I have been reading this article. 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 $n^{th}$ 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 $n^{th}$-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!

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!