Notes:Potentially invalid vertex scheme

Instance

[ilmath]\xymatrix{ a \ar@{-}[r] \ar@{-}[d] & b \ar@{-}[r] \ar@{-}[d] & c \ar@{-}[r] \ar@{-}[d] & a \ar@{-}[d] \\ d \ar@{-}[ur] \ar@{-}[r] \ar@{-}[d] & e \ar@{-}[r] \ar@{-}[d] \ar@{-}[ur] & f \ar@{-}[r] \ar@{-}[d] \ar@{-}[ur] & d \ar@{-}[d] \\ a \ar@{-}[r] \ar@{-}[ur] & b \ar@{-}[r] \ar@{-}[ur] & c \ar@{-}[r] \ar@{-}[ur] & a }[/ilmath]
The labelling scheme in question

This is done in the context of Munkres - Elements of Algebraic Topology - page 16:

• What is the geometric realisation of the complex shown to the right?
  • Surely [ilmath]\cong\mathbb{S}^1\times I[/ilmath] where [ilmath]I:\eq[0,1]\subset\mathbb{R} [/ilmath]?

  intuitively, if one folds along defd to join abca to itself, we get a rectangle. We then have to bend it round to join ada to ada, thus making a tube.

  However, we do have to identify be to eb and ditto for cf... I'm not sure what these do

  • • But in the case described the ada-to-ada join is not ambiguous, dealing with that first we see that it is very much so. Do we twist or not?

Caveat:Is this even a "valid" vertex labelling scheme?

• After playing around with a leather bookmark, if you do the ada-to-ada join with a twist in it, thus yielding a mobius band AND then try and fold it... you cannot.

Problems:

1. The left and right edges, ada are ambiguous, do we twist or not (mobius band vs tube)
2. be-to-be join
3. cf-to-cf join

"Solution"

• This should perform "more identifications than one would think" and not be a torus.