Timeline for How to determine the homotopy groups of the suspension of a space?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| S Feb 28, 2015 at 10:40 | history | suggested | Exterior | CC BY-SA 3.0 |
added TeX to answer
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| Feb 28, 2015 at 10:32 | review | Suggested edits | |||
| S Feb 28, 2015 at 10:40 | |||||
| Oct 29, 2009 at 12:27 | comment | added | Tyler Lawson | You are right, this "symmetric elements" functor preserves surjections but not coequalizers, but I think the functor I want is its 0'th left (nonabelian) derived functor. | |
| Oct 28, 2009 at 16:36 | comment | added | Charles Rezk | The functor on abelian groups A |-> (symmetric elements in A tensor A) is not right exact, and has a 0-th right derived functor Γ<sup>2</sup>. This functor is also the "2nd divided square" construction. Is Γ<sup>2</sup> G<sub>ab</sub> the closed form you want? | |
| Oct 28, 2009 at 12:36 | history | edited | Tyler Lawson | CC BY-SA 2.5 |
correction again, was screwed up between symmetric elements in the tensor square and bilinear pairings
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| Oct 24, 2009 at 1:39 | vote | accept | Skye | ||
| Oct 23, 2009 at 18:20 | history | edited | Tyler Lawson | CC BY-SA 2.5 |
corrected universal bilinear form vs set of bilinear pairings.
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| Oct 23, 2009 at 17:34 | history | edited | Tyler Lawson | CC BY-SA 2.5 |
state exact sequence precisely.
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| Oct 23, 2009 at 17:07 | history | answered | Tyler Lawson | CC BY-SA 2.5 |