We modify the proof of Lemma 3.9. We think of the corner $q$ as the origin in the upper half plane model, and take $v_{n}$ instead to be the Möbius vector field $z^{2}\partial_{z}$ on $\Sigma$. This has one higher order of vanishing:

$$v_{n}=(a_{n1}z^{\phi_{1}/\pi+1},\ldots a_{nn}z^{\phi_{n}/\pi+1})+O(z^{2}).$$

This leads to

$$\Omega(v_{1},\ldots v_{n})=z^{(\sum\phi_{k})/\pi-n+2}(\det(a_{kj})+o(1)),$$

so the excess vanishing order at $q$ is $\geq 2-n$. The index of the ordinary linearized Cauchy-Riemann operator is

$$\deg q=2\sum(\text{interior zeros})+\sum(\text{boundary zeros})+\sum(\text{excess corner zeros})+n,$$

whence $\deg q\geq 2$.

When the equality is achieved, then there is no interior or boundary zero, and $\det(a_{kj})\neq 0$ at the corner, hence the immersion claim. The argument in Cor. 3.6 shows that $v_{1},\ldots v_{n},\frac{v_{n}}{z},\frac{v_{n}}{z^{2}}$ span the real vector space of first order deformations in $W^{1,2;\pi}$. In particular, the only first order deformation which decays at $q$ are spanned by $v_{n}$ and $z^{-1}v_{n}$, namely the Möbius generators. Since the index of the ordinary Cauchy-Riemann operator is two, the cokernel must have dimension zero, namely the obstruction vanishes. ∎