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For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$: $$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Now since $^\sigma g=g$, we notice that $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ so we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc's paper, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$: $$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Now since $^\sigma g=g$, we notice that $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ so we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc, J.-J. Groupe de Brauer et arithmétique des groupes algébriques linéaires sur un corps de nombres. J. Reine Angew. Math. 327 (1981), 12–80, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$: $$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Now since $g\cdot\,^\sigma g=1$, we notice that $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ so we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc's paper, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$: $$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Now since $^\sigma g=g$, we notice that $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ so we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc's paper, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_1$ is conjugate to $P_0$ over $k$: $$P_1=g\cdot P_0\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Since $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc's paper, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.

For simplicity we write $G$ for $G^{\rm der}$ and $M_0$ for $M_0^{\rm der}$, then $G$ is a connected reductive group, and $M_0$ is a Levi subgroup of a minimal parabolic $P_0$ of $G$. Let $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k, G)]$$ be a cohomology class, represented by some 1-cocycle $c\in Z^1(k,M_0)$. Since $\xi=[c]$ is contained in the kernel, the exists $g_1\in G(\bar k)$ such that $c_\sigma=g_1^{-1}\cdot \,^\sigma g_1$ for all $\sigma\in \Gamma:={\rm Gal}(\bar k/k)$. We set $$M_1=g_1\cdot M_0\cdot g_1^{-1},\quad P_1=g_1\cdot P_0\cdot g_1^{-1},$$ then one can easily check that $M_1$ and $P_1$ are defined over $k$. Since $(P_1,M_1)$ is conjugate to $(P_0,M_0)$ over $\bar k$, we see that $P_1$ is a parabolic of $G$ and that $M_1$ is a Levi subgroup of $P_1$. By Theorem 4.13(c) $P_0$ is conjugate to $P_1$ over $k$: $$P_0=g\cdot P_1\cdot g^{-1}\quad\text{for some }g\in G(k).$$ Set $g_2=g g_1\in G(\bar k)$, then $$g_2 \cdot P_0 \cdot g_2^{-1}=P_0\,,$$ hence, $g_2\in N_G(P_0)=P_0$. Now since $g\cdot\,^\sigma g=1$, we notice that $$c_\sigma=g_2^{-1}\cdot \,^\sigma g_2\,,$$ so we see that $$\xi\in{\rm ker}[H^1(k,M_0)\to H^1(k,P_0)].$$ Since $H^1(k,P_0)= H^1(k,M_0)$, see e.g. Lemma 1.13 in Sansuc's paper, we conclude that $\xi=1$, hence $${\rm ker}[H^1(k,M_0)\to H^1(k, G)]=1.$$ A twisting argument shows that every fiber has only one element.