(Disclaimer: This question was also asked at MSE (https://math.stackexchange.com/questions/71020/can-we-collapse-omega-1-without-adding-a-dominating-real). I'm posting it here because, when I asked it, I was torn between my sense that it was appropriate for MO and my suspicion that this question is much, much easier than I'm making it; and it seems to be attracting no attention at MSE. I'm not a set theorist, so it's hard for me to judge the difficulty of the questions I'm asking; please let me know if this question is too elementary for MO, and I'll delete it.)

The question is precisely that of the title: is there a notion of forcing $\mathbb{P}$ which collapses $\omega_1$ to $\omega$ but does not add a real which dominates every real in the ground model? (Here "real" means "element of $\omega^\omega$.)

It seems like the answer should be "no," and I've attempted to prove this myself. The problem is that the easiest way to do so would be to define a dominating function in terms of an arbitrary surjection $f: \omega\twoheadrightarrow\omega_1;$ however, there seems to be no clear way to do this. My first thought was to look at the set $S_f=\lbrace n: \forall m < n, f(m) < f(n)\rbrace$. This is certainly a real, but there is no reason it should be dominating, let alone not present in the original model already; in fact, we can alter the usual collapsing poset to demand that $S_f$ be precisely the evens, or precisely the powers of 17, or in fact any infinite co-infinite subset of $\omega$.

On the other hand, looking at the poset of finite increasing functions from $\omega$ to $\omega_1$ (which builds a countable sequence cofinal in $\omega_1$, hence collapsing $\omega_1$ to $\omega$), it's unclear how this would add a dominating real; so perhaps the answer to my question is yes.