In a lecture note of Joel Bellaiche, he mentioned that when $V$ is finite then we have
$$\dim H^0(\hat{\mathbb{Z}},V)=\dim H^1(\hat{\mathbb{Z}},V)$$
where $$\hat{\mathbb{Z}}=\lim_{\longleftarrow}\mathbb{Z}/n\mathbb{Z}$$
I know that the cohomological dimension of $\hat{\mathbb{Z}}$ is $1$. Also, in Serre's local fields, we have the following description
$$H^q(\hat{\mathbb{Z}},A)=\lim_{\longrightarrow}H^q(\hat{\mathbb{Z}}/n\hat{\mathbb{Z}},A^{n\hat{\mathbb{Z}}})$$
Now, $\hat{\mathbb{Z}}$ is generated by $1$ and let $F$ be the homomorphism of ${\rm Aut}(A)$ which is given by the action of $1$ on $A$. Then $A^{n\hat{\mathbb{Z}}}=A^{F^n}$. Also, let $A'$ be the subset of elements $a$ of $A$ such that there exists $n\in\mathbb{N}$ such that $$(1+F+\cdots+F^{n-1})a=0$$ Then $$H^1(\hat{\mathbb{Z}},A)=A'/(F-1)A$$ and $H^0(\hat{\mathbb{Z}},A)$ is simply $A^{\hat{\mathbb{Z}}}$. Can we use these to build an isomorphism between $H^0(\hat{\mathbb{Z}},V)$ and $H^1(\hat{\mathbb{Z}},V)$? Can you please give me some hints?
Thanks in advance!
Since $A$ is finite and the action of $\hat{\mathbb{Z}}$ on $A$ is determined by it's topological generator $\hat{1}$, we get that $H^0(\hat{\mathbb{Z}},A)=A^F$ and $H^1(\hat{\mathbb{Z}},A)=A/(F-1)A$, where $F$ is the automorphism of $A$ corresponding to the action of $\hat{1}$ on $A$. We have the following exact sequence $$0\longrightarrow A^F\longrightarrow A\xrightarrow{\;F-1\;}A\longrightarrow A/(F-1)A\longrightarrow0$$ This shows that $$A/A^F\cong (F-1)A$$ $$\implies |A/A^F|=|(F-1)A|$$ $$\implies |A^F|=|A|/|(F-1)A|=|A/(F-1)A|\\\implies |H^0(\hat{\mathbb{Z}},A)|= |H^1(\hat{\mathbb{Z}},A)|$$
Using this and a theorem due to Tate, we can prove that for any $p$-adic representation $V$ of $\hat{\Bbb{Z}}$ we have $$\dim(H^0(\hat{\mathbb{Z}},A))=\dim(H^1(\hat{\mathbb{Z}},A))$$