Timeline for Fundamental groups of topological groups.
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2013 at 17:44 | comment | added | Todd Trimble | @OmarAntolín-Camarena Right (thinking now of groups as one-object categories), and that seems a little cleaner also because it avoids the side argument that modding $S^\infty$ out by the antipodal map is the same as modding out by a subgroup $\mathbb{Z}/2$. Thanks! | |
| Sep 11, 2013 at 17:29 | comment | added | Omar Antolín-Camarena | Can't you omit all mention of $S^\infty$ by saying that $\mathbb{Z}/2$, being Abelian, is a group object (even an internal $\mathbb{Z}/2$-vector space) in the category of groups and then apply the classifying space functor? (Very cool example, by the way.) | |
| Sep 11, 2013 at 16:46 | history | edited | Todd Trimble | CC BY-SA 3.0 |
neatened some formatting; cleared up a technical point on topology
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| Oct 22, 2010 at 1:38 | comment | added | Todd Trimble | identification of a point with its antipode = negation is the same as identifying a point $x$ with $1+x$ in the topological Boolean ring, which is the same as modding out by the subspace $\{0, 1\}$. This is one way of making precise an argument behind one of the claims in my answer. | |
| Oct 22, 2010 at 1:34 | comment | added | Todd Trimble | Whuh, you don't believe me? :-) The answer anyhow is yes; it's another way of describing the standard Milgram construction of the classifying bundle, viz., the realization of the nerve applied to the groupoid map $K\mathbb{Z}_2 \to B\mathbb{Z}_2$ (that '$B$' being our nLab notation for the 1-object category attached to the group), and that's homeomorphic to the projection $S^\infty \to \mathbb{RP}^\infty$. Incidentally, the same Lawvere theory line of argument I'm using here shows that $S^\infty$ is a topological Boolean algebra for which negation is the antipodal map, and the (cont.) | |
| Oct 21, 2010 at 23:54 | comment | added | David Roberts♦ | I know what you're trying to say, but is S^\infty actually given by applying the functor you give to Z_2?? I would prefer to say it is some model of EZ_2... But otherwise this is nice argument to show that BZ_2 can be chosen as a Z_2-module plus some topology. | |
| Oct 21, 2010 at 13:55 | comment | added | Todd Trimble | Well, if you like, but the way I said it is fine. The '1' is traditional notation for the group identity. It becomes the 0 element of a vector space structure if we assume the axiom $x^2 = 1$. | |
| Oct 21, 2010 at 13:42 | comment | added | Mariano Suárez-Álvarez | You mean $x^2=0$ as an axiom for vector spaces in categories. | |
| Oct 21, 2010 at 13:39 | vote | accept | Chris | ||
| Oct 21, 2010 at 12:46 | history | answered | Todd Trimble | CC BY-SA 2.5 |