Projective plane crosscap polygon

On the disk $D^2$ the antipodal pairing $u\mapsto -u$ pairs the two closed semicircles, so there is one edge class with two incident face-sides, and it pairs the two corners $(1,0)$ and $(-1,0)$, so there is one vertex class; the two corner sectors join under the two side-germ pairings in a single cyclic link. Hence the digon is a connected surface schema with $V=1$, $E=1$, $F=1$ whose realization $Y$ is a nonempty compact connected Hausdorff boundaryless surface; traversing the boundary circle once covers the single edge class twice in the same direction, which is the word $aa$.

The closed upper hemisphere $H=\{x\in S^2:x_3\ge 0\}$ is closed and bounded in ${\mathbb{R}}^3$, hence compact by Heine–Borel; let $P:=H\mathrm{/}\text{\hspace{0.17em}⁣}\sim$ be the quotient that identifies $x$ with $-x$ for $x$ in the equator, the standard model of $\mathbb{R}{P}^2$; $P$ is compact as the continuous image of $H$ under the quotient map.

The radial map $\phi :D^2\to H$, $\phi (u)=(u,\sqrt{1-\parallel u{\parallel }^2})$, is a homeomorphism: it is continuous because the coordinate functions and the square root are continuous, it is bijective with inverse the coordinate projection $(x_1,x_2,x_3)\mapsto (x_1,x_2)$, which is continuous, and $\parallel \phi (u)\parallel =1$ holds for every $u$ in $D^2$; on the boundary circle $\phi (-u)=-\phi (u)$, so $\phi$ conjugates the antipodal pairing of the disk boundary to the antipodal pairing of the equator and induces a continuous bijection $P\to Y$ by the characteristic property of the quotient.

The quotient $P$ is compact by step 1.2 and $Y$ is Hausdorff by step 1.1, so the continuous bijection $P\to Y$ of step 1.3 is a homeomorphism by [L4]; hence the word $aa$ realizes $\mathbb{R}{P}^2$, with $Y\cong P$.

The antipodal boundary map is a half-turn, so it preserves boundary direction. Take two small collars around paired noncorner boundary points. On each collar choose coordinates $(t,r)$, where $t$ increases in the induced boundary direction and $r\ge 0$ points inward; these give the same face orientation on both collars. The pairing identifies $(t,0)$ on the first with $(t,0)$ on the second. A chart across the seam therefore uses $(t,r)$ on the first collar and $(t,-r)$ on the second. Reflection of the second coordinate changes the sign of a plane local homology generator: the boundary of a small oriented disk is a generator in degree one, and the reflection reverses its cyclic orientation. Thus a single generator on the seam chart agrees with the face generator on one collar and with its negative on the other. Its restrictions are a consistent local section; it is their signs relative to the face orientation that differ.

The cell counts of step 1.1 give $\chi (Y)=V-E+F=1-1+1=1$ by [L5], and step 2.1 transfers this value to $\mathbb{R}{P}^2$.

Join points in the two collar interiors by a path in the open disk, and close it by crossing the seam once. Along the interior path the face orientation gives a constant generator section. Across the seam step 2.2 changes its sign relative to that section, so the resulting loop transports a generator to its negative. A global integral orientation would be preserved along every path by [L6], contradicting this loop since a generator of an infinite cyclic group is not its negative. Hence $Y\cong \mathbb{R}{P}^2$ is nonorientable. All constructions and pairings are finite; no choice axiom is used.