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Is there a group $G$ which is at the same time a (compact-Hausdorff)-ly generated weakly Hausdorff space (or short CGWH space) such that inverse and product are continuous maps and the space is not Hausdorff?

Note that the group multiplication is a map from $G \times G$ to $G$. In the sense of this question, $G \times G$ denotes the product in the category of CGWH spaces. It is not necessarily equipped with the usual product topology.

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  • $\begingroup$ Take the free CGWH group $FX$ on any non-Hausdorff CGWH space $X$. $\endgroup$
    – Tyrone
    Commented Mar 23, 2023 at 12:36
  • $\begingroup$ @Tyrone Thank you for the quick answer. Can I find the definition of a free CGWH group somewhere? My search through the internet or to come up with the correct definition by myself were not successful. $\endgroup$ Commented Mar 23, 2023 at 14:46
  • $\begingroup$ You should find the details in Lamartin's article. I can try to find a more specific reference tomorrow if you need. $\endgroup$
    – Tyrone
    Commented Mar 23, 2023 at 15:18
  • $\begingroup$ Thanks, this article is already quite detailed. $\endgroup$ Commented Mar 24, 2023 at 10:49

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To give this question the correct color, I give this proper answer. The mathematical part of this answer is purely based on Tyrone's comments.

Yes, there is a CGWH group which is not Hausdorff. This is shown in Lamartin's article in Example 2.14.

The idea is the following: The forgetful functor from CGWH groups to CGWH spaces has a left adjoint $F$. Let $X$ be any non-Hausdorff CGWH space. By definition, $FX$ is a CGWH group. There is a natural continuous map $X\to FX$. This can be seen by explicitely constructing $FX$ or by taking the adjoint to the continuous group Homomorphism $\operatorname{id}: FX \to FX$. One can show that this map is a closed embedding. Therefore, $FX$ is not Hausdorff.

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