I was looking at the proof of the Noether normalization lemma in the affine case.
To be clear : let $X$ be an affine variety , there exists a finite map $\phi : X \to \mathbb{A}^m$.
In the proof, the authors embedded $X$ in $\mathbb{A}^n$ for some $n$. Then they embedded this affine space in $\mathbb{P}^n$ ,took the closure of $X$ (call that $Y$) in this space and made a projection from a point nor in $Y$, nor in the affine space.
So, they got a map $ \psi : Y \to \mathbb{P}^{n-1}$ which is a finite morphism with the image and that takes $X$ to $\mathbb{A}^{n-1}$.
Iterating that, they got a finite morphism $\phi : Y \to \mathbb{P}^m $ that takes $X$ in $\mathbb{A}^m$, but i can't really see why the restriction to $X$ is still finite with the image.
What am i missing? Is this trivial?
Hope i've been clear about my doubt
You are right in observing that finiteness is in general not preserved when passing to an open subset of the domain. However, it is preserved under base change, so what you need to see is that $$ X = \mathbb{A}^m \times_{\mathbb{P}^m} Y, $$ then the map $X \to \mathbb{A}^m$ is just a base change of the finite $Y \to \mathbb{P}^m$.
As to why this is true: $\mathbb{A}^m \to \mathbb{P}^m$ is an open immersion, so the right hand side of the above fibered product is $\phi^{-1} (\mathbb{A}^m)$. To see equality with $X$, consider the case of one projection $\psi$ and observe that the projection $\mathbb{P}^n \setminus\{x\} \to \mathbb{P}^{n-1}$ maps the decomposition $$\mathbb{P}^{n} = \mathbb{A}^n \cup \mathbb{P}^{n-1}$$ to a similar decomposition of $\mathbb{P}^{n-1}$.