Doubt on proof of Implicit function theorem 
On The second part of the proof, where it's stated that V is open as it is the inverse image of the open set $V_0$ under the continuous mapping $y \rightarrow (0, y)$.
Let $\pi$ be this continuous mapping. Then, $\forall _{U_{open\text{ in }\mathbb{R}^n\times\mathbb{R}^p}} \pi^{-1}(U)$ is open in/relative to the domain of $\pi$, since $\forall_\epsilon \exists_\delta Dom(\pi)\cap B(v_1;\delta)\subset \pi^{-1}(B(v_0;\epsilon))$ with $\lim_{y\rightarrow v_1} \pi (y) = \pi(v_0)$.
In this situation, is $Dom(\pi)=V$ or $Dom(\pi)=V_1$, where $V$ and $V_1$ are as defined in the image? If it's the first, then I do not understand how $V= \pi^{-1} (V_0)$ is open in/relative to $\mathbb{R}^p$, even if $V_0$ is open in $\mathbb{R}^n\times\mathbb{R}^p$...
If it's the second possibility, then I understand that $\pi^{-1} (V_0)$ is open in $V_1$, and since $V_1$ is open in $\mathbb{R}^p$, $\pi^{-1} (V_0)$ is also open in $\mathbb{R}^p$. In this last possibility is that then $\pi^{-1} (V_0)\neq V$... Then how do we prove that $V$ is open in $\mathbb{R}^p$ ?

So, I think I now understand what's happening.
$Dom(\pi)=V_1$ and yet we still have $\pi^{-1} (V_0)=V$ since, $V=\{y\in V_1|(0,y)\in V_0\}$  and $\pi(y)=(0,y)$.
Silly me!
