Timeline for Why is the dual of a torus the same as its fundamental group?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
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| Sep 7, 2019 at 1:06 | comment | added | LSpice | In particular, the equality $\operatorname{Hom}(\mathbb Z^n, \mathbb Z) = \operatorname{Hom}(\mathbb Z, \mathbb Z^n)$ is a very noncanonical isomorphism. | |
| Feb 17, 2011 at 3:07 | comment | added | S. Carnahan♦ | Some of your equalities should be isomorphisms. | |
| Feb 16, 2011 at 20:50 | history | edited | Chris Gerig | CC BY-SA 2.5 |
gave more detail
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| Feb 16, 2011 at 20:42 | comment | added | Chris Gerig | The fact that $\pi_1$ is abelian is well-known and trivial to show by the product-formula... But yes there should be more to say on why the set of homotopy classes is equal to the set of continuous group homomorphisms... for this, a rather twisted way to go about it I will place in an edit. | |
| Feb 16, 2011 at 8:35 | comment | added | Dan Petersen | I don't see why the set of homotopy classes $[\mathbb{T}^n,\mathbb{R}/\mathbb{Z}]$ should be equal to the set of group homomorphisms $\mathbb{T}^n \to \mathbb{R}/\mathbb{Z}$. (Also, some argument is needed to show that $\pi_1$ is abelian, e.g. the Eckman-Hilton lemma.) | |
| Feb 16, 2011 at 6:35 | history | answered | Chris Gerig | CC BY-SA 2.5 |