This is one of my old unsolved questions when I reading Novikov's book on homology theory. I do not know how to prove it because standard triangulation, fundamental diagram, etc does not help and it should be easy to prove.

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    $\begingroup$ I think you mean "bouquet" of flowers. $\endgroup$ – Andrew Jul 27 '12 at 15:31
  • $\begingroup$ Maybe this helps. Remove a point from a torus. Then stretch open the surface until you have 2 bands attached at a piece of surface. This retracts even more onto the wedge of 2 circles. $\endgroup$ – i. m. soloveichik Jul 27 '12 at 23:46
  • $\begingroup$ No, this would not work because we only know it is open. Taking the closure of it does not admit a process of going back to the original surface. $\endgroup$ – Bombyx mori Jul 28 '12 at 5:31
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    $\begingroup$ Is the number of boundary components finite? If yes, then n-point compactification is a closed surface, which means you had a closed surface with n points removed. If the number of boundary components is infinite, I'm not even sure the claim is true... By the way, what are flowers? (I expected "circles") $\endgroup$ – user31373 Jul 28 '12 at 12:53
  • $\begingroup$ I think it is allowed to be infinite, yes, I am not sure it is true too. But Novikov is a fields medalist so this statement must not be wrong for trivial reasons. $\endgroup$ – Bombyx mori Jul 28 '12 at 14:35

This question just floated up again, so let me put some references here.

This MO question gives several different proofs that for your surface $S$, $\pi_1(S)$ is free. In fact, Lee Mosher's answer gives a direct proof that $S$ is homotopy equivalent to a graph, and hence to a bouquet of circles. You can also proceed by noting the universal cover of $S$ is contractible, and hence $S$ is homotopy equivlanent to any $K(\pi_1(S),1)$, of which the appropriate bouquet of circles is one.


I would agree with Leonid. The claim should definitely be true for open surfaces which are built from closed ones by removing a top handle (i.e. a disc), (this should follow from a handle decomposition and the classification of closed surfaces). For example (as also stated in wiki if you search for surfaces) you could take a cantor set in the sphere, and take its complement. I don't expect this to have a good homotopy type.

  • $\begingroup$ I double checked, I think it should hold in your case, since it is still connected. $\endgroup$ – Bombyx mori Jul 31 '12 at 11:20

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