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Timeline for Failure of Theorem of the Cube?

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Dec 6, 2010 at 23:31 vote accept jlk
Dec 6, 2010 at 20:45 answer added Laurent Moret-Bailly timeline score: 7
Dec 6, 2010 at 15:53 answer added t3suji timeline score: 5
Dec 6, 2010 at 6:59 history edited jlk CC BY-SA 2.5
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Dec 6, 2010 at 3:41 comment added BCnrd Not every component of $X$ necessarily has a $k$-rational point (the component group could be non-constant). Can you work out something reasonable when $X$ is a finite etale commutative group? If not, there seems little point in considering the more general case.
Dec 6, 2010 at 1:30 history edited jlk CC BY-SA 2.5
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Dec 6, 2010 at 1:18 comment added jlk @BCnrd: Thank you for the comment. @David Roberts: Thanks. Fixed.
Dec 6, 2010 at 1:15 history edited jlk CC BY-SA 2.5
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Dec 6, 2010 at 1:14 comment added BCnrd Dear jlk: It is more natural to tack on an additional tensor factor of $0^{\ast}(L)^{-1}$ where $0:G^3 \rightarrow G$ is the constant map to the origin. When working over a field this has no serious effect, but in the relative setup (which makes various Yoneda-type arguments more effective) it is clearly necessary.
Dec 6, 2010 at 0:51 comment added Eric Peterson The $m_\underline{i}$ really do have the same codomain. For example, in coordinates, $m_3(x,y,z)=z=p_3(x,y,z)$ and $m_{2,3}(x,y,z)=y \cdot z$, just as described by 'summing the coordinates whose indices are in $\underline{i}$'.
Dec 6, 2010 at 0:04 comment added David Roberts You've changed notation there a bit. Is $G=X$? And I'm not sure all your pullbacks end up over the same space - $m_i$ and $m_{ij}$ $i,j=1,2,3$ don't have the same codomain.
Dec 5, 2010 at 23:34 history asked jlk CC BY-SA 2.5