Lemma 1

Let G be a simply connected† topological group, W a symmetric connected open neighbourhood of e, H a group and f a mapping of $W^3$ into H such that

for x, y, z in W. There exists a morphism $f'$ of G into H such that $f' \mid W = f \mid W$.

For $(g, h) \in G \times H$ and U an open neighbourhood of e in W, let $A(g, h, U)$ be the set of $(gu, hf(u)) \in G \times H$ for $u \in U$. Then $(g, h) \in A(g, h, U)$ and $A(g, h, U_1) \cap A(g, h, U_2) = A(g, h, U_1 \cap U_2)$. Let $(s, t) \in A(g, h, U)$; then $s = gu$ and $t = hf(u)$ for $u \in U$; there exists an open neighbourhood $U'$ of e in W such that $uU' \in U$; then, for $u' \in U'$,

† Cf. Chapter XI of General Topology (to appear). It is proved in this chapter that if $G_1, G_2$ are connected topological groups, $\phi$ is an open continuous homomorphism of $G_1$ onto $G_2$ with discrete kernel and $G_2$ is simply connected, then $\phi$ is a homeomorphism. Recall on the other hand that a simply connected space is connected.

and hence $A(s, t, U') \subset A(g, h, U)$. It then follows that the $A(g, h, U)$ form the base of a topology on $G \times H$. We shall denote by $Y$ the set $G \times H$ with this topology and denote by $p$ the canonical projection of $Y$ onto $G$, which is open. The restriction of $p$ to $A(g, h, U)$ is a homeomorphism of $A(g, h, U)$ onto $gU$. Hence $(Y, p)$ is a covering space of $G$. Let $Y_0$ be the subgroup of $Y$ generated by $A(e, e, W)$ and let $\mathfrak{B}$ be the set of $A(e, e, U)$. Clearly $\mathfrak{B}$ satisfies conditions (GV$_1'$) and (GV$_1''$) of General Topology, Chapter III, § 1, no. 2. The set $Y_0'$ of $y \in Y_0$ such that the mappings $z \mapsto yzy^{-1}$ and $z \mapsto y^{-1}zy$ of $Y_0$ into $Y_0$ are continuous at $(e, e)$ is a subgroup of $Y_0$. Let $w \in W$. The mapping $w' \mapsto ww'w^{-1}$ of $W$ into $G$ is continuous and hence the mapping

of $A(e, e, W)$ into $Y$ is continuous at $(e, e)$. Now $f(ww'w^{-1}) = f(w)f(w')f(w^{-1})$ and therefore $(ww'w^{-1}, f(ww'w^{-1})) = (w, f(w))(w', f(w'))(w, f(w))^{-1}$.

As $w^{-1} \in W$, we see that $(w, f(w)) \in Y_0'$. Thus, $A(e, e, W) \subset Y_0'$, so that $Y_0' = Y_0$. The group $Y_0$, with the filter base $\mathfrak{B}$, therefore satisfies condition (GV$_{III}'$) of General Topology, Chapter III, § 1, no. 2. As

$Y_0$ is a topological group which is connected since $A(e, e, W)$ is connected. Then $p(Y_0)$ is an open subgroup of $G$, whence $p(Y_0) = G$ since $G$ is connected. The kernel of $p \mid Y_0$ is discrete. As $G$ is simply connected, $p \mid Y_0$ is a homeomorphism of $Y_0$ onto $G$. Therefore $Y_0$ is the graph of a morphism $f'$ of $G$ into $H$. For $g \in W$, $(g, f(g)) \in A(e, e, W) \subset Y_0$, whence $f'(g) = f'(g)$.