I am a bit puzzled about some probable implication in a paper.  

Let $\varphi:\mathbb{P}^1\_K\to \mathbb{P}^1\_K$ be a rational map, where $K$ is a number field and let $\alpha\in K$ be such that $\varphi(\alpha)\neq\infty$. Then is it true that $\alpha$ is an exceptional point (backward orbit of $\alpha$ under $\varphi$ is finite) implies $\alpha$ is a preperiodic point (forward orbit of $\alpha$ under $\varphi$ is finite)? If this is not true providing an example will be appreciated.



A question about the backward orbit and forward orbit of a rational map in 1-dimensional projective space
Let $\alpha\_0=\alpha$, and for each $n\ge1$, choose a point $$\alpha\_n\in\phi^{-1}(\alpha\_{n-1})\subset \mathbb P^1(\overline K).$$ (The map $\phi$ is surjective on $\mathbb P^1(\overline K)$.) The points $\alpha\_n$ are in the backward orbit of $\alpha$, which you have assumed is finite. Hence I can find some $m>n\ge0$ such that $\alpha\_m=\alpha\_n$. Then  

$$
\alpha = \phi^m(\alpha\_m) = \phi^m(\alpha\_n) = \phi^{m-n}\circ\phi^n(\alpha\_n)
= \phi^{m-n}(\alpha).
$$
Hence $\alpha$ is periodic.


