From the text: Let \(G\) be a topological group. By a cocycle of \(G\) we mean a continuous map \(f\colon G\to G\) such that \(f(st)=f(s)f(t)^s\), with \(a^s=sas^{-1}\), \(s,t\in G\), \(a\in G\). We denote by \(Z(G)\) the set of all cocycles. Two cocycles \(f\), \(f'\) are equivalent, written \(f\sim f'\), if there is an \(a\in G\) such that (*) \(f(s)=a^{-1}f'(s)a^s\), \(s\in G\). Cocycles \(f\), \(f'\) are locally equivalent, written \(f\underset\text{loc}\sim f'\), if there is an \(a_s\in G\) for each \(s\in G\) such that \(f(s)=a_s^{-1}f'(s)a_s^s\). Two subsets \(B(G)\), \(B_{\text{loc}}(G)\) of \(Z(G)\) are defined by \[ B(G)=\{f\in Z(G);\;f\sim 1\},\qquad B_{\text{loc}}(G)=\{f\in Z(G);\;f\underset\text{loc}\sim 1\}, \] respectively, where 1 denotes the constant function on \(G\) of value \(1_G\). A cocycle in \(B(G)\) is a coboundary and one in \(B_{\text{loc}}(G)\) is a local coboundary. Clearly a coboundary is a local coboundary: \(B(G)\subset B_{\text{loc}}(G)\). The Shafarevich-Tate set \(\text{ Ш}(G)\) is the quotient of \(B_{\text{loc}}(G)\) with respect to the equivalence (*). \(B(G)\) forms a distinguished point in \(\text{ Ш}(G)\). When \(\text{ Ш}(G)=1\), i.e. \(B_{\text{loc}}(G)=B(G)\), we say that \(G\) enjoys the “Hasse principle”.
Here, the author proves: Theorem. Let \(G\) be a profinite group. Then there is a bijection \[ \text{ Ш}(G)\approx\operatorname{Aut}_c(G)/\text{Inn}(G). \] In particular, \(\text{ Ш}(G)\) gets a group structure.
Now let \(K\) be a finite Galois extension over \(\mathbb{Q}\) and \(G_K=\text{Gal}(\overline\mathbb{Q}/K)\). Then celebrated results due to Neukirch, Ikeda, Iwasawa and Uchida tell us that there is an isomorphism \[ \operatorname{Aut}(G_K)/\text{Inn}(G_K)\approx\text{Gal}(K/\mathbb{Q}). \] Combining this with the theorem it follows that the Shafarevich-Tate group \(\text{ Ш}(G_K)\) can be embedded in the finite group \(\text{Gal}(K/\mathbb{Q})\). In particular, \(\text{ Ш}(G_\mathbb{Q})=1\), i.e., the full Galois group \(G_\mathbb{Q}=\text{Gal}(\overline\mathbb{Q}/\mathbb{Q})\) enjoys the Hasse principle.