Theorem 1

The mapping Θ is a group isomorphism from Ex$_{\tau}(G,F)$ to $H^2(G,F)$.

Let us first prove that Θ is a group homomorphism. Let $\mathscr{E}= (\Gamma , \iota , \pi )$ and $\mathscr{E}'= (\Gamma ', \iota ', \pi ')$ be $\tau$-extensions, and let $\sigma$ (resp. $\sigma '$) be a section of the mapping $\pi$ (resp. $\pi '$). We denote by $\mathscr{E} \mathscr{E}'= (\Gamma '', \iota '', \pi '')$ the product of the $\tau$-extensions $\mathscr{E}$ and $\mathscr{E}'$. We denote by $[\gamma , \gamma ']$ the image in $\Gamma ''$ of an element $(\gamma , \gamma ')$ of $\Gamma \times_G\Gamma '$ by the surjective homomorphism from the remark of VIII, p. 293. The mapping from G to $\Gamma ''$ that sends an element $g$ to $[\sigma (g), \sigma '(g)]$ is a section $\sigma ''$ of the mapping $\pi ''$. Let $g_1$ and $g_2$ be elements of G. We have the relations

We therefore have $c_{\sigma''}(g_1, g_2) =c_{\sigma}(g_1, g_2)c_{\sigma'}(g_1, g_2)$.

Let us prove that the mapping Θ is injective. Let $\mathscr{E}= (\Gamma , \iota , \pi )$ be a $\tau$-extension, and let $\sigma$ be a section of the mapping $\pi$ such that the image of $c_{\sigma}$ in $H^2(G,F)$ is trivial. There exists a mapping $a: G\rightarrow F$ such that

for all $g_1, g_2\in G$. We then define $\sigma ': G\rightarrow \Gamma$ by

for $g\in G$. Using (11), we obtain that $c_{\sigma'}$ is constant with value 1; consequently, $\sigma '$ is a group homomorphism, which proves that the $\tau$-extension $\mathscr{E}$ is semitrivial (I, §6, No. 1, p. 67, Proposition 4).

Let us prove that the mapping Θ is surjective. Let $c$ be an element of $Z^2(G,F)$. We endow the set $\Gamma = F\times G$ with the following law of composition: (12) $(f_1, g_1)(f_2, g_2) =(f_1(^{g_1}f_2)c(g_1, g_2), g_1g_2)$

for all $f_1, f_2\in F$ and $g_1, g_2\in G$. We have the relations

and

for all $f_1, f_2, f_3\in F$ and $g_1,g_2, g_3\in G$. It follows that this law is associative because $c$ is a 2-cocycle. For every $(f, g)\in \Gamma$, we have

Now, it follows from the definition of a 2-cocycle that $c(g, e) =^gc(e, e)$, from which we deduce that $(f, g)(c(e, e)^{-1}, e) = (f, g)$. We prove likewise that

The law of composition defined by (12) therefore has identity element $(c(e, e)^{-1}, e)$, and every element $(f, g)$ of Γ is invertible, with inverse

So we have endowed Γ with a group structure. Let us denote by $\iota : F\rightarrow G$ the mapping that sends $f$ to $(c(e, e)^{-1}f, e)$, by $\pi : \Gamma \rightarrow G$ the second projection, and by $\sigma$ the mapping $g\mapsto (1, g)$. The mappings $\iota$ and $\pi$ are group homomorphisms, so the triple $\mathscr{E}= (\Gamma , \iota , \pi )$ is a $\tau$-extension, $\sigma$ is a section of the mapping $\pi$, and the associated cocycle $c_{\sigma}$ is equal to $c$ because

for $g_1, g_2\in G$.