Are abelian subcategories of semisimple categories semisimple? Let $\mathcal{C}$ be a semisimple abelian category and $\mathcal{D}\subset \mathcal{C}$ an abelian subcategory. Does it follow that $\mathcal{D}$ is semisimple? I think it does: choose an arbitrary $X\in \text{Obj}(\mathcal{D})$, which then has a decomposition $X=\bigoplus_{i\in I} S_i$ where $S_i\in \text{Obj}(\mathcal{C})$ are simple. Then set
$$J:=\{j\in I\mid S_i\in \text{Obj}(\mathcal{D})\}$$
and we have a short exact sequence in $\mathcal{C}$
$$0 \to \bigoplus_{j\in J} S_j\to X \to \bigoplus_{j\in I\backslash J} S_j\to 0$$
and thus $\bigoplus_{j\in I\backslash J} S_j\in \text{Obj}(\mathcal{D})$. Then is just remains to argue that  $\bigoplus_{j\in I\backslash J} S_j$ is simple in $\mathcal{D}$. The only subobjects of $\bigoplus_{j\in I\backslash J} S_j$ in $\mathcal{C}$ are direct sums of the $S_j$ for $j\notin J$, which thus are not in $\mathcal{D}$. However, the problem is that it could happen that a direct sum of the $S_j$'s lies in $\mathcal{C}$. I think one can iterate the process as above, and that this must terminate as every object is the finite direct sum of simple objects, but I'm not sure.
Is $\mathcal{D}$ indeed semisimple, does the argument above work or is there a nicer argument? Or did I miss a counterexample?
 A: What exactly do you mean by subcategory?
If the category $\mathcal{D}$ is a full subcategory of $\mathcal{D}$, then all projections to simple sub-objects in $\mathcal{C}$ are morphisms in $\mathcal{D}$. Hence, the latter is generated by the simple objects in it and is therefore semi-simple.
However, if you do not assume that the subcategory is full, then the subcategory need not be semi-simple.
For example, the category $\mathcal{C}$ of finite dimensional vector spaces over a field $k$ is semi-simple. Let $\mathcal{D}$ be the subcategory consisting of pairs $(V,N)$ where $V$ is a finite dimensional $k$-vector space with a nilpotent endomorphism $N$. Morphisms in $\mathcal{D}$ are $k$-linear maps that commute with the nilpotent endomorphism. One checks that this is an abelian category. (Kernels etc. are exactly as before; the nilpotent endomorphisms restrict/descend.)
The category $\mathcal{D}$ is not semi-simple. The object $(k[T]/(T^2),T)$ is not a direct sum of simple objects as $(k,0)$ is a sub-object and is not a direct summand.
Edit: To clarify the "inclusion" of objects of $\mathcal{D}$ in $\mathcal{C}$ one can take the objects of $\mathcal{C}$ to be pairs $(V,N)$ as above. Morphisms in $\mathcal{C}$ ignore the $N$ and are just $k$-linear maps of $k$-vector spaces.
