# A topological group in which either multiplication or inversion doesn't pullback open sets to $F_\sigma$ sets.

In a Hausdorff topological group, both multiplication and inversion maps are continuous, so inverse image of any open set is always open under multiplication as well as inversion. Thus to find a topological group in which either multiplication or inversion doesn't always pullback open sets to $$F_\sigma$$ sets, we must only consider non-metrizable groups as in metric spaces every open set is $$F_\sigma$$.

So if we consider any group with indiscreet topology, say $$H$$, and take box product of uncountable copies of it, i.e. $$G=H^I$$ (box topology), we will get a non-metric topological group. In this group $$G$$, if we look at inversion mapping, the pullback of element of $$G$$, where every element of "uncountable tuple" (not sure how to write it precisely) is $$H$$, must not be $$F_\sigma$$. I am not sure how convincing this example is, or if there are other better examples out there. Thanks.

• A box product of indiscrete space is indiscrete again. Non-metrisable, because it's not $T_1$ e.g., but the only non-empty open set is also closed... – Henno Brandsma Apr 16 at 12:41

Take any $$G$$ that is a topological group that is not perfectly normal: there is some open set $$O$$ that is not an $$F_\sigma$$. Then $$i^{-1}[O]$$ is also not an $$F_\sigma$$ as the inversion $$i$$ is a homeomorphism from $$G$$ to itself.
So an example is trivial, in a way. E.g. take the group $$\{0,1\}^I$$ (in the usual product topology) for an uncountable index set $$I$$, with coordinatewise addition mod $$2$$ as a group operation, as a standard example of such a group.
• We have used Sierpinski topology on $X=\{0,1\}$, i.e. $\tau_X=\{\phi , \{0\}, \{0,1\}\}$ correct? But is $X$ a topological group under addition mod $2$ as operation, as inverse image of $\{0\}$ is $\{(0,0),(1,1)\}$ under addition mod 2, which is not open in $X \times X$, also $X$ is not Hausdorff. Are you implying that $G=X^I$ is a topological group without $X$ being one? Sorry if it's trivial but I can't wrap my head around it. – blabla Apr 18 at 16:10
• @blabla no the discrete topology on $\{0,1\}$ which is a topological group. – Henno Brandsma Apr 18 at 16:12