Cutting the $2$-dimensional real Projective Space

Question

I have the following question. Let $M$ be a smooth manifold which is homeomorphic to $\mathbb{R}P^{2}$. If one cuts $M$ along a non-contractible path then $M$ should be homeomorphic to a closed disc, right? Why is this so?

Ben

Answer 1

Of course you can assume $M=\mathbb{R}P^2$; if you know $\pi_1(\mathbb{R}P^2)$ you get that up to homotopy your path is the upper semicircle of $D^2$ (with the identifications on the boundary) so that cutting along it gives as result the "elimination" of those identifications, leaving a simple closed disk.

Answer 2

I'll presume you mean a simple closed path. Lano's comment gives a correct answer but there are some hidden steps in the "up to homotopy" comment. One way to reveal those hidden steps is to apply the classification of surfaces. Let $S$ be the surface you obtain from $M$ by cutting along the path (your notation conflates $M$ and $S$). Since the path is not contractible, a little homology calculation shows that the path does not separate its complement in $M$, so $S$ is connected. Another homology calculation shows that the Euler characteristic is $\chi(S)=1$. By the classification of surfaces, $S$ is a closed disc.