Sphere as a polygonal quotient

The pairing is the parameter reversal $s_1(t)\sim s_2(1-t)$, so the corner $v_0=s_1(0)$ is paired with $s_2(1)=v_0$ and the corner $v_1=s_1(1)$ with $s_2(0)=v_1$: each corner forms its own vertex class, the two sides form one edge class, and the disk interior is the one face. Hence the realization $Y$ has $V=2$, $E=1$, $F=1$, and these are the cells of its finite CW structure.

First replace the bigon by a round-disk representative. A disk homeomorphism takes its two corners to two points of the unit circle; a circle homeomorphism carrying these to antipodal points extends radially, $ru\mapsto r\phi (u)$, with radial inverse. Composing gives a disk homeomorphism taking the two corners to opposite ends of a diameter. Conjugate the side pairing by this homeomorphism; it induces a quotient homeomorphism by [L2], using the inverse to obtain the inverse quotient map. We may now cut the round representative along that diameter $\gamma$, whose interior lies in the disk interior; the pieces are two bigons, $B_1$ with sides $s_1$ and one copy of $\gamma$, and $B_2$ with sides $s_2$ and the other copy ${\gamma }^{\mathrm{\prime }}$. By the split identity of [L3] the quotient $Y$ is homeomorphic to the quotient of $B_1\bigsqcup B_2$ by the pair $s_1\sim s_2$ together with the pair $\gamma \sim {\gamma }^{\mathrm{\prime }}$; since $\mathrm{\partial }{B}_1=s_1\cup \gamma$ and $\mathrm{\partial }{B}_2=s_2\cup {\gamma }^{\mathrm{\prime }}$ and the two pairings match at the common endpoints $v_0,v_1$, they combine into a single homeomorphism $h:\mathrm{\partial }{B}_1\to \mathrm{\partial }{B}_2$ that identifies the whole boundary circle of $B_1$ with the whole boundary circle of $B_2$.

Fix homeomorphisms ${\alpha }_1:B_1\to {\bar{B}}_2(0,1)$ and $u_2:B_2\to {\bar{B}}_2(0,1)$ onto the closed unit disk, which exist because bigons are closed disks, and let $\phi ={\alpha }_1\circ h^{-1}\circ u_2^{-1}$ on the unit circle; extending $\phi$ radially by $\kappa (tu)=t\phi (u)$ and putting ${\alpha }_2=\kappa \circ u_2$ gives a homeomorphism ${\alpha }_2:B_2\to {\bar{B}}_2(0,1)$ with ${\alpha }_2\circ h={\alpha }_1$ on $\mathrm{\partial }{B}_1$. Here ${\kappa }^{-1}(tu)=t{\phi }^{-1}(u)$ and ${\alpha }_2^{-1}=u_2^{-1}\circ {\kappa }^{-1}$. Define $\mathrm{\Phi }$ on the class of $x\in B_1$ as $({\alpha }_1(x),\sqrt{1-\parallel {\alpha }_1(x){\parallel }_2^2})$ and on the class of $x\in B_2$ as $({\alpha }_2(x),-\sqrt{1-\parallel {\alpha }_2(x){\parallel }_2^2})$: the two formulas agree on the glued boundary circle because ${\alpha }_2\circ h={\alpha }_1$ there, so $\mathrm{\Phi }$ is a well-defined continuous map out of the quotient by [L2]. It is surjective onto $S_{+}^2\cup S_{-}^2=S^2$ and injective because the two formulas are homeomorphisms onto the closed upper and lower hemispheres, which meet exactly in the equator. The quotient is compact as a continuous image of the compact disk $D$ [L2], while $S^2$ is Hausdorff [L4], so $\mathrm{\Phi }$ is a homeomorphism by the compact-to-Hausdorff criterion [L4], and $Y\cong S^2$.

The single pair has opposite exponents, so by [L1] it is orientation compatible; by step 2.1 the realization $Y$ is a boundaryless surface, and the generator carried by the oriented face continues unchanged across the single edge class, its locally constant generator classes over face and vertex charts supplying a continuous generating section as in [L6]. Hence $Y$, and by step 2.1 the sphere $S^2$, is orientable.

The cell counts of step 1.1 give $\chi (Y)=V-E+F=2-1+1=2$ by [L5], and this is the Euler characteristic of $S^2$ by step 2.1. Every construction used is finite and explicit: one diameter splitting a round representative into two bigons, one radial extension of a circle homeomorphism, and two hemisphere formulas, so no choice axiom is used.