By lemma 4.9 in Dugger-Hollander-Isaksen, a hypercover is defined as an augmented simplicial object $U_\bullet\to X$ in the category of simplicial presheaves such that each $U_n$ is a coproduct of representables and each $U_n\to (cosk_{n-1} U_\bullet)_n$ is a local epimorphism.

In the answer of this post, it shows the relation of hypercover for simplicial presheaves with hypercovering in $\infty$-topos. But the definition in the question doesn't require $U_n$ to be a coproduct of representables. Is the condition that $U_n$ being a coproduct of representables not essential for the definition of hypercover of simplicial presheaves?


1 Answer 1


The condition that $U_n$ being a coproduct of representables is not essential for the definition of hypercover of simplicial presheaves. In fact, in Jardine's Local Homotopy Theory, a hypercover is just defined to be a local trivial fibration (more general class than DHI's), which one can prove to be equivalent (need quite some work, using Lemma 4.8 of Jardine's book) to DHI's definition in their specific form; also, the hypercover descent in DHI is equivalent to Jardine's descent.

  • $\begingroup$ Does saying $U_n\to (cosk_{n-1} U_\bullet)_n$ is a local epimorphism implies that $U_n$ is coproduct of representables? $\endgroup$
    – Nicky
    Sep 15, 2019 at 21:14
  • $\begingroup$ No. Each $U_n$ being a coproduct of representables is an additional condition. DHI requires this I think is because this is more intuitive about coverings (a covering sieve is a sub presheaf of a representable one, which is also a local epimorphism). What I said in the answer is that DHI's and Jardine's hypercover gives the same descent result which I think is your main concern. See [mathoverflow.net/questions/340015/… for a relevant exposition. $\endgroup$
    – Lao-tzu
    Sep 15, 2019 at 21:22
  • $\begingroup$ Can you give some more details on the proof? I don't see how the Lemma you referenced would help prove that both definitions give equivalent notions of descent. $\endgroup$ Jan 3, 2022 at 22:56

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.