to a closed square in R 2 , and ζ 1 , . . . , ζ 4 the boundary edges as in (1) and (2). At this point we do not assume any of the reciprocality conditions. We consider the modulus A standard method now shows that there exists a weakly admissible function realizing M 1 . More precisely, let (ρ j ) be a minimizing sequence of admissible functions. Then, after passing to a subsequence, ρ j converges to ρ ∈ L 2 (Q) weakly in L 2 . Moreover, by Mazur's lemma [34,Page 19], there exists a sequence (ρ k ) of convex combinations of the ρ j ; Now it follows by Fuglede's lemma [34, Page 131] that for almost every γ in Q. In particular, for almost every γ joining ζ 1 and ζ 3 in Q, so We would now like to define the function u by integrating the minimizing function ρ over paths. This is possible although some technicalities arise. Denote by 0 the family of paths in Q that have a subpath for which (7) does not hold. Then mod( 0 ) = 0.
We will be working with paths that do not belong to the exceptional family 0 . For instance, we show in Lemma 4.3 that the upper gradient inequality (6) holds for the function u, weak upper gradient ρ, and all paths γ outside 0 . Since ρ is integrable on such paths γ , it follows that u will be absolutely continuous there. The subpath property in the definition of 0 is given to guarantee that paths outside 0 can be concatenated succesfully.
Define u as follows: For x ∈ Q, first assume there exists such that some subpath γ x of γ joins ζ 1 and x. Then define where the infimum is taken over all possible γ and γ x . If u(x) cannot be defined this way for x ∈ Q, let where E is the set of points y for which u(y) is already defined.

Lemma 4.1 The function u
Proof We have to show that for every x ∈ Q and every > 0 there exists y ∈ B(x, ) such that u(y) is defined by (9). First notice that B(x, ) ∩ Q contains a non-trivial continuum G. Therefore, by Proposition 3.5, there exists a family of paths joining ζ 1 and G in Q, such that mod( ) > 0. Then Fuglede's lemma guarantees that for some γ a ∈ (7) holds for all subpaths of γ a . Let F be a non-trivial component of |γ a | ∩ B(x, ). Then, applying Proposition 3.5 and Fuglede's lemma again gives a path γ b joining F and ζ 3 such that (7) holds for all subpaths of γ b . Now we can define γ by concatenating a suitable subpath of γ a with γ b . Then γ joins ζ 1 and ζ 3 and |γ | intersects B(x, ). Moreover, all subpaths of γ satisfy (7). Therefore, u(y) can be defined by (9) for all y ∈ |γ |.
Proof The argument is similar to the previous lemma. Proposition 3.5 gives path families 1 and 2 of positive modulus joining ζ 1 and |γ |, and ζ 2 and |γ |, respectively. Moreover, Fuglede's lemma gives paths γ a ∈ 1 \ 0 and γ b ∈ 2 \ 0 . Now γ can be defined by concatenating γ a , γ b , and a suitable subpath of γ . That γ / ∈ 0 follows because γ a , γ b and γ all have the same property.

Lemma 4.3
The function ρ is a weak upper gradient of u in Q. In fact, (6) holds (with ρ) for all rectifiable paths γ / ∈ 0 .
Proof Let x and y ∈ Q. Since we only require the upper gradient inequality outside a set of modulus zero, we may assume that there is a rectifiable path γ / ∈ 0 joining x and y in Q. Then, by Lemma 4.2, u(x) and u(y) are defined by (9). We may assume that u(y) > u(x). Then, by the definition of u, where the infimum is taken as in (9).
We need the following auxiliary result to prove further properties for u.
Proof First, let y ∈ |η| and 0 < r < diam |η|/2. Assume that u(z) L at some z ∈ B(y, r/2). Then, by the definition of u, there exists a curve α joining B(y, r/2) and Q\B(y, 2r ) such that u L everywhere on α. Moreover, Lemma 3.3 implies that for every r/2 < s < r some continuum C(s) ⊂ S(y, s) intersects both |η| and α. That is, there are a s , b s ∈ C(s) such that . Therefore, Proposition 3.5 and Lemma 3.1 show that for almost every such s there are rectifiable curves in C(s) joining a s and b s . Furthermore, for almost every such s, the upper gradient inequality gives Integrating from r/2 to r and applying Proposition 3.2, we have Now for every δ > 0 and every y ∈ |η|\E there exists r y < δ such that (10) holds for every r < r y . By the 5r -covering lemma, among all such balls B(y, r ) we can find a finite or countable subcollection {B j = B(y j , r j )} such that the balls B j are pairwise disjoint and |η|\E ⊂ j B(y j , 5r j ).
Applying (10) in all B j and summing the estimates gives By the disjointness of the balls B j , the sum on the right can be estimated from above by where N 5δ (|η|) is the closed 5δ-neighborhood of |η|. Since η is rectifiable, this integral converges to zero when δ → 0. Combining the estimates gives the claim.
Proof Suppose to the contrary that u(x 0 ) 1+3 for some > 0 and x 0 ∈ Q. Then, by the definition of u, we find a curve α in the interior of Q such that u 1 + 2 everywhere on α. Proposition 3.5 shows that Given η : By the upper gradient inequality and the absolute continuity of u on η , 0 < t 0 < 1 and η ρ ds .
Combining with Lemma 4.4, we conclude that ρχ E / is weakly admissible by (11). On the other hand, the function is weakly admissible for (ζ 1 , ζ 3 ; Q) by the definition of u. But now (12) gives This contradicts the minimizing property of ρ. The proof is complete.
We next establish a maximum principle. Let ⊂ X be open. We denote

Lemma 4.6 Let ⊂ X be open. Then
Proof To prove the second equality, let x 0 ∈ ∩ Q and u(x 0 ) = m. Then there is x ∈ ∩ Q such that u(x) m + is defined by (9). Moreover, there exists a path γ x joining ζ 1 and x such that u m + 2 on |γ x |. But |γ x | must intersect ∂ * . The second equality follows. The proof of the first equality is similar to the proof of Lemma 4.5. Let M = sup y∈∂ * u(y) 1, and suppose there is δ > 0 such that u(x) M + 2δ for some x ∈ ∩ Q. Then, by the definition of u, we can choose a curve α in ∩ Q such that u > M + δ on α. Applying Proposition 3.5, we see that mod(α, ζ 1 ; Q) > 0. Arguing as in the proof of Lemma 4.5, we see that On the other hand, ρχ Q\E is weakly admissible for the minimizing problem, because u M on ∂ * . This contradicts the minimality of ρ.

Continuity of u
Let u be the function defined in Sect. 4. In this section we show that u is continuous, assuming conditions (2) and (3). The rest of this section is devoted to the proof of Theorem 5.1. We say that D ⊂ X is a half-annulus, if D is homeomorphic to defined in polar coordinates. The boundary of D consists of inner and outer circles, and the two flat components. (2) and (3), and fix x ∈ X and R > 0. Moreover, let r < R/2 and assume that D is a half-annulus with inner circle T r ⊂ B(x, r ), outer circle T R ⊂ X \B(x, R), and flat components I and J . Then

Lemma 5.2 Suppose X satisfies
where depends on r , R, κ and x.
Proof By condition (3), On the other hand, by (2), The lemma follows by combining the estimates.

Remark 5.3
For future reference, we note that Lemma 5.2 holds if the assumptions are replaced by assumption (4). See Sect. 15 for further details.

Lemma 5.4 Suppose X satisfies (13). Then u is continuous in
Moreover, we require that D does not intersect ζ 4 or ζ 3 . Then the boundary circle T of D satisfies T ⊂ X \B(x, R) for some R > 0. Let r < R/2, and choose another topological disk D ⊂ B(x, r ) containing x, with boundary circle T . Then the two boundary circles and ∂ Q bound a half-annulus A in Q. Denote by T R and T r the circular boundary components (the restrictions of T and T , respectively), and by I, J ⊂ ∂ Q the flat components. Moreover, But u = 0 on ζ 1 , so the first estimate above holds true without ζ 1 on the last term. Also, if ζ 1 intersects the boundary of γ , then γ intersects ζ 1 . So also the second estimate holds without ζ 1 . In other words, Since ρ is a weak upper gradient of u, it follows that δ r γ ρ ds for almost every γ ∈ (I, J ; A). Consequently, we have On the other hand, by (13) we have We conclude that δ r → 0 as r → 0, showing that u is continuous at x.
Lemma 5.5 Suppose X satisfies (13). Then u is continuous in ζ 3 and equals 1 there.
Proof Let x ∈ ζ 3 . In view of Lemma 4.5, it suffices to show that Without loss of generality, x / ∈ ζ 4 . If (14) does not hold, there exists > 0 and a sequence of points We choose a topological closed disk D such that x ∈ int D . Moreover, we require that D does not intersect ζ 4

. Then the boundary circle
By the definition of u and Lemma 3.1, there is a simple path η / ∈ 0 joining ζ 1 and D in Q such that We may assume that η does not intersect ζ 3 , since otherwise (15) violates (8). Now |η|, T , T and ζ 3 bound a half-annulus A with flat boundary components I ⊂ |η| and J ⊂ ζ 3 . We claim that when r is small enough there exists a path γ ∈ (I, J ; A)\ 0 satisfying γ ρ ds < .
Indeed, otherwise we would have But this contradicts (13) when r is small enough, so (16) holds.
Concatenating γ with a subpath of η and applying (15) and (16) now gives a path γ / ∈ 0 joining ζ 1 and ζ 3 in Q such that This contradicts (8), and so (14) holds. The proof is complete.
Continuity of u in the interior of Q is proved using the methods above. However, the proof is more technical and we need an auxiliary lemma. Lemma 5.6 Suppose X satisfies (13), and fix x ∈ int Q. Moreover, suppose there is a simple, rectifiable path γ : Mapping |γ | to a segment in R 2 if necessary, we can choose D so that |γ | separates D into two components D 1 and D 2 . Also, since γ ρ ds < ∞, we can choose D small enough such that γ ρχ D ds < .
It then follows from the definition of u and Lemma 4.2 that Let r < R/2, and choose another topological disk D ⊂ B(x, r ) with boundary circle T , such that x ∈ D . Again, we can choose D such that |γ | separates D into two components D 1 ⊂ D 1 and D 2 ⊂ D 2 . We control the oscillation of u in D 1 and D 2 separately. Since the estimates are identical, we only consider the case D 1 . Now D 1 contains a half-annulus A 1 bounded by T , T , and |γ |. The flat boundary components are I, J ⊂ |γ |. Then, if This contradicts (13). Applying the same argument to A 2 , we conclude that We conclude that u is continuous at x.
Lemma 5.7 Suppose X satisfies (13). Then u is continuous in int Q.
Proof Fix x ∈ int Q and let > 0. Choose a topological disk D ⊂ Q containing x, with boundary circle T . Moreover, let r > 0 and let D ⊂ B(x, r ) ⊂ D be another disk containing x. Denote the boundary circle of D by T r . Then, by the definition of u, there exists a rectifiable path γ / ∈ 0 joining ζ 1 and D such that for every y ∈ |γ |. Moreover, by Lemma 3.1, we find a simple path γ / ∈ 0 joining ζ 1 and D with |γ | ⊂ |γ |. By Lemma 5.6, u is continuous on |γ |. We would like to repeat the argument used in the previous lemmas, applying the maximum principle, (18) and Lemma 4.6 in the domain bounded by T , T r , and |γ |. But this domain is not a half-annulus, so Lemma 4.6 does not apply directly.
To correct this, notice that by the uniform continuity of u on |γ | and (18) there is a neighborhood V of |γ | ∩ D such that for all y ∈ V . We choose simple paths I and J in V connecting T and T r such that |γ | separates I and J in V . Now I , J , T and T r bound a half-annulus A, with flat boundary components I and J . As before, the maximum principle and (19) for every η joining I and J in A. Applying (20) to all such paths, together with the weak upper gradient property of ρ, gives This contradicts (13) when r is small enough. We conclude that u is continuous in x.

Level sets of u
In this section we examine the properties of the level sets of u, and in particular show that almost every level set is a rectifiable curve. This helps us define the conjugate function v in the next section.
Proposition 6.1 Suppose that X satisfies (2) and the minimizer u satisfies the conclusions of Theorem 5.1. Then for H 1 -almost every t the level set u −1 (t) is a simple rectifiable curve |γ t | joining ζ 2 and ζ 4 .
We will later show that u is the real part of a homeomorphism, so in particular u −1 (t) is a simple curve for all 0 < t < 1. The rest of this section is devoted to the proof of Proposition 6.1.
Recall our notation Proof First, we have since (t − s) −1 ρ is weakly admissible. The reverse inequality also holds, since if there was an admissible function g such that would be weakly admissible for (ζ 1 , ζ 3 ; Q) (because u = 0 in ζ 1 and u = 1 in ζ 3 ), and This contradicts the minimizing property of ρ. Therefore, the first equality in (21) holds. To prove the second equality, we denote Let δ > 0, and Then ρ δ is weakly admissible for (ζ 1 , ζ 3 ; Q), and If I s,t < (t − s)M 1 , then the right term is strictly smaller than M 1 when δ > 0 is small enough. This contradicts the minimizing property of ρ. Similarly, if I s,t > (t − s)M 1 , we get a contradiction by the above argument, replacing A s,t with Q\A s,t . Lemma 6.3 Suppose 0 < s < t < 1, and that u satisfies the conclusions of Theorem 5.1. Then A s,t and u −1 (t) are connected and simply connected sets connecting ζ 2 and ζ 4 in Q. Moreover, the sets This violates the maximum principle, Lemma 4.6, and so A s,t must be simply connected. The same argument shows that u −1 (t) is simply connected. Next, suppose W is a connected component of A s,t . We claim that W has to intersect both ζ 2 and ζ 4 . Notice that by the maximum principle, W has to intersect either ζ 2 or ζ 4 . We lose no generality by assuming that W intersects ζ 2 . To show that W also intersects ζ 4 , suppose to the contrary that this was not the case.
Then, by Lemma 3.3, there is a continuum Y ⊂ ∂ * W separating W and ζ 4 , where ∂ * is as in Lemma 4.6. Now, if s < u(x) < t at some point x ∈ Y , there is a neighborhood B of x such that s < u < t everywhere on B. This contradicts the definition of W . Therefore, u only takes values s and t on Y . But Y is connected, so u is constant on Y . On the other hand, Y and ζ 2 bound a domain in Q that includes W , and the maximum principle implies that u equals either t or s everywhere in this domain This is a contradiction, since W ⊂ A s,t belongs to this domain. We conclude that W intersects ζ 4 . Now let V 1 and V 2 be disjoint connected components of A s,t . Then, since both separate ζ 1 and ζ 3 , there exists x ∈ Q\A s,t such that A s,t separates x from both ζ 1 and ζ 3 . This contradicts the maximum principle, Lemma 4.6. We conclude that A s,t is connected. To show that u −1 (t) is connected and connects ζ 2 and ζ 4 , it suffices to notice that the same holds for A t−1/j,t+1/j and express u −1 (t) as the intersection.
The remaining claims can be proved by applying the maximum principle as in the previous paragraphs. We leave the details to the reader.
To prove Proposition 6.1, we recall the compactness property of a family of paths with bounded length, and lower semicontinuity of path length under uniform convergence. The first property follows from the Arzela-Ascoli theorem, while the second property is a simple consequence of the definition of path length.
Then the paths γ j can be reparametrized so that the sequence of the reparametrized paths has a subsequence converging uniformly to a rectifiable path γ : The following differentiation result will be frequently applied, see [21, Theorem 3.22] for the proof. Suppose A ⊂ Q is a Borel set. Moreover, suppose φ : A → [0, ∞] is Borel measurable and integrable, and ψ : A → R Borel measurable. Define whenever t ∈ (a j , b j ) and |b j − a j | → 0. Lemma 6.5 Suppose ϕ is defined as above. Then the differential ϕ (t) < ∞ exists for almost every t ∈ R and defines a measurable function such that B ϕ (t) dt ϕ(B) (22) for all Borel sets B ⊂ R. If moreover ϕ(B) = 0 whenever H 1 (B) = 0, then equality holds in (22).
Towards the proof of Proposition 6.1, we first show that almost every level set of u has finite 1-measure and contains a rectifiable path as in the statement of the proposition. Proof We apply Lemma 6.5 with ψ = u, φ = 1, and choose 0 < t < 1 such that ϕ (t) exists. Then H 2 (u −1 (t)) = 0, so u −1 (t) does not have interior points.
Let h > 0 such that [t − h, t + h] ⊂ (0, 1). By Lemma 6.3, A t−h,t−h/2 and A t−h/4,t−h/8 contain simple paths α and α , respectively, both joining ζ 2 and ζ 4 . Let D h be the Jordan domain bounded by α, α , β β , where β is a subpath of ζ 2 and β is a subpath of ζ 4 . Then Hence, by Lemma 6.2, Here we use notation M s,t introduced before Lemma 6.2. Combining with (2), we have On the other hand, by (23), where h is the length of a shortest path γ h joining β and β in D h . Notice that γ h is simple, since otherwise we could find a shorter path inside |γ h | with the same property.
Combining the estimates, we have We take a sequence h j → 0. Then, by (24) and our choice of t, Hence, by Lemma 6.4, there is a subsequence of the simple paths (γ h j ) converging uniformly to a rectifiable pathγ t,− . Moreover, by Lemma 3.1, |γ t,− | contains a simple rectifiable path γ t joining ζ 2 and ζ 4 in u −1 (t). This proves the second claim in the lemma. We found the pathγ t,− as a limit of paths converging "from left". With the same argument, replacing t −h/q by t +h/q everywhere, we get a sequence of simple rectifiable paths converging uniformly to a rectifiable pathγ t,+ . Thus γ t,+ is a limit of paths converging "from right". Notice that both |γ t,− | and |γ t,+ | are subsets of u −1 (t). The first claim in the lemma follows if we can show that Let (γ − k ) and (γ + k ) be the sequences of simple paths constructed above, such that γ − k →γ t,− and γ + k →γ t,+ uniformly as k → ∞, and let k be the domain bounded by |γ − k |, |γ + k |, ζ 2 and ζ 4 . Then u −1 (t) = ∩ k k . Since u −1 (t) does not have interior points, it follows that for every such that x + k ∈ |γ + k | for every k and x + k → x, or both. Thus, by the uniform convergence of the paths γ − k and γ + k , x ∈ |γ t,− | or x ∈ |γ t,+ |. We conclude that (25) holds. The proof is complete.
Proof of Proposition 6.1 Again, we apply Lemma 6.5 with ψ = u, φ = 1, and choose 0 < t < 1 such that ϕ (t) exists and the claims of Lemma 6.6 hold. So u −1 (t) contains a simple rectifiable path γ t . We need to show that u −1 (t) does not contain points outside |γ t |.
To prove this, it is convenient to use Euclidean coordinates. In other words, we now think of d as a metric in R 2 . Then we may assume that Q = [−1, 1] 2 and moreover that |γ t | = {0} × [−1, 1]. Suppose there is a point a ∈ u −1 (t)\|γ t |. Then we may assume that a ∈ int Q removing, if necessary, at most countably many values of t for which u −1 (t) contains a non-trivial subcontinuum of ∂ Q.
Next let I be the line segment {−1/4} × [−1, 1]. Then, when h is small enough and u −1 (t − h) contains a simple path γ t−h , this path together with I bounds a simply connected domain U , a ∈ ∂U , whose boundary is the union of a subcurve J 1 of |γ t−h |, the subsegment J 0 = [−1/2, −1/4] × {0} of |η|, and two subsegments J 2 and J 3 of I . Now we slightly modify U in order to have a Jordan domain V ⊂ U to which condition (2) can be applied. Since u is continuous and u = t on J 0 , we can choose a simple path η in U , depending on h, close to J 0 as follows: u t − h/2 on J 0 = |η | and J 0 , J 1 , J 2 and J 3 bound a Jordan domain V such that a ∈ ∂ V . What is important to us is that J 0 can be chosen so that there exists a constant c > 0 not depending on h such that whenever γ is a path connecting J 2 and J 3 in V , then (γ ) c. Also, since V ⊂ A t−h,t and for h small enough, we have where A does not depend on h. Applying (2) and (26) shows that On the other hand, since u = t − h on J 1 and u t − h/2 on J 0 , the function 2h −1 ρ is weakly admissible for (J 0 , J 1 ; V ). Thus, by (27) we have Notice that there is > 0 not depending on h such that dist(I, |γ t |) . We claim that the function is weakly admissible for (ζ 1 , ζ 3 ; Q). To see this, let γ : [0, 1] → Q be a rectifiable path, γ / ∈ 0 , such that γ (0) ∈ ζ 1 and γ (1) ∈ ζ 3 . Denote by 0 < T < 1 the largest number such that u(γ (T )) = t − h.
On the other hand, if γ (S) ∈ V for some S > T , then since a subpath of γ joins I and γ t in A t−h,t . We conclude that ρ is indeed weakly admissible. Now, by (28) and Hölder's inequality, We conclude that when h is small enough, This contradicts the minimizing property of ρ. The proof is complete.

Conjugate function v
In this section we construct a "conjugate function" v for our minimizing function u and prove continuity. Then f = (u, v) is the desired QC map in Q if X is reciprocal; this will be shown in the next sections. We note that the conjugate function is easier to find if X is 1-reciprocal. Indeed, if we construct v as u but replacing ζ 1 and ζ 3 with ζ 2 and ζ 4 , respectively, then f = (u, v) : int Q → (0, 1) 2 is a conformal homeomorphism. In the general case of κ-reciprocal X we have to work more to find v. The idea behind the construction is that v should be defined integrating the minimizer ρ over the level sets of u in a suitable way.

Recall the notation
Lemma 7.1 Suppose 0 < t < 1, and that the minimizer u satisfies the conclusions of Theorem 5.1. Then for every > 0 there exists h > 0 such that Proof If the claim is not true, then for some 0 < t < 1 and > 0, is non-empty for all h. But the sets F h are nested and compact, and by the continuity of u. This is a contradiction since the intersection of the sets F h cannot be empty.
We denote Recall that, by Proposition 6.1, the set F has full 1-measure in (0, 1). Now let x = γ t (T ) ∈ U and denote when x ∈ U \U . That 0 v M 1 follows from Lemmas 6.2 and 7.1. Also, notice that ζ 1 ∪ ζ 3 ⊂ U .
The following proposition allows us to extend v to all of Q. Recall the notation ∂ * from Lemma 4.6. (2) and (3). Let V be a connected component of Q\U . Then v is constant on ∂ * V .

Proposition 7.2 Suppose X satisfies
Proof Let V be a connected component of Q\U . We will argue by contradiction, assuming that v is not constant on ∂ * V . First, notice that there exists 0 < t 0 < 1 such that V ⊂ u −1 (t 0 ), by Proposition 6.1. Therefore, Lemma 6.2 implies that ρ(x) = 0 for almost every Then, by the definition of v we can find a radius r > 0 and points x ∈ B(a, r ) Then we furthermore find 0 < < r and h > 0 such that We denote Then combining (30) and (32) gives Now choose points p ∈ B(a, r ) ∩ V , q ∈ B(b, r ) ∩ V , and a simple path η joining p and q in V . Moreover, choose δ > 0 small enough such that N δ (|η|) ⊂ V . By condition (3) we can choose r > 0 small enough to begin with so that there are Borel functions g 1 and g 2 such that g 1 is admissible for (ζ 1 , B(a, 2r ); Q), g 2 is admissible for (ζ 1 , B(b, 2r ); Q), and We now define a function g by setting This definition of g is motivated by the fact that g is weakly admissible for (ζ 1 , ζ 3 ; Q). Indeed, let γ ∈ (ζ 1 , ζ 3 ; Q)\ 0 . If |γ | intersects B(a, 2r ) then γ g 1 ds 1. Similarly, if |γ | intersects B(b, 2r ) then γ g 2 ds 1. If |γ | intersects |η|, then γ δ −1 χ V ds 1. Otherwise γ either passes through A t−h,t or A s−h,s , in which case γ ρχ / h ds 1. Now, for 0 < m < 1, also the function is weakly admissible for (ζ 1 , ζ 3 ; Q). By the minimizing property of ρ, we have Differentiating in (35) with respect to m and setting m = 0 then gives To conclude the proof, we show that this is a contradiction. Recall that ρ(x) = 0 for almost every x ∈ V . In particular, Also, Hölder's inequality, (34) and the minimizing property of ρ give Therefore, combining the definition of g with (33), (37) and (38) gives This contradicts (36). The proof is complete. By Proposition 7.2, we can extend v to all of Q: if V is a connected component of Q\U and x ∈ V , then v(x) = v(y), where y is any point on ∂ * V . Fix a ∈ U . Moreover, let μ, r > 0, x, y ∈ B(a, r ) ∩ U , and suppose v(x) v(y) − 7μ. Using the notation of Proposition 7.2 for v(x) and v(y), we then conclude that (33) holds. This time we choose r > 0 small enough and a Borel function g 1 which is admissible for (ζ 1 , B(a, r ); Q), such that (34) holds with g 2 removed from the estimate. Then we define g = h −1 ρχ + g 1 , and conclude as above that g is weakly admissible for (ζ 1 , ζ 3 ; Q). This then leads to a contradiction precisely as in the proof of Proposition 7.2. So we conclude that, when r > 0 is small enough, v(y) v(x) + 7μ. Interchanging the roles of x and y, and recalling the definition of v, we then have We conclude that v|U is continuous at a, and furthermore that v is continuous at every b ∈ Q by the discussion above. The claims of the proposition now follow directly from the definition and continuity of v, Lemma 6.2, and Proposition 7.2.
We orient X so that winding around ∂ Q starting from ζ 1 and ending at ζ 4 defines positive orientation.

Change of variables with f = (u, v)
In order to prove quasiconformality, we need to establish analytic properties for f . In this section we prove a change of variables formula by employ-ing decompositions of the rectangle [0, 1] × [0, M 1 ] and the corresponding preimages.
We decompose the interval [0, 1] × [0, M 1 ] as follows. We first choose k 0 ∈ Z and 2 −1 < m 1 1 such that Let k ∈ N, k −k 0 + 2, and consider the rectangles where 0 i 2 k − 1, 0 j 2 k+k 0 − 1. Then two rectangles either coincide or have disjoint interiors, and the union of all the rectangles covers all of [0, 1] × [0, M 1 ]. We denote and byQ(i, j, k) the interior of Q(i, j, k). Also, when (i, j, k) is fixed, and 0 s < t 1, we use the notation A s,t (i, j, k) = A s,t ∩Q(i, j, k).

Lemma 8.1 Suppose X satisfies (2) and (3). Then we have
for every (i, j, k) as above.
Proof Fix k −k 0 + 2. We claim that for every (i, j). Suppose to the contrary that (42) does not apply for some (i, j). Setting we get a set function for which Lemma 6.5 can be applied. Since (42) does not hold, Lemma 6.5 shows that there exists a set G ⊂ (i2 −k , (i + 1)2 −k ) of positive 1-measure such that for every t ∈ G In particular, for some t ∈ G the level set u −1 (t) is a simple curve γ t . By the definition of Q(i, j, k), we find a = γ t (T ) and contradicting (43) (the detailed proof of the last inequality involves the argument used in the proof of Proposition 7.2 and is left to the reader). We conclude that (42) holds. Now we can apply (39) and (42) to all (i, j) to estimate This gives (41). The proof is complete.
Applying Lemma 8.1 gives the desired change of variables formula. (2) and (

Proposition 8.2 Suppose X satisfies
Proof By Monotone Convergence, we may assume that g is bounded. Let first g k be of the form where a j 0 and R j is a rectangle of the form (40) for every j, such that the rectangles have disjoint interiors. Then (41) gives In general, the bounded Borel function g can be approximated in

Now it suffices to show that
The proposition follows from (45), the definition of E, and Dominated Convergence.
To prove (45), notice that | f (Q\E)| = 0. Hence, applying (41) again shows that, given > 0, the set f (Q\E) can be covered by rectangles R of the form (44) So (45) follows. The proof is complete.

Invertibility of f
In this section we show that the map f : Q → [0, 1] × [0, M 1 ] is a homeomorphism. In particular, v is then defined by (29) in every x ∈ Q.
The rest of this section is devoted to the proof of Proposition 9.1. We first show that f −1 (z) does not contain non-trivial continua for any z Suppose F ⊂ f −1 (z) is a non-trivial continuum. There exists a non-trivial We first give an estimate resembling a lower modulus bound. Let τ : [0, 1] → Q be a simple path such that τ (T ) ∈ E if and only if T = 0 and τ (T ) ∈ F if and only if T = 1. If there exists a ∈ F ∩ int Q, then we choose τ so that τ (1) = a. Otherwise F is a simple curve in ∂ Q, and we choose τ so that τ (1) is not a boundary point of F in ∂ Q.

Lemma 9.2 Suppose X satisfies
Proof We may assume that there exists L ∈ N such that R = 2 L r . Let j ∈ N, j L. Now denote Here Then Therefore, either If (47) occurs, then, since β s / ∈ 0 , (6) is satisfied with γ j . In other words, We claim that if (48) occurs, then Suppose for the moment that (50) holds. Then, applying (49) and (50), we have Summing over j and recalling L = log 2 R/r yields (46). It remains to prove (50), assuming that (48) Now we find 0 < a, b < 1 and points x a , x b ∈ U such that u −1 (a) and u −1 (b) are simple paths γ a and γ b joining ζ 2 and ζ 4 , and From now on the argument proceeds exactly as the proof of Proposition 7.2, so we only recall the main points. We choose a small h > 0, and define a weakly admissible function for (ζ 1 , ζ 3 ; Q) as follows. First, near γ a and γ b , we apply the function ρ the same way as in the definition of v(x a ) and v(x b ). Then, near γ j we use the function ψ/ h. Finally, in Q\(B(x a , ) ∪ B(x b , )) we apply (3) to construct an admissible function p for (∂ Q, B(x a , ) ∪ B(x b , ); Q) such that the integral of p 2 is small. We take the sum g of these functions, and test the minimizing property of ρ for M 1 = mod(ζ 1 , ζ 3 ; Q) with the function (1 − α)ρ + αg. Taking → 0 and α → 0, we arrive at (50). Then applying the coarea inequality, Proposition 3.2 to the distance function ψ (recall that |β s | ⊂ ψ −1 (s)), we have On the other hand, ϕ extends to a set function with differential ϕ (s) at almost every s, and δ 0 ϕ (s) ds ϕ((0, δ)) by Lemma 6.5. Therefore, Combining (46), (51) and (52), we have Furthermore, applying Hölder's inequality and Proposition 8.2, we have But this is a contradiction when r → 0. The lemma follows.
Proof of Proposition 9. 1 We know that f is continuous and surjective by Corollary 7.4. We use the notation of Sect. 7; Then for every t ∈ F the conjugate function v is increasing on |γ t | by construction, so f −1 (t, s) is a continuum for all s. But then f −1 (t, s) has to be a point by Lemma 9.3. We conclude that if z = (t, s), t ∈ F, then f −1 (t) is a point. It remains to prove that the same property holds when t / ∈ F. Fix such t, and z = (t, s).
We now claim that f −1 (z) contains a point x z with the following property: for every > 0 the x z -component V of f −1 B(z, ) contains points a, b ∈ U such that u(a) < t and u(b) > t. Indeed, we know that the set u −1 (t) separates A = {u < t} and B = {u > t} in Q. Consider the set Recall that u −1 (t) does not contain interior points by Proposition 7.2 and Lemma 9.3. Therefore C is non-empty and also separates A and B in Q, since A\C and B\C are open and disjoint in Q. We conclude that C contains a continuum joining ζ 2 and ζ 4 , so in particular v takes all values between 0 and M 1 in C. The claim follows.
Notice that by Proposition 7.2 and Lemma 9.3, U is dense in Q. Suppose there exists a point x 0 ∈ f −1 (z), x 0 = x z , and let > 0. Then by the density of U , the x 0 -component W of f −1 B(z, ) contains a point a 0 ∈ U such that u(a 0 ) < t or u(a 0 ) > t. Without loss of generality, assume that u(a 0 ) < t. Connecting a 0 to x 0 in W and a to x z in V , we may assume that u(a 0 ) = u(a) = t 0 < t.
We now have components V and W of f −1 B(z, ), and points a ∈ V and a 0 ∈ W , such that Recall that the restriction of f to |γ t 0 | is injective. In particular, a and a 0 can be connected by a subcurve η of |γ t 0 | such that f (η) ⊂ B(z, ). So we conclude that in fact V = W . But this is a contradiction when is small enough, since f −1 (z) is not connected by Lemma 9.3. The proof is complete.

Variational modulus
In the next section we prove perhaps the most intricate property of our map f , showing that under the reciprocality assumption the function Cρ is a weak upper gradient of f when C is large enough. To accomplish this, we now introduce a modified version of the conformal modulus called variational modulus, and prove a reciprocality result connecting the variational modulus to conformal modulus. The variational modulus appears, though implicitly, in the work of Gehring [23] in Euclidean space, where it coincides with conformal modulus.

We say that a Borel function H 0 is
and define the variational modulus mod( ) by where the infimum is taken over all V -admissible functions H . (2) and (3). Then

Lemma 10.1 Suppose X satisfies
Proof We first prove Let u 0 be the minimizing function in Q 0 constructed exactly as u in Sect. 4, with minimizing weak upper gradient ρ 0 . Moreover, let H be V -admissible for , and γ t = u −1 where A s,t = {x ∈ Q 0 : s < u 0 (x) < t} as before. By Lemma 6.5 and the V -admissibility of H , h exists for almost every 0 < t < 1, and moreover Recall also that Q 0 Therefore, (54) follows from (55) by applying Hölder's inequality and minimizing over H .
To conclude the proof, we show that We claim that the function ρ 0 · mod(ζ 0 1 , ζ 0 3 ; Q 0 ) −1 is V -admissible for . This immediately implies (56). Let γ , and g ∈ F (γ ) be as in (53). Then, g is in particular weakly admissible for (ζ 0 1 , ζ 0 3 ; Q 0 ). Applying the minimizing property of ρ 0 , we have Subtracting the left integral from both sides of the inequality and letting s → 0, we have verifying our claim.
We can apply the variational modulus to estimate conformal modulus in the rectangles Q(i, j, k) defined in Sect. 8. Recall that assuming reciprocality means that we assume the conditions (1), (2) Then  (1) and (2). The main content of the lemma is the second inequality in (58), as we will see in Section 11.
11 Regularity of f By Proposition 9.1, our map f : Q → [0, 1] × [0, M 1 ] is a homeomorphism, assuming (2) and (3). In this section we show that if we also assume condition (1), then we have one of the modulus inequalities required for quasiconformality.
Upper gradients for maps are defined similarly to upper gradients of functions. We say that a Borel function g 0 is an upper gradient of a map F : for every a, b ∈ Y and every locally rectifiable path γ joining a and b in Y . If moreover Y is equipped with locally finite H 2 -measure, then g is a weak upper gradient of F if there exists an exceptional set of modulus zero such that (66) holds for every γ / ∈ . Furthermore, if g ∈ L 2 (Y ) is a weak upper gradient of F, then there exists an exceptional set of modulus zero such that if h 0 is a Borel function in Z , then for every γ / ∈ . See [34] for the proof of this property and more information on upper gradients and absolute continuity. Proposition 11.1 Suppose X is κ-reciprocal. Then 2000κ 1/2 ρ is a weak upper gradient of f . Remark 11.2 Proposition 11.1 and Lemma 4.3 imply in particular that f belongs to the Newtonian Sobolev space N 1,2 (Q, R 2 ). See [34] for the theory of Sobolev spaces in metric measure spaces.
Before proving Proposition 11.1, we apply it to prove the modulus inequality discussed above.
Taking infimum over admissible functions g gives the claim.
Proof of Proposition 11.1 We use the notation R(i, j, k) and Q(i, j, k) introduced in Sect. 8. We fix k and denotê In other words,Q(i, j, k) is the preimage under f of a rectangleR(i, j, k) with the same center as R(i, j, k), so thatR(i, j, k) is a scaled copy of R(i, j, k) with scaling factor 3, except when Q(i, j, k) intersects ∂ Q. We will consider four subsets ofQ(i, j, k) together with their boundaries. We denote j + 1, k).
Then the union of the sets P forms a topological annulus around Q(i, j, k), except when Q(i, j, k) intersects ∂ Q.
The rectangles f (P (i, j, k)) have two opposite sides three times as long as the two other sides; we say that the long boundary curves of P are the preimages of the longer sides. We denote by the family of rectifiable paths joining the long boundary curves of P (i, j, k) in P (i, j, k). Then Lemma 10.2 gives mod( ) 3κ for all . Therefore, we can choose a weakly admissible function ν (i, j, k) : We now define σ k : Q → [0, ∞], Notice that if x ∈ int Q(i, j, k) for some (i, j), then there are at most 8 triples (i , j , ) such that x ∈ P (i , j , k). Therefore, applying (68) and Lemma 8.1, we see that In particular, the sequence (σ k ) is bounded in L 2 (Q), so there exists a subsequence (σ k n ) converging weakly to σ ∈ L 2 . Furthermore, by Mazur's lemma, there exists a sequence (σ n ) of convex combinations of the functions σ k n converging to σ strongly in L 2 . Notice that (69) then holds with σ k replaced by σ and for every (i, j, k). Now, if ⊂ Q is open in Q, then we can take a "Whitney decomposition" of f ( ) and cover it with the sets R(i, j, k) ⊂ f ( ) with disjoint interiors (and varying k). Then is covered by the corresponding sets Q(i, j, k) ⊂ , and applying Lemma 8.1 and (69) with σ gives Since this holds for every open ⊂ Q, we conclude that for H 2 -almost every x ∈ Q. So the proposition follows if we can show that 64σ is a weak upper gradient of f . First, notice that since the functions ν (i, j, k) are weakly admissible for the path families (i, j, k), we can choose an exceptional setˆ of zero modulus such that the following holds: whenever γ contains a subpathγ in (i, j, k)\ˆ , then γ ν (i, j, k)χ P (i, j,k) ds 1.
On the other hand, the triangle inequality gives Therefore, applying (71) and recalling the bounded overlap of the setŝ Q(i, j, k), we have Theorem 12.1 follows from Proposition 9.1, Corollary 11.3, and Corollary 12.3 below. In this section we prove Sobolev regularity for the inverse map f −1 . This leads to the last modulus inequality in the definition of quasiconformality, finishing the proof of Theorem 12.1.
We formulate the next results in slightly greater generality than what is needed to prove Theorem 12.1. Notice that the results can be applied to our map f , thanks to Proposition 9.1 and Corollary 11.3.

Proposition 12.2 Suppose X satisfies
for every path family in , then (2κ K J F −1 ) 1/2 is a weak upper gradient of F −1 .

Corollary 12.3 Suppose X and F are as in Proposition 12.2. Then
for every path family in .
Proof Let be a path family in , and let h be admissible for . Then, by Proposition 12.2 and (67), is weakly admissible for F , and moreover by (72) Minimizing over h gives (74). The proof is complete.
The rest of this section is devoted to the proof of Proposition 12.2. The basic idea for the proof is classical in QC mapping theory, see [57,Theorem 31.2]. However, here we replace the classical geometric conditions by the reciprocality condition (2).
We say that a continuous function w : → R is ACL, if w is absolutely continuous on H 1 -almost every line segment parallel to the coordinate axes. Notice that if w is ACL, then it has partial derivatives at almost every point, defining the gradient ∇w.
For the rest of this section, we denote H = F −1 . Let a ∈ X , and denote H a (y) = dist (H (y), a).
We will use the following fact, cf. [34,Theorems 7.1.20 and 7.4.5] for the proof: If there exists a Borel function g ∈ L 2 ( ) such that, for every a ∈ X , H a is ACL and |∇ H a (y)| g(y) for almost every y ∈ , then g is a weak upper gradient of H .
Proof of Proposition 12.2 In view of the previous discussion it suffices to show that H a is ACL and that the function satisfies (75) for every a ∈ X . Notice that by assumption it suffices to consider the restriction of H to an arbitrary cube Q ⊂ , and that by scaling and translating if necessary we may assume Q = [0, 1] 2 . We denote Recall from Lemma 6.5 that exists and is finite for almost every 0 < t < 1. We fix such a t, and let with H 1 (E) < . We will prove where C may depend on t but not on E. This suffices for the ACL-property since H is a homeomorphism and since we can apply the same argument to the horizontal segments. We may assume that E is a Borel set. Furthermore, since H (E) is an increasing limit of compact subsets, we may assume that E is compact. Now there exists δ > 0 and a covering and By Lemma 6.4, for every α > 0 there exists ν < δ such that For now we choose α = 1/2. Also, choosing ν smaller if necessary we may assume that Now, the moduli of j (ν) and j (ν) are easy to calculate. Combining with assumption (73), we then have By condition (2), Moreover, by (78), the constant function 2/ (H ( Combining the estimates, we get Summing over j and applying Hölder's inequality yields Recalling the disjointness of the interiors of the segments I j and that Lδ < , we see that the right hand term is bounded by (77) follows. We conclude that H is ACL.
To conclude the proof we have to show that H = F −1 satisfies (75) with the function g in (76). Let y = (y 1 , y 2 ) ∈ , and We denote E t = {t} × [y 2 − δ, y 2 + δ]. Let a ∈ X . Then, since H is ACL, so is H a = dist(·, a). Now Notice that choosing α arbitrarily small in (78) and showing (79) with this sharper bound yields Combining with (80), Hölder's inequality and Lemma 6.5, we have Dividing both sides by 4δ 2 , taking δ → 0 and applying the Lebesgue differentiation theorem then gives for almost every y ∈ . Repeating the argument gives the same estimate for ∂ 1 H a . Combining the estimates, we conclude (76).
13 Existence of QC maps f 0 : X → R 2 and → S 2 Theorem 12.1 shows the existence of QC maps on subsets of a reciprocal space X . In this section we finish the proof of Theorem 1.4 by showing the existence of a QC map on the whole space X . This is done by exhausting X with a sequence of subsets for which Theorem 12.1 can be applied, and then using normal family arguments.
Recall that if (F j ) is a sequence of K -QC maps F j : U → V j between planar domains containing 0 and 1 such that F j (0) = 0 and F j (1) = 1 for every j, then (F j ) is a normal family and there exists a subsequence (F j ) converging locally uniformly to a K -QC map F, cf. [57, 20.5,21.3,37.2]. Also, notice that if F and G are K 1 -and K 2 -QC homeomorphisms, respectively, and if the composition F • G is a well-defined homeomorphism, then F • G is K 1 K 2 -QC; this follows from the definition of quasiconformality. Theorem 1.4 is a direct consequence of the following.

Theorem 13.1 Suppose X is κ-reciprocal.
Then there is a 512 · 10 18 κ 6 -QC homeomorphism from X onto either R 2 or D.
Proof Let {X j }, X j ⊂ X j+1 for all j, be an exhaustion of X by open topological squares such that the closures X j are closed topological squares. Moreover, fix a, b ∈ X 1 , a = b. By Theorem 12.1 and the Riemann mapping theorem, there exists for every j ∈ N a 8 · 10 6 κ 2 -QC homeomorphism f j : X j → B j , where B j = B(0, r j ) ⊂ R 2 is a disk, normalized such that f j (a) = 0 and f j (b) = 1. We denote the inverse map by h j = f −1 j : B j → X j . Now fix k ∈ N, and let Then g k j is 64 · 10 12 κ 4 -QC for all j k. Moreover, g k j (0) = 0 and g k j (1) = 1. Thus (g k j ) ∞ j=k is a normal family, so there exists a subsequence (g k j k ) converging locally uniformly to a 64 · 10 12 κ 4 -QC map g k : B k → R 2 . It follows that Taking a diagonal subsequence ( f ), we see that ( f |X k ) converges for every k ∈ N to a 512 · 10 18 κ 6 -QC map. We conclude that the pointwise limit map f : X → R 2 is 512 · 10 18 κ 6 -QC. Applying the Riemann mapping theorem if necessary, we see that the image f (X ) can be chosen to be either R 2 or D.
We now consider the case where Y is homeomorphic to the Riemann sphere S 2 . The reciprocality conditions now easily generalize; we assume H 2 (Y ) < ∞, that (1) and (2)  Now mod( ) → 0 as → 0 by condition (3). Then also mod( f ) → 0 by the quasiconformality of f . Applying Proposition 3.5, we conclude that f (X ) does not have boundary in R 2 , and f extends continuously, mapping y 0 to ∞ on the Riemann sphere. Moreover, the extension is 512 × 10 18 κ 6 -QC. The proof is complete.

Minimizing dilatation
In this section we prove Theorem 1.5. Note that the corresponding result also holds when Y is homeomorphic to S 2 ; this follows from the proof given below.
The constant 2 in Theorem 1.5 is not best possible. In fact, in view of Example 2.2 and the results of Behrend [8] (see also [5] and [7]) on the area ratios of symmetric convex bodies, it is natural to ask if the sharp constant is π/2, or even if there always exists a QC map f 0 satisfying Both inequalities would be best possible by Example 2.2. The results mentioned above together with the arguments in this section guarantee that there exists a QC map satisfying the first inequality in (81), and also there exists a QC map satisfying the second inequality. However, we do not know if a single map satisfies both inequalities.
In the proof of Theorem 1.5, we apply certain differentiability properties of Sobolev maps with values in metric spaces, together with the measurable Riemann mapping theorem and John's theorem on convex bodies. Instead of relying on the measurable Riemann mapping theorem, one could reprove it with the methods used in this paper.
We now begin the proof of Theorem 1.5. We will not consider the case X ⊂ R N separately since it follows from the general arguments below. We assume f is QC, and denote h := f −1 : → X.
We will use some Lipschitz analysis. The following lemma is a special case of a statement concerning metric-valued Sobolev functions. See [34,Theorem 8.1.49] for the proof. Lemma 14.1 There exist measurable, pairwise disjoint sets G j , j = 0, 1, 2, . . ., covering , such that |G 0 | = 0 and h|G j is j-Lipschitz continuous for all j = 1, 2, . . ..
Recall that every metric space Z isometrically embeds to the Banach space L ∞ (Z ). Fix j 1. Then h|G j can be extended to a Lipschitz map By Kirchheim [36,Theorem 2], h j is metrically differentiable; for almost every x ∈ R 2 there exists a seminorm M D(h j , x) on R 2 such that We denote by |M D(h j , x)| the operator norm  D(h j , x) is a non-zero norm for almost every x ∈ .
Proof The first claim follows from [34,Proposition 6.3.22]. Towards the second claim, recall that J h denotes the volume derivative of h. Then, by Proposition 8.2, J h (x) > 0 for almost every x ∈ . Now (82) and a density point argument shows that M D(h j , x) has to be a non-zero norm at almost every x ∈ . By Lemma 14.2, we can define in a field G = G h of norms which are non-zero at almost every point, as follows. Let if x is a point of metric differentiability for h j for which Lemma 14.2 (ii) holds, and G x = 0 otherwise.
We would like to apply the measurable Riemann mapping theorem in order to make the distortion of h smaller. To this end, recall that the unit ball in a norm in R 2 is a symmetric convex body in the Euclidean plane. In the points x where G x is a non-trivial norm, denote the unit ball by Let E x be an ellipse, E x ⊂ C x , whose Lebesgue measure is maximized among all such ellipses. We can now define an ellipse fieldG by setting G x = E x whenever defined, andG x = B(0, 1) otherwise. Also, it follows from the construction that the complex dilatation associated to the ellipse fieldG is measurable. Thus the measurable Riemann mapping theorem gives a QC homeomorphism ν : → such that for almost every x ∈ there exists some r x > 0 so that We denote Also, let C x = Dν (C x Combining John's theorem and the previous construction, we have Also, for some R x > 0. Then, by (85), where H 2 M D is the Hausdorff measure with respect to the norm M D (H j , x).
x is proved in [36,Lemma 6]. Also, recalling (83), we have Combining the estimates, we have This together with (84) and the proof of Corollary 11.3 gives the inequality mod( ) 2 mod(H ) for every path family in R 2 .
For the reverse inequality, first notice that On the other hand, (86) and (87) imply The proof of Theorem 1.5 is complete.

Existence of QC maps under mass upper bound
In this section we prove Theorem 1.6. In other words, we show that the mass upper bound (4) implies the three conditions of reciprocality. We prove each condition separately in Lemma 15.1, Propositions 15.5 and 15.8, respectively.
Proof Define when r d(y, x) R, and g = 0 elsewhere. Then g is admissible for We denote T = log 2 R/r . Then, applying (4) yields Now we notice that the continuity of the energy minimizer u holds under condition (4). We need a slight modification of Proposition 3.2. Proof The proof of Theorem 5.1 shows that it suffices to establish (13). Let x ∈ X and R > 0 such that X \B(x, R) = ∅. For 0 < r < R, consider the family r of all rectifiable paths separating B(x, r ) and X \B(x, R) in X . Then (13) follows if we can show that
The lemma is proved by slightly modifying the standard proof for the Hardy-Littlewood maximal function on doubling spaces. More precisely, one can follow the proof given in [27, Theorem 2.2] step by step, but when the doubling property is used there we apply our current definition of the maximal function instead.
Proof Fix Q, the boundary paths ζ 1 , . . . , ζ 4 , and the minimizer ρ as in Sect. 4. Recall We would like to show that M 2 = mod(ζ 2 , ζ 4 ; Q) C/M 1 . In view of Lemma 15.4, it is sufficient to show that the function C 1 (Mρ)/M 1 is admissible for M 2 , for large enough C 1 depending only on the constant C U in (4). Let γ be a rectifiable path in Q joining ζ 2 and ζ 4 , and > 0. We may assume that γ is simple. Then, testing the modulus of (ζ 1 , ζ 3 ; Q) with the function as in the proof of Proposition 7.2, we notice that Here N (γ ) is the closed -neighborhood of |γ | as before. We now choose a covering of N (γ ) by balls B(x j , 5 ) such that each x j ∈ |γ | and the balls B(x j , ) are pairwise disjoint. By (91) and (4), we have We conclude that 961C U (Mρ)/M 1 is admissible for M 2 , as desired.
To conclude the proof of Theorem 1.6, we show that the lower bound (2) follows from (4). The proof is an application of Proposition 3.2 and the following estimate for the minimizer u.  u(B(x, r ))) Proof Applying Proposition 3.2 with the distance function from x shows that H 1 (S(x, s)) < ∞ for almost every r < s < 2r . Similarly, the upper gradient inequality for u and ρ holds for every path η with |η| ⊂ S(x, s), for almost every s. Fix such s, and let E j (s) be a connected component of S(x, s) such that E j (s) separates X . Then E j (s) contains a curve γ j that bounds a domain U j (s). Moreover, B(x, r ) ⊂ ∪ j U j (s), so u(B(x, r ))) j diam u(U j (s)). Combining the estimates, we have Integrating over s and applying Proposition 3.2 again gives the desired estimate.
We need a version of the coarea inequality for our minimizer u. We will follow the proof of Proposition 3.2 given in [2, Proposition 3.1.5], replacing the Lipschitz condition assumed there with Lemma 15.6. Proposition 15.7 Suppose X satisfies (4). Let g : Q → [0, ∞] be a Borel function. Then the function t → u −1 (t) g dH 1 is measurable, and Proof Fix ∈ Z, and denote E = {x ∈ Q : 2 < g(x) 2 +1 }. Then, for j ∈ Z, let It then suffices to show that Let > 0 and choose a finite or countable covering of E j by balls B i = B(x i , r i ), x i ∈ E j , such that 2r i < for every i, and We denote λB i = B(x i , λr i ). Notice that removing ∂ Q does not affect the left side of (92), so we may and will assume that 2B i ⊂ Q for every i. Now, by Lemma 15.6, On the other hand, by (4) and (94), Combining the estimates yields We now define Integrating over t and taking the integral inside the sum then yields On the other hand, by the definition of -content, for every 0 < t < 1. Combining the estimates with (95) yields Mρ dH 2 (measurability follows by standard real analysis). Letting → 0, (93) follows by monotone convergence.
Proof By Proposition 15.7 applied to the constant function 1, we know that H 1 (u −1 (t)) < ∞ for almost every t. Also, by Lemmas 15.3 and 6.3, u −1 (t) is connected for all t. Therefore, by Proposition 3.1, u −1 (t) contains a simple path γ t joining ζ 2 and ζ 4 in Q for almost every t. Now, if g is admissible for mod(ζ 2 , ζ 4 ; Q), then γ t g dH 1  Minimizing over g gives the claim.

Existence of QS maps
In this section we prove Corollary 1.7 as an application of Theorems 1.4 and 1.6. Recall that Corollary 1.7 is proved in [11] using different methods. Theorem 1.6 can be seen as a generalization of Corollary 1.7. In [11] another generalization of Corollary 1.7 is given for quasisymmetric (QS) maps in general, possibly fractal, topological spheres. Wildrick [59] extended Corollary 1.7 to surfaces homeomorphic to R 2 .
for distinct points x 1 , x 2 , x 3 . We use the chordal distance d(a, b) = |a−b| in S 2 . We have now defined the concepts in the statement of Corollary 1.7. The method in the proof of Corollary 1.7 assuming Theorem 1.4 is nowadays standard in QC mapping theory and can be found in [29]. The argument is repeated here for completeness. We need three facts. First, if E and F are disjoint continua in S 2 and dist(E, F) T min{diam E, diam F}, 0 < T < ∞, This estimate is proved integrating a given admissible function over suitably chosen concentric circles intersecting both E and F, and then integrating over the radius, cf. [57,Section 10], and [29, Section 3]. Metric measure spaces satisfying (97) are called Loewner spaces, see [29]. Secondly, if X is λ -linearly locally contractible, then it satisfies the so-called L LC-conditions for all λ > λ , cf. [11]: (1) if B(x, r ) is a ball in X and a, b ∈ B(x, r ), then there exists a continuum E ⊂ B(a, λr ) joining a and b.
(2) if B(x, r ) is a ball in X and a, b ∈ X \B(x, r ), then there exists a continuum F ⊂ X \B(a, r/λ) joining a and b.
Finally, applying Proposition 3.2 and linear local contractibility as in Remark 3.4, we get the lower bound C L r 2 H 2 (B(x, r )) whenever r diam(X ), for measures of balls, where C L depends only on λ . Combining with (4), we see that a space X satisfying the conditions of Corollary 1.7 is Ahlfors 2-regular. In particular, X is then a doubling metric space.
Proof of Corollary 1.7 By Theorem 1.6 and the proof of Theorem 13.2, there exists a 2-quasiconformal map f : Y → S 2 . Then by Remark 16.2 it suffices to show that f −1 is η-QS with η depending only on C U and λ . Moreover by Ahlfors regularity and a theorem of Väisälä, see [27,Theorem 10.19], it suffices to show that f −1 is weakly QS, i.e., that (96) holds with t = 1.
We first choose points a 1 , a 2 , a 3 ∈ Y such that We denote b j = f (a j ). Precomposing f −1 with a Möbius transformation, if necessary, we may then assume that for all i, j. Now take distinct points y 1 , y 2 , y 3 ∈ S 2 such that |y 1 − y 2 | |y 1 − y 3 |.
We denote x k = f −1 (y k ), k = 1, 2, 3. Then, by triangle inequality, for at least two indices j. Among them we can then choose one of the indices, say j = 1, such that also The L LC-conditions now guarantee the existence of a continuum where C U is the constant in (4). Combining (100) and (101) gives 8λ 2 exp(16C U /φ (16)).
We conclude that f −1 is QS.

Concluding remarks
We briefly discuss the absolute continuity properties of QC maps between X and R 2 . It follows from Proposition 8.2, and the fact that planar QC maps satisfy Condition (N ), that every QC map f : X → R 2 satisfies condition (N ). One could hope for Condition (N ) to hold also for the inverse. Then it would follow from Lemma 14.1 that a reciprocal X is always countably 2-rectifiable. However, we show that this is not the case in general.

Proposition 17.1
There exists a reciprocal X ⊂ R 3 that is not countably 2-rectifiable. In fact, X satisfies (4).
Proof We only briefly describe the construction of X and leave the details to the interested reader. We choose a self-similar Cantor set C ⊂ [0, 1] 3 such that for all x ∈ C and 0 < r < 1, cf. [39, pp. 65-67]. Then, we construct a "tree" consisting of tubes that follow the construction of the set C. More precisely, each new step in the construction of C corresponds to a branching of the tree such that several tubes grow from every already existing tube. We can arrange the tubes such that the limiting set X includes the whole set C so that X is not rectifiable, and such that there is no overlapping so X is homeomorphic to R 2 . Also, we can choose the area of each tube to be as small as we wish. Therefore, combining with (102) we can guarantee that the mass upper bound (4) holds. Reciprocality then follows from Theorem 1.6.
Our discussion is related to the so-called inverse absolute continuity problem for QS maps: if f : X → R 2 is QS, does f satisfy condition (N )? See [24,25,31,58]. There are several similar unsolved problems in QC mapping theory, see [3] for an overview. From Theorem 1.4 and the fact that planar QC maps preserve sets of measure zero, it follows that the answer is affirmative if X is reciprocal. This fact can be also proved directly employing condition (1), as we now demonstrate. (1), and let f : X → R 2 be QS. If E ⊂ X, then H 2 (E) = 0 if and only if | f (E)| = 0.

Proposition 17.2 Suppose X satisfies
Proof That | f (E)| = 0 implies H 2 (E) = 0 is well-known to hold even without assumption (1) by the works of Gehring, Väisälä and Tyson, see [56]. Suppose H 2 (E) = 0 and | f (E)| > 0. Since f is QS, by [56] we know that there exists K 1 such that mod( ) K mod( f −1 ) for every path family in R 2 . Moreover, an examination of the proof given there shows that in fact where mod( ) is defined as mod( ) except for the definition of admissibility; we say that a Borel function g is admissible for mod( ) if g is admissible for and g = 0 almost everywhere on the set f (E). Combining the estimates and minimizing over admissible functions, we conclude mod( ) −2 .
The same estimate holds with replaced by . Now, by (103), But this contradicts (1) when > 0 is small enough. The proof is complete.
The following related question immediately arises from [24] and [25]. Notice that a QS f is automatically QC in the sense of the metric definition. The answer to Question 17.3 is affirmative if in addition X satisfies (4); this follows from [29] and also from Theorems 1.4 and 1.6. To finish, we discuss the three conditions in the definition of reciprocality. Although used only once in the proof of Theorem 1.4, we feel that the most important of the conditions is (1). For instance, it is the failure of (1) that prevents the existence of a QC map in Example 2.1. (1), or some modification of it, imply (2) and/or (3)?

Question 17.4 Does condition
It is not difficult to give examples of spaces that do not satisfy (3), but we do not know if such an example satisfying (1) exists. It can be proved that if X satisfies (1), then mod({x}, E; Q) = 0 for every Q ∈ X , x ∈ Q and every compact E ⊂ Q\{x}. Concerning condition (2), it seems that this condition should hold in great generality. Question 17.5 Does condition (2) hold for all X ?