Quasiregularity almost everywhere (removability) The three equivalent definitions of quasiregular mapping that I am using are these ones:

Let $U\subset\mathbb{C}$ be an open set and $K < \infty$. Then:
  
  
*
  
*A mapping $g:U\to\mathbb{C}$ is $K$-quasiregular if and only if $g = f\circ\phi$ for some $K$-quasiconformal map $\phi:U\to\phi(U)$ and for  some holomorphic map $f:\phi(U) \to g(U)$.
  
*A continuous mapping $g:U\to\mathbb{C}$ is $K$-quasiregular if and only if $g$ is locally $K$-quasiconformal except at a discrete set of
  points.
  
*A mapping $g:U\to \mathbb{C}$ is $K$-quasiregular if and only if for every $z\in U$ there exist neighbourhoods of $z$ and $g(z)$
  denoted by $N_z$ and $N_{g(z)}$ respectively, a $K$-quasiconformal
  mapping $\psi:N_z\to \mathbb{D}$ and a conformal mapping
  $\varphi:N_{f(z)}\to\mathbb{D}$ such that $(\varphi\circ g\circ
 \psi^{-1})(z) = z^d$, for some $d\geq 1$.
  

I am looking for some result or proof to deal with the following:
I have a continuous mapping $g:\mathbb{C}\to\mathbb{C}$ that is quasiregular on the whole plane except in two Jordan curves. How can I ensure that $g$ is quasiregular on the whole plane?
 A: Lemma.
From the first definition of quasiregularity, $g$ is locally $K$-quasiconformal at the discrete set of points that are preimages by $\phi$ of critical points of $f$.
The statement of the following theorem is taken from Branner B., Fagella N.-Quasiconformal Surgery in Holomorphic Dynamics (2014) and for a proof it refers to Prop. 4.2.7 and 4.9.9 in J. H. Hubbard - Teichmüller Theory and Applications to Geometry, Topology and Dynamics, Vol. 1. Matrix Editions (2006).
Theorem.
If $\Gamma$ is a quasiarc (the image of a straight line under a quasiconformal mapping) and $\phi:U\to V$ a homeomorphism that is $K$-quasiconformal on $U\setminus\Gamma$, then $\phi$ is $K$-quasiconformal on $U$, and hence $\Gamma$ is quasiconformally removable. In particular, points, lines and smooth arcs are quasiconformally removable.
Answer of the doubt:
Let $\gamma_1$ and $\gamma_2$ be the curves of the question.


*

*If $z\in (\gamma_1\cup \gamma_2)\setminus(\gamma_1\cap\gamma_2)$, we will use the regularity at $\mathbb{C}\setminus(\gamma_1\cup \gamma_2)$. Since $g$ is regular at $\mathbb{C}\setminus(\gamma_1\cup \gamma_2)$, it is there locally quasiconformal except at a discrete set of points. As the zeros of a holomorphic function are isolated, the lemma ensures that $g$ is locally quasiconformal at $\mathbb{C}\setminus(\gamma_1\cup \gamma_2)$ except at a set of isolated points. Because of that, given a point $z\in\gamma_1\cup \gamma_2$, there exists a neighbourhood $N_z$ such that $g$ is quasiconformal at $N_z\setminus \Gamma$, where $\Gamma := N_z\cap (\gamma_1\cup \gamma_2)$. By the theorem, $g$ is quasiconformal at $N_z$ and hence quasiregular.

*If $z\in\gamma_1\cap\gamma_2$, then we have a point in a neighbourhood $N_z$ so that $g_{|N_z}$ is quasiregular and we can apply an analogous argument of the first case with that point.

