It is enough to show that $f(tx) = tf(x)$ for all $x \in \mathbf{R}^n$ and all $t \in \mathbf{R}$. The argument is the same as that of Proposition 5 of Chapter V, § i, no. 3 if we replace $x$ by $x$ and $\mathbf{R}$ by $\mathbf{R}^m$.

By the same reasoning as in Proposition 6 of Chapter V, § i, no. 4 we show that the homomorphism extending $f$, if it exists, is unique. Next, the subgroup $G_1$ of $G$, generated by $f(A)$, is commutative; for if $x$ and $y$ are any two points of $A$, then $\frac{1}{2} x, \frac{1}{2} y$ and $\frac{1}{2} (x + y)$ belong to $A$, therefore

which shows that $f \left( \frac{1}{2} x \right)$ and $f \left( \frac{1}{2} y \right)$ commute; therefore so do $f(x) = (f \left( \frac{1}{2} x \right))^2$ and $f(y) = (f \left( \frac{1}{2} y \right))^2$, which proves the assertion.

Let $a_1, a_2, \ldots, a_n$ be $n$ non-zero vectors contained in $A$ and proportional to the basis vectors of the parallelootope, and for each index $i$ let $D_i$ be the line through $o$ and $a_i$, that is, the set of points $ta_i$ as $t$ runs through $\mathbf{R}$. Let $A_i$ be the set of all $t \in \mathbf{R}$ such that $ta_i \in A$; then $A_i$ is an interval of $\mathbf{R}$ containing $[0, 1]$, and the function

is defined and continuous on $A_i$ and satisfies the relation

whenever $t, t'$ and $t + t'$ all belong to $A_i$. By Proposition 6 of Chapter V, § 1, no. 4, there exists a continuous homomorphism $\bar{f}_i$ of $\mathbf{R}$ into $G$ which extends $f_i$. Since $\mathbf{R}^n$ is the direct sum of the subgroups $D_i$, we can define a homomorphism $\bar{f}$ of $\mathbf{R}^n$ into the abelian group $G_1$ by the rule $\bar{f}(x) = \prod_{i=1}^n \bar{f}_i(t_i)$ where $x = \sum_{i=1}^n t_i a_i$; $\bar{f}$ is an extension of $f$ since, if $x \in A$, all the components $x_i$ of $x$ on the lines $D_i$ also belong to $A$, by reason of the choice of the $a_i$; moreover, $\bar{f}$ is continuous on $\mathbf{R}^n$, because it is continuous on each of the lines $D_i$, and $x_i$ is a linear function of $x$ (and therefore continuous).

for each pair of points $x, y$ such that $x \in V, y \in V$ and $x + y \in V$. Then there exists a unique continuous homomorphism of $\mathbf{R}^n$ into $G$ which agrees with $f$ at all points of a neighbourhood $W$ of $o$.

Take $W$ to be an open box with centre $o$, contained in $V$, and apply Proposition 2 to $W$.

We shall see later that this property of $\mathbf{R}^n$ extends to a larger class of topological groups, the "simply connected" groups.

Let $\overline{f}$ be the continuous homomorphism of $\mathbf{R}^n$ into $G$ which agrees with $f$ at all points of a neighbourhood of $o$; $\overline{f}(\mathbf{R}^n)$ contains, by hypothesis, a neighbourhood of the identity element of $G$ and therefore (Chapter III, § 2, no. 1, Corollary to Proposition 4) is an open subgroup of $G$; moreover $\overline{f}$ is a strict morphism of $\mathbf{R}^n$ onto $\overline{f}(\mathbf{R}^n)$, by Chapter III, § 2, no. 8, Proposition 24.

A suitably chosen local isomorphism $f$ of $\mathbf{R}^n$ with $G$ extends to a strict morphism of $\mathbf{R}^n$ onto an open subgroup of $G$ (Proposition 2, Corollary 2) and therefore onto $G$ itself, since $G$ is connected. It follows that $G$ is isomorphic to a quotient group $\mathbf{R}^n/H$ of $\mathbf{R}^n$: $H$ is discrete, otherwise there would exist points $x \neq o$ of $H$ arbitrarily close to $o$ and such that $f(x) = f(o)$, contrary to the hypothesis that $f$ is a local isomorphism. The theorem is therefore a consequence of Theorem 1 of § 1, no. 1.

Let $f$ be a continuous homomorphism of $\mathbf{R}^m$ into $\mathbf{T}^n$. We shall show that there is a linear mapping $u$ of $\mathbf{R}^m$ into $\mathbf{R}^n$ such that the homomorphisms $x \to f(x)$ and $x \to \varphi(u(x))$ coincide at all points of a neighbourhood of $o$ in $\mathbf{R}^m$; the proposition will then follow from Corollary 1 to Proposition 2 of no. 2. Now let $V$ be a neighbourhood of $o$ in $\mathbf{R}^n$ such that $\varphi$, restricted to $V$, is a local isomorphism of $\mathbf{R}^n$ with $\mathbf{T}^n$, and let $\psi$ be the inverse of $\varphi$, defined on $\varphi(V)$. Since $f$ is continuous, $V' = \overline{f}^{-1}(\varphi(V))$ is a neighbourhood of $o$ in $\mathbf{R}^m$; the mapping

restricted to $V'$, is a continuous mapping of $V'$ into $\mathbf{R}^n$ such that $\psi(f(x+y)) = \psi(f(x)) + \psi(f(y))$ for each pair of points $x, y$ in $\mathbf{R}^m$ such that $x \in V', y \in V'$ and $x + y \in V'$; hence (no. 2, Corollary 1 to Proposition 2) this mapping coincides with a well-determined continuous homomorphism $u$ of $\mathbf{R}^m$ into $\mathbf{R}^n$ at all points of a neighbourhood $W$ of o in $\mathbf{R}^m$. By Proposition 1 (no. 1) $u$ is a linear mapping of $\mathbf{R}^m$ into $\mathbf{R}^n$; for all $x \in W$ we have therefore $f(x) = \varphi(u(x))$, which completes the proof.

Equivalently (§ 1, no. 1), if $u(e_i) = \sum_{j=1}^n a_{ij} e_j$, the $a_{ij}$ must be integers such that $\det(a_{ij}) = \pm 1$. In particular, for $n = 1$: