Closure of image of algebraic variety

Question

Suppose $A$ is algebraic variety and $f$ is a regular map from $A$ to grassmannian. Let us consider closure of $f(A)$ in Zariski and analitical topology.

Is it true that these closures coincide?

Answer 1

Yes.
Actually this results from a very general theorem: given a morphism $f:X\to Y$ of algebraic varieties over $\mathbb C$, the subset $f(X)\subset Y$ has the same closure in the Zariski topology as in the classical topology of $Y$.
This is Proposition 7, page 12 in Serre's 1956 article GAGA article , which after 59 years remains the best source for this kind of questions.