\section{Symmetric Connectivity Criterion}
\label{sec:sym_conn_criterion}

In what follows, we fix the translation subgroup of the planar group $p1$ by the two primitive vectors
\begin{equation}
    \mathbf e_1=(1,0),\qquad \mathbf e_2=(0,1), 
\end{equation}
and all occurrences of connected mean path-connected.

Let the (half-open) unit cell and the $2\times 2$ supercell be
\begin{equation}
    B=[0,1)\times[0,1),\qquad
A=[0,2)\times[0,2). 
\end{equation}
Let
\begin{equation}
    \pi:\mathbb R^2\to \mathbb T^2:=\mathbb R^2/\mathbb Z^2 
\end{equation}
be the quotient (covering) map.

For a path-connected subset $C\subset\mathbb T^2$, the full preimage $\pi^{-1}(C)\subset\mathbb R^2$ may have
several path-connected components. For any such component $\widetilde C$, define its stabilizer
\begin{equation}
    H(\widetilde C)=\{g\in\mathbb Z^2\mid \widetilde C+g=\widetilde C\}. 
\end{equation}

\begin{lemma}\label{lem:stabilizer_well_defined}
Let $C\subset\mathbb T^2$ be path-connected. If $\widetilde C_1,\widetilde C_2$ are any two path-connected components
of $\pi^{-1}(C)$, then there exists $k\in\mathbb Z^2$ such that $\widetilde C_2=\widetilde C_1+k$, and moreover
\begin{equation}
    H(\widetilde C_1)=H(\widetilde C_2). 
\end{equation}
Hence $H(\widetilde C)$ depends only on $C$, and we may write it as $H(C)$.
\end{lemma}

\begin{proof}
Pick $x_1\in\widetilde C_1$ and set $\bar x:=\pi(x_1)\in C$. Since $\pi(\widetilde C_2)=C$, there exists
$x_2\in\widetilde C_2$ with $\pi(x_2)=\bar x$. Then $x_2-x_1\in\mathbb Z^2$; write $k:=x_2-x_1$ so that
$x_2=x_1+k$.

The translation $T_k(x)=x+k$ satisfies $\pi\circ T_k=\pi$, hence $T_k(\pi^{-1}(C))=\pi^{-1}(C)$.
Therefore $T_k(\widetilde C_1)$ is path-connected, contained in $\pi^{-1}(C)$, and contains $x_2$. By maximality of the component $\widetilde C_2$ containing $x_2$, we get $T_k(\widetilde C_1)\subset \widetilde C_2$. Applying the same argument to $T_{-k}$ yields the reverse inclusion, hence $\widetilde C_2=T_k(\widetilde C_1)=\widetilde C_1+k$.

Finally, for any $g\in\mathbb Z^2$,
\begin{equation}
    (\widetilde C_1+k)+g=\widetilde C_1+k
\iff
\widetilde C_1+g=\widetilde C_1, 
\end{equation}
so $g\in H(\widetilde C_1)$ iff $g\in H(\widetilde C_2)$. Thus $H(\widetilde C_1)=H(\widetilde C_2)$.
\end{proof}

\begin{lemma}
\label{lem:two_p1_invariant_components_intersect}
Let $E,F\subset\mathbb R^2$ be nonempty, $\mathbb Z^2$-invariant subsets, i.e.
\begin{equation}
    E+(m,n)=E,\qquad F+(m,n)=F,\qquad \forall (m,n)\in\mathbb Z^2. 
\end{equation}
If both $E$ and $F$ are path-connected, then $E\cap F\neq\varnothing$.
\end{lemma}

\begin{proof}
Let $\pi:\mathbb R^2\to\mathbb T^2$ be the quotient map. We first show $\pi(E)\cap\pi(F)\neq\varnothing$.

Pick $p\in E$. Since $p+\mathbf e_1\in E$ and $E$ is path-connected, there exists a path
\begin{equation}
    \gamma_x:[0,1]\to E,\qquad \gamma_x(0)=p,\ \gamma_x(1)=p+\mathbf e_1. 
\end{equation}
Set $\alpha:=\pi\circ\gamma_x$, which is a loop in $\mathbb T^2$ based at $\pi(p)$.
Likewise, pick $q\in F$. Since $q+\mathbf e_2\in F$ and $F$ is path-connected, there exists a path
\begin{equation}
    \gamma_y:[0,1]\to F,\qquad \gamma_y(0)=q,\ \gamma_y(1)=q+\mathbf e_2, 
\end{equation}
and set $\beta:=\pi\circ\gamma_y$, a loop in $\mathbb T^2$ based at $\pi(q)$.

Lift $\alpha$ to $\widetilde\alpha$ with $\widetilde\alpha(0)=p$. By uniqueness of path lifting,
$\widetilde\alpha=\gamma_x$, hence $\widetilde\alpha(1)-\widetilde\alpha(0)=\mathbf e_1$. Similarly, the lift of
$\beta$ starting at $q$ satisfies $\widetilde\beta(1)-\widetilde\beta(0)=\mathbf e_2$. Thus, under the standard
identification $\pi_1(\mathbb T^2)\cong\mathbb Z^2$, the loops $\alpha,\beta$ represent the classes $(1,0)$ and $(0,1)$.

By intersection theory on surfaces (e.g. the mod-$2$ intersection number; see Sec. 2.4 of \citet{guilleminDifferentialTopology1974}), the mod-$2$ intersection number of two loops depends only on their homotopy (equivalently homology) classes, and the two coordinate generators $(1,0)$ and $(0,1)$ have mod-$2$ intersection equal to $1$. Hence $\alpha$ and $\beta$ cannot be disjoint, so $\alpha([0,1])\cap\beta([0,1])\neq\varnothing$. Consequently,
\begin{equation}
    \pi(E)\cap\pi(F)\neq\varnothing. 
\end{equation}

Now take $\bar z\in\pi(E)\cap\pi(F)$. Choose $e\in E$ and $f\in F$ with $\pi(e)=\pi(f)=\bar z$.
Then $e-f\in\mathbb Z^2$. Let $t:=e-f\in\mathbb Z^2$, so $f+t=e$. Since $F$ is $\mathbb Z^2$-invariant, $f+t\in F$,
hence $e\in E\cap F$. Therefore $E\cap F\neq\varnothing$.
\end{proof}

\begin{theorem}
\label{thm:2x2_gamma_connectivity}
Let $S\subset\mathbb R^2$ be $\mathbb Z^2$-invariant. Fix $A=[0,2)\times[0,2)$, $B=[0,1)\times[0,1)$ and
$\Gamma \subset B$ as above. Assume that every path-connected component of
$S\cap A$ intersects $\Gamma$. Then $S$ is path-connected.
\end{theorem}

\begin{proof}
Let $C:=\pi(S)\subset\mathbb T^2$, and write the decomposition into path-connected components
$C=\bigsqcup_{j\in J} C_j$. Since $S$ is $\mathbb Z^2$-invariant, one has the identity
\begin{equation}
    S=\pi^{-1}(C), 
\end{equation}
because if $\pi(x)\in C$ then $\pi(x)=\pi(y)$ for some $y\in S$, hence $x-y\in\mathbb Z^2$ and thus $x\in S$.

Fix $j\in J$, and choose any lift component $\widetilde C_j\subset\pi^{-1}(C_j)$. By translating $\widetilde C_j$
by some integer vector (which yields another lift component of the same $C_j$ by Lemma~\ref{lem:stabilizer_well_defined}),
we may assume
\begin{equation}
    \widetilde C_j\cap B\neq\varnothing. 
\end{equation}
Consider $\mathbf e_1$. If $\mathbf e_1\notin H(\widetilde C_j)$, then $\widetilde C_j$ and $\widetilde C_j+\mathbf e_1$
are two distinct (hence disjoint) path-connected components of $\pi^{-1}(C_j)\subset S$.
Since $\widetilde C_j\cap B\neq\varnothing$, we have $(\widetilde C_j+\mathbf e_1)\cap (B+\mathbf e_1)\neq\varnothing$,
so $(\widetilde C_j+\mathbf e_1)\cap A\neq\varnothing$. Let $D$ be any path-connected component of
$(\widetilde C_j+\mathbf e_1)\cap A$. Then $D$ is a path-connected component of $S\cap A$ (it cannot connect inside $A$
to any other lift component because distinct lift components are disjoint). Moreover, $D\subset B+\mathbf e_1$, hence
$D\cap\Gamma=\varnothing$ for all choices of $\Gamma\in\{\Gamma_1,\Gamma_2,\Gamma_1\cup\Gamma_2\}$, since $\Gamma\subset\partial B$
and $B+\mathbf e_1$ is disjoint from $\partial B$.
This contradicts the hypothesis that every path-connected component of $S\cap A$ intersects $\Gamma$.
Therefore $\mathbf e_1\in H(\widetilde C_j)$.

The same argument with $\mathbf e_2$ in place of $\mathbf e_1$ shows $\mathbf e_2\in H(\widetilde C_j)$.
Hence $H(\widetilde C_j)$ contains $\mathbf e_1$ and $\mathbf e_2$, and thus
\begin{equation}
    H(\widetilde C_j)=\mathbb Z^2. 
\end{equation}
By Lemma~\ref{lem:stabilizer_well_defined}, this implies $H(C_j)=\mathbb Z^2$.

Assume for contradiction that $|J|\ge 2$, and pick two distinct components $C_{j_1},C_{j_2}$.
Choose lift components $\widetilde C_{j_1}\subset\pi^{-1}(C_{j_1})$ and $\widetilde C_{j_2}\subset\pi^{-1}(C_{j_2})$.
By Step~1, both satisfy $H(\widetilde C_{j_\ell})=\mathbb Z^2$, hence each $\widetilde C_{j_\ell}$ is a $\mathbb Z^2$-invariant
path-connected subset of $\mathbb R^2$. Then Lemma~\ref{lem:two_p1_invariant_components_intersect} yields
$\widetilde C_{j_1}\cap \widetilde C_{j_2}\neq\varnothing$, which contradicts the fact that
$\pi(\widetilde C_{j_1})\subset C_{j_1}$ and $\pi(\widetilde C_{j_2})\subset C_{j_2}$ with $C_{j_1}\cap C_{j_2}=\varnothing$.
Therefore $|J|=1$, i.e. $C$ is path-connected.

Since $C$ is path-connected and $H(C)=\mathbb Z^2$, Lemma~\ref{lem:stabilizer_well_defined} implies that
$\pi^{-1}(C)$ has only one lift component, hence $\pi^{-1}(C)$ is path-connected. Using $S=\pi^{-1}(C)$, we conclude
that $S$ is path-connected.
\end{proof}