# How to prove every non-compact, connected 2 dimensional surface is homotopical to a bouquet of flowers?

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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I think you mean "bouquet" of flowers. – Andrew Jul 27 '12 at 15:31
yeah............. – Bombyx mori Jul 27 '12 at 17:35
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. – i. m. soloveichik Jul 27 '12 at 23:46
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. – Bombyx mori Jul 28 '12 at 5:31
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") – user31373 Jul 28 '12 at 12:53

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.