Timeline for Removing a simplicial subset from a simplicial set
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2013 at 20:25 | answer | added | Anonymous | timeline score: 0 | |
| Apr 15, 2013 at 15:42 | comment | added | John Wiltshire-Gordon | @Chris Schommer-Pries: Good point, but I'd like a combinatorial answer. In particular, if there are finitely many simplices before, I'd like finitely many after. | |
| Apr 15, 2013 at 14:03 | comment | added | Chris Schommer-Pries | The singular simplicial set of the space $|X| \setminus |A|$ has the property you ask for, namely its geometric realization is homotopy equivalent to $|X| \setminus |A|$. Perhaps you want some additional constraints? | |
| Apr 15, 2013 at 11:00 | answer | added | Tim Porter | timeline score: 1 | |
| Apr 15, 2013 at 1:34 | history | edited | John Wiltshire-Gordon | CC BY-SA 3.0 |
Clearer explanation
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| Apr 15, 2013 at 1:31 | comment | added | John Wiltshire-Gordon | Dear Professor May, I meant geometrically realizing A and X first and then performing subtraction in the category of spaces. | |
| Apr 15, 2013 at 1:27 | comment | added | Peter May | John, what precisely do you mean by X\A and taking its `geometric realization'. It seems you mean the simplices of X not in A, which as you observe is not a simplicial set, so its geometric realization has no obvious meaning. You are desperately trying to avoid X/A. In your equivariant example, presumably G is a group acting simplicially and A is the subcomplex of simplices with nontrivial stabilizer. Then you might replace X\A by the maximal G-free subcomplex of X, which maybe makes sense. | |
| Apr 14, 2013 at 23:07 | history | asked | John Wiltshire-Gordon | CC BY-SA 3.0 |