By viewing the domain of the polygon as a strip with extra boundary punctures, we produce a holomorphic vector field $v_{n}$ as the $\mathbb{R}$-translation vector field. However, unlike in the strip case, at the degree one self intersection corners $v_{n}$ does not typically have the required decay to be admitted as a deformation vector field. Indeed, by thinking about such a corner point as the origin in the upper half plane model of $\Sigma$ with local coordinate $z$, then $zv_{n}$ decays at the corner, but not necessarily $v_{n}$ itself.

Now $v_{1}\wedge\ldots v_{n}$ is a section of $\Lambda^{n}TX$ with boundary value on $\Lambda^{n}TL$, and $\Omega(v_{1},\ldots v_{n})$ is a holomorphic function on $\Sigma$. We assume from now on that it is not identically zero. The index of the ordinary Cauchy-Riemann operator is

Invoking (23) this is computable from the vanishing orders of $\Omega(v_{1},\ldots v_{n})$:

The interior and boundary vanishing orders are non-negative. Since the $v_{k}$ are holomorphic near the corners without correction, the proof of Lemma 3.5 shows that the excess vanishing order at the $CF^{0}(L,L^{\prime})$ corner and the $q$ corner are both non-negative. At the degree one self intersection corners, the excess vanishing order of $\Omega(v_{1},\ldots v_{n-1},zv_{n})$ is nonnegative by the same previous arguments, so $\Omega(v_{1},\ldots v_{n-1},v_{n})$ itself has excess vanishing order $\geq-1$. Hence $\deg q-k+1\geq n-k+1,$ namely $\deg q\geq n$.

When the equality is achieved, then all bounds are saturated. In particular, $\Omega(v_{1},\ldots v_{n})$ can only vanish at the corners, so $u:\Sigma\to X$ is an immersion up to boundary. At the corners, the same arguments in Corollary 3.6 shows the failure of immersion is minimal.

If $v$ is the deformation vector field corresponding to an arbitrary kernel element of the extended linearized operator, then after subtracting off a constant linear combination of $v_{1},\ldots v_{n-1}$, we may assume $v$ is tangent to $\Sigma$ at any chosen point on $\partial\Sigma$. The same argument in Corollary 3.6 shows $v$ is tangent to the image of $\Sigma$. The immersion property allows us to lift $v$ to the domain $\Sigma$. There is no room to deform the complex structure of $\Sigma$, nor is there any automorphism of $\Sigma$, so in fact $v$ vanishes identically. This shows that $v_{1},\ldots v_{n-1}$ span all first order deformations. But $\deg q=n$ implies that the index of the extended linearized operator is

Thus the cokernel dimension is zero, namely the obstruction space vanishes. Consequently, the moduli space of such holomorphic polygons is smooth. ∎