Lemma 2.4.4 from Kempf's Algebraic Varieties I have a question about the following lemma in the book.
 
Its proof starts by saying that without loss of generality we can assume that $X$ and $Y$ are affine, and $k[X]$ is a finitely generated module over $k[Y]$. Why is this justified? In particular, since cover by affines is not necessarily disjoint, why does solving it locally suffice?
Furthermore, they follow it with the an assertion that in fact it suffices to consider the case when $X$ is irreducible affine and $Y = f(X)$.  I fail to see why finiteness of morphism is preserved in this case?
Definition of a finite morphism is given as follows: a morphism $f: X \rightarrow Y$ is finite if there is an open cover $Y = \cup_i U_i$by affines $U_i$ whose inverse images are affine and $k[f^{-1}(U_i)]$ is finitely generated as $k[U_i]$ module.
Any help is appreciated. Thanks.
 A: The reason we can make these reductions is that finiteness is preserved by base change, and both steps of this reduction (restricting to some subset) can be described by base change. Here's how to make the reduction steps:
First, reducing to $Y$ irreducible: If $Z_2$ is irreducible, then $f(Z_2)$ is also irreducible, and it is contained in an irreducible component $Y_0$ because irreducible components are maximal with respect to inclusion among irreducible subsets. By restricting to $Y_0$, aka base changing along the closed immersion $Y_0\to Y$, we get to the situation $f|_{f^{-1}(Y_0)}: f^{-1}(Y_0)\to Y_0$, where $Y_0$ is irreducible and $f|_{f^{-1}(Y_0)}$ is again finite, being the base change of a finite morphism. Thus it suffices to treat the case where $Y$ is irreducible. 
If we reduce to the case where $Y$ is affine, this will get us to the case where $X$ is affine as well, as finite morphisms are affine: the preimage of any affine subset under an affine map is again affine. In order to reduce to the case of $Y$ affine, it suffices to note that we can tell if $f(Z_1)\subset f(Z_2)$ is a proper inclusion or not if $(f(Z_1)\cap U)\subset (f(Z_2)\cap U)$ is proper for every open subset $U$ in some cover of $Y$ (think about why this is - how do you know an inclusion is proper? You find a point in one but not in the other...). As restricting to $U\subset Y$ is the base change by the open immersion $U\to Y$, we see that finiteness (and therefore affineness) is again preserved. As every variety admits an open cover of affine varieties, this means we can reduce to the case of $X,Y$ affine.
By performing these reductions in sequence, we get what we want: $X,Y$ affine, $f$ finite, and $Y$ irreducible.
