MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Lemma of Higher Topos Theory, the statement of the lemma requires that the fibers are not only nonempty but contractible. However, in the proof, I don't see where contractibility is directly used, only the fact that the fibers are nonempty. There is one other place where contractibility is mentioned, "Since the boundary of this simplex maps entirely into the contractible kan complex $S_t$, it is possible to extend $f'$ to $X(n+1)$." However, I don't see how contractibility directly factors in, since that would only attest to the uniqueness of the extension. The existence of the extension comes from the fact that the inclusion $\partial \Delta^n \times \Delta^1 \subseteq X(n+1)$ is left anodyne and $S_t$ is a nonempty Kan complex and the fact that the map f' factors through the inclusion of $S_t$.

Please correct me if I'm wrong. Also, there is a relevant post on meta where I first asked if this question is appropriate, and I was greenlighted by Anton.

share|cite|improve this question
up vote 7 down vote accepted

The inclusion $\partial \Delta^n \times \Delta^1 \subseteq X(n+1)$ isn't any kind of anodyne extension, though. It's formed by attaching an n-simplex to $\partial \Delta^n \times \Delta^1$ with boundary $\partial \Delta^n \times 0$. So extending a map to $S_t$ from $\partial \Delta^n \times \Delta^1$ to $X(n+1)$ is exactly the same as extending a map from $\partial \Delta^n$ to $\Delta^n$, and being able to do this for all $n$ is exactly the same as $S_t$ being a contractible Kan complex (since the maps $\partial \Delta^n \to \Delta^n$ as $n$ varies form generating cofibrations for the Kan model structure).

share|cite|improve this answer
Could you elaborate a little bit on the contractibility part? I still don't see how it factors in, I guess. Is being contractible and kan equivalent to being trivially fibrant? – Harry Gindi Dec 29 '09 at 4:46
Yeah: X being a Kan complex means X -> * is a fibration, and X being contractible means it is also a weak equivalence, hence an acyclic fibration. – Reid Barton Dec 29 '09 at 4:50
Oh, then it's all clear now. Thank you. – Harry Gindi Dec 29 '09 at 4:52

Whenever I get confused about quasicategories I think back to ordinary categories. A left fibration between (nerves of) ordinary categories is the same as an opfibration with groupoid fibers, and a trivial fibration between ordinary categories is the same as an equivalence that is surjective on objects. Clearly an opfibration in groupoids with nonempty fibers need not be an equivalence unless the fibers are contractible.

share|cite|improve this answer
Yeah, I wasn't thinking model categorically, so the point I was missing was "If you're kan and contractible, then you're trivially fibrant", which is true for the "obvious reason". – Harry Gindi Dec 30 '09 at 7:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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