Let $M$ be an abelian monoid. For sake of simplicity we shall assume that in $M$ the cancellation law holds true. With this last assumption we define the group completion $G$ of $M$ as $$G:=M\times M/\sim$$ where $(a,b)\sim (a',b')$ if and only if $a+b'=a'+b$. It has been quite surprizing find out, reading the paper [Completions and Fibrations for Topological Monoids](https://www.jstor.org/stable/2154549?seq=1#page_scan_tab_contents) - Paulo Lima-Filho, that in general, if $M$ is a topological abelian monoid then its group completion $G$, endowed with the quotient topology, is not a topological group. Can anyone help me to provide an example of this fact?



Group completion of topological monoids
A suitable counterexample can be constructed as follows. 


Let $(e\_n)\_{n\in\omega}$ be the standard orthonormal basis of the Hilbert space $\ell\_2$. For every $n\in\mathbb N$ consider the linear hull $L\_n$ of the vectors $e\_1,\dots,e\_{n}$ in $\ell\_2$. On the union $L^\infty:=\bigcup\_{n=1}^\infty L\_n$ consider the strongest topology that induces the Euclidean topology on each space $L\_n$. It is well-known (and easy to see) that the space $L^\infty$ is a non-metrizable $k\_\omega$-space (i.e., the direct limit of a sequence of compact Hausdorff spaces). Observe that the vector $e\_0$ is orthogonal to the subset $L^\infty$ in $\ell\_2$.


In the Hilbert space $\ell\_2$ consider the submonoid $S:=\bigcup\_{n=0}^\infty(n e\_0+L\_n)$. Now consider the product $S\times S$ and the map $$q\_S:S\times S\to S-S=\mathbb Z+\bigcup\_{n=1}^\infty L\_n,\;\;q\_S:(x,t)\mapsto x-y.$$ It is easy to see that the quotient topology on $S-S$ is not metrizable -- it coincides with the topology of the direct limit of the sequence $(\mathbb Z+L\_n)\_{n=1}^\infty$. 


Finally consider the topological monoid $M:=\ell\_2\times S$. We claim that $M$ is a required counterexample. Indeed, consider the subgroup $M-M\subset \ell\_2\times\ell\_2$ and the map $q:M\times M\to M-M$, $q:(x,y)\mapsto x-y$.


Since the space $M\times M$ is second-countable, the set $M-M$ endowed with the quotient topology is a sequential $\aleph\_0$-space (see Theorem 11.3 on p.494 [here](https://www.sciencedirect.com/science/article/pii/B9780444865809500136)) and hence $M-M$ has countable $cs^\*$-character. Assuming that $M-M$ is a topological group, we can apply Theorem 1 from [this paper](https://arxiv.org/pdf/0911.4075.pdf) of Banakh and Zdomskyy and conclude that $M-M$ is either metrizable or contains an open $k\_\omega$-subgroup. But none of these two conditions applies to $M-M$: this space contains a topological copy of the $k\_\omega$-space $L^\infty$ and hence is not metrizable and contains a copy of the Hilbert space $\ell\_2$, so cannot contain an open (and hence closed) $k\_\omega$-subgroup.


