In an abelian category, an object $G$ is a generator iff for any nonzero object $X$, $\operatorname{Hom}(G, X) \neq 0$ In an abelian category, an object $G$ is a generator iff for any nonzero object $X$, $\operatorname{Hom}(G, X) \neq 0$
Would you give me a proof or some references?
I could show the “only if” part.
I don’t know that at which I use the abelian-ness of the category.
In the proof of “only if” part, I didn’t use it.
 A: Here is a proof under the additional hypothesis that $G$ is projective; I don't know either a proof or a counterexample without this assumption. (Edit: In the comments Jeremy Rickard gives the example of $\mathbb{Z}/2\mathbb{Z}$ in the category of $\mathbb{Z}/4\mathbb{Z}$-modules).
Let $G$ be a weak generator, meaning that if $\text{Hom}(G, X) = 0$ then $X = 0$ (so $\text{Hom}(G, -)$ reflects zero objects), which is also projective. We want to show that $G$ is a generator, meaning that $\text{Hom}(G, -)$ is faithful, or equivalently that if $f : X \to Y$ is a morphism such that $\text{Hom}(G, f) = 0$ then $f = 0$ (so $\text{Hom}(G, -)$ reflects zero morphisms).
We have that $f = 0$ iff $\text{im}(f) = 0$. Since $G$ is projective, $\text{Hom}(G, -)$ is exact, so preserves kernels and cokernels, and hence preserves images. . This means that if $f : X \to Y$ is a morphism, then
$$\text{Hom}(G, \text{im}(f)) \cong \text{im}(\text{Hom}(G, f)).$$
So we have equivalences
$$\text{Hom}(G, f) = 0 \Leftrightarrow \text{im}(\text{Hom}(G, f)) = 0 \Leftrightarrow \text{Hom}(G, \text{im}(f)) = 0 \Leftrightarrow \text{im}(f) = 0 \Leftrightarrow f = 0$$
which is the desired result.
Without the hypothesis that $G$ is projective you get only that $\text{Hom}(G, -)$ preserves kernels and so reflects monomorphisms (using $\text{ker}(f)$ instead of $\text{im}(f)$ above).
