I am reading an article (on second order characterizability) which at some point in a proof states that by forcing with $\mathbb P=\{f:\alpha\to\{0,1\},\alpha\in\omega_1\}$ we do not add subsets to $\omega$ and we preserve cardinals. This due to the forcing being $\omega_1$-closed (I suppose this guarantees that there are no new subsets in $\omega$) and $\omega_2$-c.c. (I suppose this preserves the cardinals).
I do not understand, however, how exactly those two properties ($\omega_1$-closed and $\omega_2$-c.c.) give those facts in the forcing extension. I'd appreciate it some guidance in this regard.
If you are familiar with the proof that c.c.c. forcings preserve cardinals, then the same thing holds here. By $\omega_2$.c.c. we have that no cardinals above $\omega_2$ are collapsed. To see why, suppose that $\dot f$ is a name for function from $\kappa$ to $\lambda$, with $\kappa<\lambda$ cardinals in the ground model with $\lambda$ regular. For each $\alpha<\kappa$ there is an antichain of size $<\omega_2$, $D_\alpha$ such that each condition $p\in D_\alpha$, $p$ decides the value of $\dot f(\check\alpha)$.
This means that the set $F_\alpha=\{\beta\mid\exists p\in D_\alpha: p\Vdash\dot f(\check\alpha)=\check\beta\}$ is in the ground model and has size at most $\aleph_1$ there. But this also means that the range of $\dot f$ is necessarily a subset of $\bigcup F_\alpha$, which has cardinality $\kappa\cdot\aleph_1<\lambda$, therefore the range of $\dot f$ has to be bounded in $\lambda$.
In the case that $\lambda$ is singular, it's easy to observe that if it is collapsed, then some regular cardinal below $\lambda$ is also collapsed.
The closure gives us the other direction, given any function whose domain is a subset of $\omega_1$, we can extend the conditions and decide more and more information about it, until we decide all the values of the function. This means that the function was in the ground model. So no cardinals were collapsed above $\omega_2$ or below $\omega_3$. So no cardinals have been collapsed.