This is mostly a reformulation of the results on finite Galois extensions proved in More on Algebra, Section \ref{more-algebra-section-ramification}. The surjectivity of the map $D \to \text{Aut}(\kappa(\mathfrak m)/\kappa)$ is More on Algebra, Lemma \ref{more-algebra-lemma-one-orbit-geometric-galois}. This gives the first exact sequence. \medskip\noindent To construct the second short exact sequence let $\Lambda$ be the set of finite Galois subextensions, i.e., $\lambda \in \Lambda$ corresponds to $L/L_\lambda/K$. Set $G_\lambda = \text{Gal}(L_\lambda/K)$. Recall that $G_\lambda$ is an inverse system of finite groups with surjective transition maps and that $G = \lim_{\lambda \in \Lambda} G_\lambda$, see Fields, Lemma \ref{fields-lemma-infinite-galois-limit}. We let $B_\lambda$ be the integral closure of $A$ in $L_\lambda$. Then we set $\mathfrak m_\lambda = \mathfrak m \cap B_\lambda$ and we denote $P_\lambda, I_\lambda, D_\lambda$ the wild inertia, inertia, and decomposition group of $\mathfrak m_\lambda$, see More on Algebra, Lemma \ref{more-algebra-lemma-galois-inertia}. For $\lambda \geq \lambda'$ the restriction defines a commutative diagram $$ \xymatrix{ P_\lambda \ar[d] \ar[r] & I_\lambda \ar[d] \ar[r] & D_\lambda \ar[d] \ar[r] & G_\lambda \ar[d] \\ P_{\lambda'} \ar[r] & I_{\lambda'} \ar[r] & D_{\lambda'} \ar[r] & G_{\lambda'} } $$ with surjective vertical maps, see More on Algebra, Lemma \ref{more-algebra-lemma-compare-inertia}. \medskip\noindent From the definitions it follows immediately that $I = \lim I_\lambda$ and $D = \lim D_\lambda$ under the isomorphism $G = \lim G_\lambda$ above. Since $L = \colim L_\lambda$ we have $B = \colim B_\lambda$ and $\kappa(\mathfrak m) = \colim \kappa(\mathfrak m_\lambda)$. Since the transition maps of the system $D_\lambda$ are compatible with the maps $D_\lambda \to \text{Aut}(\kappa(\mathfrak m_\lambda)/\kappa)$ (see More on Algebra, Lemma \ref{more-algebra-lemma-compare-inertia}) we see that the map $D \to \text{Aut}(\kappa(\mathfrak m)/\kappa)$ is the limit of the maps $D_\lambda \to \text{Aut}(\kappa(\mathfrak m_\lambda)/\kappa)$. \medskip\noindent There exist canonical maps $$ \theta_{\lambda, can} : I_\lambda \longrightarrow \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda)) $$ where $n_\lambda = |I_\lambda|/|P_\lambda|$, where $\mu_{n_\lambda}(\kappa(\mathfrak m_\lambda))$ has order $n_\lambda$, such that $\theta_{\lambda, can}(\tau \sigma \tau^{-1}) = \tau(\theta_{\lambda, can}(\sigma))$ for $\tau \in D_\lambda$ and $\sigma \in I_\lambda$, and such that we get commutative diagrams $$ \xymatrix{ I_\lambda \ar[r]_-{\theta_{\lambda, can}} \ar[d] & \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda)) \ar[d]^{(-)^{n_\lambda/n_{\lambda'}}} \\ I_{\lambda'} \ar[r]^-{\theta_{\lambda', can}} & \mu_{n_{\lambda'}}(\kappa(\mathfrak m_{\lambda'})) } $$ see More on Algebra, Remark \ref{more-algebra-remark-canonical-inertia-character}. \medskip\noindent Let $S \subset \mathbf{N}$ be the collection of integers $n_\lambda$. Since $\Lambda$ is directed, we see that $S$ is multiplicatively directed. By the displayed commutative diagrams above we can take the limits of the maps $\theta_{\lambda, can}$ to obtain $$ \theta_{can} : I \to \lim_{n \in S} \mu_n(\kappa(\mathfrak m)). $$ This map is continuous (small detail omitted). Since the transition maps of the system of $I_\lambda$ are surjective and $\Lambda$ is directed, the projections $I \to I_\lambda$ are surjective. For every $\lambda$ the diagram $$ \xymatrix{ I \ar[d] \ar[r]_-{\theta_{can}} & \lim_{n \in S} \mu_n(\kappa(\mathfrak m)) \ar[d] \\ I_{\lambda} \ar[r]^-{\theta_{\lambda, can}} & \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda)) } $$ commutes. Hence the image of $\theta_{can}$ surjects onto the finite group $\mu_{n_\lambda}(\kappa(\mathfrak m)) = \mu_{n_\lambda}(\kappa(\mathfrak m_\lambda))$ of order $n_\lambda$ (see above). It follows that the image of $\theta_{can}$ is dense. On the other hand $\theta_{can}$ is continuous and the source is a profinite group. Hence $\theta_{can}$ is surjective by a topological argument. \medskip\noindent The property $\theta_{can}(\tau \sigma \tau^{-1}) = \tau(\theta_{can}(\sigma))$ for $\tau \in D$, $\sigma \in I$ follows from the corresponding properties of the maps $\theta_{\lambda, can}$ and the compatibility of the map $D \to \text{Aut}(\kappa(\mathfrak m))$ with the maps $D_\lambda \to \text{Aut}(\kappa(\mathfrak m_\lambda))$. Setting $P = \Ker(\theta_{can})$ this implies that $P$ is a normal subgroup of $D$. Setting $I_t = I/P$ we obtain the isomorphism $I_t \to \lim_{n \in S} \mu_n(\kappa(\mathfrak m))$ from the surjectivity of $\theta_{can}$. \medskip\noindent To finish the proof we show that $P = \lim P_\lambda$ which proves that $P$ is a pro-$p$-group. Recall that the tame inertia group $I_{\lambda, t} = I_\lambda/P_\lambda$ has order $n_\lambda$. Since the transition maps $P_\lambda \to P_{\lambda'}$ are surjective and $\Lambda$ is directed, we obtain a short exact sequence $$ 1 \to \lim P_\lambda \to I \to \lim I_{\lambda, t} \to 1 $$ (details omitted). Since for each $\lambda$ the map $\theta_{\lambda, can}$ induces an isomorphism $I_{\lambda, t} \cong \mu_{n_\lambda}(\kappa(\mathfrak m))$ the desired result follows.