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



Recovering an abelian category from the Ext of its simple objects
Here's a counterexample that appears in nature.


Fix a prime $p$ and a field $k$ of characteristic $p$, and let
$G=C\_{p^{n}}$ be a cyclic group of order $p^{n}$ (where $n\geq1$ if
$p$ is odd, and $n\geq2$ if $p=2$).


Then the category $\operatorname{mod}kG$ of finitely generated
$kG$-modules has only one simple module: the trivial module $k$.


As graded $k$-algebras
$$\operatorname{Ext}^{\ast}\_{kG}(k,k)=H^{\ast}(G,k)\cong k[s,t]/(s^{2}),$$
with $s$ and $t$ in degree $1$ and $2$ respectively, independent of
$n$.


But $\operatorname{mod}kG$ determines $n$, as $p^n$ is the length of an
indecomposable projective module.


So, for example, $\operatorname{mod}\mathbb{F}\_{2}C\_{4}$ and
$\operatorname{mod}\mathbb{F}\_{2}C\_{8}$ are not equivalent, but have
the same $\operatorname{Ext}$-algebra.


Alternatively, this generalizes to looking at
$\operatorname{mod}k[x]/(x^{m})$ for any field $k$ and varying values
of $m\geq3$.


So a slightly smaller counterexample is
$\operatorname{mod}k[x]/(x^{3})$ and $\operatorname{mod}k[x]/(x^{4})$.


