Lemma 1

Let G be a topological group and let V be a connected neighbourhood of the identity element $e$ of G. For every neighbourhood U of $e$ in G and every $x\in V$, there exist an integer $n\in \mathbf{N}$ and elements $u_1, . . . , u_n$ in U such that $u_1. . . u_n=x$ and $u_1. . . u_k\in V$ for every integer $k$ such that $1\leqslant k\leqslant n$.

One may suppose that U is open and contained in V. Let A denote the set of $x\in V$ satisfying the condition of the lemma. If $x\in A$ and if $y\in xU\cap V$, then $y\in A$, whence AU $\cap V\subset A$; this shows that A is open in V. Let $x\in V$ be such that $xU^{-1}\cap A=\not\emptyset$; then one has $x\in$ AU$\cap V$, hence $x\in A$. Consequently, if $x\in V$ and $x\notin A,xU^{-1}\cap A =\emptyset$ and A is closed in V. Since $e\in A$ and V is connected, it follows that A = V.

Let us now prove the proposition. Let $j$ be the mapping $g\mapsto$ $(g, f(g))$ of V into $G\times G'$ and let H be the subgroup of $G\times G'$ generated by $j(V)$.

Let U be a neighbourhood of $e$ in G, contained in V. Let $x\in V$; by lemma 1, there exist $u_1, . . . , u_n\in U$ such that $x=u_1. . . u_n$ and such that $u_1. . . u_k\in V$ for every integer $k$ such that $1\leqslant k\leqslant n$. By induction, one has $f(u_1. . . u_k) =f(u_1). . . f(u_k)$ for every integer $k,1\leqslant k\leqslant n$. In particular, $j(x)$ belongs to the subgroup generated by $j(U)$. It follows that H is generated by $j(U)$.

Let $\mathscr{B}$ be the set of subsets of H of the form $j(U)$, where U is a neighbourhood of $e$ in V. Let us show that there exists a unique topology on H, compatible with its group structure, for which $\mathscr{B}$ is a base of the filter of neighbourhoods of the identity element. For this, it is enough to prove that the set $\mathscr{B}$ satisfies conditions (GV$'_I$), (GV$'_{II}$) and (GV$'_{III}$) of III, p. 4.

Let therefore U be a neighbourhood of $e$ in G, contained in V; there exists a neighbourhood $U'$ of $e$ such that $U'\cdot U'\subset U$. For every pair $x, y$ of points of $U'$, one has $xy\in V$ and $f(xy) =f(x)f(y)$. Hence $j(U')\in \mathscr{B}$ and $j(U')\cdot j(U')\subset j(U)$. This shows that condition (GV$'_I$) is satisfied.

The set $U''= V\cap U^{-1}$ is then a neighbourhood of $e$ in V and, for $x\in U''$, one has $x^{-1}\in U$ and $f(x^{-1}) =f(x)^{-1}$. Hence $j(U'')^{-1}\subset j(U)$, which shows condition (GV$'_{II}$).

Finally, let us fix a neighbourhood U of $e$ in V such that $U = U^{-1}$ and $U^3\subset V$. Let W be a neighbourhood of $e$ in U and let $h= (g, f(g))$ be an element of $j(U)$. There exists a neighbourhood $W'$ of $e$ contained in W such that $gW'g^{-1}\subset W$. Then $j(W')\in \mathscr{B}$ and $hj(W')h^{-1}\subset j(W)$, for one has $f(gxg^{-1}) =f(g)f(x)f(g^{-1})$, for $x\in W'$.

Let $h\in H$. Since U is a neighbourhood of $e$ contained in V$,j(U)$ generates H; since U is symmetric, there exist elements $u_1, . . . , u_n$ in U such that $h=j(u_1). . . j(u_n)$. By induction on $n$, there exists a neighbourhood $W'$ of $e$ contained in W such that $hj(W')h^{-1}\subset j(W)$. Consequently, condition (GV$'_{III}$) is satisfied.

Let us then endow the group G with this topology.

Let us denote by $p: H\rightarrow$ G the restriction to H of the first projection $G\times G'\rightarrow G$. This is a group homomorphism. For every neighbourhood U of $e$ contained in V, the set $\overset{-1}{p}(U)$ contains the neighbourhood $j(U)$ of the identity element of H; hence $p$ is a continuous homomorphism of topological groups. For every neighbourhood U of $e$ contained in V, one has $p(j(U)) = U$, hence $p$ is an open mapping. Since G is connected, $p$ is surjective. Its kernel is discrete because it meets $j(V)$ only in the identity element. It then follows from Corollary 3 (I, p. 100) that the G-space $(H, p)$ is a covering.

Let $\varphi$ be the restriction to H of the second projection $G\times G'\rightarrow G'$. If $g\in V$, one has $(g, f(g)) =j(g)\in j(V)$ and $\varphi (g, f(g)) =f(g)$, so that $\varphi (y) =f(p(y))$ for $y\in j(V)$.

Suppose moreover that $G'$ is a topological group and that the mapping $f$ is continuous at $e$. Then the homomorphism $\varphi$ is continuous at the identity element of H, and hence is continuous (TG, III, p. 15, Prop. 23).