those properties of $u$ which can be proved without any use of the reciprocality conditions.

The primary technical difficulty we must deal with is that, without assuming the reciprocality conditions, we do not know [*a priori*]{} that the function $u$ is continuous. However, it is shown in Lemma 4.6 of [@Raj:16] that $u$ satisfies a maximum and a minimum principle. To state it, we use the following notation. For an open set $\Omega \subset X$, or a relatively open set $\Omega \subset Q$, let $$\partial_* \Omega = (\partial \Omega \cap Q) \cup (\overline{\Omega} \cap (\zeta_1 \cup \zeta_3)) .$$ Then we have the following.

\[lemm:maximum\_principle\] Let $\Omega \subset X$ be open. Then $\sup_{x \in \Omega \cap Q} u(x) \leq \sup_{y \in \partial_*\Omega} u(y)$ and $\inf_{x \in \Omega \cap Q} u(x) \geq \inf_{y \in \partial_*\Omega} u(y)$.

Lemma \[lemm:maximum\_principle\] allows us to establish topological properties for the closures of sets of the form $u^{-1}([s,t])$.

\[prop:connect\] For all $s, t \in [0,1]$, $s \leq t$, the set $\overline{u^{-1}([s,t])}$ is connected and intersects both $\zeta_2$ and $\zeta_4$.

Let $E = \overline{u^{-1}([s,t])}$. To prove the first claim, suppose that $E$ is not connected. Then there is an open set $U \subset X$ such that $$\label{sussa}
        U 