The quotient space of a closed orientable surface by a $\Bbb Z_2$-action

Question

Let $M_g$ be a closed orientable surface, embedded in $\Bbb R^3$ such that the $y$-axis intersects it in $2g + 2$ points and $M_g$ is invariant under the 180$^\circ$ rotation around the $y$-axis. Thus, the 180$^\circ$ rotation around the $y$-axis defines a $\Bbb Z_2$-action on $M_g$ with $2g+2$ fixed points. I want to show that the quotient space $M_g/\Bbb Z_2$ is homeomorphic to the $2$-sphere $S^2$. But, since this action is not free, I cannot apply some well-known theorems about quotient manifolds. How can we check this?

Answer 1

Draw a convincing picture.

This image shows the involution you are describing. What is a fundamental domain for it? Well it looks like the top/bottom half of the image as an open subset, and then these have to be glued together along the intersection of the image with the horizontal slice coming into and out of your screen. This slice cuts the handles as well, so when the gluing is carried out, each is sewn shut into a disk. Since the top piece is a disk, and every handle in the surface has been sewn shut, this space is a topological sphere.