Why Grothendieck's Homotopy Hypothesis is so difficult?

Recall that Grothendieck's Homotopy Hypothesis states that the homotopy category of weak globular $n$-groupoids (as in https://arxiv.org/pdf/1009.2331.pdf) is equivalent to the category of homotopy $n$-types.

It's also well known that strict globular groupoids are not enough since $\Pi_3 (S^2)$ cannot be rectified to a strict $3$-groupoid due to the non-vanishing of the Whitehead bracket.

On the other side, it's well known that every quasi-category can be rectified into a strict $(\infty, 1)$-category. More generally, there's Berger-Moerdjik result that says that a bunch of algebras over operads can be rectified. In particular, $A_{\infty}$-algebras can be rectified and also homotopy coherent diagrams (Vogt's theorem).

In view of these observations, I have the following questions:

1) By considering $\Pi_3 (S^2)$ as a quasi-category (i.e., extend it by degenerated simplices), we can rectify it. Why doesn't it contradict my first observation? (EDIT : As Yonatan mentioned in the answer below, only the level $0$ and $1$ can be rectified. Therefore there's no contradiction.)

2) Why can't we rectify weak globular groupoids? By Vogt's theorem, homotopy coherent diagrams can be rectified. So, what fails if one views a weak globular groupoid as a homotopy coherent diagram and, then, rectify it?

Further comments about the intuition of why one cannot rectify weak globular groupoids would be also of great utility.

• What is a strict $(\infty, 1)$-category? If you mean something like a category strictly enriched over strict $\infty$-groupoids then there's no way your rectification result holds. If not, there's no contradiction. – Qiaochu Yuan May 7 '17 at 7:49
• The problem is not the $(\infty,1)$-categories themselves but functors between them. Even if you rectify the categories you must still consider the functors in a weak sense, with all possible homotopy coherent data. I think this is explained well in the introduction of Lurie's Higher Topos Theory. By the way, the homotopy hypothesis per se isn't about globular sets, it is about an existence of some notion of $(\infty,1)$-category where the subcategory of groupoids would be equivalent to homotopy types. Globular groupoids is just one possible model. – Anton Fetisov May 7 '17 at 11:50
• @AntonFetisov: You can rectify any ∞-category to a bifibrant simplicial category, and the space of functors between such categories is equivalent to the space of functors between the original ∞-categories. – Dmitri Pavlov May 7 '17 at 12:53
• @QiaochuYuan Well, I was thinking about Lurie's comment in HTT page 6 and 7, where he says that a category enriched over $\infty$-groupoids with a coherent associative composition can be rectified into an ordinary category enriched over simplicial objects. I think that he means something like Cordier nerve construction (and its adjoint) or the rectification of categories enriched over $A_{\infty}$-spaces. But I'm not sure if the latter is really true. – user40276 May 7 '17 at 21:44
• @user40276: okay, in that case there's no contradiction with the observation that $\Pi_3(S^2)$ can't be rectified to a strict $3$-groupoid, because the $3$-truncation of a simplicial category isn't a strict $3$-category. – Qiaochu Yuan May 7 '17 at 22:33

First, let me remark that your question does not seem to be about the homotopy hypothesis, but about rectification. More specifically, the homotopy hypothesis concerns the question of whether (some version of) weak globular groupoids is equivalent to, say, Kan complexes, while the rectification problem you mention asks whether weak globular groupoids are the same as strict globular groupoids. The confusion then seems to arise from the fact that on the side of Kan complexes, there are some aspects which can be rectified. For example, we can rectify a Kan complex into a strict simplicial groupoid. One can then maybe rephrase the question as: given that we believe the homotopy hypothesis, how can it be that Kan complexes can be rectified while weak globular groupoids cannot? The answer to this question is simply that Kan complexes cannot be rectified either. Indeed, if one replaces a Kan complex by a simplicial groupoid, the mapping spaces of this simplicial groupoid will still be Kan complexes. This can be informally described as saying that we have rectified the levels of objects and morphisms, but we have left unrecitifed the level of homotopies between morphisms, homotopies between homotopies etc. all the way up. Note that by only rectifying the 0 and 1 dimensional levels you can still avoid the counterexample $S^2$. On the other hand, that's the maximum you can do. For example, $S^2$ cannot be modeled by a strict 2-category enriched in simplicial sets: indeed, the group of self equivalences of an identify morphism in any simplicially enriched strict 2-category is a simplicial abelian group, but the corresponding group of automorphism for $S^2$ is an $\mathbb{E}_2$-group whose $\mathbb{E}_2$-structure does not refine to an $\mathbb{E}_{\infty}$-structure.
• Now about your example of $S^2$, so this group of automorphism is strictly associative? In other words, only the commutativity holds up to homotopy, right? But is this a more general phenomenon? In other words, in the examples where I can't strictly stuff, are there some $E_n$ constraints implicitly making all the problem? You see, since I can rectify $A_{\infty}$-algebras and, more generally, I can rectify $A_{\infty}$-categories to DG-categories, it seems that there are some constraints not related to associativity implicitly somewhere. – user40276 May 7 '17 at 22:06
• To some extent yes, it is the higher commutativity that is the problem. For example, every monoidal groupoid is equivalent to a strict monoidal groupoid (i.e., a monoidal groupoid in which the associativity constraints hold "on the nose"), but not every braided monoidal groupoid is equivalent to one in which the commutativity constraints hold on the nose. Indeed, if the braiding $T_x: x \otimes x \to x \otimes x$ of an object $x \in X$ is not the identity then $X$ cannot be strictified. That's another way of thinking of what happens with the $S^2$ example. – Yonatan Harpaz May 8 '17 at 18:55
• Ok, thanks for the response. I'm aware that braided monoidal categories can't be strictified, but what exactly is the extent in "To some extent yes,..."? In other words, is there a precise theorem stating that the rectification can always be turned into a $E_n$ kind of obstruction? – user40276 May 25 '17 at 0:31