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I have been reading Asok and Doran - $\mathbb A^1$-homotopy groups, excision, and solvable quotients. In the Example 2.17, page 1155, there is a claim that for $m>1$ and $n\geq 1$, $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$.

Let me explain the notation quickly:

$\mu_n$ is the group scheme of $n^\text{th}$ roots of unity acting on $\mathbb{A}^m-0$ by diagonal action. So $\mathbb{A}^m-0/\mu_n$ is a smooth scheme. $\mathbb{G}_m/\mathbb{G}_m^n$ is the cokernel of the $n^\text{th}$-power map $\mathbb{G}_m\xrightarrow{(-)^n} \mathbb{G}_m$ in the category of Nisnevich sheaves of abelian groups.

My question is how to see the isomorphism $\pi_0^{\mathbb{A}^1}(\mathbb{A}^m-0/\mu_n)\simeq\mathbb{G}_m/\mathbb{G}_m^n$?

Comments are most welcome!

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    $\begingroup$ Here's a short sketch of the relevant steps: $\mathbb{A}^m-0/\mu_n$ is an approximation of the classifying space of $\mu_n$. For $m>1$ the universal covering $\mathbb{A}^m-0$ is $\mathbb{A}^1$-simply-connected, then the map $\mathbb{A}^m-0/\mu_n\to B_{et}\mu_n$ (which classifies the $\mu_n$-covering) induces an isomorphism on $\pi_0$. The sheaf of connected components of $B_{et}\mu_n$ is $H^1_{et}(-,\mu_n)\cong \mathbb{G}_m/\mathbb{G}_m^n$. $\endgroup$ Commented May 9, 2021 at 18:24
  • $\begingroup$ @MatthiasWendt: Thanks a lot for your comments. For $m=2$, $\mathbb{A}^2-0$ is not $\mathbb{A}^1$-simply connected as $\pi_1^{\mathbb{A}^1}(\mathbb{A}^2-0)\simeq K_2^{MW}$. For $m>2$, could you please indicate how to see that the map $\mathbb{A}^m-0/\mu_n\to B_{et}\mu_n$, induces isomorphism on $\pi_0$. Also, I suppose for $m=2$ there should be a separate argument, then? $\endgroup$ Commented May 10, 2021 at 9:30
  • $\begingroup$ You're absolutely right, sorry about the $m=2$ mistake. Maybe I should try to give more detailed explanations in an answer, not a series of comments. Might take me some time though... $\endgroup$ Commented May 10, 2021 at 17:45
  • $\begingroup$ @MatthiasWendt: That will be great, thanks! $\endgroup$ Commented May 11, 2021 at 3:46

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Over a field of characteristic coprime to $n$, the classifying space ${\rm B}_{\rm et}\mu_n$ is $\mathbb{A}^1$-local. In this situation, any $\mu_n$-covering $\widetilde{X}\to X$ is an $\mathbb{A}^1$-covering space and provides a classifying morphism $X\to {\rm B}_{\rm et}\mu_n$. From the Kummer sequence $\mu_n\to \mathbb{G}_{\rm m}\to \mathbb{G}_{\rm m}$ we get a fiber sequence ${\rm B}_{\rm et}\mu_n\to {\rm B}\mathbb{G}_{\rm m}\to {\rm B}\mathbb{G}_{\rm m}$. This implies that the sheaf of $\mathbb{A}^1$-connected components is $\mathbb{G}_{\rm m}/\mathbb{G}_{\rm m}^n$ (the Nisnevich sheaf quotient) which we can alternatively write as the Nisnevich sheafification of ${\rm H}^1_{\rm et}(-,\mu_n)$. (This is in the Morel-Voevodsky paper.)

Now turn to $\mathbb{A}^m\setminus\{0\}$. The $\mu_n$-covering $\mathbb{A}^m\setminus\{0\}\to\mathbb{A}^m\setminus\{0\}/\mu_n$ is classified by a map $\mathbb{A}^m\setminus\{0\}/\mu_n\to {\rm B}_{\rm et}\mu_n$ and the more precise claim is that the induced map on $\pi_0^{\mathbb{A}^1}$ is an isomorphism of Nisnevich sheaves.

For surjectivity, we consider an embedding $\mathbb{G}_{\rm m}\to\mathbb{A}^m\setminus\{0\}$ as $i$-th component. This is a $\mu_n$-equivariant map if we take the componentwise action on $\mathbb{A}^m\setminus\{0\}$. Therefore, we get an induced map on quotient schemes $\mathbb{G}_{\rm m}\to \mathbb{A}^m\setminus\{0\}/\mu_n$. On sheaves of $\mathbb{A}^1$-connected components we get induced maps $$ \mathbb{G}_{\rm m}\to \pi_0^{\mathbb{A}^1}\mathbb{A}^m\setminus\{0\}/\mu_n\to \mathbb{G}_{\rm m}/\mathbb{G}_{\rm m}^n $$ such that the composition is the natural quotient map. This proves the surjectivity.

We check injectivity locally in the Nisnevich topology. For that, let $R$ be a henselian local ring. A morphism ${\rm Spec}R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ is represented by a tuple $(x_1,\dots,x_m)$ with $x_i\in R$ such that at least one of the $x_i$ is not in the maximal ideal of $R$. Two such tuples $(x_1,\dots,x_m)$ and $(y_1,\dots,y_m)$ give the same morphism if the tuples $(x_1^n,\dots,x_m^n)$ and $(y_1^n,\dots,y_m^n)$ agree. Now we want to show that any map ${\rm Spec} R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ (corresponding to a tuple $(x_1,\dots,x_m)$) factors through a coordinate embedding $\mathbb{G}_{\rm m}\to\mathbb{A}^m\setminus\{0\}/\mu_n$ used before. One entry $x_i$ is not contained in the maximal ideal, assume it's the first. Then $(x_1,t\cdot x_2,\dots,t\cdot x_m)$ provides a naive $\mathbb{A}^1$-homotopy from the given tuple to $(x_1,0,\dots,0)$. So up to $\mathbb{A}^1$-homotopy, any map ${\rm Spec} R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ factors through a coordinate embedding $\mathbb{G}_{\rm m}\to \mathbb{A}^m\setminus\{0\}$. If such a homotopy class maps to the base point of $\mathbb{G}_{\rm m}/\mathbb{G}_{\rm m}^n$ then by the identifications above, it must lift through the $\mu_n$-covering $\mathbb{G}_{\rm m}\xrightarrow{n} \mathbb{G}_{\rm m}$. But since we have a morphism from the covering $\mathbb{G}_{\rm m}\to \mathbb{G}_{\rm m}$ to the covering $\mathbb{A}^m\setminus\{0\}\to \mathbb{A}^m\setminus\{0\}/\mu_n$, the map ${\rm Spec}R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ lifts (up to $\mathbb{A}^1$-homotopy) through a map ${\rm Spec}R\to \mathbb{A}^m\setminus\{0\}$. But the target there is $\mathbb{A}^1$-connected, meaning that the original map ${\rm Spec} R\to \mathbb{A}^m\setminus\{0\}/\mu_n$ must be null-homotopic, showing injectivity.

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