Proposition 5

Let us keep the notation and the hypotheses of Proposition 4.

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 $X'/\overset{-1}{p}(e)$.