I'm doing a course on elliptic curves, and I'm stuck on this.
Here are the definitions:
Let $G$ be a group and let $A$ be a $G$-module (that is, a $\mathbb{Z}[G]$-module). We have the "cochains" $$C^1(G,A)=\{\text{maps }: \; G \rightarrow A\},$$ and "cocycles" $$Z^1(G,A)=\{(a_\sigma)_{\sigma \in G}\; | \; a_{\sigma \tau}=\sigma a_\tau+a_\sigma \},$$ and "coboundaries" $$ B^1(G,A)=\{(\sigma b-b)_{\sigma \in G}\}.$$ we have $$C^1(G,A) \supset Z^1(G,A) \supset B^1(G,A).$$ We then define the 1st cohomology group $$H^1(G,A)=\frac{Z^1(G,A)}{B^1(G,A)}.$$ For any normal subgroup $H \leq G$, we also have the inflation map $$ \text{inf}: H^1\left(\frac{G}{H}, A^H\right)\rightarrow H^1(G,A).$$
Let $K$ be a perfect field. Then $G=\text{Gal}(\overline{K}/K)$ is a topological group with basis around the identity given by the subsets $\text{Gal}(\overline{K}/L)$ where $[L:K]< \infty$. (I think we also want $L/K$ to be Galois, but the lecturer didn't say this.)
We then impose that the stabiliser of any $a \in A$ is an open subgroup of $G$, and that the "cochains" are continuous w.r.t. the discrete topology for $A$.
Then the lecturer claims that $$H^1(\text{Gal}(\overline{K}/K),A)=\lim_{\rightarrow}H^1\left(L/K, A^{\text{Gal}(\overline{K}/L)}\right).$$
I can't see why this is true. I'm really struggling to get a hold of the RHS. Here the direct limit is taken over finite extensions $L$ of $K$ with respect to the inflation map.
Edit: here is my attempt at proving $$Z^1(\text{Gal}(\overline{K}/K),A)=\lim_\rightarrow \; Z^1\left( \text{Gal}(L/K),A^{\text{Gal}(\overline{K}/L)}\right).$$
I think what I want to show is that a cocyle $(a_\sigma)\in Z^1(\text{Gal}(\overline{K}/K),A)$ is induced by a cocyle in $Z^1( \text{Gal}(L/K),A^{\text{Gal}(\overline{K}/L)})$ for some $[L:K]<\infty$.
Now, by the continuity we can find a finite extension of $K$, $L_1$, such that the coset of $\text{Gal}(\overline{K}/L_1)$ is contained in the preimage of $0=a_1$. Since this coset contains $1$, it is actually just $\text{Gal}(\overline{K}/L_1)$.
By the condition about stabilisers, we can find $L_2$ such that a coset of $\text{Gal}(\overline{K}/L_2)$ fixes $a_1$, and again this coset is just $\text{Gal}(\overline{K}/L_2)$.
Take $L$ to be the Galois closure of the compositum $L_1L_2$.
Suppose $\sigma, \sigma' \in \text{Gal}(\overline{K}/K)$ such that $\sigma|_L=\sigma'|_L$. Then $\sigma^{-1} \sigma'=\tau \in \text{Gal}(\overline{K}/L)$. So $$a_\sigma'=a_{\sigma \tau}=\sigma a_\tau+a_\sigma =a_\sigma.$$
Thus $a_\sigma$ depends only on $\sigma|_L$.
So for $\tau \in \text{Gal}(\overline{K}/L)$, $a_{\tau \sigma}=a_\sigma$. So $\tau(a_\sigma)=a_{\tau \sigma}-a_\tau=a_\sigma$. Thus each element of $\{a_\sigma: \sigma \in \text{Gal}(\overline{K}/K)\}$ is fixed by $\text{Gal}(\overline{K}/L)$, as required.
First of all, in the definition of the profinite topology, it doesn't matter whether or not you restrict to $L$ that are Galois over $K$. (The one set of subgroups is cofinal in the other.)
As for your main question, have you tried checking the analogous statement for $C^1$, $Z^1$, or $B^1$? (If you do, then the statement about $H^1$ that you want should follow from the fact that passing to direct limits is exact.)