Let $M$ to be a compact connected surface that is not homeomorphic to the sphere $\mathbb S^2$. Then, I want to show that there exists some $x \in M$ such that $M \setminus \{x \}$ is homotopy equivalent to a bouquet of circles (wedge sum of finite $\mathbb S^1$'s copies).
Some ideas:
I know that a compact connected surface must be homeomorphic to either a connected sum of copies of $\mathbb P^2$ or $\mathbb T^2$ given it is not homeomorphic to $\mathbb S^2$.
I think $\mathbb T^2$ minus one point is homotopy equivalent to a bouquet of two circles: $\mathbb S^1 \lor \mathbb S^1$.
I also saw somewhere that $\mathbb P^2$ minus one point is homeomorphic to $\mathbb S^1$.
Since connected sum is commutative and $\mathbb T^2 \# \mathbb P^2$ can be written as $\mathbb P^2 \# \mathbb P^2 \# \mathbb P^2$, $M$ is either homeomorphic to $\#_{i = 1}^n \mathbb P^2$ or $\mathbb T^2$.
For the second case, I can just claim that $M$ minus any point is homeomorphic to a bouquet of two circles, so it is homotopy equivalent to a bouquet of two circles, if I am not wrong.
However, for the first case, I am not sure how to proceed.
Best Answer
$M$ may be constructed by taking a polygon and identifying edges in such a way that all vertices become identified with each other.
If you remove a vertex from the center of the polygon, then you can retract onto the perimeter of the polygon, which is a bouquet of circles.
Let $X$ denote the unit disk in $\mathbb{R}^2$ with the origin removed. Let $C$ denote the unit circle. Working in polar co-ordinates, we give an explicit homotopy from the identity on $X$ to a map $X\to C$ which restricts to the identity on $C$: $$h_t(r,\theta)=((1-t)r+t,\theta).$$
Now we obtain $M\backslash \{x\}$ from $X$ by partitioning $C$ into arcs and identifying pairs of arcs in some way. The homotopy $h_t$ remains the same. Now $C$ modulo the identifications will be a bouquet of circles.