Proof. (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and

Let $(x_{1}, \cdots, x_{k}) \in E^{k}$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_{1}^{k} \subset S$. In which case,

by assumption. Thus $I^{-1}T(x_{1}, \cdots, x_{k}) \in B_{\sigma}(E; F)$.

In addition, for any $S_{1} \in \sigma$ and entourage $E(S_{2}, U)$ of $B_{\sigma}(E; F)$ where $S_{2} \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_{1} \cup S_{2}$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^{k}) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^{k}) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^{k}_{\sigma}(E; B_{\sigma}(E; F))$.

It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^{k}, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.

On the other hand, let $S_{1} \in \sigma$ and $E(S_{2}, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_{2} \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_{1} \cup S_{2}$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.

(2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then

Thus (2) holds for all $k \in \natp$.$\square$