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I have the following situation: $\mathcal{M}$ is a semi simple, indecomposable module category over a semisimple, rigid monoidal category $\mathcal{C}$ with finitely many irreducible objects and irreducible unit object. Let $N$ be an object in $\mathcal{M}$.

In my exercise it says that $\mathrm{Hom}(-, N)$ is left exact under those conditions. Why is that?

P. Schulze
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  • What is a semisimple, indecomposable category? Do you mean that $N$ is semisimple and indecomposable? Since $N$ is semisimple then its every submodule is a direct summand. So $N$ is simple because it is indecomposable. – freakish Feb 27 '19 at 14:35
  • I asked myself the same thing, but I think the definitions consider two different things: The semi simplicity of the category means, that every object can be written as a direct sum of simple objects, while for the category to be indecomposable means that the category itself can't be written as a direct sum of two non-trivial categories. – P. Schulze Feb 28 '19 at 13:13
  • I am not sure anymore... maybe you are right... – P. Schulze Feb 28 '19 at 13:16
  • Wow, I've never heard about the concept of indecomposable category. Do you have some reference so I could learn more about it? – freakish Feb 28 '19 at 13:25
  • https://arxiv.org/abs/math/0111139 Definition 7 – P. Schulze Feb 28 '19 at 13:26
  • @freakish lets say that $N$ is simple. Does it then follow that $\mathrm{Hom}(-, N)$ is left exact? – P. Schulze Feb 28 '19 at 13:46
  • My mistake, the $\text{Hom}(N,-)$ is exact which easily follows from $N$ being simple. But I don't think that $\text{Hom}(-,N)$ needs to be in general. But let me ask this: so $\mathcal{M}$ is a (full?) subcategory of $\text{Mod}(R)$ for some ring $R$? Such that all objects in $\mathcal{M}$ are semisimple (even though $R$ does not have to be) and $\mathcal{M}$ is indecomposable? That sounds to weak. Or is $\mathcal{M}$ just $\text{Mod}(R)$ in which case $R$ has to be semisimple. – freakish Feb 28 '19 at 14:24
  • Maybe I should write that more explicitly in the question, let me edit it. @freakish is it more clear now? – P. Schulze Feb 28 '19 at 14:56

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An equivalent characterization of $\mathcal{M}$ being semisimple is that every $N\in \mathcal{M}$ is injective (see e.g. R is semisimple if and only if every R module is projective for the dual statement). $\operatorname{Hom}(-,N)$ being exact is equivalent to $N$ being injective.

Julian Kuelshammer
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  • thanks! do you have any reference? Maybe one where explicitly categories are used? I didn't finde anything categorical, just modules. – P. Schulze Feb 27 '19 at 14:18
  • @Polly What definition of semisimple are you using? (another related stackexchange question is https://math.stackexchange.com/q/2055190/15416 ) – Julian Kuelshammer Feb 27 '19 at 14:34
  • In the paper I am working in no definition is given. I always worked with this one: https://ncatlab.org/nlab/show/semisimple+abelian+category – P. Schulze Feb 28 '19 at 09:45