Let $Idem = Idem^{(\infty)}$ be the walking idempotent [1], and let $Idem^{(n)}$ be its n-skeleton. Note that $Idem$ has one nondegenerate simplex in each dimension. Let $\iota\_n^m: Idem^{(n)} \to Idem^{(m)}$ be the inclusion. Lurie has shown [2] the following:


* If $X$ is a quasicategory, and if $Idem^{(3)} \xrightarrow f X$ is a map, then there exists a map $Idem \xrightarrow g X$ such that $g\iota\_1^\infty = f\iota\_1^3$.


That is, although a homotopy-coherent idempotent involves infinitely many pieces of coherence data corresponding to the infinitely many nondegenerate simplices of $Idem$, nevertheless all of this data is guaranteed to exist once the first 3 have been found. However, the resulting coherence data might be different from the original data in dimensions 2 and 3 [3].


The proof is a bit involved, and I have not studied it in detail, but what I do understand seems to suggest an affirmative answer to the following


**Question:** Is the inclusion $Idem^{(n)} \to Idem$ homotopy cofinal for certain $n$ (i.e. is it cofinal in the $\infty$-categorical sense -- depending on how one defines this, it may be necessary to Joyal-fibrantly replace $Idem^{(n)}$ before the question makes sense)? As noted in the comments, this is definitely not true for $n$ odd, nor is it true for $n=0$.  Somehow it seems unlikely for $n=2$; hence the title question, which asks this for $n=4$.


I don't believe the inclusion $Idem^{(n)} \to Idem$ is left or right anodyne, so any proof will encounter some complications.


One idea would be to use the usual map $N \to Idem^{(n)}$, where $N \subseteq \mathbb N$ is the graph on which the poset $\mathbb N$ of natural numbers is free -- the 0-cells are natural numbers, and there is a 1-cell from $n$ to $n+1$ for each $n$. For the inclusion $N \to \mathbb N$ is a categorical equivalence, and it's easy to show that $\mathbb N \to Idem$ is homotopy cofinal using Quillen's Theorem A (since the relevant slice categories are just ordinary 1-categories). By composition, $N \to Idem$ is homotopy cofinal, but this factors through $Idem^{(n)}$. By a cancellation property of cofinality, in order to show that $Idem^{(n)} \to Idem$ is homotopy cofinal, it will suffice to show that $N \to Idem^{(n)}$ is homotopy cofinal, which sounds straightforward, since $N$ and $Idem^{(n)}$ are finite-dimensional. But I'm not sure the relevant slice objects are still finite-dimensional...


[1] That is, $Idem$ is the category with one object and one non-identity morphism $i$, satisfying the equation $i^2 = i$. In this question I identify 1-categories with their nerves, which are quasicategories.


[2] HTT 4.4.5.20 in the current version. This does not appear in the published version of HTT. It appears in older versions of HA as 7.3.5.14, but was moved to HTT when Lurie [rewrote the section on idempotents in HTT](https://mathoverflow.net/questions/266384/is-the-category-idem-filtered).


[3] For example, consider the inclusion $Idem^{(3)} \to \widetilde{Idem^{(3)}}$ given by a Kan fibrant replacement such as $Ex^\infty$. This map is nontrivial on homology, so it cannot actually be extended to a map $Idem \to \widetilde{Idem^{(3)}}$ since this would amount to a trivialization. Thus the extension produced by Lurie's theorem must be *changing* the image of the 3-cell of $Idem^{(3)}$, at least. In particular, it's *not* the case that all quasicategories have the right lifting property with respect to $\iota\_3^\infty$.



Is the inclusion of its 2-skeleton into the walking idempotent homotopy cofinal?
I don't know what's wrong with the following computation, but the answer is clearly *no*: if there were a cofinal functor from a finite simplicial set to $Idem$, then any $\infty$-category with finite colimits would have split idempotents, which is not the case (witness finite spaces).


 Somewhat surprisingly, this seems to work for even $n>0$, even for $n=2$! That is,



> 
> **Claim:** Let $n \in \mathbb N$. Then the inclusion $Idem^{(n)} \to Idem$ is homotopy cofinal (equivalently, since everything is self-dual: co-cofinal) if and only if $n$ is positive and even.
> 
> 
> 


**Proof:** We have seen that this can't happen when $n=0$ or $n$ is odd. Otherwise, we verify the hypotheses of the Joyal-Lurie version of Quillen's Theorem A, i.e. we check that the simplicial set $X^{(n)} = Idem^{(n)} \times\_{Idem} Dec(Idem)$ is weakly contractible. Here $Dec(Idem)$ is the decalage contruction $Dec(C)\_n = C\_{n+1}$ where we forget the 0th degeneracy. So in terms of simplices, we have $X^{(n)}\_m = Idem^{(n)}\_m \times\_{Idem\_m} Idem\_{m+1}$. For $m \geq n+1$, an $m$-simplex in $X^{(n)}\_m$ is a string of morphisms in $Idem$ (each either $i$ or $1$) of length $m+1$, such that among the last $m$ morphims in the string, all but at most $n$ are $1$. Any such simplex is a degeneracy of a simplex obtained by deleting one of the copies of $1$. That is, $X^{(n)}$ is $n$-skeletal. Now, the $n$-skeleton of $X^{(n)}$ agrees with that of $X^{(\infty)}$, which is weakly contractible. Therefore, since $n \geq 2$, we have $\pi\_1(X^{(n)}) = \tilde H\_{\leq n-1}(X^{(n)}) = 0$. So it will suffice to show that $H\_n(X^{(n)}) = 0$. There are two nondegenerate simplices of degree $n$: the string $1,i,\dots,i$ (a $1$ followed by $n$ $i$'s) and the string $i,i,\dots,i$ (a string of $n+1$ $i$'s). The boundaries of these (remember that $n$ is even and we are omiting the $\partial\_0$ term of the boundary map) are $1,i,\dots,i$ and $i,i,\dots,i$ (where now we have one fewer term in each string) respectively, which are linearly independent. Thus there are no nondegenerate cycles and $H\_n(X^{(n)}) = 0$ as desired. 


