Soft sheaves adapted to $f_!$ I'm reading Gelfand-Manin, Homological Algebra. I understand that the class of soft sheaves is sufficiently large, because every injective sheaf is soft. Now to see that this class is adapted to $f_!$, I have to show that every acyclic complex of soft sheaves is mapped by the functor $f_!$ to an acyclic complex. 
For this, I'd say it's enough to just show that if 
$0\rightarrow \mathcal{F}\rightarrow \mathcal{G}\rightarrow\mathcal{H}\rightarrow 0$ 
is an exact sequence of soft sheaves, then $0\rightarrow \mathcal{f_!F}\rightarrow \mathcal{f_!G}\rightarrow\mathcal{f_!H}\rightarrow 0$ is also exact.
But in the book, they say the above is only enough because of an exercise from earlier which says that in an exact sequence of sheaves $0\rightarrow \mathcal{F}\rightarrow \mathcal{G}\rightarrow\mathcal{H}\rightarrow 0$ , if $\mathcal{F}$ and $\mathcal{G}$ are soft, then so is $\mathcal{H}$.
Can somebody tell me where this exercise is needed in the proof of the adaptedness of soft sheaves to $f_!$ ?
 A: Well, to show that $f_!$ takes acyclic complexes of soft sheaves to acyclic complexes, we need to split up an acyclic complex of soft sheaves
$$0 \to \mathcal F_1 \xrightarrow{\phi_1} \mathcal F_2 \xrightarrow{\phi_2} \mathcal F_3 \xrightarrow{\phi_3} \mathcal F_4 \xrightarrow{\phi_4} \dots$$
into short exact sequences, namely
$$0 \to \mathcal F_1 \xrightarrow{\phi_1} \mathcal F_2 \xrightarrow{\phi_2} \text{im}(\phi_2) \to 0,$$
$$0 \to \text{im}(\phi_2) \to \mathcal F_3 \to \text{coker} (\phi_2) \to 0,$$
$$0 \to \text{coker}(\phi_2) \xrightarrow{\phi_3} \mathcal F_4 \xrightarrow{\phi_4} \text{im}(\phi_4) \to 0,$$ and so on.
Showing that $f_!$ preserves the acyclicity of our complex is equivalent to showing that it preserves these short exact sequences. This does not immediately follow from the assertion that $f_!$ preserves short exact sequences of soft sheaves, however, as we need to know that $\text{im}(\phi_2)$, $\text{coker}(\phi_2)$, $\text{im}(\phi_4)$, etc. are soft. This is where the earlier exercise is needed.
A: It is not needed in the proof, it just makes it easier. Because of that exercise you only need to prove that $f_!$ is right exact.
