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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

• CommentRowNumber12.
• CommentAuthorTim Campion
• CommentTimeFeb 25th 2020
• (edited Feb 25th 2020)

Somehow, I can’t find 7.3.5.14 in Higher Algebra, nor can I find any discussion of this coherence issue anywhere else in HA. Presumably it was removed in some update to HA (or else I’m just colossally failing). Does anybody know where I can find an exposition of this (i.e. the fact that an incoherent idempotent plus a finite number of coherences automatically lifts to a coherent idempotent) which isn’t phrased in type theory like Mike’s paper or blog post? It’s reminiscent of HTT 5.2.7.4, which basically shows that this coherence statement is solvable for idempotent endofunctors.

• CommentRowNumber13.
• CommentAuthorTim Campion
• CommentTimeFeb 25th 2020
• (edited Feb 25th 2020)

I figured it out – when Lurie rewrote the idempotents section in HTT, he seems to have moved this there – it’s HTT 4.4.5.20 now.

The fun thing about this coherence question is that it tells us the following: the walking idempotent $Idem$ is a compact object in the $\infty$-category of $\infty$-categories – even though it has infinitely many nondegenerate simplices as a simplicial set! In particular, writing $Idem$ as the sequential colimit of its skeleta, we see that the identity map factors through some stage of the colimit, $\infty$-categorically. More precisely, I think what we can say is that $Idem$ is a retract of a Joyal-fibrant replacement of its 3-skeleton.

Now, I believe it’s shown that if $X$ is a finite simplicial set, or just Joyal-equivalent to a finite simplicial set, then any $\infty$-category with an initial object and pushouts has $X$-indexed colimits. We know that not every $\infty$-category with an initial object and pushouts admits splitting of idempotents (e.g. finite spaces, by the Wall finiteness obstruction). Therefore, $Idem$ is not Joyal-equivalent to a finite simplicial set, Therefore $Idem$ is an example of a compact object in $Cat_\infty$, the $\infty$-category of $\infty$-categories, which cannot be obtained as a finite colimit of $\Delta[n]$’s, even though every compact object is a retract of a finite colimit of $\Delta[n]$’s. The point is that this is analogous to the phenomenon in the $\infty$-category $Gpd_\infty$ of spaces, where not every compact object is a finite colimit of contractible spaces, even though every compact object is a retract of such – as seen via the Wall finiteness obstruction.

That is, if you’re looking for examples of the Wall phenomenon in $Gpd_\infty$, it’s a bit exotic. But if you look for examples of it in $Cat_\infty$, you have this great, familiar example in the form of $Idem$.

Relatedly, the fact that the skeleta of $Idem$ can’t be equivalent to $Idem$ implies that there must exist examples of inequivalent coher-ifications of homotopy idempotents of all orders, as Mike asked about here.

Do these arguments show that the inclusion of the 3-skeleton into $Idem$ is $\infty$-categorically cofinal and co-cofinal?

• CommentRowNumber14.
• CommentAuthorMike Shulman
• CommentTimeMar 12th 2020

Nice observation. I don’t know about finality.

• CommentRowNumber15.
• CommentAuthorDmitri Pavlov
• CommentTimeMar 12th 2020

2. This is possibly a silly question, but what do we know about the weak homotopy type of $\mathsf{Idem}$?

Lurie mentions in HTT 4.4.5.3 that $\mathsf{Idem}$ is precisely the classifying space of the “Boolean monoid” $\mathbb{B}=(\{0,1\},\text{OR},1)$, and thus clearly we have

\begin{aligned} \pi_0(\mathsf{Idem}) &\cong \pi_0(\mathbf{B}\mathbb{B})\\ &\cong *, \end{aligned}

but also

\begin{aligned} \pi_1(\mathsf{Idem}) &\cong \pi_1(\mathbf{B}\mathbb{B})\\ &\cong \mathrm{K}_0(\mathbb{B})\\ &\cong 0. \end{aligned}

What about $\pi_n(\mathsf{Idem})$ for $n\geq2$?

• CommentRowNumber17.
• CommentAuthorHurkyl
• CommentTimeSep 19th 2021

HTT 4.4.5.10 says that Idem is weakly contractible; its weak homotopy type is the point.

3. Ah, perfect! Thanks, Hurkyl! :)