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Let $S^3 \rightarrow S^2$ be the Hopf fibration. Can we recover $\pi_2(S^2)$ of $S^2$ from the simplicial set $X : \Delta^{op} \rightarrow \text{Set}$, $$ X(n) = \pi_0 (S^3 \times_{S^2} \cdots \times_{S^2} S^3) $$ $X$ is $\pi_0$ of the Čech nerve?

$S^3 \rightarrow S^2$ is not a good cover, but it does have $\pi_2, \pi_1, \pi_0 = 0$. So another way of thinking about this is whether $S^3 \rightarrow S^2$ acts enough like a good cover to recover $\pi_2(S^2)$.

In general, I am interested in categories like the category of schemes, which lacks good covers with which to tell the homotopy type $\pi_n(\text{Spec}(\mathbb{Z}))$. In absence of such good covers, one might hope to take some kind of limit over all of the covers, which better and better approximate a contractible one. This would be the next best thing to having a good cover, in the sense of Čech cohomology.

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    $\begingroup$ Unfortunately, no. The map $S^3 \to S^2$ is a principal $S^1$-bundle, and this allows one to prove that there is a homeomorphism $S^3 \times_{S^2} S^3 \cong S^3 \times S^1$ as spaces over $S^2$. Inductively you can then show that $X(n) \cong \pi_0(S^3 \times (S^1)^n) = 0$. $\endgroup$
    – Tyler Lawson
    Commented Sep 22, 2020 at 18:24
  • $\begingroup$ This is not to say that the Čech nerve is bad -- far from it. But $\pi_0$ just doesn't recover enough information. $\endgroup$
    – Tyler Lawson
    Commented Sep 22, 2020 at 18:26
  • $\begingroup$ @TylerLawson. If I used a cover of $S^2$ which had $\pi_3$ equal to $0$, then would this work? $\endgroup$
    – user30211
    Commented Sep 22, 2020 at 19:04
  • $\begingroup$ Perhaps only a contractible cover would be able to tell the $\pi_2$? $\endgroup$
    – user30211
    Commented Sep 22, 2020 at 19:04
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    $\begingroup$ Here's one way to think about this. The homotopy colimit of the two truncation is the two truncation of the homotopy colimit. So the problem is that you need the spaces $X(n)$ to be two-connected in order to compute $\pi_2$ how you want to. I don't see a great way to do this-- since even if you replaced the hopf fibration with one that has a contractile total space, you would get $X(2) = \Omega S^2$ which is not sufficiently connected. $\endgroup$
    – Phil Tosteson
    Commented Sep 22, 2020 at 20:34

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