Smooth morphisms vs. submersions

Question

This question is almost a duplicate of that question, which has a good answer. The difference is that I ask for references rather than proofs. By a reference I mean a reference to a book, or to a paper, or to an arXiv preprint. The answerer does not know references; see his comment.

Let $f\colon X\to Y$ be a surjective morphism of algebraic varieties over $\Bbb C$. Consider the following assertions:

Assertion 1. If the morphism $f$ is smooth, then it is a submersion, that is, for any $\Bbb C$-point $x\in X({\Bbb C})$, the linear map of the tangent spaces $d_x\colon T_x(X)\to T_{f(x)} Y$ is surjective.

Assertion 2. If the varieties $X$ and $Y$ are smooth, then the morphism $f$ is smooth if and only if it is a submersion.

I am asking for references for these two assertions. Many thanks in advance!

Answer 1

Assertion 2 is stated as Proposition 10.4 page 270 in Hartshorne's book. In fact, assertions 1 is equally proved in Hartshorne proposition 10.4 (though it is not in the statement of the proposition). It suffices to notice that the kernel of: $$ \Omega_{Y/\mathbb{C}} \longrightarrow \Omega_{X/ \mathbb{C}} \longrightarrow \Omega_{X/Y} \longrightarrow 0$$ is a torsion-sheaf as:

$\bullet$ $\Omega_{X/Y}$ is locally free of rank $\dim X - \dim Y$ (definition of smoothness)

$\bullet$ $\Omega_{X/\mathbb{C}}$ (resp. $\Omega_{Y/\mathbb{C}}$) is generically of rank $\dim X$ (resp. $\dim Y$) because $X$ is integral (resp. $Y$ is integral).

Then dualize this exact sequence an oberve that dual of a torsion sheaf is zero.