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In the following, we work with additive categories.

We say that a category is weakly idempotent complete if all the epimorphisms which admit a section have kernel. It's equivalent to the dual statement: all the monomorphisms which admit a retract have cokernel.

A stronger notion is the notion of idempotent complete. A category is idempotent complete if each morphism in $\{f:A \rightarrow A \;|\; f^2=f\}$ has a kernel or equivalently a cokernel. As suggested by the name of these notions, an idempotent complete category is weakly idempotent.

We know that an abelian category is idempotent complete.

We have the following examples :

• We consider the category $K_{vect} ^{ ^{\ge n}}$ of vector spaces over a field $K$ with no non-nul vector space of smaller dimension than $n$. Then $K_{vect} ^{\ge n}$ is not weakly idempotent complete since the trivial projection $K^{n+1} \twoheadrightarrow K^n$ has a section but no kernel.

• We consider the category $K_{vect} ^ {^{\equiv 0 [2]}}$ of vector spaces over a field $K$ with pair dimension or infinite dimension. Then $K_{vect} ^ {^{\equiv 0 [2]}}$ is weakly idempotent complete. But it's not idempotent complete since the projector $K^{2} \stackrel{\begin{pmatrix}Id & 0 \\ 0 & 0\end{pmatrix}}{\rightarrow} K^2$ has no kernel or cokernel.

So this is the question :

Is someone knows an example of an idempotent complete category which is not abelian?

(here we don't need to have Grothendieck's axioms)

Thanks you, Timothée

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    $\begingroup$ I think the most interesting example is Deligne-Milne category $Rep(GL_t)$ for integral parameter $t$. jmilne.org/math/Books/DMOS.pdf $\endgroup$ – Denis T. Jun 14 '20 at 13:14
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    $\begingroup$ Free abelian groups. Divisible abelian groups. Abelian groups with no elements of order $101$. More generally, if you take an additive subcategory of an abelian category that is closed under taking direct summands, then it will be idempotent complete, and although it may be abelian, it will typically not be. $\endgroup$ – Jeremy Rickard Jun 14 '20 at 13:30
  • $\begingroup$ Could you post your answer as an answer and not as a comment ? If you do this I could accept the answers :) $\endgroup$ – MoreauT Jun 14 '20 at 15:23
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    $\begingroup$ Very often people post answers in comments when they are hesitant about whether a question is on-topic here: they don't necessarily want to encourage more such questions, but also they don't want to leave the OP empty-handed. Of course I'm not sure that's Jeremy's thinking, but just to let you know this sometimes happens. $\endgroup$ – Todd Trimble Jun 14 '20 at 15:33
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The OP asked for responses as an answer. I have made this answer CW, and encourage everyone to make a big list. I start with the answer that comes up in my work, and then compile the answers from the comments. (As this answer is CW, feel free to reorganize, remove this intro, etc.)

  1. Project modules over a ring $R$. Finitely generated projective modules over $R$.
  2. Free abelian groups. (Example 1. for $R = \mathbb{Z}$.)
  3. Vector bundles over a fixed base $M$ (Example 1. for $R = \mathcal{O}(M)$.)
  4. Divisible abelian groups.
  5. Abelian groups with no elements of order $101$.
  6. Deligne- Milne category $Rep(GL_t)$ for $t$ integral. https://www.jmilne.org/math/Books/DMOS.pdf
  7. Smooth manifolds.
  8. The homotopy category of any idempotent complete additive ∞-category is still idempotent complete, and they are typically not abelian. For example the homotopy category of spectra, the derived category of any Grothendieck abelian category, or the category $DM(S)$ of Voevodsky for $S$ a locally Noetherian scheme... If you want small examples you can take compact objects there instead.
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  • $\begingroup$ Smooth manifolds? The OP said to work with additive categories. $\endgroup$ – Todd Trimble Jun 14 '20 at 21:41
  • $\begingroup$ @ToddTrimble Well, it is CW :) Anyway, I added it just because my favourite example of idempotent completion is to start with the category of open domains in R^n and smooth maps... $\endgroup$ – Theo Johnson-Freyd Jun 15 '20 at 5:00
  • $\begingroup$ Hey, it's one of mine too! mathoverflow.net/questions/162552/… $\endgroup$ – Todd Trimble Jun 15 '20 at 11:35

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