$\def\sO{\mathcal{O}}
\def\supp{\operatorname{Supp}}
\def\sI{\mathcal{I}}
\def\sC{\mathcal{C}}
\def\colim{\operatorname{colim}}$I am studying complex spaces using Grauert, Remmert, Coherent Analytic Sheaves. In Chapter 2, §5.3 (p. 60), there is this result (where $(X,\sO_X)$ and $(Y,\sO_Y)$ are complex spaces):
In the proof, I don't understand why equality $$ \supp f_*(\sO_X)=f(X) $$ is true. I know how to see $(\supset)$. Namely, one uses the result that for a ring-valued diagram $F:\sI\to\mathsf{Ring}$, one has $\colim F=0$ if and only if $Fi=0$ for some $i\in\sI$. But how one does see $(\subset)$? If $y\in\supp f_*(\sO_X)$, then $f_*(\sO_X)_y\neq 0$, and by the categorical lemma, it follows that $\sO_X(f^{-1}(V))\neq 0$ for all open neighborhoods $V\subset Y$ of $y$, i.e., $f^{-1}(V)\neq\varnothing$ for these neighborhoods. After this, I don't know how to proceed.
EDIT: In Chapter 1, §1.4 there is this statement:
Actually, the argument above works to prove the identity $\supp f_*(\sO_X)=\overline{f(X)}$, so it suffices to show that $f(X)$ is closed. But why should $f(X)$ be closed at all? I guess it has to do somehow with the $\sO_Y$-coherence of $f_*(\sO_X)$.
Suppose we have $y \in Y \setminus f(X)$. Then by Hausdorffness and the open mapping theorem we find an open neighbourhood $x \in U \subseteq Y$ s.t. $U \cap f(X)= \emptyset$. Now we have $f_*(\mathcal{O}_X)(U)=\mathcal{O}_X(f^{-1}(U))=\mathcal{O}_X(\emptyset) = 0$.
As $x \in U$ we can conclude $f_*(\mathcal{O}_X)_x = 0$ and $x$ is not in the support of $f_*(\mathcal{O}_X)$.