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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.

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  • $\begingroup$ 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... $\endgroup$ – Henno Brandsma Apr 16 at 12:41
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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.

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  • $\begingroup$ 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. $\endgroup$ – blabla Apr 18 at 16:10
  • $\begingroup$ @blabla no the discrete topology on $\{0,1\}$ which is a topological group. $\endgroup$ – Henno Brandsma Apr 18 at 16:12
  • $\begingroup$ oh ok then. Thanks. $\endgroup$ – blabla Apr 18 at 16:16

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