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Let $S$ be a closed compact surface, $p\in S$ and $X=S-\{p\}$. Show that X admits a bouquet of circles as deformation retract. How many circles?

I'm starting to study algebraic topology and I can't even begin to solve this question I need some hints to start to solve it.


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Do you know that any closed compact surface come from gluing sides of a polygon? – user27126 Nov 6 '12 at 3:45
@Sanchez do you mean every surface is triangularizable? – user42912 Nov 6 '12 at 3:47
@user42912 Sanchez means something more like – A Walker Nov 6 '12 at 4:05
Dear user, This question is closely related, and the answers and pictures there may help. Regards, – Matt E Nov 6 '12 at 4:56
up vote 2 down vote accepted

I'll answer this for the case of the torus; hopefully the proof of the general case will be made clear by analogy.

We may view the torus as the quotient of a square, identified opposite edges with the appropriate gluing. Thus the punctured torus can be thought of as a punctured square, which deformation retracts onto its boundary.

For the torus, though, this boundary can be expressed in terms of the two loops that give generate the fundamental group of $\pi_1(T)$. Thus $T$ deformation retracts to the wedge of two circles. This fact can be verified purely geometrically as well.

In general, this method shows how the circles appear (as generators of $\pi_1$ of the surface), and should give you a handle on their number.

If you want another data point, consider the punctured sphere, which obviously deformation retracts to a point (i.e. the wedge of 0 circles).

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thank you, it helps me a lot. I'm using the classification theorem to prove the question. It miss only how to prove that for example a torus of g-genus can have as a deformation retract a wedge sum of 2g circles. I know how to do this in specific surfaces, but I don't know how to generalize. Do you have any idea? thank you again – user42912 Nov 14 '12 at 23:42
Once you know that the punctured $4g$-gon deformation retracts to the boundary, think for a bit about what the boundary is. Each side corresponds to a loop starting and ending at our fixed basepoint. So navigating the boundary corresponds to navigating loop $a$, then $b$, then $a$ (backwards), etc. We can view this path abstractly, and it consists of $2g$ distinct loops, each of which is a circle and each of which intersect at a unique common point (our basepoint). Note: this way we have not relied upon the embedding of $M_g$ into $\mathbb{R}^3$ in any way. – A Walker Nov 15 '12 at 7:09

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