A hypercovering in an $\infty$-topos $E$ is a simplicial object $U \in {\rm Fun}(\Delta^{\rm op},E)$ such that each map $U_n \to ({\rm cosk}_{n-1} U)_n$ is an effective epimorphism. Theorem of Higher Topos Theory shows that if $E$ is hypercomplete, then all hypercoverings are effective, i.e. ${\rm colim}(U)$ is terminal.

The notion of hypercovering makes sense also for semisimplicial objects $U \in {\rm Fun}(\Delta_s^{\rm op},E)$, with $\Delta_s$ the subcategory of injective maps in $\Delta$. I believe the argument of Lemma in HTT shows that the underlying semisimplicial object of a simplicial object $U$ is a semisimplicial hypercovering if and only if $U$ is a simplicial hypercovering, and since the inclusion $\Delta_s \subseteq \Delta$ is homotopy cofinal (Lemma in HTT) their colimits also coincide. However, not every semisimplicial object underlies any simplicial object. On the other hand, we can left Kan extend a semisimplicial object from $\Delta_s$ to $\Delta$ to obtain a simplicial object with the same colimit, but this operation doesn't preserve the property of being a hypercovering. This exhausts my ideas for reducing the semisimplicial case to the simplicial one; I've also inspected the proof of, but it's not clear to me how to generalize it to the semisimplicial case, since it seems to use the degeneracies (even though, intriguingly, the way it uses them is to reduce one of the main questions to the semisimplicial case, in!).

So is a semisimplicial hypercovering in a hypercomplete $\infty$-topos necessarily effective?

  • 3
    $\begingroup$ I don't know if it answers your question, but appendix A of SAG has a much-expanded section on semisimplicial hypercoverings. $\endgroup$ May 30, 2019 at 12:39
  • 3
    $\begingroup$ Ah, there it is in A.5.3.3, thanks! If you post that as an answer I'll accept it. $\endgroup$ May 30, 2019 at 12:46

1 Answer 1


You should check out appendix A of SAG, which has a much-expanded section on semisimplicial hypercoverings. In particular, take a look at A.5.3.3.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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