Timeline for Can the loops in the definition of the fundamental group be considered injective?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 11, 2021 at 0:27 | comment | added | Arshak Aivazian | Sorry, I did not understand you and thought that you mean that you already know the answer to the question that I would like to ask. Thank you! | |
| Nov 11, 2021 at 0:23 | comment | added | Jeremy Brazas | No, i only meant what I said. I did not mean to say this stronger statement. | |
| Nov 11, 2021 at 0:13 | comment | added | Arshak Aivazian | Do you mean that there is no subspace $\mathbb{R}^3$ (1) homotopically equivalent to the Griffiths twin-con and (2) whose fundamental group is realized on injective loops? | |
| Nov 10, 2021 at 22:34 | comment | added | Jeremy Brazas | You are correct that it is a difficult open problem to decide if the fundamental group of every subset of $\mathbb{R}^3$ is torsion-free. It is still the case that the Griffiths twin-cone is an example of a Peano continuum in $\mathbb{R}^3$ where injective paths are unhelpful for understanding $\pi_1$. If you reformulate your question to something just about $\mathbb{R}^3$ it is probably best to just ask a new question. | |
| Nov 10, 2021 at 22:30 | vote | accept | Arshak Aivazian | ||
| Nov 10, 2021 at 22:30 | comment | added | Arshak Aivazian | Thank you, now I see that as $\mathrm{C}$ should have been limited to only subspaces $\mathbb{R}^3$. I am very embarrassed for such a protracted bringing the question to the correct form, so I will accept this answer for now. But with the approval of the senior site contributors, I would like to edit the question again and leave the appropriate "disclaimer". | |
| Nov 10, 2021 at 21:58 | history | answered | Jeremy Brazas | CC BY-SA 4.0 |