Deleting a non-contractible embedded loop from the real projective plane Suppose I have a loop $\gamma : [0,1] \rightarrow \mathbb{R}P^2$ in the real projective plane without self-intersections that is not contractible, that is, $[\gamma]$ is a generator of the fundamental group of $\mathbb{R}P^2$. Is it true that the complement $\mathbb{R}P^2 \setminus \text{Im}(\gamma)$ is homeomorphic to an open disk? This is easy to verify for simple examples, but I do not have an intuition about more complicated loops in $\mathbb{R}P^2$.
My naive idea would be to show that a small tubular neighbourhood of $\text{Im}(\gamma)$ is homeomorphic to a Möbius band, and that deleting a Möbius band from $\mathbb{R}P^2$ leaves an open disk. But I cannot prove either of these statements, and even if I could, I am not sure how to go from deleting a tubular neighbourhood to deleting the loop. I wouldn't be surprised if there were some neat homology-based argument for this, but alas I am not comfortable enough with homology to come up with one.
 A: For convenience I will prove that the result of deleting a small open neighbourhood of $\gamma$ is a closed disk. It is not hard to see that this is equivelant.
This can be shown by combining 3 ingredients, Mayer-Vietoris (MV), and classification of surfaces (COS) (the most general vesrion, possibly with boundary, punctures not neccesarily orientable), and the classification of 1-dimensional disk bundles over $S^1$ (CO1DB) (there is two, the trivial bundle and the Moebius band). All homology groups are taken with integer coeffecients.
1.
The first step is two show that $\gamma$ is non-separating. Suppose not for a contradiction. Then we have a decomposition of $\mathbb{RP}^2$ into two connected components $A,B$ along $\gamma$. Applying MV we have a short exact sequence
$$ 0 \rightarrow \mathbb{Z} \rightarrow H_{1}(A) \oplus H_{1}(B) \rightarrow \mathbb{Z}_2 \rightarrow 0 .$$
This implies that $b_{1}(A) + b_{1}(B) =1$. WLOG lets say $b_{1}(B)=0$. By COS, the only connected surface with one $S^1$-boundary component having $b_{1}=0$ is a disk, so $B$ is a disk. But that implies that $[\gamma]$ is trivial in $\pi_{1}(\mathbb{RP}^2)$-a contradiction.
2.
Concluding that the complement is a disk. Since, $\mathbb{RP}^2 \setminus \gamma$ has one connected component, deleting a small open neighbourhood $N(\gamma)$ of it gives a connected surface with one boundary component, homeomorphic to $S^1$. I call this boundary curve $\tau$, and set $U = \mathbb{RP}^2 \setminus N(\gamma)$. Furthermore, by CO1DB $N(\gamma)$ is homeomorphic to a Moebius band (because the boundary of $N(\gamma)$ is connected the trivial bundle cannot appear). In particular, $b_{1}(N(\gamma))=1$. So applying $MV$ for the decomposition along $\tau$ gives that $b_{1}(U)=0$. By COS, $U$ is a closed disk.
*The use of COS throughout may be reduced the following Lemma.
Lemma (consequence of COS). Let $V$ be a connected surface, possibly non-orientable, with one $S^1$ boundary component and without punctures (equivelantly compact as a topological space). Suppose that $b_{1}(V)=0$, then $V$ is a closed disk.
