Let $C$ be an abelian category, assume for simplicity that $C$ is enriched over $Vect_k$ (vector spaces over $k$) for some fixed field $k$.

Suppose also that $C$ is both Artinian and Noetherian, so that for any object $X$ there is a sequence of objects $0=X_0 \hookrightarrow \ldots \hookrightarrow X_n = X$ with $X_i/X_{i-1}$ simple. Finally suppose that $C$ has enough injective/projective objects so that $\operatorname{Ext}_C$ can be defined.

Given $C$, we build a new category $S$, enriched over graded $k$-vector spaces, in the following way:

- The objects of $S$ are the
*simple*objects of $C$ - If $X,Y\in Ob(S)$ then $\operatorname {Hom}_S(X,Y) = \bigoplus_{n\geq0}\operatorname{Ext}^n_C(X,Y)$
- Compositions of morphisms are defined using the natural maps $\operatorname{Ext}^n_C(X,Y)\otimes \operatorname{Ext}^m_C(Y,Z)\to\operatorname{Ext}^{n+m}_C(X,Z)$

My question is: Can we recover $C$ from $S$ (say up to equivalence)?

Assuming the answer is "yes", I guess that there is an analogue for when $C$ is only enriched over $Ab$, maybe if we redefine $S$ so that $\operatorname{Hom}_S(X,Y)=\operatorname{Hom}_{D(C)}(X,Y)$ or something