are going to use later. First, we observe that whether a linear morph is upward only depends on the upwardness of the initial and final drawings of the morph.

\[obs:upward-stays-upward\] Let $\Gamma_0$ and $\Gamma_1$ be two strictly-upward straight-line drawings of a rooted tree $T$. Then the linear morph $\langle \Gamma_0,\Gamma_1 \rangle$ is upward.

Assume that the morph $\langle \Gamma_0,\Gamma_1 \rangle$ happens between the time instants $t=0$ and $t=1$. For any $t\in [0,1]$, denote by $\Gamma_t$ the drawing of $T$ in $\langle \Gamma_0,\Gamma_1 \rangle$ at time $t$.

Consider any edge $(p(v),v)$ of $T$. Since $\Gamma_0$ and $\Gamma_1$ are strictly-upward, it follows that $y_{\Gamma_0}(p(v))>y_{\Gamma_0}(v)$ and $y_{\Gamma_1}(p(v))>y_{\Gamma_1}(v)$. Hence, at any time instant $t\in [0,1]$ of the morph $\langle \Gamma_0,\Gamma_1 \rangle$, we have $y_{\Gamma_t}(p(v))=(1-t)\cdot y_{\Gamma_0}(p(v)) + t \cdot y_{\Gamma_1}(p(v)) > (1-t)\cdot  y_{\Gamma_0}(v) + t \cdot y_{\Gamma_1}(v) = y_{\Gamma_t}(v)$. It follows that the drawing $\Gamma_t$ is strictly-upward, and hence that $\langle \Gamma_0,\Gamma_1 \rangle$ is an upward morph.

The planarity of a morph can not be ensured as simply as its upwardness. However, if the initial and final drawings of the morph satisfy some further conditions, it turns out that planarity is actually guaranteed; this is similar to a lemma by Da Lozzo [*et al.*]{} [@DBLP:conf/gd/LozzoBFPR18].

