Timeline for Homotopy domination of a wedge of two polyhedra
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Mar 29, 2021 at 16:53 | comment | added | Ben Wieland | My previous comment doesn't work for that particular pair of groups. Also, it is too complicated, I guess because I didn't want to commit to a particular number of spheres or worry about a very simple space having a retract. Just take $X_1=S^2\vee S^2$ and $X_2=S^3/Q_8$. The wedge has a retract with nontrivial obstruction, but neither space does. $X_1$ doesn't because it is simply connected and $X_2$ doesn't because it is rationally irreducible. Or $X_2=S^3/(\mathbb Z/23)$ and maybe more spheres for $X_1$. | |
| Mar 25, 2021 at 17:07 | comment | added | Ben Wieland | Probably there is an example using the Wall finiteness obstruction. There are pairs of groups (probably including $(\mathbb Z/2,\mathbb Z/2)$) where the individual groups have no finiteness obstruction, but the free product does. Thus there exists a finitely dominated complex $A$ which is not finite whose fundamental group is a free product, but which is not a wedge of spaces with fundamental group the two factors, because those pieces $A_1,A_2$ would necessarily be finite, and thus $A_1\vee A_2$ would be finite. Something like $X_1=X_2=\mathbb RP^2\vee S^3$. | |
| Dec 8, 2017 at 4:50 | vote | accept | M.Ramana | ||
| Nov 27, 2017 at 5:54 | history | edited | M.Ramana | CC BY-SA 3.0 |
compacta has been changed to polyhedra
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| Nov 15, 2017 at 11:12 | history | edited | M.Ramana | CC BY-SA 3.0 |
added 60 characters in body; edited title
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| Nov 13, 2017 at 17:46 | answer | added | Nicholas Kuhn | timeline score: 6 | |
| Nov 13, 2017 at 16:13 | answer | added | Tyrone | timeline score: 5 | |
| Nov 12, 2017 at 12:27 | history | asked | M.Ramana | CC BY-SA 3.0 |