Proposition 1

Let $(u_\lambda)_{\lambda \in L}$ and $(v_\mu)_{\mu \in M}$ be two summable families of elements of $\mathbf{A}[[\mathbf{I}]]$. Then the family $(u_\lambda v_\mu)_{(\lambda, \mu) \in L \times M}$ is summable and we have

Let $(\alpha_{\lambda, \nu})_{\nu \in \mathbf{N}^{(I)}}$ (resp. $(\beta_{\mu, \nu})_{\nu \in \mathbf{N}^{(I)}}$) be the family of coefficients of $u_\lambda$ (resp. $v_\nu$). For each $\nu \in \mathbf{N}^{(I)}$ there exists only a finite number of pairs $(\nu_1, \nu_2) \in \mathbf{N}^{(I)} \times \mathbf{N}^{(I)}$ such that $\nu_1 + \nu_2 = \nu$, hence only a finite number of pairs $(\lambda, \mu) \in L \times M$ such that the coefficient of $X^\nu$ in $u_\lambda v_\mu$ is $\neq 0$. Hence the family $(u_\lambda v_\mu)_{(\lambda, \mu) \in L \times \mathscr{J}}$ is summable. Now the formula (2) follows from the associativity of the sum (Gen. Top., III, p. 265, formula (2)).

In $\mathbf{A}[[\mathbf{I}]]$ the product is an associative and commutative composition law. We may therefore speak of a multipliable family of elements of $\mathbf{A}[[\mathbf{I}]]$ and of the product of a multipliable family (Gen. Top., III, p. 262, remark 3).