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Covering map from $p\colon X \to S_1 \vee S_1$.

X=

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I know infinite graphs with four edges incident at each vertex can be 2-oriented, but I don't think this would help me.

I also need to find out $\pi_1(X)$ afterwards. So I can't use this in the above.

Bob
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  • If you take $X$, and identify all the corners as one point, don't you get the wedge of $10$ circles? It should be possible to map five of them to each of the two circles in the figure-eight $S^1\vee S^1$ without any problem, unless I'm missing something. – Arthur Nov 21 '16 at 15:10
  • Because there are ten edges in $X$ that after the identification start at and end at the same point. – Arthur Nov 21 '16 at 15:14

1 Answers1

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Label the edges in $X$ with either an $a$ if they are an 'outer' edge or a $b$ if they are an 'inner' edge. Map each edge to the corresponding edge of $S^1_a\vee S^1_b$ in the obvious way (orientation doesn't matter). This is clearly continuous and surjective.

  • The preimage of a small neighbourhood of the wedge point is five disjoint copies of a space that looks like the letter 'X'.

  • The preimage of any other small open neighbourhood looks like five disjoint copies of an open interval.

Hence the map satisfies the local homeomorphism property and is a covering space.

Dan Rust
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  • Identifying the edges would change the homotopy type. Are you aware of the result that if you quotient a connected graph out by a maximal sub-tree then it gives a wedge of circles? Such a quotient can be formed as a homotopy equivalence. So, find a maximal sub-tree and work out what is left after quotienting out by it. – Dan Rust Nov 21 '16 at 16:00
  • Look at a small ball around any of the vertices - it looks like four half open intervals glued together at their closed ends, i.e. a letter 'X'. I'm not sure if I can picture what you mean by 'the lower edge, the 2 diagonals which are above everything else, and the 2 right hand edges', but you should find that a maximal sub-tree has four edges (there are many maximal sub-trees). You seem to have given something with five edges, which would not be a tree. – Dan Rust Nov 21 '16 at 16:06
  • You mean the edges $[1,3]$, $[3,5]$ and $[2,4]$? Now that's only three edges, and that isn't a tree (as it's not connected). How about just $[1,2],[1,3],[1,4],[1,5]$? – Dan Rust Nov 21 '16 at 16:13
  • Homeomorphisms let you wiggle things around.... – Dan Rust Nov 21 '16 at 16:19
  • Sorry I'm a bit hard pressed for time at the moment and that question needs a bit more attention given to it than what I've given here. Good luck. – Dan Rust Nov 21 '16 at 16:22