$(u,v)$ in $\Gamma_1$. More in general, for any $t\in [0,1]$, the point $q_t=(1-t) \cdot q_0 + t \cdot q_1$ belongs to the segment representing $(u,v)$ in $\Gamma_t$.

Now consider any three points $q_0$, $r_0$, and $s_0$ of $\Gamma_0$, which are moved to three points $q_1$, $r_1$, and $s_1$ of $\Gamma_1$ by $\langle \Gamma_0,\Gamma_1 \rangle$, respectively. Suppose that $q_i$ is to the left (to the right) of the line through $r_i$ and $s_i$, for each $i=0,1$. Then Corollary 7.2 in [@DBLP:journals/siamcomp/AlamdariABCLBFH17] ensures that $q_t$ is to the left (resp. to the right) of the line through $r_t$ and $s_t$, for any $t\in [0,1]$. The applicability of Corollary 7.2 from [@DBLP:journals/siamcomp/AlamdariABCLBFH17] to the morph $\langle \Gamma_0,\Gamma_1 \rangle$ is a consequence of the fact that $\langle \Gamma_0,\Gamma_1 \rangle$ is *unidirectional*, that is, all the node trajectories are parallel (indeed, each node $v\in V(T)$ moves along a horizontal line, given that $y_{\Gamma_0}(v)=y_{\Gamma_1}(v)$).

We prove the planarity of $\langle \Gamma_0,\Gamma_1 \rangle$. Suppose, for a contradiction, that two edges $(u,p(u))$ and $(v,p(v))$ of $T$ cross during $\langle \Gamma_0,\Gamma_1 \rangle$. For the sake of the simplicity of notation, let $u'=p(u)$ and $v'=p(v)$. Then there exists a point $q_0$ that belongs the segment $\overline{u_0u'_0}$ and such that, for 