I have the following equivalence between two categories: $F: {}_{R}\mathsf{M} \to {}_{S}\mathsf{M}$, with $G$ inverse. I have to prove that if $X$ is a generator in the first category, then $F(X)$ is in the second one.
I use the following equivalent definition of generator: $X$ is generator if and only if for every $M_1, M_2$ in ${}_{R}\mathsf{M}$, if $f \in \operatorname{Hom}_R(M_1, M_2)$, $ f \neq 0$ then there is $g: X \to M$ such that $fg \neq 0$.
I consider $f':N_1 \to N_2$ in ${}_{S}\mathsf{M}$, $f' \neq 0$, so $G(f') \neq 0$ in ${}_{R}\mathsf{M}$. Then because $X$ is generator I can find $g:X \to G(N_1)$ with $G(f')g \neq 0$.
Now applying $F$, $FG(f')F(g)\neq 0$. To demonstrate that $F(X)$ is a generator I have to prove that there is a $g':F(X) \to N_1$ such that $f'g' \neq 0$. I've done the following, and I'd like to know if it is right.
I used the fact that $FG \cong 1_S$ so there is a natural isomorphism $\zeta$. So if $\zeta_{N_1}: FG(N_1) \to N_1$ and the same for $N_2$, the diagram commutes and I have the following identity
$$ FG(f')=\zeta_{N_2}^{-1}f'\zeta_{N_1}. $$
So $FG(f')F(g)=\zeta_{N_2}^{-1}f'\zeta_{N_1}F(g) \neq 0$ and this implies $f'\zeta_{N_1}F(g) \neq 0$ because $\zeta_{N_2}$ is isomorphism. So I found $g'=\zeta_{N_1}F(g): F(X) \to N_1$ such that $f'g' \neq 0$.
I'd like to know if my proof is correct. Thanks!
Yes, your proof is absolutely correct.