An orientation-reversing homeomorphism need not be quasiconformal

Let $T(z)=\bar{z}$, ${\mathrm{\Omega }}^{\mathrm{\prime }}=T(\mathrm{\Omega })$, and ${\mathrm{\Gamma }}^{\mathrm{\prime }}=T\mathrm{\Gamma }$. Given any Borel density $\sigma$ on ${\mathrm{\Omega }}^{\mathrm{\prime }}$, set $\rho =\sigma \circ T$ on $\mathrm{\Omega }$. By [F2], each rectifiable path has the same density length before and after reflection, while nonrectifiable paths have infinite length on both sides. Thus ${\mathrm{\ell }}_{\rho }(\mathrm{\Gamma })={\mathrm{\ell }}_{\sigma }({\mathrm{\Gamma }}^{\mathrm{\prime }})$. By [F3], the two densities also have equal area, with $0<A<\mathrm{\infty }$ on one side exactly when it holds on the other. Since $T$ is an involution, this is a bijection of the admissible density classes; taking the defining supremum gives $\lambda ({\mathrm{\Gamma }}^{\mathrm{\prime }})=\lambda (\mathrm{\Gamma })$ and hence $\mu ({\mathrm{\Gamma }}^{\mathrm{\prime }})=\mu (\mathrm{\Gamma })$, including empty, zero, and infinite cases.

Every quadrilateral family and its reflected image therefore satisfy the two-sided modulus inequalities with constant $K=1$. But [F4] shows that $T$ reverses the local orientation, so it fails the orientation-preserving clause in the library's geometric definition.

The map is smooth and belongs to $W_{\mathrm{l}\mathrm{o}\mathrm{c}}^{1,2}$, but [F4] gives $\mathrm{\mid }{f}_{\bar{z}}\mathrm{\mid }=1$ and $\mathrm{\mid }{f}_z\mathrm{\mid }=0$. For every finite $K\ge 1$, the analytic definition has $k=(K-1)\mathrm{/}(K+1)<1$ and would require $1\le k\cdot 0=0$, which is impossible. Thus the modulus bounds alone do not imply either orientation-preserving geometric or analytic quasiconformality.