Let $F/K$ be a finite extension of number fields, let $v$ be a prime of $K$ that is totally split in $F$. Call $w_1,\dots, w_n$ the primes of $F$ above $v$. Let $T$ be a $Gal(\bar{K}/K)$-module. Then we have the following maps:
$$H^1(F,T)\overset{\oplus res}{\longrightarrow}\oplus_{i=1}^n H^1(F_{w_i},T)$$ $$H^1(K,T)\overset{res}{\longrightarrow}H^1(K_v,T)$$
I want to close the diagram with two vertical arrows: the first is clearly $res: H^1(K,T)\to H^1(F,T)$, while the second is also called $\oplus res H^1(K_v,T)\to \oplus_{i=1}^n H^1(F_{w_i},T)$. But now, since $v$ is split in $F$, we have that $K_v=F_{w_i}$ so this last map seems to be a sort of $\oplus id$.
The point is that I know that these maps depend somehow on embeddings, so I tried to make things clear choosing decomposition groups $D_{v}$ and $D_{w_i}$ and work with them. Now, we can suppose without loss of generality that $D_v=D_{w_1}$, so that each $D_{w_i}$ is a conjugate of $D_{w_1}$ with respect to an element of $Gal(F/K)$.
It would be awesome if someone could help me understanding the meaning of the map $ H^1(D_v,T)\to \oplus_{i=1}^n H^1(D_{w_i},T)$ that should make the previous diagram commutative. This is not really a sum of restrictions, but rather a sum of conjugations via elements of $Gal(F/K)$?
Edit: Ok, let’s try again with something more focused on your setting. I’ll ignore the $H^1$ in cohomology – as previously, it’s much easier to work with an injective resolution $I$.
Let’s choose a decomposition group $D_v$ above $v$. This choice defines a place $w$ of $F$, and the decomposition group $D_w=D_v \cap G_K$. Since $v$ is totally split, we actually have $D_v=D_w$ but we’ll ignore it. On the other hand, I’m assuming that $F/K$ is Galois.
It’s obvious that you compute the restriction from $G_K$ to $D_w$ by restricting to $G_F$ and localizing afterwards; or by localizing first to $D_v$ and then restricting.
Now let’s consider another decomposition group $D’_v$: it can be written as $gD_vg^{-1}$ for some $g \in G_K$, and defines the place $w’=g \cdot w$ of $F$, and the corresponding decomposition group over $F$ is $D’_{w’}=G_F \cap D’_v=gD_wg^{-1}$.
So we have an isomorphism $x \in I^{D_v} \longmapsto g \cdot x\in I^{D’_{v’}}$ (preserving the restriction from $I^{G_F}$), and if $F/K$ is Galois, a corresponding isomorphism $I^{D_w} \rightarrow I^{D’_{w’}}$. The latter does not preserve the restriction from $K$: indeed, it acts as multiplication by $g$ on $I^{G_K}$.
Now, pick a place $w_0$ of $F$ above $v$, and $D_v$ a decomposition group above $v$ that induces the place $w_0$. For each place $w|v$ of $F$, choose $g_w$ such that $g_w \cdot w_0=w$.
Recall that we’re interested in linking the morphisms: $I^{G_K} \subset I^{D_v}$ and $I^{G_F} \rightarrow \bigoplus_{w |v}{I^{g_wD_v g_w^{-1} \cap G_F}}$ given by the natural inclusions.
The natural map which fits the bill is $I^{D_v} \rightarrow \bigoplus_{w|v}{I^{g_w D_v g_w^{-1} \cap G_F}}$ whose projection at $w$ is multiplication by $g_w$.
How do we see the $Gal(F/K)$-action on $\bigoplus_{w|v}{I^{g_wD_vg_w^{-1} \cap G_F}}$?
Well, note that $I^{D_v \cap G_F}$ is naturally a $D_v/D_v \cap G_F$-module – thus it’s a $D_vG_F/G_F=Gal(F/K)_{w_0}$-module.
Then it shouldn’t be very hard to prove that $\bigoplus_{w|v}{I^{g_w D_v g_w^{-1} \cap G_F}}=\bigoplus_{w|v}{g_w I^{D_v \cap G_F}}$ is naturally the induced module of $I^{D_v \cap G_F}$ from $Gal(F/K)_{w_0}$ to $Gal(F/K)$.
Moreover, you can also see that
Old answer: Let $G$ be a profinite group, $g \in G$, $H \leq G$ be a closed subgroup. I’m considering cohomology of discrete modules.
There are restriction maps $H^{\star}(G,M) \rightarrow H^{\star}(H,M)$ and $H^{\star}(G,M) \rightarrow H^{\star}(gHg^{-1},M)$. I claim that there’s a natural isomorphism between $H^{\star}(H,M)$ and $H^{\star}(gHg^{-1},M)$ identifying these maps.
Proof: choose an injective resolution $I^{\cdot}$ of $M$ as a $G$-module. Then the restriction to $H$ is the image in cohomology of $I^G \subset I^H$. The restriction to $gHg^{-1}$ is the image in cohomology of $I^G \rightarrow I^{gHg^{-1}}$. The isomorphism is given by $x \in I^H \longmapsto g \cdot x \in I^{gHg^{-1}}$.
This tells us that the restriction map $H^{\star}(G,M) \rightarrow H^{\star}(H,M)$ does not depend on the conjugacy class of $H$ in $G$.
Or, in other words, if $H$ is not actually a subgroup of $G$ but another abstract profinite group, with a continuous injective morphism $H \rightarrow G$ defined only up to inner automorphism of $G$, the restriction map $H^{\star}(G,M) \rightarrow H^{\star}(H,M)$ is well-defined nonetheless.
In particular, if $L/K$ is a field extension, and $G=Gal(K_{sep}/K)$, $H=Gal(L_{sep}/L)$, you can check that the maps $H \rightarrow G$ that we could use are conjugates of one another by an element of $G$, so the restriction $H^{\star}(K,M) \rightarrow H^{\star}(L,M)$ is independent of the embedding.