 Solution

Let $\alpha =({\omega }_1{)}^M$ and takeThis formula is upward absolute between transitive models: if the smaller model contains such an $f$, the assumed absoluteness of “function”, domain, range, and $\omega$ shows that the same witness works in the larger model.

The generic union $g=\bigcup G$ is a total map $\omega \to \alpha$, because the conditions deciding each input form a dense set. For every $\beta <\alpha$, the conditions putting $\beta$ somewhere in the range are also dense, so $g$ is surjective. Thus $M[G]\models {\phi }_1(\alpha )$. But $M/\models {\phi }_1(\alpha )$ because $M$ regards $\alpha$ as its first uncountable ordinal. Hence ${\phi }_1$ is not downward absolute between $M$ and $M[G]$.