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Let $A, B$ two rings. I know that $G \in \operatorname{mod}-A$ is a generator for $\operatorname{mod}-A$ if and only if $\operatorname{Hom}(G,-)$ is a faithful functor from $\operatorname{mod}-A$ to the category of abelian groups, which means that for any $f \in \operatorname{Hom}(X,Y)$, with $X,Y \in \operatorname{mod}-A$, there exists a $g \in \operatorname{Hom}(G,X)$ such that $fg \neq 0$.

I have to prove that, if $F$ is an equivalence from $\operatorname{mod}-A$ to $\operatorname{mod}-B$ then $G$ is a generator for $\operatorname{mod}-A$ if and only if $F(G)$ is a generator for $\operatorname{mod}-B$.

Can someone help me?

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This has nothing to do with rings. If $F : C \to D$ is any equivalence of categories, then $G \in C$ is a generator iff $F(G) \in D$ is a generator. This is simply because the notion of a generator is defined in the language of categories, hence is invariant under equivalence. This is a general principle. If you are not familiar with it, you can also prove it directly: If $G \in C$ is a generator, then $\hom(G,-)$ is faithful. Choose some quasi-inverse $F^{-1}$ of $F$. Since $F^{-1}$ is also faithful, it follows that $\hom(G,-) \circ F^{-1} = \hom(G,F^{-1}(-)) \cong \hom(F(G),-)$ is faithful. Thus, $F(G)$ is a generator. So actually for this direction we only need that $F$ has a right adjoint which is faithful.

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