Proposition 9

Let $G$ be a commutative group, and let $\mathfrak{T}_1, \mathfrak{T}_2$ be two Hausdorff topologies compatible with the group structure of $G$. Suppose that $\mathfrak{T}_1$ is finer than $\mathfrak{T}_2$ and that there is a fundamental system of neighbourhoods of $o$ for $\mathfrak{T}_1$ which are closed for $\mathfrak{T}_2$. Let $G_1, G_2$ be the completions of $G$ with respect to the topologies $\mathfrak{T}_1, \mathfrak{T}_2$ respectively, and let $f : G_1 \to G_2$ be the continuous homomorphism which extends the identity mapping of $G$ (no. 3, Proposition 5). Then $f$ is injective.

Suppose that $G$ is written additively. Let $\mathfrak{U}_1$ be the uniformity on $G$ corresponding (no. 1) to the topology $\mathfrak{T}_1$: it will suffice to show that if $\mathfrak{F}$ and $\mathfrak{F}'$ are two minimal Cauchy filters (Chapter II, § 3, no. 2) with respect to $\mathfrak{U}_1$, which converge in $G_2$ to the same point $a$, then $\mathfrak{F} = \mathfrak{F}'$ (Chapter II, § 3, no. 7). For this, it is enough to show that $\mathfrak{F} \cap \mathfrak{F}'$ is a Cauchy filter with respect to $\mathfrak{U}_1$. Let $V$ be a neighbourhood of $o$ in $G$ with respect to $\mathfrak{T}_1$, such that $V$ is closed in $\mathfrak{T}_2$, and let $W$ be a symmetric neighbourhood of $o$ in $G$ with respect to $\mathfrak{T}_1$, such that $W + W \subset V$. By hypothesis, there is a $W_d$-small set $M$ (resp. $M'$) in $\mathfrak{F}$ (resp. $\mathfrak{F}'$); if $x \in M$ and $y \in M$ we have $y - x \in W$, i.e. $y \in x + W$. If $\overline{W}$ and $\overline{V}$ are the closures of $W$ and $V$ in $G_2$, it follows that $y \in x + \overline{W}$, and therefore, since $a$ is in the closure of $M$, that $a \in x + \overline{W}$ for each $x \in M$. Similarly, $a \in x' + \overline{W}$ for each $x' \in M'$, and hence $x - x' \in \overline{W} + \overline{W}$; but since $(x, y) \to x + y$ is a continuous mapping of $G_2 \times G_2$ into $G_2$, we have $\overline{W} + \overline{W} \subset \overline{W} + \overline{W} \subset \overline{V}$. It follows that if $x \in M$ and $x' \in M'$, then $x - x' \in \overline{V} \cap G = V$, because $V$ is closed in $\mathfrak{T}_2$; and this completes the proof.