Let's say that a limit diagram $\bar p:K^\triangleleft\to\def\sC{\mathscr{C}}\sC$ is a weakly contractible limit if the simplicial set $K$ is weakly contractible (in that $K\to*$ is a weak homotopy equivalence).

I want to say for an $\infty$-category $\sC$ and an object $x\in\sC_0$ that the forgetful functor $\sC_{/x}\to\sC$ creates (i.e., preserves and reflects) weakly contractible limits. Is there a reference that already has this result?

The nLab shows here that the limit of a diagram $p:K\to\sC_{/x}$ coincides with the limit of the corresponding diagram $p/x:K^\triangleright\to\sC$, so this reduces the problem to showing that the inclusion $K\hookrightarrow K^\triangleright$ is an initial map (using the nLab's convention, and thus is dual to the notion of a "cofinal map" in the sense of Lurie's Higher Topos Theory). By Proposition of Higher Topos Theory, this is equivalent to showing that $K\hookrightarrow K^\triangleright$ is left anodyne.

So perhaps an equivalent question is: if $K$ is a weakly contractible simplicial set, then is $K\hookrightarrow K^\triangleright$ left anodyne? Is there an easy way to see this?


1 Answer 1


You can also use Quillen's theorem A : to prove that $C \to D$ is initial, it suffices to show that for every $d\in D$, $C \times_D D_{/d}$ is weakly contractible.

In the case where $C\to D$ is fully faithful, this is always the case for $d$ in the image of $C$ as this pullback has a terminal object. So here we are left with proving that this is the case for the cone point, which I'm going to call $\infty$. For that one, as it is terminal in $K^\triangleright$, we see that $K \times_{K^\triangleright} (K^\triangleright)_{/\infty} \simeq K$ is weakly contractible, by assumption.

  • $\begingroup$ I am not sure whether usual proofs of Thm A use some version of this. In Kerodon, this is placed before Thm A: kerodon.net/tag/02KS $\endgroup$
    – Z. M
    Commented Nov 19, 2022 at 12:55
  • 3
    $\begingroup$ @Z.M I am relatively confident that they don't. I know at least one proof which doesn't, and browsing through Kerodon, it also seems to be that the proof in Kerodon doesn't either. (from Kerodon's system for tracking where tags are used, it also seems to indicate that 02KS is not used anywhere but 02KT, which itself is used nowhere) $\endgroup$ Commented Nov 19, 2022 at 14:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.