uniqueness of local Artin map

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This problem defines a map with the same properties as local Artin map and asks you to prove they are equal. I'm having problems with b) and c). Is the first part of b) comes from Galois correspondence? And how is that [$L_{\sigma}:K]=[L:K]$? I believe $Frob_K$ is the generator of $Gal(K^{nr}/K)$. So shouldn't $[L_{\sigma}:K]$ be smaller than $[L:K]$? Please let me know where I'm wrong. As to c) I have no clue.

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Hint for (b): We know that $\mathrm{Gal}(L^{\mathrm{nr}}/L_\sigma)$ is generated by $(\sigma,\mathrm{Frob}_K)$. First of all, by the Galois correspondence, you want to check $L_\sigma=L_\tau$, i.e., $(\sigma,\mathrm{Frob}_K)$ and $(\tau,\mathrm{Frob}_K)$ generate the same subgroup of $\mathrm{Gal}(L/K)\times\mathrm{Gal}(K^{\mathrm{nr}}/K)$, then $\sigma=\tau$.

Moreover, the degree $[L_\sigma:K]$ is simply the index of $\mathrm{Gal}(L^{\mathrm{nr}}/L_\sigma)$ in $\mathrm{Gal}(L^{\mathrm{nr}}/K)$, i.e., the index of the subgroup generated by $(\sigma,\mathrm{Frob}_K)$ in $\mathrm{Gal}(L/K)\times\mathrm{Gal}(K^{\mathrm{nr}}/K)$. Thus, both questions are now a group theory problem: prove the size of this quotient is $[L:K]$.

Warning: to be precise, you need to consider the closure of the subgroup generated by $(\sigma,\mathrm{Frob}_K)$.

Hint for (c): We know $$K^\times/N_{L_{s_K(a)}/K}L_{s_K(a)}^\times\cong\mathrm{Gal}(L_{s_K(a)}/K)\cong (\mathrm{Gal}(L/K)\times\mathrm{Gal}(K^{\mathrm{nr}}/K))/(\sigma,\mathrm{Frob}_K),$$ whereas $$K^\times/\langle N_{L/K}\mathcal O_L^\times,aN_{L/K}\varpi\rangle\cong\langle \mathcal O_K^\times,\varpi_K\rangle/\langle N_{L/K}\mathcal O_L^\times,aN_{L/K}\varpi\rangle.$$ Moreover, there are isomorphisms $\mathrm{Gal}(K^{\mathrm{nr}}/K)\cong\langle\varpi_K\rangle$ and $\mathrm{Gal}(L/K)\cong\mathcal O_K^\times/N_{L/K}\mathcal O_L^\times$. Under these isomorphisms, $(\sigma,\mathrm{Frob}_K)$ corresponds to $aN_{L/K}\varpi$. You should track all of these isomorphisms, thinking about which uses $s_K$ and which just uses Galois theory, and why they are compatible, which should just reduce to the axioms 1, 2, and 3.