# When does one have $f_\ast(Im(a))=Im(f_\ast a)$?

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.

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Taking the cohomology of $0\to K\to F\to E$, you can see that $f_*(E)/Im(f_*a)$ is the kernel of $R^1f_*K \to R^1f_*F$. – Cantlog Oct 16 '13 at 14:15
So, if $R^1f_\ast K=0$ then $f_\ast E=Im(f_\ast a)$, but not conversely? – Brenin Oct 17 '13 at 7:27
@Brenin: this is an equivalence if $R^1f_*F=0$. – Cantlog Oct 17 '13 at 10:51
@JeskoHüttenhain: my comment does not fully answer the question. It would need an example showing that $R^1f_*K=0$ is not a necessary condition. – Cantlog Oct 17 '13 at 10:56
Ah, jesus, I mixed up $F$ and $E$. I'll remove my comment. – Jesko Hüttenhain Oct 17 '13 at 11:11