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Let $\mathsf{A}$ be an abelian category. I understand that if the homotopy category $\mathsf{K}(\mathsf{A})$ is abelian, then $\mathsf{K}(\mathsf{A})$ is semi-simple. (There's a proof of this in another question of mine: 1.)

Now, I think that this should imply that $\mathsf{A}$ itself is semi-simple. (The analogous fact for $\mathsf{D}(\mathsf{A})$ is true. 2) But I can't seem to prove it.

Let $0\to A \to B\to C\to 0$ be an exact sequence in $\mathsf{A}$. If it is also exact when seen in $\mathsf{K}(\mathsf{A})$, then $A\to B$ is a split monomorphism in $\mathsf{K}(\mathsf{A})$ and, since $\mathsf{A}$ embeds fully faithfully in $\mathsf{K}(\mathsf{A})$, it is also split in $\mathsf{A}$, finishing the proof. However it is not clear to me why the sequence is also exact in $\mathsf{K}(\mathsf{A})$ or even why $A\to B$ is still a monomorphism in $\mathsf{K}(\mathsf{A})$.

Gabriel
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  • Any full subcategory of a semisimple category that is closed under retracts (i.e. direct summands in the additive context) is semisimple, no? – Zhen Lin Jun 02 '21 at 22:42
  • Dear @ZhenLin, what do you mean by "closed under retracts"? That if a morphism in the subcategory has a retract, then this retract is in the subcategory? If so, I agree that $\mathsf{A}$ is closed under retracts (since it embeds fully faithfully) but I don't see why this implies that it is semisimple – Gabriel Jun 03 '21 at 08:33
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    No, I mean what I said: closed under direct summands. – Zhen Lin Jun 03 '21 at 09:26
  • That is, if $A\oplus B$ is in the subcategory, then so is $A$? If so, why does that implies that $\mathsf{A}$ is semisimple? – Gabriel Jun 03 '21 at 09:38
  • Sorry, I think I was assuming that any object in the subcategory that is a simple object in the big category is also a simple object in the subcategory. – Zhen Lin Jun 03 '21 at 10:26
  • No problem :) Do you happen to know any other way of proving the result in the question? – Gabriel Jun 03 '21 at 11:37
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    The proof that $A$ is abelian if $D(A)$ is was given in the linked you provided. It actually shows that if $A$ fully embed in an abelian triangulated category, then $A$ is semi-simple. So it works with $K(A)$ mutatis mutandis. – Roland Jun 03 '21 at 17:13

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