Remark 1

— We are going to show that the mapping $f \mapsto f^b$ is a strict morphism (GT, III, §2, No. 8) of $\mathscr{K}(X)$ onto $\mathscr{K}(X/H)$.

a) The mapping is continuous: it suffices to prove that, for every compact subset $K$ of $X$, the restriction of $f \mapsto f^b$ to $\mathscr{K}(X, K)$ is a continuous mapping of $\mathscr{K}(X, K)$ into $\mathscr{K}(X/H, \pi(K))$ (TVS, II, §4, No. 4, Prop. 5); since $H$ operates properly in $X$, the set $P$ of $\xi \in H$ such that $K\xi$ intersects $K$ is compact; one concludes from (3) that $\sup_{x \in K} |f^b(\pi(x))| \leq \beta(P) \sup_{x \in K} |f(x)|$, and this proves our assertion.

b) Let $K'$ be a compact subset of $X/H$. Let us choose a compact subset $K$ of $X$ such that $\pi(K) = K'$, and let us show that the restriction of $f \mapsto f^b$ to $\mathscr{K}(X, K)$ is a strict morphism of $\mathscr{K}(X, K)$ onto $\mathscr{K}(X/H, K')$. It suffices to construct a right inverse for this restriction (GT, III, §6, No. 2, Prop. 3). Now, by Lemma 1 of No. 1 (whose notations we adopt), one obtains such an inverse by composing the following mappings: $\alpha)$ the mapping $f' \mapsto f' \circ \pi$ of $\mathscr{K}(X/H, K')$ into the set $E$ of functions of $\mathscr{K}^1(X)$ whose support is contained in $KH$; $\beta)$ the mapping of $E$ into $E$ that, to every $g \in E$, makes correspond the function equal to $g/u^1$ on $KH$, and to 0 on $X - KH$; $\gamma)$ the mapping of $E$ into $\mathscr{K}(X)$ that, to every function $h \in E$, makes correspond $uh$.

c) This established, if $V$ is a convex neighborhood of 0 in $\mathscr{K}(X)$, then $V \cap \mathscr{K}(X, K)$ is a convex neighborhood of 0 in $\mathscr{K}(X, K)$, therefore $V^b \cap \mathscr{K}(X/H, K')$ is a convex neighborhood of 0 in $\mathscr{K}(X/H, K')$ by b), therefore $V^b$ is a neighborhood of 0 in $\mathscr{K}(X/H)$ (TVS, II, §4, No. 4). This completes the proof.