Isomorphism of Forcing Posets

Question

Let $\text{Col}(\alpha, \beta)$ be the forcing poset that collapses $\beta$ to $\alpha$, let $Q$ be any old forcing poset, with $|Q| = \lambda$.

Why is $Q \times \text{Col}(\omega, \lambda) \cong \text{Col}(\omega,\lambda)$?

Answer 1

This is a consequence of Proposition 10.20 in Kanamori's The Higher Infinite (p. 129):

Suppose that $P$ is a separative partial order such that $|P|\leq|\alpha|$ and$$\Vdash_P\exists f(f\colon\omega\to\alpha\text{ is surjective}\land f\notin\check V).$$ Then there is an injective, dense embedding of a dense subset of $\text{Col}(\omega,\{\alpha\})$ into $P$.

Note that $\text{Col}(\omega,\lambda)$ has cardinality $\lambda$, and therefore $Q\times\text{Col}(\omega,\lambda)$ has the same cardinality. Both forcings add a new surjection from $\omega$ onto $\lambda$, and therefore both have a dense subset isomorphic to a dense subset of $\text{Col}(\omega,\lambda)$.

It follows that the Boolean completions are isomorphic, and so both forcings are equivalent.