The FHHF theorem, stated in Vakil’s FOAG(The Rising Sea: Foundations Of Algebraic Geometry Notes, see here for the available electronic versions), is as follows:
1.6.I. IMPORTANT EXERCISE (THE FHHF THEOREM). This result can take you far, and perhaps for that reason it has sometimes been called the Fernbahnhof (FernbaHnHoF) Theorem, notably in [Vak1, Exer. 1.6.I]. Suppose F: $\mathscr{A} \rightarrow \mathscr{B}$ is a covariant functor of abelian categories, and $\mathrm{C}^{\bullet}$ is a complex in $\mathscr{A}$. Show that
(a) ( $\mathrm{F}$ right-exact yields $\mathrm{FH}^{\bullet} \longrightarrow \mathrm{H}^{\bullet} \mathrm{F}$ ) If $\mathrm{F}$ is right-exact, describe a natural morphism $\mathrm{FH}^{\bullet} \rightarrow \mathrm{H}^{\bullet} \mathrm{F}$. (More precisely, for each $i$, the left side is $F$ applied to the cohomology at piece $i$ of $\mathrm{C}^{\bullet}$, while the right side is the cohomology at piece $i$ of $\mathrm{FC}^{\bullet}$.
(b) (F left-exact yields $\mathrm{FH}^{\bullet} \longleftarrow \mathrm{H}^{\bullet} \mathrm{F}$ ) If $\mathrm{F}$ is left-exact, describe a natural morphism $\mathrm{H}^{\bullet} \mathrm{F} \rightarrow \mathrm{FH}^{\bullet}$.
(c) (F exact yields $\mathrm{FH}^{\bullet} \stackrel{\sim}{\longrightarrow} \mathrm{H}^{\bullet} \mathrm{F}$ ) If $\mathrm{F}$ is exact, show that the morphisms of (a) and (b) are inverses and thus isomorphisms.
The abstract nonsense way to do it is as in the thread Prove the FHHF theorem using as much abstract non-sense as possible
Can it be proved using the Freyd-Mitchell embedding theorem? I took two embeddings taking $\mathscr{A}, \mathscr{B}$ into categories of $R$ and $S$ modules respectively. But it seems that the functor $F$ cannot be some kind “transferred”.
The main reason I wanna to do this is trying to gain some practice in the embedding theorem.
Thank you in advanced for any help.
