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Jan 5, 2010 at 3:12 comment added Mariano Suárez-Álvarez You can do this construction for the category generated by the graph (when you say that the boundary "leaves exactly one point [out]" you really want to say "it replaces a pair of adjacent arrows by their composition")
Jan 5, 2010 at 3:00 comment added Ilya Nikokoshev The nerve construction should work though in any case, it's a standard one.
Jan 5, 2010 at 2:59 comment added Ilya Nikokoshev It takes the chain 0-1-2 to an alternating sum of 1-2, 0-2 and 0-1. I'm not entirely sure about the construction though. Also, you need to allow points to map not only to the vertices, but also to the edges, I think.
Jan 5, 2010 at 2:51 comment added Qiaochu Yuan I don't see how the boundary operator as you defined it takes n-chains to (n-1)-chains. Could you be more explicit?
Jan 5, 2010 at 2:24 history edited Ilya Nikokoshev CC BY-SA 2.5
fixes refs
Jan 5, 2010 at 2:18 history edited Ilya Nikokoshev CC BY-SA 2.5
expanded
Jan 5, 2010 at 2:04 history edited Ilya Nikokoshev CC BY-SA 2.5
correction
Jan 5, 2010 at 1:58 history answered Ilya Nikokoshev CC BY-SA 2.5