Theorem 1

Let X, Y be two filtered groups whose filtrations $(X_n), (Y_n)$ consist of normal subgroups; let $u : X \to Y$ be a homomorphism compatible with the filtrations.

(i) Suppose that the filtration $(X_n)$ is exhaustive. For $\mathrm{gr}(u)$ to be injective, it is necessary and sufficient that $\bar{u}^{-1}(Y_n) = X_n$ for all $n \in \mathbf{Z}$.

(ii) Suppose that one of the following hypotheses holds: (α) X is complete and Y Hausdorff; (β) Y is discrete. Then, for $\mathrm{gr}(u)$ to be surjective, it is necessary and sufficient that $Y_n = u(X_n)$ for all $n \in \mathbf{Z}$.

(i) To say that the mapping $\mathrm{gr}_n(u)$ is injective means that

This is obviously the case if $(Y_{n+1}) = X_{n+1}$. Conversely, if for all $n$, we deduce by induction on $k$ that $X_{n-k} \cap \bar{u}^{-1}(Y_{n+1}) \subset X_{n+1}$ for all $n \in \mathbf{Z}$ and all $k \geq 0$. As the filtration $(X_n)$ is exhaustive, we see that, for all $n$, $X_n$ is the union of the $X_{n-k}$ ($k \geq 0$), hence $\bar{u}^{-1}(Y_{n+1}) \subset X_{n+1}$ for all $n$ and therefore $X_{n+1} \subset \bar{u}^{-1}(Y_{n+1})$, which completes the proof.

(ii) To say that the mapping $\mathrm{gr}_n(u)$ is surjective means that

This is obviously the case if $Y_n = u(X_n)Y_{n+1}$ for all $n \in \mathbf{Z}$. Let $n$ be an integer and $y$ an element of $Y_n$; we shall define a sequence $(x_k)_{k \geq 0}$ of elements of $X_n$ such that $x_k \in X_n$, $x_{k+1} \equiv x_k \pmod{X_{n+k}}$ and $u(x_k) \equiv y \pmod{Y_{n+k}}$ for all $k \geq 0$. We shall take $x_0$ equal to the identity element of X, which certainly gives $u(x_0) \equiv y \pmod{Y_n}$. Suppose that an $x_k \in X_n$ has been constructed such that $u(x_k) \equiv y \pmod{Y_{n+k}}$; then $(u(x_k))^{-1}y \in Y_{n+k}$; the hypothesis implies that there exists $t \in X_{n+k}$ such that $u(t) \equiv (u(x_k))^{-1}y \pmod{Y_{n+k+1}}$ and hence $u(x_k t) \equiv y \pmod{Y_{n+k+1}}$; it is sufficient to take $x_{k+1} = x_k t$ to carry out the induction. This being so, if Y is discrete, there exists $k \geq 0$ such that $Y_{n+k} = \{e'\}$ (the identity element of Y), whence $u(x_k) = y$ and hence in this case it has been proved that $u(X_n) = Y$, for all $n$. Suppose now that X is complete and Y Hausdorff. As $x_h^{-1} x_k \in X_{n+k}$ for $h \geq k \geq 0$, $(x_k)$ is a Cauchy sequence in $X_n$; as

X_n is closed in X and hence complete, this sequence has at least one limit x in X_n. By virtue of the continuity of u, u(x) is the unique limit of the sequence (u(x_k)) in Y, Y being Hausdorff. But the relations u(x_k) \equiv y (\text{mod. } Y_{n+k+1}) show that y is also a limit of this sequence, whence $u(x) = y$ and it has also been proved that $u(X_n) = Y_n$.