The condition is necessary by Proposition 3 of TG, III, p. 6. Conversely, suppose that $p$ induces a homeomorphism of an open neighbourhood V of the identity element $e$ of G onto a neighbourhood $V'$ of the identity element of $G'$. The mapping $p$ is then continuous (TG, III, p. 15, Prop. 23) and open (TG, III, p. 16, Prop. 24). The image H of $p$ is a subgroup of $G'$ containing V, hence open and closed in $G'$ (TG, III, p. 7, Corollary). Let N be the kernel of $p$; one has $N\cap V =\{e\}$, therefore N is discrete (loc. cit., Prop. 5). Consequently, $p$ makes G a principal covering of H, with group N (I, p. 100, Cor. 3 of Th. 1). Since H is open and closed in $G'$, the $G'$-space $(G, p)$ is a covering.

If $G'$ is a topological group and if the mapping $f$ is continuous, such a homomorphism $\varphi$ is continuous.

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).

The group G is connected. Let H$,\varphi$ and $p$ be as in proposition 2. Since G is simply connected, H is a trivializable covering of G (I, p. 124, def. 3); since H is connected and nonempty, the mapping $p$ is an isomorphism of topological groups. Put $h=\varphi \circ p^{-1}$; the mapping $h$ is a group homomorphism and there exists an open neighbourhood U of the identity element $e$ of G contained in V such that $f|U =h|U$. It follows from lemma 1 that $f$ and $h$ coincide on V. In other words, the mapping $h$ extends the mapping $f$.

If $G'$ is a topological group and if the mapping $f$ is continuous, the homomorphism $\varphi$ is continuous, hence so is $h$.

Let us prove the uniqueness of such an extension. The set of points of G where two homomorphisms of G into $G'$ coincide is a subgroup of G. Since the group G is connected, every subgroup of G containing a neighbourhood of the identity element is equal to G (TG, III, p. 8, prop. 6). The uniqueness of $h$ follows.

Let G and $G'$ be locally connected topological groups, let V and $V'$ be connected neighbourhoods of the identity element of G and of $G'$ respectively, and let $f: V\rightarrow V'$ be a homeomorphism which is a local isomorphism of G onto $G'$. By corollary 1, there exists a unique continuous group homomorphism $\varphi : G\rightarrow G'$ extending $f$ and a unique continuous group homomorphism $\varphi ': G'\rightarrow$ G extending $f^{-1}$. The homomorphisms $\varphi '\circ \varphi$ and Id$_G$ coincide on a neighbourhood of $e$ in G, hence are equal, since G is connected. Analogously, $\varphi \circ \varphi '=$ Id$_{G'}$, whence the corollary.

Let $F(V,\mathbf{r})$ be the quotient group of the free group F(V) by the smallest normal subgroup containing the elements $xyz^{-1}$, where $(x, y, z)\in V\times$ $V\times V$ and $xy=z$ (A, I, p. 86). Let us denote by $f: V\rightarrow F(V,\mathbf{r})$ and $g: F(V,\mathbf{r})\rightarrow G$ the canonical mappings. By construction, the mapping $f$ is a local homomorphism of G into $F(V,\mathbf{r})$. By corollary 1, there exists a unique group homomorphism $\overline{f}: G\rightarrow F(V,\mathbf{r})$ extending $f$. Since the group $F(V,\mathbf{r})$ is generated by $f(V)$, the homomorphism $\overline{f}$ is surjective. For $x\in V$, one has $g(\overline{f}(x)) =g(f(x)) =x$; since V generates G$,g\circ \overline{f}=$ Id$_G$, which proves that $\overline{f}$ is injective. It is therefore an isomorphism.

where $\sim$ denotes the relation of strict homotopy. The law of composition of the group $\pi_1(X, e)$ is the homomorphism obtained by composing the canonical isomorphism $\pi_1(X, e)\times \pi_1(X, e)\rightarrow \pi_1(X\times X,(e, e))$ (III, p. 297, corollary) and the homomorphism $\pi_1(m,(e, e))$ of $\pi_1(X\times X,(e, e))$ into $\pi_1(X, e)$. The group $\pi_1(X, e)$ is commutative.

Let $\mu:\pi_1(X, e)\times \pi_1(X, e)\rightarrow \pi_1(X, e)$ be the group homomorphism obtained by composing the canonical isomorphism of $\pi_1(X, e)\times \pi_1(X, e)$ onto $\pi_1(X\times$ $X,(e, e))$ and the homomorphism $\pi_1(m,(e, e))$. It is to be proved that one has

for all $\gamma ,\gamma '\in \pi_1(X, e)$. Taking account of Remark 2 of III, p. 297, one has:

Let $c\in \Omega_e(X)$ be a loop of class $\gamma$; let us denote by $m_1: X\rightarrow X$ the mapping defined by $m_1(x) =m(x, e)$; then $\mu(\gamma , \varepsilon_e)$ is the class of $m_1\circ c$. By definition of a Hopf space, the mappings pointed at $e$, $m_1$ and Id$_X$, have the same pointed homotopy class at $e$. Hence the loops at $e$, $m_1\circ c$ and $c$, are strictly homotopic (III, p. 296, Cor. 1 of Prop. 3) and one has $\mu(\gamma , \varepsilon_e) =\gamma$. One proves analogously that one has $\mu(\varepsilon_e, \gamma ') =\gamma '$. Relation (1) then follows from relation (2).

a) The covering $X'$ is Galois and the group of its automorphisms is commutative.

b) There exists a continuous law of composition $m': X'\times X'\rightarrow X'$ and only one such that one has $p\circ m'=m\circ (p, p)$ and $m'(e', e') =e'$. Endowed with this law of composition, the pointed space $(X', e')$ is a Hopf space.

c) Endowed with the law of composition induced by $m'$, the fibre $X'_e$ is a group with identity element $e'$. The mapping of $\pi_1(X, x)$ into $X'_e$ given by $\gamma \mapsto e'\cdot \gamma$ is a surjective group homomorphism with kernel $p_*(\pi_1(X', e'))$. The mapping $g\mapsto g(e')$ is an isomorphism of the group Aut$_X(X')$ onto the group $X'_e$.

a) The group $\pi_1(X, e)$ is commutative (prop. 3). Since the covering $X'$ of X is connected, it is then a Galois covering and the group Aut$_X(X')$ is isomorphic to the quotient group $\pi_1(X, e)/p_*(\pi_1(X', e'))$ (III, p. 312, corollary 4 to proposition 2); this group is commutative.

b) Let us denote by $q: X'\times X'\rightarrow X$ the mapping $m\circ (p, p)$. The required mapping $m'$ is a continuous lifting of $q$ to $X'$ such that $m'(e', e') =e'$. The space $X'\times X'$ is locally arcwise connected; it is therefore enough, in order to prove the existence of such a continuous lifting, to verify that $q_*(\pi_1(X'\times X',(e', e')))$ is contained in $p_*(\pi_1(X', e'))$ (III, p. 308, prop. 1). By prop. 3, the mapping $\pi_1(m,(e, e))$ identifies with the law of composition of the group $\pi_1(X, e)$ when one identifies $\pi_1(X\times X,(e, e))$ with $\pi_1(X, e)\times \pi_1(X, e)$. The group $q_*(\pi_1(X'\times X',(e', e')))$ is therefore equal to the group $p_*(\pi_1(X', e'))$, whence the existence of $m'$. Since $X'\times X'$ is connected, the uniqueness of $m'$ follows from I, p. 34, corollary 1 of prop. 11.

Let us prove that the mapping $m'$ endows the pointed space $(X', e')$ with a structure of Hopf space. Let $m_1: X\rightarrow X$ be the mapping $g\mapsto m(g, e)$ and let $\sigma_1: X\times \mathbf{I}\rightarrow X$ be a homotopy pointed at $x$ joining $m_1$ to Id$_X$. Put $\tau =\sigma_1\circ (p$, Id$_{\mathbf{I}}): X'\times \mathbf{I}\rightarrow X$. The mapping $m'_1: X'\rightarrow X'$ defined by $h\mapsto m'(h, e')$ lifts the mapping $\tau (\cdot ,0)$; let $\tau ': X'\times \mathbf{I}\rightarrow X'$ be the lifting of $\tau$ which is a homotopy with origin $m'_1($III, p. 301, prop. 2). The mapping $t\mapsto \tau '(e', t)$ is a lifting to $X'$ of the constant mapping $t\mapsto e$; since $\tau '(e',0) =e'$, we therefore have $\tau '(e', t) =e'$ for every $t\in \mathbf{I}$. The mapping $\tau '(\cdot ,1)$ is then a lifting to $X'$ of the mapping $p$ which maps $e'$ to $e'$. Since $X'$ is connected, Id$_{X'}$ is the unique X-morphism of $X'$ into itself which fixes the point $e'$; we therefore have $\tau '(\cdot ,1) =$ Id$_{X'}$. This shows that $\tau '$ is a homotopy pointed at $e'$ which joins the mapping $m'_1$ to the mapping Id$_{X'}$. Analogously, there exists a homotopy pointed at $e'$ joining the mapping $m'_2:h\mapsto m'(e', h)$ to the mapping Id$_{X'}$. This proves that the pointed space $(X', e')$, endowed with the law of composition $m'$, is a Hopf space.

c) Since $X'$ is a connected covering of X, the orbital mapping of $e'$ induced by the operation of $\pi_1(X, e)$ on $X'_e$ is surjective and induces a bijection of $\pi_1(X, e)/p_*(\pi_1(X', e'))$ onto $X'_e($III, p. 305, Theorem 1).

Let $c$ and $d$ be loops in X at $e$, and let $\gamma$ and $\delta \in \pi_1(X, e)$ be their classes. Let $c'$ and $d'$ be the paths with origin $e'$ in $X'$ which lift $c$ and $d$. We have $e'\cdot \gamma =c'(1)$ and $e'\cdot \delta =d'(1)$ (III, p. 304). By Proposition 3 of IV, p. 374, the loop $m\circ (c, d)$ is strictly homotopic to the loop $c*d$; hence we have $e'\cdot (\gamma \delta ) =e'\cdot (m\circ (c, d))$. Now, the path $m'\circ (c', d')$ is a lift with origin $e'$ of the path $m\circ (c, d)$; hence we have

The mapping $\gamma \mapsto e'\cdot \gamma$ is therefore a group homomorphism of $\pi_1(X, e)$ into the set $X'_e$ endowed with the law of composition induced by $m'$. Consequently, $X'_e$ is a group for the law of composition induced by $m'$, and the orbital mapping of $e'$ is an isomorphism of the quotient group $\pi_1(X, e)/p_*(\pi_1(X', e'))$ onto $X'_e$.

The last part of assertion c) then results from Corollary 3 of III, p. 311.

a) If $m$ is an associative ( resp. commutative) law of composition, the same is true of $m'$.

b) If $e$ is a right-hand ( resp. left-hand) identity element for the law $m$, then $e'$ is a right-hand ( resp. left-hand) identity element for the law $m'$.

c) If X is a topological group, $m$ its law of composition and $e$ its identity element, the law of composition $m'$ endows $X'$ with a group structure compatible with the topology of $X'$, of which $e'$ is the identity element. The mapping $p: X'\rightarrow X$ is a group homomorphism whose kernel is discrete and contained in the center of $X'$.

a) Suppose that the law $m$ is associative. Then the mappings from $X'\times X'\times X'$ into X which map $(h_1, h_2, h_3)$ to $m(p(h_1), m(p(h_2), p(h_3)))$ and $m(m(p(h_1), p(h_2)), p(h_3))$ respectively are equal. The mappings $(h_1, h_2, h_3)\mapsto m'(h_1, m'(h_2, h_3))$ and $(h_1, h_2, h_3)\mapsto m'(m'(h_1, h_2), h_3)$ from $X'\times X'\times X'$ into $X'$ are continuous liftings of them which coincide at the point $(e', e', e')$. Since $X'\times X'\times X'$ is connected, they are equal (I, p. 34, Corollary 1 of Prop. 11), which shows that the law $m'$ is associative.

Suppose now that the law $m$ is commutative; the mappings $(h_1, h_2)\mapsto m'(h_1, h_2)$ and $(h_1, h_2)\mapsto m'(h_2, h_1)$ are continuous liftings to $X'$ of the mapping $(h_1, h_2)\mapsto m(p(h_1), p(h_2))$ which coincide at the point $(e', e')$. Since $X'\times X'$ is connected, they are equal (loc. cit.) and the law $m'$ is commutative.

If $e$ is a right (resp. left) identity element for the law $m$, the mapping $h\mapsto m'(h, e')$ (resp. $h\mapsto m'(e', h)$) from $X'$ into $X'$ is an X-morphism of coverings which coincides with Id$_{X'}$ at the point $e'$, hence at every point of $X'$, since $X'$ is connected (loc. cit.), whence b).

Let us finally prove c). Suppose that X is a topological group. From the foregoing, the law $m'$ is associative and $e'$ is an identity element for it. Let $i: X\rightarrow X$ denote the mapping $g\mapsto g^{-1}$. It is continuous (TG, III, p. 1) and the homomorphism $\pi_1(i, e):\pi_1(X, e)\rightarrow \pi_1(X, e)$ is none other than the mapping $\gamma \mapsto \gamma^{-1}($IV, p. 374, prop. 3). Consequently, the subgroup $(i\circ p)_*(\pi_1(X', e'))$ is equal to $p_*(\pi_1(X', e'))$. By prop. 1 of III, p. 308, there therefore exists a continuous mapping $i': X'\rightarrow X'$ such that $p\circ i'=i\circ p$ and $i'(e') =e'$. The mappings $h\mapsto m'(h, i'(h))$ and $h\mapsto$ $m'(i'(h), h)$ of $X'$ into $X'$ are liftings to $X'$ of the constant mapping with image $e$ of $X'$ into X. They are therefore constant and their image is $e'=m'(e', e')$. Thus every element $h$ of $X'$ is invertible, with inverse $i'(h)$, which shows that $X'$, endowed with the composition law $m'$, is a group. By construction of the law $m'$, the mapping $p: X'\rightarrow X$ is a group homomorphism. Since the mappings $m'$ and $i'$ are continuous, the group structure of $X'$ is compatible with its topology (TG, III, p. 1). The fibre $\overset{-1}{p}(e)$ is a discrete subgroup of $X'$ which is contained in the center of $X'($I, p. 100, cor. 3) and X is isomorphic to the quotient topological group $X'/\overset{-1}{p}(e)$.

a) The space $\widetilde{G}$ is simply connected and simply connected by arcs. Endowed with the mapping $p$, the pointed space $(\widetilde{G},\widetilde{e})$ is a universal covering of the pointed space $(G, e)$.

(b) The kernel N of $p$ is a discrete subgroup of $\widetilde{G}$, contained in the center of $\widetilde{G}$. The group homomorphism $N\rightarrow$ Aut$_G(\widetilde{G})$ which, to $n\in N$, associates the right translation in $\widetilde{G}$, is a group isomorphism. The homomorphism of $\pi_1(G)$ into N which, to $\gamma \in \pi_1(G)$, associates the unique element $n$ of N such that $\widetilde{e}\cdot \gamma =n$, is a group isomorphism.

(c) If $G'$ is a topological group, with identity element $e'$, and if $p': G'\rightarrow$ G is a group homomorphism which makes $G'$ into a covering of G, the unique continuous mapping $u:\widetilde{G}\rightarrow G'$ such that $u(\widetilde{e}) =e'$ and $p'\circ u=p$ is a group homomorphism. Endowed with the mapping $u,(\widetilde{G},\widetilde{e})$ is a universal covering of $(G', e')$.

By IV, p. 342, Theorem 1, there exists a covering $(\widetilde{G}, p)$ of G, simply connected and simply connected by arcs, Galoisian with group $\pi_1(G)^{\circ}$, and a point $\widetilde{e}$ of $\overset{-1}{p}(e)$ such that the pointed covering $(\widetilde{G},\widetilde{e})$ is a universal covering of $(G, e)$.

By IV, p. 375, Prop. 4 and IV, p. 377, Prop. 5, there exists on the space $\widetilde{G}$ a unique group structure compatible with its topology for which $p$ is a group homomorphism and $\widetilde{e}$ an identity element. It follows from Prop. 4 that the group $\widetilde{G}$ satisfies assertions (a) and (b).

Let us prove assertion (c). Let $G'$ be a topological group and let $p': G'\rightarrow G$ be a group homomorphism which makes $G'$ into a covering of G. Let $e'$ be the identity element of $G'$. The existence and uniqueness of a mapping $u:\widetilde{G}\rightarrow G'$ such that $p=p'\circ u$ and $u(\widetilde{e}) =e'$ follow from the fact that $(\widetilde{G},\widetilde{e})$ is a universal covering of the pointed space $(G, e)$. Since the mappings $p$ and $p'$ are surjective group homomorphisms, $u$ is a group homomorphism. Endowed with the mapping $u,\widetilde{G}$ is a covering of $G'($I, p. 81, Prop. 7), hence $(\widetilde{G},\widetilde{e})$ is a universal covering of $(G', e')$ (IV, p. 345, Cor. 2 of Theorem 1).

Under the hypotheses of the proposition, we shall say, by abuse, that $\widetilde{G}$ is a universal covering of G.