Möbius line Thom space as a projective-plane quotient

The disk bundle $D(L)=[0,1]\times [-1,1]\mathrm{/}((0,t)\sim (1,-t))$ is a Möbius band and its sphere bundle $S(L)$ is its single boundary circle. Realize ${\mathbb{R}\mathbb{P}}^2$ as the disk with antipodal boundary points identified. Removing from it a smaller concentric open disk leaves the closed annulus with its outer circle antipodally identified, which is again a Möbius band: the outer identification reverses the inward transverse direction on one traversal, which is the defining twist. Since the removed disk is complementary to that Möbius band, ${\mathbb{R}\mathbb{P}}^2$ is obtained from $D(L)$ by attaching a closed disk along $\mathrm{\partial }D(L)$, and collapsing that attached disk turns the pushout into $D(L)\mathrm{/}\mathrm{\partial }D(L)$, which is $\mathrm{Th}(L)$ by [F1].

On a disk pair $D_1\subset D_2$ of concentric Euclidean disks, the radial map that sends $D_1$ to the center of $D_2$ and rescales the annulus $D_2\setminus D_1$ onto $D_2$ minus its center, while fixing the complement of $D_2$, is continuous, injective off $D_1$, and onto; it therefore descends to a continuous bijection $D_2\mathrm{/}{D}_1\to D_2$ from the compact quotient to the Hausdorff disk, hence a homeomorphism. Applying this inside the projective plane to a disk containing the attached disk exhibits $\mathrm{Th}(L)\cong {\mathbb{R}\mathbb{P}}^2\mathrm{/}{D}^2\cong {\mathbb{R}\mathbb{P}}^2$ with the basepoint corresponding to the centre of the complementary disk. The trivial line over $S^1$ instead has Thom space $\mathrm{\Sigma }(S_{+}^1)$ by [F2]: its two boundary circles are collapsed to the same basepoint, whereas the unreduced suspension $\mathrm{\Sigma }{S}^1\cong S^2$ keeps its two suspension points distinct.