Let $P$ be a separative partial order such that $\left| P \right| \leq \left| \alpha \right|$ and $$\Vdash_P\exists f(f\colon\omega\to\alpha\text{ is surjective}\land f\notin\check V).$$ I want to prove that for any $p \in P$ there is a maximal antichain below p of cardinality $\left| \alpha \right|$. In Proposition 10.20 in Kanamori's "The Higher Infinite" (p. 129) this is proved as follows: If $\left| \alpha \right| = \omega$, this follows from $\Vdash f \notin \check V$ since any condition must have incompatible extensions. If $\left| \alpha \right| > \omega$, this follows from the collapse of $\left| \alpha \right|$ to $\omega$ and the consequent failure of the $\left| \alpha \right|-c.c$.
I see that in the case of $\left| \alpha \right| = \omega$ it is enough to prove that any condition have incompatible extensions. But how does this follows from $\Vdash f \notin \check V$? And how does the failure of the $\left| \alpha \right|-c.c$ follow from the collapse of $\left| \alpha \right|$ to $\omega$ in the case of $\left| \alpha \right| > \omega$? I know this is true if $\left| \alpha \right|$ is regular.