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May 8, 2014 at 14:56 comment added Lev Borisov Yes, you are right. Thank you for the correction!
May 8, 2014 at 14:53 comment added FreeQuark Thank you for your answer! I am not versed in homological algebra, but I think I understood the latter argument. I think that in your list of special edges, the last ones should be $[(1,1,0),(1,1,1)]$ and $[(1,1,1),(1,1,0)]$? Otherwise one the faces of $[0,1]^3$ does not contain any of the edges above. Also, do you think this proof generalizes to higher dimensions (I am particularly interested in $\mathbb{Z}^4$). It seems to depend only on the choice of a list of segments like the ones you wrote.
May 4, 2014 at 21:21 history answered Lev Borisov CC BY-SA 3.0