Transfer the inner automorphism on the Weil group to the multiplicative group of local fields Let $K$ be a $p$-adic field, $W_K$ its Weil group. By local class field theory (CFT), there exists a canonical morphism $\mathfrak{a}_K:W_K\rightarrow K^\times$, which induces an isomorphism $W_K^{ab}\rightarrow K^{\times}$, and it satisfies a series of functorial properties. More precisely, if $L/K$ is a finite extension, then $W_L\subset W_K$ and the CFT isomorphism passes this morphism to the norm map $N_{L/K}:L^{\times}\rightarrow K^{\times}$. And for the natural embedding $K^{\times}\rightarrow L^{\times}$, the CFT isomorphism passes it to the group-theoretic transfer map ("Ver").
Now I met the following fact: if $L/K$ is finite Galois, then $Gal(L/K)=W_K/W_L$. Pick $\sigma\in Gal(L/K)$ and a lift of $\sigma$ in $W_K$, denoted by $\tilde{\sigma}$. Since $W_L$ is now a normal subgroup of $W_K$, we may consider the conjugate action of $\tilde{\sigma}$ on $W_L$, namely
$$W_L\rightarrow W_L, \tau\mapsto \tilde{\sigma}\tau\tilde{\sigma}^{-1}.$$
This induces a homomorphism $W_L^{ab}\rightarrow W_L^{ab}$. Now we pass this map to $L^{\times}\rightarrow L^{\times}$ via the CFT isomorphism. The claim is: this give exactly the action of the Galois group element: $x\mapsto \sigma(x)$.
I found this fact seemingly intuitive, but I didn't find it on any standard textbook. Finally, I saw this on Tate's famous article "Number Theoretic Background" (1.5). However, Tate's article goes much further, and I didn't really understand it (it works in a much wider and axiomatic/functorial framework). So I didn't understand his proof of this fact. But I guess this is rather standard for veterans.
Can anybody explain its background/idea for me? Or any reference (at the level of textbooks)? Thanks a lot in advance!
 A: This is actually a fact about the Artin reciprocity isomorphism $r_{L/K}: Gal(L/K)^{ab} \rightarrow K^{\times}/N_{L/K}L^{\times}$, for all the finite Galois extensions $L/K$ of $p$-adic fields.
The precise statement is this: if $M/L$ is a finite Galois extension of $p$-adic fields both finite Galois over a $p$-adic field $K$, then, for any $s \in Gal(M/K)$, $r_{M/L}(s \cdot s^{-1})=s \circ r_{M/L}$.
To prove this, I think we can only come back to the definition of the reciprocity homomorphism.
Short version: read again local class field theory and assume everything is Galois over a certain fixed base field $F$. Show that everything commutes to conjugation by $G_F$.
Here’s the long version.
Note: Thus I’ll use group cohomology, which will always be the Tate cohomology for finite groups.
First, $r_{M/L}$ is the composition of the identification $i: Gal(M/L)^{ab} \rightarrow H^{-2}(Gal(M/L),\mathbb{Z})$ and the cup-product with the fundamental class $u_{M/L} \in H^2(Gal(M/L),M^{\times})$ mapping $H^{-2}(Gal(M/L),\mathbb{Z})$ isomorphically to $H^0(Gal(M/L),M^{\times})$, and then of the identification $j: H^0(Gal(M/L),M^{\times}) \cong L^{\times}/N_{M/L}M^{\times}$.
Thus, to prove out claim, we need to show the following:

*

*Conjugating $Gal(M/L)^{ab}$ by some $s \in Gal(M/K)$ corresponds through $i$ to the standard action by conjugation on cohomology (*).

*The cup product commutes to the standard action by conjugation on the cohomology.

*Through the identification $j$, conjugation by $s \in Gal(M/K)$ on cohomology amounts to applying $s$ to $L^{\times}$.

*$u_{M/L}$ is invariant under conjugation by $Gal(M/K)$.

(*) if $G$ is a finite group and $H \triangleleft G$, and $M$ is a $G$-module, then, for all $g \in G$, we can define a morphism of complexes of cochains for the $H$-module $M$ by $c(\cdot) \longmapsto g \cdot c(g^{-1}\cdot g)$. This products a natural action of $G$ (the “standard action by conjugation”) on $H^{\cdot}(H,M)$.


*is thus an instance of functoriality for the cup-product on group cohomology and therefore holds.

We can next restate 3 as follows in more generality: let $M$ be a module over a finite group $G$ and $H \triangleleft G$. Under the identification $H^0(H,M) \cong M^H/N_HM$, conjugation on cohomology by $g \in G$ corresponds to applying $g: M^H \rightarrow M^H$, and this is a formal verification.
In the same way, we can rephrase 1: let $G$ be a finite group and $H \triangleleft G$, under the identification $H^{-2}(H,\mathbb{Z}) \cong H^{ab}$, conjugation by $g \in G$ on cohomology amounts to conjugation by $g$ on $H^{ab}$.
To prove this, we need to spell out what that isomorphism is: we have an exact sequence $0 \rightarrow I \rightarrow \mathbb{Z}[H] \rightarrow \mathbb{Z} \rightarrow 0$ of $G$-modules, with $\mathbb{Z}[H]$ induced over $H$ (the second map is the degree $h \longmapsto 1$), so that we have a $G$- isomorphism $H^{-2}(H,\mathbb{Z}) \cong H^{-1}(H,I)$.
Now, a formal verification shows that we have a $G$-isomorphism $H^{-1}(H,I) \cong I^{N_H=0}/\sum_h{(h-1)I}$. Finally, we can check that $h-1 \longmapsto h$ defines an isomorphism $I^{N_H=0}/\sum_h{(h-1)I} \rightarrow H^{ab}$ respecting conjugation by $g \in G$.
All that remains is 4, the really interesting point – and obviously, we have no choice but go back to the definition.
Let $M’$ be the unramified extension over $L$ of same degree as $M/L$, it’s clearly Galois over $K$. Let $N=M \cdot M’$, it’s Galois over $L$ and $K$.
It’s well-known that the images of the injective inflation maps $H^2(Gal(M/L),M^{\times}) \rightarrow H^2(Gal(N/L),N^{\times})$ and $H^2(Gal(M’/L),M’^{\times}) \rightarrow H^2(Gal(N/L),N^{\times})$ are the same, that both maps commute with conjugation by $Gal(N/K)$ (note the change of groups, although it doesn’t really matter), and that the images of $u_{M/L}$ and $u_{M’/L}$ in this bigger group are the same.
Therefore, we may assume that $M/L$ is unramified. But then the valuation map (which is $Gal(M/K)$-equivariant) $M^{\times} \rightarrow \mathbb{Z}$ induces an isomorphism in $Gal(M/L)$-cohomology. Moreover, by considering the action on the residue field, $Gal(M/L)$ is central in the $Gal(M/K)$, so that the $Gal(M/K)$-action on $H^2(Gal(M/L),\mathbb{Z})$ is trivial, which concludes.
