Suppose $f: X \to Y$ is a holomorphic map of connected compact complex manifolds. Is there an easy way to see that $f(X)$ is analytic (i.e. Zariski closed) in $Y$? It is clear that $f(X)$ is closed, because images of compact sets are compact, and in a Hausdorff space, compact implies closed. But this is for the "ordinary" topology, not the Zariski topology. In [1], they proof
Proper Mapping Theorem: For any proper holomorphic map [of arbitrary complex spaces] $f: X\to Y$ the image set $f(X)$ is an analytic set in $Y$.
The proof is first noting that $f(X) = \operatorname{Supp} f_* \mathcal O_X$. Then by the Direct Image Theorem, $f_* \mathcal O_X$ is a coherent sheaf on $Y$, so its support is analytic. $\square$
But proving that $f_* \mathcal O_X$ is coherent is quite a lot of work (in fact this is one of the main endeavors of [1], besides introducing analytic spaces), so I wonder if this might be obtained simpler, if we restrict ourself to compact manifolds.
My application is actually to show that if the Jacobian of $f$ has maximal rank $\geq \dim Y$ at some point $p \in X$, then $f$ is surjective. The rank assumption will ensure that $f(X)$ contains a small ball, which is Zariski dense. So if $f(X)$ is analytic, it has to be everything.
[1] Grauert, Remmert; Coherent Analytic Sheaves