Theorem 1

The restriction of the exponential mapping of $\hat{A}(X)$ to $\hat{L}(X)$ is a bijection of $\hat{L}(X)$ onto a closed subgroup of the Magnus group $\Gamma(X)$.

We write $A(X) = A, A^n(X) = A^n, \hat{A}(X) = \hat{A}, L^n(X) = L^n, \hat{L}(X) = \hat{L}, \Gamma(X) = \Gamma$. Let B be the algebra $A \otimes A$ with the graduation of type $\mathbf{N}$ defined by $B^n = \sum_{i+j=n} A^i \otimes A^j$. Let $\hat{B} = \prod_{n \geq 0} B^n$ be the associated complete filtered algebra (Commutative Algebra, Chapter III, § 2, no. 12, Example 1). The coproduct $c : A \to A \otimes A$ defined in § 3, no. 1, Corollary 1 to Theorem 1 is graded of degree 0 and hence extends by continuity to a homomorphism $\hat{c} : \hat{A} \to \hat{B}$ given by

We also define continuous homomorphisms $\delta'$ and $\delta''$ of $\hat{A}$ into $\hat{B}$ by

By Corollary 2 to Theorem 1 of § 3, no. 1, $L^n$ is the set of $a_n \in A^n$ such that $c(a_n) = a_n \otimes 1 + 1 \otimes a_n$. It follows that $\hat{L}$ is the set of $a \in \hat{A}$ such that

Let $\Delta$ be the set of $b \in \hat{A}$ of constant term equal to 1 and satisfying the relation

in other words, the set of $b = \sum_{n \geq 0} b_n$ such that $b_n \in \mathbf{A}^n$ for all $n \geq 0$, $b_0 = 1$ and $c(b_n) = \sum_{i+j=n} b_i \otimes b_j$ for $n \geq 0$. The latter characterization shows that $\Delta$ is a closed subset of $\Gamma$; as $\hat{\epsilon}$, $\delta'$ and $\delta''$ are ring homomorphisms and every element of $\delta'(\hat{\mathbf{A}})$ commutes with every element of $\delta''(\hat{\mathbf{A}})$, the restrictions to $\Gamma$ of the mappings $c$ and $\delta'\delta''$ are group homomorphisms and $\Delta$ is a subgroup of $\Gamma$.

By Proposition 1 of no. 1, the exponential mapping of $\hat{\mathbf{A}}$ is a bijection of the set $\hat{\mathbf{A}}^+$ of elements of $\hat{\mathbf{A}}$ with no constant term onto $\Gamma$. Let $a \in \hat{\mathbf{A}}^+$ and $b = \exp a$. As $\hat{\epsilon}$ is a continuous ring homomorphism,

The relations

are proved similarly and, as $\delta'(a)$ commutes with $\delta''(a)$ (no. 1, Remark 1),

Therefore $a$ satisfies (3) if and only if $b$ satisfies (4), which proves the theorem.