Suppose that $G$ is a regular feebly compact Moore quasitopological group. Must $G$ be a topological group? This was previously posted [here on MathSE](https://math.stackexchange.com/questions/2817752/a-question-on-quasitopological-group) also.


A semitopological group $G$ is a group $G$ with a topology such that the product map of $G\times G$ into $G$ is separately
continuous. If G is a semitopological group and the inverse map of $G$ onto itself associating $x^{-1}$ with
arbitrary $x\in G$ is continuous, then $G$ is called a quasitopological group.
If $G$ is a quasitopological group and the product map of $G\times G$ into $G$ is jointly continuous, then $G$ is called a topological group.


A space is feebly compact if every
family of locally finite non-empty open sets is finite.


Note that if $G$ is Tychonoff, the answer to this question is true,since $G$ will be Cech complete.



A question on quasitopological group
Yes. The paper [KKM] contains more general results. In particular, by Corollary 1 each regular semitopological group which is a cover semi-complete Baire space is a topological group. It remains to note that each $p-\sigma$-fragmentable space (see [Bou] for the definition) (in particlar, each $p$-space, so each Moore space) is cover semi-complete [KKM] and a (quasi)regular feebly compact space is Baire. 


*References*


[Bou] Ahmed Bouziad, *[Every Cech-analytic Baire semi-topological group is a topological group](https://doi.org/10.1090/S0002-9939-96-03384-9)*, Proc. Amer. Math. Soc. **124** (1996), 953-959.


[KKM] P. S. Kenderov, I. S. Kortezov, W. B. Moors *[Topological games and topological groups](https://doi.org/10.1016/S0166-8641(99)00152-2)*, Topology Appl. **109** (2001) P.157-165. (I have a preprint, to which I referred). 


