Let $A$ be an algebra and $I$ an ideal of $A$. Now, let $ \mathcal{C}$ be a subcategory of $Mod A$ where the objects are $A-$modules M such that $MI=0.$ Show that there exists an equivalence of categories between $\mathcal{C}$ and $Mod A/I$.
Let $M$ in $\mathcal{C}$, since $M$ is an $A-$ module we can define a operation $M \times A/I \to M$ given by $m(a+I) = ma$.
This makes sense, because if $(a+I) = (b+I)$ then $a-b \in I$ and $MI =0$, so $m(a-b) = 0 \Rightarrow ma = mb.$
So we can define a functor $F: \mathcal{C} \to Mod A/I$ given by $F(M_\mathcal{C}) = M_\alpha$. Here, I mean $M_{\mathcal{C}}$ is the $A-$ module $M$ such that $MI=0$ and $M_\alpha$ is the same $M$ with the structure of $A/I-$módule.
For any morphism $f:M_{\mathcal{C}} \to N_{\mathcal{C}}$, we can define $F(f) = f: M_\alpha \to N_\alpha$.
It seems easy to show that this functor is full, faithfull and dense. It's kind of ''Identity'', let me explain.
If $F(f) = F(g)$ then $f(m) = g(m), \forall m \in M_\alpha \Rightarrow f=g.$ (faithfull)
If we have $h: M_\alpha \to N_\alpha$ then we can take the same morphism $h: M_\mathcal{C} \to N_{\mathcal{C}}$ and we have $F(h) = h$ (full)
We have $F(M_\mathcal{C}) = M_\alpha \simeq M_\alpha$ (dense).
My professor didn't like this solution, is there something wrong???
Also, there is another way to solve this question, I need to find a functor $G: Mod A/I \to \mathcal{C}$ s.t $FG \simeq 1_{Mod A/I}$ and $GF \simeq 1_\mathcal{C}$. I'd appreciate if you give me a hint about this functor $G$.
Maybe it is useful if you put it in the following terms: there is a projection map $\pi : A\to A/I = B$, so it induces a functor $\pi^*: {}_B \mathsf{Mod} \to {}_A\mathsf{Mod}$ by pullback: if $\rho' : B\to \operatorname{End}(M)$ defines a $B$-modules structure on $M$, then precomposition with $\pi$ gives you $\rho : A\to \operatorname{End}(M)$. The functor is the identity on morphisms.
This functor gives an isomorphism between ${}_B\mathsf{Mod}$ and the full subcategory $\mathcal C_I$ of ${}_A\mathsf{Mod}$ consisting of those $A$-modules annhiliated by $I$. The inverse $F$ takes an $A$-module $(M,\rho)$ such that $IM=0$ and assigns it the structure map $\rho' : A/I\to \operatorname{End}(M)$ induced by $\rho$. By construction, $F\pi^*(M,\rho') = (M,\rho')$ and $\pi^*F(M,\rho) = (M,\rho)$, and both functors are the identity on morphisms, so $F\pi^*$ and $\pi^*F$ are the identity (and not only naturally isomorphic to it.)