Is a quotient of a complete group always complete?

Question

Let $\: \langle G,\cdot,\mathcal{T}\hspace{0.01 in} \rangle \:$ be a $\big($$\text{T}_0$$\big)$ topological group. $\;\;$ Let $H$ be a closed normal subgroup of $G$.

Set $\;\; \mathbf{G} \: = \: \langle G,\cdot,\mathcal{T}\hspace{0.01 in} \rangle \;\;$, $\;\;$ and assume that $\mathbf{G}$ is complete with respect to the two-sided uniformity.

Does it follow that $\: \mathbf{G}/H \:$ is complete? $\;\;$ (with respect to its two-sided uniformity)

If no, what if one also assumes that $\mathbf{G}$ is abelian?

Answer 1

No.

In the positive direction, Bourbaki proves in Topologie Générale, Chapitre IX, §9, Proposition 4 the following:

Let $G$ be a metrizable topological group. If $H$ is a closed normal subgroup then $G/H$ is metrizable. If $G$ is complete (in one of the one-sided uniformities) then so is $G/H$.

This entails a positive answer for metrizable abelian groups, as the left, right and two-sided uniformities coincide.

On the other hand, in Espaces Vectoriels Topologiques, Chapitre IV, §4, Exercices 9 et 10, they give among many other things a (somewhat involved) construction of a complete topological vector space whose quotient by a closed subspace is not complete. The linear structure is of no importance for the topological/uniform considerations, so this applies to topological abelian groups as well.

Worse, still: Susanne Dierolf proved in Über Quotienten vollständiger topologischer Vektorräume, Manuscripta Mathematica 17, Nr 1 (1975), 73–77 that every topological vector space arises as a quotient of a complete and Hausdorff topological vector space.