$S-\{p\}$ admits a bouquet of circles as deformation retract.

Question

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.

Thanks

Answer 1

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).