# Dimension image of morphism of projective varieties

Let $$f: \mathbb{P}^n \to \mathbb{P}^m$$ be a rational map. Then there exists $$U \subset \mathbb{P}^n$$ open so that $$f_{|U}$$ is a morphism. What can we say about the dimension of $$\overline{f(U)}$$? We have that $$\dim\mathbb{P}^n = \dim U$$ and $$\dim f(U)= \dim\overline{f(U)}$$. Can I conclude that $$n=\dim U \geq \dim\overline{f(U)}$$ by the surjectvity of $$f : U \to f(U)$$ ? I think that's not true because $$U$$ in general is not a projective variety and $$f$$ need not be a morphism on $$\overline{U}$$. How can I proceed? Can I find more information assuming that $$U \subset \mathbb{A}^n$$?

Thanks for the help!