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• CommentRowNumber1.
• CommentAuthorMike Shulman
• CommentTimeJan 27th 2010

At first Zoran's reply to my query at structured (infinity,1)-topos sounded as though he were saying "being idempotent-complete" were a structure on an (oo,1)-category rather than just a property of it. That had me worried for a while. It looks, though, like what he meant is that "being idempotent" is structure rather than a property, and that makes perfect sense. So I created idempotent complete (infinity,1)-category.

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeJan 27th 2010

Sorry for the typo, I was originally planning to say the thing in two steps, but then decided it is better to just quote the reference and the phrase was left...

• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeJun 11th 2010

Lurie defines a simplicial set $Idem$ such that an idempotent in an $(\infty,1)$-category $C$ is the same as a functor $Idem\to C$. Is this $Idem$ the same as the nerve of the free 1-category containing an idempotent?

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeDec 4th 2014

Suppose I have an $(\infty,1)$-category $C$ and an idempotent in its homotopy 1-category. Can it be lifted to a “coherent” idempotent in $C$ itself? Is there an obstruction theory?

1. @Mike: the closest result I know of is Lemma 7.3.5.14 of Higher Algebra. More or less, it says that if e: X -> X is an morphism, then e can be enhanced to a coherent idempotent if and only there is an equivalence e -> e^2 making the obvious maps e^2 -> e^3 determined by the two choices of parentheses commute up to homotopy.
• CommentRowNumber6.
• CommentAuthorDylan Wilson
• CommentTimeDec 5th 2014

cf. also Warning 1.2.4.8 for a counterexample if we try to weaken this criterion. (though there are no counterexamples if C is stable.)

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeDec 5th 2014

Awesome, together those answer the question completely, thanks! I’ve recorded these facts on the page.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeDec 8th 2014

I have added pointer to Mike’s HoTT wrapup here

• CommentRowNumber9.
• CommentAuthorMike Shulman
• CommentTimeAug 16th 2017

Apparently the answer to the question asked (over 7 years ago) in #3 above is “if it isn’t, it should have been”. Lurie has now changed the definition of $Idem$ to be the nerve of the free category containing an idempotent; see here and here. We should update the definition on the lab.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeAug 16th 2017

Updated idempotent complete (infinity,1)-category with the new, simpler definitions (and updated page references).

• CommentRowNumber11.
• CommentAuthorMike Shulman
• CommentTimeMay 3rd 2018

updated link to point to my paper in addition to blog post