Additive exact functors preserve homology of modules If $C$ is a chain complex of modules over a ring $R$ and $F: Mod_R \to Mod_S$ an additive exact functor then:

If $F$ is covariant then $H_n(FC)\cong FH_n(C)$.

Can anyone give my an idea or an idea to prove this statement? Thanks. 
 A: I just need this result and let me fill the proof as stated by @Zavosh in the comments: we'll prove image, kernel, and quotient is preserved under the exact functor.
Let's declare some notations, let $C=(C_n,\partial_n)_{n\in \mathbb N}$ be a complex, and $F$ be the exact covariant functor, with obtained complexes: $FC=(F(C_n),F(\partial_n))_{n\in \mathbb N}$ (This is a complex since $F$ is a functor and the composition of boundary maps factor through $F$.) Let $Z_n=\ker(\partial_n)\subset C_n$, and $B_n=Im(\partial_{n+1})\subset C_n$, and we want to show $$H_n(FC)\cong F(H_n(C))$$
We shall prove it in the following steps:
(i) Image is preserved by $F$
This only needs $F$ is a functor, since by a commutative diagram, we have
$$Im(F(\partial_{n+1}))=F(\partial_{n+1})(F(C_{n+1})))=F(\partial_{n+1}(C_{n+1}))=F(Im(\partial_{n+1}))=F(B_n)\tag{1}$$
(ii) Kernel is preserved by $F$
Since the functor $F$ is exact, by applying $F$ to the short exact sequence:
$$0\to Z_{n}\xrightarrow{i_n} C_n\xrightarrow{\partial_n} B_{n-1}\to 0$$
we still get short exact sequence:
$$0\to F(Z_{n})\xrightarrow{F(i_n)} F(C_n)\xrightarrow{F(\partial_n)} F(B_{n-1})\to 0$$
The exactness tells you that 
$\ker(F(\partial_n))\cong F(Z_n)=F(\ker(\partial_n)) \tag{2}$
(iii) Quotient is preserved by $F$
Applying $F$ to the Short exact sequence:
$$0\to B_{n}\xrightarrow{j_n} Z_n\xrightarrow{\pi_n} H_{n}(C)\to 0$$
where $j_n$ is the inclusion map, gives you
$$0\to F(B_{n})\xrightarrow{F(j_n)} F(Z_n)\xrightarrow{F(\pi_n)} F(H_{n}(C))\to 0$$
The exactness tells you that $F(Z_n)/F(B_n)\cong F(H_n(C))\tag{3}$
(iv) Conclusion 
Thus by steps above, we have $$H_n(FC)=\frac{\ker(F(\partial_n))}{Im(F(\partial_{n+1}))}\stackrel{by (1)(2)}{\cong}\frac{F(\ker(\partial_n))}{F(Im(\partial_{n+1}))}=\frac{F(Z_n)}{F(B_n)}\stackrel{by (3)}{\cong}F(H_n(C))$$
