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.
How to prove every non-compact, connected 2 dimensional surface is homotopical to a bouquet of flowers?
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algebraic-topology
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1I think you mean "bouquet" of flowers. – 2012-07-27
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0Maybe 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. – 2012-07-27
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0No, 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. – 2012-07-28
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3Is 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") – 2012-07-28
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0I 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. – 2012-07-28
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0When I was in Moscow they used a bouquet of flowers to represent $\bigwedge S^{1}$. – 2012-07-28
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0This is equivalent to proving the fundamental group is free. – 2012-08-05