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Has anyone formally calculated the étale homotopy type of $\text{Spec}(\mathbb{Z}) \cup \{ \text{place}_{\infty} \}$?

According to arithmetic topology, $\text{Spec}(\mathbb{Z}) \cup \{ \text{place}_{\infty} \}$ is formally analogous to $S^3$, so I predict that this is the étale homotopy type. We should have $\pi_3(\text{Spec}(\mathbb{Z})\cup \{ \text{place}_{\infty} \}) \cong \mathbb{Z}$, $\pi_2(\text{Spec}(\mathbb{Z}) \cup \{ \text{place}_{\infty} \}) \cong 0$, and $\pi_1(\text{Spec}(\mathbb{Z}) \cup \{ \text{place}_{\infty} \}) = 0$.

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    $\begingroup$ And $\pi_1=0$ ? $\endgroup$
    – David Roberts
    Commented Sep 22, 2020 at 20:53
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    $\begingroup$ What is your definition of etale homotopy type after adding the infinite place? $\endgroup$
    – naf
    Commented Sep 23, 2020 at 5:34
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    $\begingroup$ What topos corresponds to $\operatorname{Spec}(\mathbb{Z})\cup\{\infty\}$? I know of a couple of proposals, but they're both in a rather embrional stage... $\endgroup$ Commented Sep 23, 2020 at 6:14
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    $\begingroup$ This question has already 5 upvotes but still doesn't make sense. Please clarify what you mean by $\operatorname{Spec} \mathbf{Z} \cup \{{\rm place}_\infty\}$ or its etale homotopy type. $\endgroup$ Commented Sep 24, 2020 at 13:15
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    $\begingroup$ I'm okay with a broad scope, I was more concerned with the set of possible precise formulations of your question being empty. $\endgroup$ Commented Sep 25, 2020 at 7:29

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