The inverse of a quasiconformal map is quasiconformal with the same dilatation

By [F1], $g=f^{-1}$ is analytically K-quasiconformal independently of any area equality or inverse-null assumption. Both maps are orientation-preserving, belong locally to $W^{1,2}$, and are differentiable almost everywhere by [F2].

We prove area equality for either map $u$. On an interior rectangle, choose countable dense sets of good ACL horizontal and vertical levels. Their subrectangles form a basis with rectifiable Jordan image boundaries. These image boundaries have area zero: divide each finite-length arc into pieces of length at most $\delta$, cover by squares of side $2\delta$, and let the total cost $C\delta (\mathrm{length}+\delta )$ tend to zero. Let $D$ be such a closed rectangle and $T⋐u(D^{\circ })$. Smooth mollifications $u_n$ converge uniformly on a neighborhood of $D$, with derivatives converging in $L^2$; determinants $J_n$ converge to $J_u$ in $L^1$, and since $J_u\ge 0$, the integrals of $(J_n{)}_{-}$ tend to zero. Uniform convergence makes the straight homotopy of the boundary images avoid every $y\in T$ for large $n$. The positively oriented rectifiable Jordan contour $u(\mathrm{\partial }D)$ has winding one at $y$: triangulate $D$ into a singular2-chain, transport it by $u$, and apply naturality of the pair boundary map to its positive local orientation class. Excision identifies that class with the positive generator of $H_2({\mathbb{R}}^2,{\mathbb{R}}^2\setminus \{y\})$; the connector is an isomorphism because ${\mathbb{R}}^2$ is contractible, giving the positive punctured-plane H1 generator. Degree-one Hurewicz and the winding/degree identification give analytic winding one. This uses Long exact sequence of a pair, Naturality of the pair long exact sequence, Excision for singular homology, The first Hurewicz map is abelianization, For loops in C times, the winding number about 0 equals the circle degree, Winding number identifies the fundamental group of C times with the integers; translating by $-y$ and a nonzero complex scaling normalizes the contour basepoint to1 without changing its integral. Thus $u_n(\mathrm{\partial }D)$ also has winding one about $T$.

For a regular value $y$ of $u_n$ away from its boundary, its fibre in $D$ is finite. Delete disjoint small disks around these preimages. Round the four rectangle corners in neighborhoods avoiding its finite fibre, and apply C1 Stokes [F4] to the pulled-back closed angular form $(-vdu+udv)\mathrm{/}(2\pi (u^2+v^2))$, after translating by $y$. This gives boundary winding as the sum of the small-circle windings. The corner-arc integrals tend to zero, since the form is bounded there and their lengths tend to zero; passage to the limit recovers the rectangle formula. Interpolation to the invertible derivative on each small circle makes each term $\mathrm{sgn}{J}_n$ there. Hence a regular $y\in T$ has signed fibre count1 and at least one positive-Jacobian preimage. Sard [F2] makes the exceptional target values null. Use the smooth local inverse theorem [F4] and a countable rational-basis cover to partition the open set $\{J_n>0\}\cap D^{\circ }$ into disjoint Borel inverse-branch pieces. Nonnegative change of variables on each branch and countable additivity give ${\int }_D(J_n{)}_{+}\ge \mathrm{\mid }T\mathrm{\mid }$. Therefore ${\int }_D{J}_n\ge \mathrm{\mid }T\mathrm{\mid }-{\int }_D(J_n{)}_{-}$; passage to the limit and compact exhaustion of $u(D^{\circ })$ give ${\int }_{D^{\circ }}{J}_u\ge \mathrm{\mid }u(D^{\circ })\mathrm{\mid }$. The opposite lower area inequality is [F2]. On each chosen basis rectangle, its subrectangles from the same dense good levels, together with the whole rectangle, form a generating pi-system. Both restricted measures are finite and agree there, so [F4] applies with the constant whole-rectangle exhaustion. Countably many such rectangles cover the domain; disjointizing that cover extends equality to all relatively compact Borel sets. Exhaustion gives the area formula wherever needed. In particular $u$ sends null Borel sets to null sets. Apply this argument to both $f$ and the already-regular $g$. This supplies N and inverse-N without assuming either.

If $f_z$ vanished on a positive-area Borel set $E$ inside a compact exhaustion, the Beltrami inequality would make $J_f=0$ there. Step 3.1 gives $\mathrm{\mid }f(E)\mathrm{\mid }=0$, while the null-set property of $g$ would force $\mathrm{\mid }E\mathrm{\mid }=0$, a contradiction. Thus $f_z\mathrm{\ne }0$ a.e. and $Df$ is nonsingular a.e. The non-differentiability set of $g$ has null preimage under $g$ by the same null-set property, so almost every source point is a common differentiability point. At such a point the chain rule [F3] for $g\circ f=\mathrm{id}$ gives $0=(g_w\circ f)f_{\bar{z}}+(g_{\bar{w}}\circ f)\overline{f_z}$ and the corresponding z equation equals1. Division yields ${\mu }_g(f(z))=-f_{\bar{z}}(z)\mathrm{/}\overline{f_z(z)}=-{\mu }_f(z)f_z(z)\mathrm{/}\overline{f_z(z)}$.

The modulus of the unimodular factor in step 4.1 is one. Null-set preservation in both directions transports the coefficient equality and its essential bounds, so $\mathrm{\parallel }{\mu }_g{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }{\mu }_f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $K_g=K_f$. Step 1.1 already supplies inverse Sobolev regularity, so the coefficient calculation does not circularly assume it. Repeating the result for $g$ proves both directions of the final equivalence.