Image of a complete topological group under open and surjective map is complete?

Question

A uniform space $X$ is complete if every Cauchy filter in $X$ is convergent. Here we do not require $X$ to be Hausdorff.

Question.

Let $G$ be a complete topological group and let $H$ be a topological group. Suppose that $q:G\twoheadrightarrow H$ is an open and surjective homomorphism of topological groups.

Is $H$ complete topological group?

Motivation.

Let $X$ be a topological vector space and let $q:X\twoheadrightarrow X/Y$ be the quotient with respect to some linear subspace $Y \subseteq X$. Then $q$ is open and surjective. Actually every open surjective linear map with domain $X$ is of this form. It is standard result in functional analysis that if $X$ is Banach, then $X/Y$ is a seminormed and complete space (if $Y$ is closed, then $X/Y$ is a Banach space).

Answer 1

The answer is no. This is already false for topological vector spaces, since every (Hausdorff) topological vector space arises as the quotient of a complete one, see Jochen Wengenroth's answer to this question; see also the original article by Susanne Dierolf.