Let $(f,\tilde{f}): (X, \mathscr{O}_X) \rightarrow (Y, \mathscr{O}_Y)$ be a holomorphic map between complex spaces, such that $f_*(\mathscr{O}_X)$ is $\mathscr{O}_Y$-coherent. Define $\mathscr{I} = \mathscr{A}n(f_*(\mathscr O_X))$ to be its annihilator.
Apparently, then $\mathscr Ker(\tilde{f}) = \mathscr I$; why? It looks like something that should be straightforward, but I believe I fail to see how exactly $\tilde{f}$ sends any $g_y \in \mathscr O_y, y \in Y,$ to $f_*(\mathscr O_X)$.
The equality is equivalent to $\tilde{f}(g_y) = 0 \in f_*(\mathscr O_X) \Longleftrightarrow g_y \cdot f_*(\mathscr O_X)_y = 0$, but I do not see how this has to be the case.
PS: I failed to find a good way to typeset Ker and An.
As anticipated, this is indeed straightforward. The point is simply the following:
$f_*(\mathscr{O}_X)$ is $\mathscr{O}_Y$-coherent, implying it is an $\mathscr{O}_Y$-algebra. But what makes it into an $\mathscr{O}_Y$-algebra is precisely the ring-homomorphism $\tilde f$, by defining $ab :=\tilde f(a)b, a\in \mathscr{O}_Y, b \in f_*(\mathscr{O}_X)$. From this, the statement easily follows.