**Lemma 6.4.** For any $r > 0$ and any $\mathbf{n} \in H^1(B_r^2, \mathbb{S}^2)$ with constant value at the boundary, if

$$ \frac{1}{2} \int_{B_r^2} |\nabla \mathbf{n}|^2 \, d\mathcal{H}^2 < 4\pi $$

then $\deg(\mathbf{n}, B_r^2) = 0$.

*Proof.* By applying the area formula, we obtain

$$ \int_{B_r^2} |\partial_{x_1} \mathbf{n} \times \partial_{x_2} \mathbf{n}| \, d\mathcal{H}^2 = \int_{\mathbf{n}(B_r^2)} \mathcal{H}^0(\mathbf{n}^{-1}(y)) \, d\mathcal{H}^2(y) \geq \mathcal{H}^2(\mathbf{n}(B_r^2)). $$

On the other hand, we have $|\partial_{x_1} \mathbf{n} \times \partial_{x_2} \mathbf{n}| \leq |\partial_{x_1} \mathbf{n}| |\partial_{x_2} \mathbf{n}| \leq |\nabla \mathbf{n}|^2 / 2$. Therefore, there holds

$$ \frac{1}{2} \int_{B_r^2} |\nabla \mathbf{n}|^2 \, d\mathcal{H}^2 \geq \mathcal{H}^2(\mathbf{n}(B_r^2)). $$

If the left-hand side is $< 4\pi$, then $\mathbf{n}$ is not surjective and so $\deg(\mathbf{n}, B_r^2) = 0$ (see e.g. [19, Property 1]). $\square$

*Proof of Proposition 6.1.* Arguing as in the proof of Proposition 5, and using that the boundary conditions $g_{\varepsilon|U_+}$ are non orientable, one shows that

$$ E_{\varepsilon}(Q_{\varepsilon}) \geq C (|\log \varepsilon| - 1) $$

for any $\varepsilon$, $L$ and $r$, so $\mathcal{S}_{\text{line}} \neq \emptyset$. By Lemma 6.3, there exists $L_*$ such that $\mathcal{S}_{\text{line}} \subseteq \Omega \setminus U_0$ if $L \geq L_*$. Set

$$ (6.9) \qquad L^* := \max \left\{ L_*, \frac{M}{\pi s_*} \right\} $$

where $M$ is given by Lemma 6.3, and let $L \geq L^*$. The proposition follows once we show that $\mathcal{S}_{\text{pts}} \cap U_{-L/2} \neq \emptyset$.

By applying Lemma 6.3, Theorem 1 and Corollary 4.12, we deduce that $Q_{\varepsilon_n} \to Q_0$ in $H^1(U_{-\delta}, \mathbf{S}_0)$ for every $\delta > 0$, where $Q_0: U_0 \to \mathcal{N}$ is a locally minimizing harmonic map. Passing to the limit as $\varepsilon \to 0$ in Lemma 6.3, we see that

$$ (6.10) \qquad \frac{1}{2} \int_{U_0} |\nabla Q_0|^2 \leq M $$

for an $L$-independent constant $M$. In particular, $Q_0 \in H^1(U_0, \mathcal{N})$. An average argument, combined with Lemma 6.10, shows that there exists $-L < s < -L/2$ such that $Q_0 \in H^1(\{s\} \times B_r^2, \mathcal{N})$ and

$$ \frac{1}{2} \int_{\{s\} \times B_r^2} |\nabla Q_0|^2 \, d\mathcal{H}^2 \leq \frac{4M}{L}. $$

Due to Lemma 3.4, we find a lifting of $Q_0|_{\partial U_s}$, i.e. a map $\mathbf{n}_0 \in H^1(\partial U_s, \mathbb{S}^2)$ which satisfies (3.1) and $|\nabla Q_0|^2 = 2s_* |\nabla \mathbf{n}_0|^2$ a.e. Then, we have

$$ (6.11) \qquad \frac{1}{2} \int_{\{s\} \times B_r^2} |\nabla \mathbf{n}_0|^2 \, d\mathcal{H}^2 \leq \frac{2M}{s_* L}. $$

Combining (6.11) with (6.9), we deduce

$$ \frac{1}{2} \int_{\{s\} \times B_r^2} |\nabla \mathbf{n}_0|^2 \, d\mathcal{H}^2 \leq 2\pi. $$
Moreover, $\mathbf{n}_0$ takes a constant value on the boundary of $\{s\} \times B_r^2$, since $Q_0$ does. Then, by Lemma 6.4, $\deg(\mathbf{n}_0, \{s\} \times B_r^2) = 0$. On the other hand, $\deg(\mathbf{n}_0, \partial U_s \cap \Gamma_0) = 0$ since $\mathbf{n}_0$ takes a constant value on $\partial U_s \cap \Gamma_0$, and $\deg(\mathbf{n}_0, \Gamma_-)$ can be computed explicitly thanks to (6.4). This yields

$$ \deg(\mathbf{n}_0, \partial U_s) = \deg(\mathbf{n}_0, \Gamma_-) = \pm 1, $$

so the map $Q_{0|\partial U_s}$ is homotopically non-trivial and $\mathcal{I}_{pts} \cap U_s \neq \emptyset$. $\square$

**Acknowledgements.** The author would like to thank his Ph.D. advisor, professor Fabrice Bethuel, for constant support and helpful advice. He is also grateful to professor Giandomenico Orlandi, professor Arghir Zarnescu, Mark Wilkinson and Xavier Lamy for inspiring conversations and interesting remarks.

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