Equality of fields being deduced from isomorphism of Galois groups I'm reading and attempting to understand a proof of the local Kronecker-Weber theorem in https://arxiv.org/abs/math/0606108 (page 17).
Let $K$ be a local field and $\sigma \in W(K^{LT}/K)$ with $\sigma|_{\hat{K}} = \operatorname{Frob}_K^{n}$ (i.e. $v(\sigma) = n > 0$). Let $L = K(\mu_n)$ and extend $\sigma$ to an element of $W(K^{\operatorname{ab}}/K)$. Let $E_\sigma \subset K^{\operatorname{ab}}$ be its fixed field. We have that $E_\sigma \cap K^{\operatorname{ur}} = L$ and $E_\sigma/L$ is totally ramified Galois. By definition $\operatorname{Gal}(K^{\operatorname{ab}}/E_\sigma) \cong \widehat{\Bbb Z}$ by $\sigma \mapsto 1$ and $\operatorname{Gal}(K^{\operatorname{ur}}E_\sigma/E_\sigma) \cong \operatorname{Gal}(K^{\operatorname{ur}}/L) \cong \widehat{\Bbb Z}$, so the Galois groups of $K^{\operatorname{ab}}$ and $K^{\operatorname{ur}}E_\sigma$ over $E_\sigma$ are isomorphic. It is immediately concluded from this isomorphism that $K^{\operatorname{ab}} = K^{\operatorname{ur}}E_\sigma$.

I don't understand how this isomorphism of Galois groups implies the equality $K^{\operatorname{ab}} = K^{\operatorname{ur}}E_\sigma$.

This would of course not usually be the case (e.g. $\operatorname{Gal}(\Bbb Q(\sqrt 2)/\Bbb Q)$ and the Galois group of literally any other quadratic field).
Hopefully some number theorist could clarify this for me!
 A: Two proofs (Following TokenToucan's comment):
Conceptual proof:
We have $E_\sigma \subset K^{\text{ur}}E_\sigma \subset K^{\text{ab}}.$ By Galois theory we have
$$K^{\text{ur}}E_\sigma = (K^{\text{ab}})^{\operatorname{Gal}(K^{\text{ab}}/K^{\text{ur}}E_\sigma)}$$
and
$$\operatorname{Gal}(K^{\text{ab}}/K^{\text{ur}}E_\sigma) = \ker\lbrace \operatorname{Gal}(K^{\text{ab}}/E_\sigma) \to \operatorname{Gal}(K^{\text{ur}}E_\sigma/E_\sigma),\  \sigma \mapsto \sigma|_{K^{\text{ur}}E_\sigma}\rbrace.$$
This map is bijective by assumption, so the kernel is trivial and $K^{\text{ab}} =K^{\text{ur}}E_\sigma.$
Cheat proof through $\widehat{\Bbb Z}$: Both Galois groups are isomorphic to $\widehat{\Bbb Z}$ and the SES
$$1 \to \operatorname{Gal}(K^{\text{ab}}/K^{\text{ur}}E_\sigma) \to \operatorname{Gal}(K^{\text{ab}}/E_\sigma) \to \operatorname{Gal}(K^{\text{ur}}E_\sigma/E_\sigma) \to 1$$
implies that $\widehat{\Bbb Z}$ is isomorphic to a proper quotient of itself, but $\widehat{\Bbb Z}$ is residually finite (since it is profinite) and topologically finitely generated (it is a procyclic group) so it is Hopfian and hence not isomorphic to a proper quotient of itself. In particular, $\operatorname{Gal}(K^{\text{ab}}/K^{\text{ur}}E_\sigma) = 1$.
