evaluating gives identity on sets from the basic model

Question

I have a very basic question about forcing and generic set: why for $a\in M$ from the ground model do we have $$\dot{a}[G]=a$$ ? Where it is used in this equality here that $a\in M$ ?

Answer 1

Supposedly you're talking about $\check a$, the canonical name for $a\in M$.

In this case, the answer is by a simple $\in$-induction: Suppose that $a\in M$, and for all $b\in a$, $\check b[G]=b$. Since $\check a=\{(1_\Bbb P,\check b)\mid b\in a\}$ and $1_\Bbb P\in G$, we get that $\check a[G]=\{\check b[G]\mid b\in a\}$, but as we already assumed $\check b[G]=b$, so $\check a[G]=\{b\mid b\in a\}=a$.