Let $G$ be a left topological group, i.e. a topological space with group operation such that left multiplication $L\_g : x \mapsto gx$ is continuous (but right multiplication and inversion are not required to be). Assume also $G$ to be a topological manifold. Does this imply that $G$ is a topological group?


EDIT 1: as pointed out by Yves Cornulier, the aswer is no if $G$ is not required to be connected. Hence, I would like to know if the statement still holds when $G$ is assumed to be connected.


EDIT 2: as discussed in the comments, we are also assuming $G$ to be paracompact, with countable atlas say.



Is a left topological group which is a manifold a topological group?
Here is a counter-example with $G$ homeomorphic to $\mathbb R^2$. Let $f:\mathbb R\to\mathbb R$ be a discontinous additive homomorphism (constructed using a Hamel basis of $\mathbb R$ over $\mathbb Q$). Define a group operation $\*$ on $\mathbb R^2$ by
$$
 (x,y)\*(x',y') = (x+x'e^{f(y)},y+y') .
$$
This groups is a semidirect product of $\mathbb R$ and $\mathbb R$ with respect to the action where $y\in\mathbb R$ acts on $\mathbb R$ by multiplication to $e^{f(y)}$. The formula is continuous in $(x',y')$ but not in $(x,y)$.


