Equivalent definitions of dominant rational map Let $f: X \dashrightarrow Y$ be a rational map of irreducible varieties $X$ and $Y$.
Definition 1. A rational map $f: X \dashrightarrow Y$ is dominant if for some open dense subset $U \subset X$ where $f$ is defined the image $f(U)$ is dense in $Y$.

Show that the following definition is equivalent:
Definition 2: A rational map $f: X \dashrightarrow Y$ is dominant if for some open dense subset $U \subset X$ where $f$ is defined, $f(U)$ contains an open dense subset of $Y$.

For $(1 \implies  2)$ I want to say that since continuous images of irreducible spaces are irreducible, $f(U)$ is irreducible. Then take
$V \subset Y$ open and take $V \cap f(U)$. This is open in $f(U)$ hence dense in $f(U)$ and, therefore, dense in $Y$. I would like to claim that $V \cap f(U)$ is open in $Y$ also but $f(U)$ is not necessarily open as far as I know. I'm totally stuck on $(2 \implies 1)$.
 A: For $2 \implies 1$, you just have to show that if $f(U)$ contains an open dense subset, then $f(U)$ is dense. For this, the openness doesn't matter. Any subset that contains a dense subset is itself dense.
The harder part is $1 \implies 2$. Part of the reason this is hard is because it is false as stated! Or to be more precise, there is an extra hypothesis which you did not state: the field needs to be algebraically closed.
Example: Work over the real numbers $\mathbb{R}$. Let $X = Y = \mathbb{R}$ and let $f:X \to Y$ be given by $f(x) = x^2$. The image of $f$ is $[0,\infty) \subset Y$, a dense subset (any infinite subset would be Zariski dense). But $f(X)$ doesn't contain any nonempty Zariski open subsets (because a nonempty Zariski open subset has finite complement, and $f(X)$ already has infinite complement).
So to prove $1 \implies 2$ you will need to take the hypothesis that the field $k$ is algebraically closed, and you will have to use that somehow. Often, and in this case, that means using some theorem that has algebraic closedness as a hypothesis. Can you think of a theorem, with a hypothesis of algebraic closedness, regarding regular images of varieties, such as the image of $f(U)$?
Hint 1:

 Try using Chevalley's theorem.

Hint 2:

 By Chevalley's theorem, the image $f(U)$ is constructible, meaning it is a finite union of differences of (closed) subvarieties of $Y$: $f(U) = (W_1-Z_1) \cup \dotsb \cup (W_k-Z_k)$ for (possibly reducible) varieties $W_i,Z_i \subseteq Y$. Closure commutes with finite unions (why?), so the closure of $f(U)$ is (contained in) $W_1 \cup \dotsb \cup W_k$. For this to be equal to $Y$, one of the $W$'s must equal $Y$, say $W_1=Y$ (why?). Now, what can you say about $W_1-Z_1$?

A: You just take $V\subset f(U)$ to be the dense open in $Y$, so $W=f^{-1}(V)$ is an open in $X$,and the map is defined on $W$, such that $ f(W)=U$ is dense, which is the first definition.
