with the $K_1$ case, see below) the simple choice $\gamma(s)=s$ would not have been possible. The boundary conditions on $\gamma''$, which give continuity of $Dv$ across vertical boundaries, are only a matter of convenience.

In the rest of the proof, we estimate the energy of the function we constructed. First, we easily check that $v$ is continuous and obeys the boundary conditions. By construction, $Dv=B_2\in K_2$ in the sets $\omega_1$ and $\omega_5$. A short computation shows that $|Dv-B_2|=|\partial_1v_1-1|\le c \alpha h^2/\ell^2$ in $\omega_3$. In order to estimate the energy in $\omega_2$ and $\omega_4$, we compute $$\label{eqlinearize}
  \begin{split}
 \min_{Q\in \operatorname{SO}(2)} |Dv-QA_2|
\le& \min_{\varphi\in\mathbb{R}}
|\partial_1v_1-\cos\varphi|+|\partial_2v_2-(1-\alpha)\cos\varphi|\\
&
+|\partial_1v_2-\sin\varphi|+|\partial_2v_1+(1-\alpha)\sin\varphi|\\
  \le& |\partial_1 v_1-1|+|\partial_2 v_2-(1-\alpha)| \\
&+ 
|\partial_2 v_1+(1-\alpha)\partial_1  v_2| + c |\partial_1 v_2|^2    \,,
  \end{split}$$ where we have chosen $\varphi$ such that $\sin\varphi=\partial_1v_2$ if $\partial_1v_2\in[-1,1]$, $\varphi=0$ otherwise. Inserting the definition of $v$, one sees that the second and third term vanish, the remainder can be estimated by $$\begin{aligned}
1
\mathrm{dist}(Dv,K_2)\le& 
c\alpha \frac{h^2}{\ell^2} \|\gamma''\|_{L^\infty}
+ c\alpha\frac{h^2}{\ell^2} \|\gamma'\|_{L^\infty}^2
+ c \Bigl(\alpha \frac{h}{\ell} \|\gamma'\|_{L^\infty}\Bigr)^2\\
\le&
c\alpha \frac{h^2}{\ell^2}\quad \text{ in }\omega_2\cup\omega_4\,,\end{aligned}$$ and therefore, $$ \int_{\omega_2\cup\omega_4}\operatorname{dist}^2(Dv(x,y),K_2)\, dxdy  \leq
 c\alpha^2\frac{h^5}{\ell^3}.$$ A simple computation shows that $|Dv-\Id|\le c\alpha$ on $\omega$.

We finally turn to the interfacial part of the energy. In each 