On Mumford's Red Book (end of section 7 of chapter 2, pg 126, 127), there is an alternative proof of Noether's Normalization Lemma that goes like this:
For an affine variety $X$ over an algebraically closed field $k$, we consider $X\subset \mathbb{A}^n_k$ and embed $\mathbb{A}^n_k\subset\mathbb{P}^n_k$ as the complement of the hyperplane $X_0=0$. Then we consider $\overline{X}$, the closure of $X$ in $\mathbb{P}^n_k$ and take $L$, a maximal linear subspace of $\mathbb{P}^n_k\setminus \mathbb{A}^n_k$.
First question: Why there exists such a maximal subspace?
Let $l_0=X_0$, $l_1,\dots,l_r$ be the generators of $L$ and set $W_i=\text{Spec }k[\frac{x_0}{l_i},\dots,1,\dots,\frac{x_n}{l_i}]$, an open affine subset of $\mathbb{P}^n_k$.
Now, define $\pi_1:\mathbb{P}^n_k\setminus L\rightarrow \mathbb{P}^r_k$, such that $\pi^{-1}(V_i)=W_i$ and $\pi_1^{*}(\frac{y_k}{y_i})=\frac{l_k}{l_i}$, where $V_i=\text{Spec }k[\frac{y_0}{y_i},\dots,\frac{y_r}{y_i}]$ are the affine opens of $\mathbb{P}^r_k$. By a previous lemma of Mumford, $\pi=\pi_1 |Z$ is a finite morphism.
Then he claims that $\pi^{-1}(V_0)=X$.
Second question: why?
And, finally, $\pi'=\pi|X:X\rightarrow V_0=\mathbb{A}^r_k$ is finite and surjective: it is finite because it is the restriction of a finite morphism, but...
Third question: why is $\pi'$ surjective?
Thank you very much.