Suppose we are given two algebraic varieties $X, Y$ over an arbitrary field $\mathbb{k}$, a rational map $$\pi \colon X \mathrel{-\,}\rightarrow Y$$
and $Z$ is the generic fiber of $\pi$.
Is it always true that $X$ is birational to $Y \times Z$? If yes, how can I prove this? If not, by any chance is it true for the particular case, namely: $$\pi \colon Sym^2(\mathbb{P^2}) \mathrel{-\,}\rightarrow Grass(\mathbb{P}^{1},\mathbb{P^2})$$ where $\pi$ maps two points in general positions to the line passing through them? As far as I know, the generic fiber of $\pi$ is $Sym^2(\mathbb{P}^1)$ which is birational to $\mathbb{P}^2$ so that means there is a birational equivalence $$Sym^2(\mathbb{P^2}) \sim \mathbb{P^2} \times \mathbb{P^2}$$