Let $L/K$ be a Galois extension of a number field and $E/K$ be an elliptic curve. let $E(K)$ be an Mordel-Weil group of $E/K$.
I encountered a notation $\hat{E(K)}$, which is the completion with respect to the topology defined by the subgroups of finite index. (related: https://math.stackexchange.com/users/985494/brauermaninobstruction)
The paper I am reading(link, the bottom of p214) claims there is a natural surjection homomorphism from$\hat{E(K)}$ to $\hat{H^0}(Gal(L/K),E(L))$. Here, $\hat{H^0}$ denotes Tate-cohomology group (https://en.wikipedia.org/wiki/Tate_cohomology_group).
I have no idea how to construct surjection from $\hat{E(K)}$ to $\hat{H^0}(Gal(L/K),E(L))$.
To sum up, why(how) can we make surjective homomorphism from from $\hat{E(K)}$ to $\hat{H^0}(Gal(L/K),E(L))$?
P.S. Before I notice a comment from Qiaochu Yuan, I made a mistake. I mistook Tate-cohomology $\hat{H^0}(Gal(L/K),E(L))$ for usual cohomology $H^0(Gal(L/K),E(L))$.
It's easier to see 'the other away around.' Namely, if the Galois extension $L/K$ is finite, then $T$, the trace/norm of $E(L)$ in $E(K)$, is a subgroup of finite index in $E(K)$ (because it contains $[L:K] E(K)$). Hence the natural map $$\hat H^0(L/K, E(L)) \stackrel{\text{def}}{=} E(K)/T \to {E(K)}^\wedge/T^\wedge $$ is an isomorphism.
(The superscript wedges indicate completion. Sorry for the change of notation - it was disturbing me to have hats being used both for Tate cohomology and for completion on the same line.)
Addendum:
In answer to the question below - why is $G/H \to G^\wedge/H^\wedge$ an isomorphism.
Suppose that $$\cal C =\{C_i\mid i\in I\},$$ is set of normal, cofinite, cofinal subgroups of $G$. (By 'cofinal', I mean that any normal subgroup of finite index $H$ contains some $C_i$.) The intersection $C_{i_1}\cap \cdots \cap C_{i_n}$ of any finite collection of $C_i$ is also normal and cofinite [as is the product $C_{i_1} \cdots C_{i_n}$]. So if $\cal C$ also contains all such, then one can calculate completion $G^\wedge$ using the $C_i$: $$ G^\wedge = \varprojlim_{i\in I} G/C_i.$$
In particular (the point), if $H$ is a normal subgroup of finite index in $G$, then we can in fact calculate $G^\wedge$ with $\cal C$ consisting of subgroups $C_i$ of $H$ ($C_i$ normal and of finite index in $G$).
Now, given $i_1, \cdots, i_n$ a corresponding basis open set $U$ in $G^\wedge$ has component $g_iC_i $ in the $i$th place, where the $g_i $ form a compatible system (for the inverse limit). In fact, if one writes $C = C_{i_1}\cap \cdots C_{i_n}$, then there exists a $g\in G$ such that
$$g C = g_{i_k} C$$ for $k=1,\cdots,n$. Thus, pretending that the $i_1,\cdots, i_n$ are the 'first' elements in $I$, $U$ is of the form
$$ U = \{(gC_1,\cdots, gC_n, g_{i}C_i, \cdots)\mid i\not= i_1,\cdots i_n, \text{ and } g C_i \cap C= g_i C\cap C_i\}.$$
In other words, G is dense in $G^\wedge$, so that $G \twoheadrightarrow G^\wedge / W$, for any open normal subgroup $W$ of $G^{\wedge}$.
Now $H^\wedge$ is such a subgroup, and the above applies: namely, if $g\in H$, then the equation $$g_i \equiv g \pmod {C\cap C_i},$$ forces $g_i$ to be an element of $H$, since, by construction/fiat, the $C$ and $C_i$ are all subgroups of $H$.
Therefore $$ G/H \to G^{\wedge}/H^{\wedge}$$ is an isomorphism.