I study the book 3264 & All That Intersection Theory in Algebraic Geometry by Eisenbud & Harris and I'm a little bit confused on the proof of Kleiman’s theorem on pages 21:
Theorem 1.7 (Kleiman’s theorem in characteristic $0$). Suppose that an algebraic group $G$ acts transitively on a variety $X$ over an algebraically closed field of characteristic $0$, and that $A \subset X$ is a subvariety.
(a) If $B \subset X$ is another subvariety, then there is an open dense set of $g \in G$ such that $gA$ is generically transverse to $B$.
(b) More generally, if $\varphi:Y \to X$ is a morphism of varieties, then for general $g \in G$ the preimage $\varphi^{-1}(gA)$ is generically reduced and of the same codimension as $A$.
Proof: (a) This is the special case $Y = B$ of (b).
(b) Let the dimensions of $X, A, Y$ and $G$ be $n, a, b$ and $m$ respectively. If $x \in X$, then the map $G \to X, g \mapsto gx$ is surjective and it's fibers are the cosets of the stabilizer of $x$ in $G$. Since all these fibers have the same dimension, this dimension must be $m -n$. Set
$$\Gamma=\{(x,y,g) \in A \times Y \times G \vert gx= \varphi(y) \}.$$
Because $G$ acts transitively on $X$, the projection $\pi: \Gamma \to A \times Y$ is surjective. Its fibers are the cosets of stabilizers of points in $X$, and hence have dimension $m - n$. It follows that $\Gamma$ has dimension
$$\dim \Gamma =a+b+m-n$$
On the other hand, the fiber over $ g$ of the projection $\Gamma \to G$ is isomorphic to $\varphi^{-1}(gA)$. Thus either this intersection is empty for general $g$, or else it has dimension $a + b - n$, (???, see Q_1) as required.
Since $X$ is a variety it is smooth at a general point. Since $G$ acts transitively, all points of $X$ look alike, so $X$ is smooth. Since any algebraic group in characteristic $0$ is smooth (see for example Lecture 25 of Mumford [1966]), the fibers of the projection to $A \to Y$ are also smooth, so $\Gamma$ itself is smooth over $A_{\text{sm}} \times Y_{\text{sm}}$. Since field extensions in characteristic $0$ are separable, the projection $(\Gamma\setminus\Gamma_{\text{sing}}) \to G$ is smooth over a nonempty open set of $G$ (???), where $\Gamma_{\text{sing}}$ is the singular locus of $\Gamma$. ...
two steps I not understand.
Q_1: which intersection is meant in text in "...Thus either this intersection is empty for general $g$..."?
Q_2: why the fact that field extensions in characteristic $0$ are separable implies that the projection $(\Gamma\setminus\Gamma_{\text{sing}}) \to G$ is smooth over a nonempty open set of $G$?
The text referenced in question 1 is thinking about the intersection of the preimage of $gA$ inside $A\times Y \times G$ with the graph $\Gamma\subset A\times Y\times G$. All that's going on here is that if you have a map $f:X\to Y$ and a subset $Z\subset X$, you get that $(f|_Z)^{-1}(Y)=f^{-1}(Y)\cap Z$.
Question 2 is just giving you a reminder of the main idea of the proof for generic smoothness of a morphism in characteristic zero.
In fact, if $X$ is smooth over a field, the result gets better: there exists a dense open subset $V\subset Y$ so that $f|_{f^{-1}(V)}$ is smooth. See this handout from Vakil, items 3.1 and 3.3, or Vakil's Rising Sea, items 25.3.1 and 25.3.3 for instance. I'm sure it's in other sources as well, but this is where I first learned it.
Anyways, the proof (of the first statement) works by examining what happens at the generic points via the fraction fields - since the extension $K(Y)\to K(X)$ is of transcendence degree $n=\dim(X)-\dim(Y)$, it's separably generated by $n$ elements. So we get that our map is smooth at the generic point by examining the stalk of the sheaf of differentials, and as smoothness is local on the source, on some dense open.
The way we use this here is that we can break $f:\Gamma\setminus\Gamma_{sing}$ in to a disjoint integral smooth subvarieties and $G$ in to it's connected components. Running the upgraded result on each of these, we obtain the fact that $f:\Gamma\setminus\Gamma_{sing}\to G$ is smooth on a dense open subset of the form $f^{-1}(V)$ for $V\subset G$ open.