Let $f:X\to Y$ be a morphism of schemes, and $a:F\to G$ a morphism of coherent sheaves on $X$, with image $Im(a)=E\subseteq G$. I am trying to understand under which conditions one has $$f_\ast(Im(a))=Im(f_\ast a).$$
First of all, the displayed formula does not always hold: e.g. take any surjective $a$ such that $f_\ast a$ is not surjective.
[Aside. In the situation I am dealing with, I have the condition $R^1f_\ast(E)=0$ but, sadly, I do not believe it helps.]
Instead, if $K=\ker\,(F\twoheadrightarrow E)$, the condition $R^1f_\ast K=0$ might help, I guess. I try to explain why: by factoring $a$ through the image, we can write it as $F\to E\hookrightarrow G$, so we have $$Im\,(f_\ast F\overset{f_\ast a}{\longrightarrow}f_\ast G)=Im\,(f_\ast F\to f_\ast E\hookrightarrow f_\ast G),$$ and this equals $f_\ast E$ when $f_\ast F\to f_\ast E$ stays surjective, i.e. $R^1f_\ast K=0$.
I would like to know how much of the above is correct. I can summarize everything in the question:
Given a morhism $a:F\to G$ and the exact sequence $0\to K\to F\to E=Im(a)\to 0$, is it true that $f_\ast E=Im(f_\ast a)$ if and only if $R^1f_\ast K=0$?
Thank you for any help with this.