Some notation:
For an $A$-module $T$, denote by $\operatorname{Gen}T$ the full subcategory of $\operatorname{Mod}A$ consisting of epimorphic images of modules in $\operatorname{Add}T$.
The following theorem is what I'm trying to prove, but I'm stuck on a very small detail in proving $(1)\Rightarrow (2)$
The following are equivalent for a torsion class $\mathcal{T}$ in $\operatorname{Mod}A$, where $A$ is just a ring with $1$.
- For every $M \in \operatorname{Mod}A$, there is a sequence $$M \xrightarrow{\phi} B \rightarrow C \rightarrow 0$$ where $\phi$ is a left $\mathcal{T}$-approximation and $\operatorname{Ext}_A^1(C, \mathcal{T})=0$.
- There is a finendo quasitilting $A$-module $T$ such that $\operatorname{Gen}T = \mathcal{T}$
First of all, let $A=M$ and $T=B\oplus C$. Then $\operatorname{Gen}T = \mathcal{T}$ (I have shown this, it's not too hard). What I need to show then to complete the proof is that $\operatorname{Ext}_A^1(B, \operatorname{Gen}T)=0$. The way it is done in the proof (it's not done in any detail) is by constructing a short exact sequence $$0 \rightarrow \overline{A} \xrightarrow{\overline{\phi}}B \rightarrow C \rightarrow 0$$ over $\overline{A}=A/\operatorname{Ann}T$. It is here that I am stuck, as I cannot show $\overline{\phi}$ to be mono.
Now, to construct $\overline{\phi}$: If $a\in \operatorname{Ann}T$, then $\phi(a) = a\phi(1) = 0$ because $\operatorname{Ann}T=\operatorname{Ann}B$. So by the projection $A \xrightarrow{\pi} \overline{A}$, the map $\overline{\phi}$ exists where $\overline{\phi}(a+\operatorname{Ann}T) = \phi(a)$.
Showing that $\overline{\phi}$ is a monomorphism proved to be a challenge for me. One way to show it is by showing that $\operatorname{Ker}\phi = \operatorname{Ann}T$, but I cannot show the inclusion $\operatorname{Ker}\phi \subseteq \operatorname{Ann}T$. Suppose that $\phi(a)=0=a\phi(1)$, then I cannot really see that $a$ annihilates $B$ and therefore $T$.
Any help on this step would be greatly appreciated, as I have done every other step of the proof in detail, but this one eludes me.