Let $F : \mathcal{C} \to \mathcal{D}$ be an exact functor between abelian categories $\mathcal{C}, \mathcal{D}$. Prove that $F$ is faithful $\iff$ $F(C) \neq 0$ for every non-zero object $C$ of $\mathcal{C}$
The implication $\Rightarrow$ is simple: if $F(C)=0$ with $C=0$ then $$Hom_{\mathcal{C}}(C,C)\to Hom_{\mathcal{D}}(0,0)$$ is not injective: $id_C\mapsto 0$. I'm not able to prove the other implication. I need that the functor is additive. In fact if $f\neq 0$ then $Im(f)\neq 0$ then $F(Im(f))=Im(F(f))$ (it follows from the exactness of F) which is $\neq 0$ so $F(f)\neq 0$. But in general i can't know that it's additive (i believe it's true if $F$ is an isomorphism of categories)
A map $f: A \to B$ fits into an exact sequence $$0\to \ker f \to A \stackrel{f}\to B \to \operatorname{coker} f \to 0.$$ Under the exact functor $F$ this goes to an exact sequence $$0\to F(\ker f) \to F(A) \stackrel{F(f)}\to F(B) \to F(\operatorname{coker} f) \to 0.$$ The middle map $F(f)$ is zero if and only if the maps $F(\ker f)\to F(A)$ and $F(B) \to F(\operatorname{coker}f)$ are isomorphisms. Likewise, if $f$ was not the zero map, then what we know is that the maps $\ker f \to A$ and $B \to \operatorname{coker} f$ are not both isomorphisms.
So, it is enough to show that if a map $g$, which we may suppose to be injective or surjective (that is, $g$ is a monomorphism or is an epimorphism), is not an isomorphism, then $F(g)$ is not an isomorphism either. Our assumption that $F(C)$ is not zero if $C$ is not zero fits in here: if $g$ is injective but not surjective, let $C = \operatorname{coker} g$; while if $g$ is surjective but not injective, let $C = \ker g$. It follows that $F(g)$ is not an isomorphism (it has non-zero cokernel or non-zero kernel, respectively).