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Dualize a noncollinear finite point set in the real projective plane to projective lines. Their intersection vertices and consecutive line segments form an embedded graph. A vertex incident to $k$ lines has graph degree $2k$ and corresponds to a primal line containing $k$ points. Ordinary lines correspond to degree-four vertices. The faces are polygons with at least three sides.

The quotient map $q:S^2\to X$ is continuous and surjective by the quotient topology, so $X$ is compact as a continuous image of a compact space. To prove the Hausdorff property explicitly, let $[u]\ne [v]$ and set $d=\min (\Vert u-v\Vert ,\Vert u+v\Vert )>0$. Choose $r<d/3$. The saturated open setsare disjoint by the triangle inequality. Each contains both antipodes of every one of its points, so $q^{-1}(qU)=U$ and similarly for $V$. Consequently $qU,qV$ are disjoint open quotient neighborhoods separating the classes. Thus the quotient is compact and Hausdorff. It is the real projective plane; equivalently the metric on the antipodal sphere quotient is $d_X([u],[v])=\min (\Vert u-v\Vert ,\Vert u+v\Vert )$ and induces this topology.

An ordinary line contains exactly two points of the configuration. We give a cubic covering from few ordinary lines by using point-line duality, a defect count and cubic interpolation. Cubics may be reducible: a line with equation $L=0$ is contained in the cubic $L^3=0$, and a conic can be multiplied by a linear factor.

Pass to the real projective plane, which preserves collinearities and ordinary-line counts. If all points are collinear, one cubic already suffices. Otherwise dualize each point $p=[a:b:c]$ to the line $p^{\ast }:aX+bY+cZ=0$. The dual lines form an embedded graph $\Gamma$: its vertices are their intersections and its edges are the consecutive segments on each projective line. A vertex where $k$ dual lines meet has graph degree $2k$ and corresponds to a primal line containing $k$ points.

Let $t_k$ count vertices incident to exactly $k$ dual lines, and let $f_j$ count faces with $j$ sides. In particular $t_2<5n$. If $v,e,f$ are the graph's numbers of vertices, edges and faces, thenThe Euler characteristic of the projective plane is one, so $v-e+f=1$. Substituting the counts gives the Euler defect identity for a projective line arrangementThus both sums measuring departure from a degree-six triangular grid are $O(n)$.

Call an edge a good dual-arrangement edge if both endpoints have degree six and both adjacent faces are triangles; call it bad otherwise. Count all edge incidences at the defective vertices and faces. Since $k\le 4(k-3)$ for $k\ge 4$ and $j\le 4(j-3)$ for $j\ge 4$,This controls actual bad edges, not merely the number of defective vertices.

Strengthen the local condition: an edge is a safe dual-arrangement edge if every edge on every path of length at most two from either endpoint is good. The bounded-radius propagation of edge defects says that only $O(n)$ edges fail this test. Indeed, trace from an unsafe edge to the first bad edge. Every intervening edge is good, so the intermediate vertices have degree six. Reversing such paths offers only a bounded number of choices, at most a fixed constant times the bad-edge count. Every edge belongs to exactly one dual line, so the pigeonhole principle gives a dual line $p^{\ast }$ containing only $O(1)$ unsafe edges.

The algebraic input is cubic propagation along a triangular strip: a consecutive strip of safe edges on $p^{\ast }$ has all of its crossing dual lines represented by primal points on a single cubic. Here is the mechanism, rather than an appeal to the desired covering theorem. The triangles on either side extend two cells outward into three indexed line families. After dualizing back, their points $a_i,b_j,c_k$ satisfy collinearities whenever $i+j+k=0$. Adjacent three-by-three blocks are the eight-point cubic completion for two triples of lines configuration. A nonzero homogeneous cubic can be fitted to nine starting points because its coefficient space has dimension ten; each successive block has eight points already on that cubic and forces its remaining point onto it. Continuing along the strip keeps every crossing point on the same cubic. One concrete seed is $a_{-1},a_0,a_1,a_2,b_{-3},b_{-2},b_{-1},c_1,c_2$. The first overlapping block forces $c_3$, the next forces $b_{-4}$, and the next forces $c_4$; another block then forces $a_{-2}$. Alternating these completion steps extends the $b$ and $c$ families for the full length of the strip.

For clarity, the completion fact has a short algebraic proof. Let $F=L_1{L}_2{L}_3$ and $G=M_1{M}_2{M}_3$, with the two line triples meeting in nine distinct points, and let a cubic $H$ vanish at all but $L_3\cap M_3$. On $L_1$, both $H$ and $G$ have the same three zeros, so $H-\lambda G$ is divisible by $L_1$ for a suitable scalar $\lambda$. Its quadratic quotient vanishes at the three intersections on $L_2$, hence is divisible by $L_2$. The remaining linear quotient vanishes at the two known intersections on $L_3$, hence is a multiple of $L_3$. Thuswhich also vanishes at the ninth point. This is the line-triple case of the Cayley-Bacharach theorem. In the strip construction, degree-six vertices and the two-cell neighbourhood ensure that the local nine points are distinct; the initial nine-point interpolation and repeated completion provide the required cubic.

It remains to account for every point, including defects and degeneracies. If some dual line has at most two distinct intersection vertices, all other dual lines pass through one of them. Dualizing back covers $P$ by at most two primal lines, hence by cubics, and we are done. Otherwise every dual line has at least three vertices, and the local strip construction applies.

On the chosen $p^{\ast }$, cut the cyclic edge sequence at its $b=O(1)$ unsafe edges, adding one arbitrary cut if necessary. There are at most $b+1$ safe runs, each covered by a cubic via the strip argument. Intersection vertices incident to cut or unsafe edges number at most $2b+2$. At each such exceptional vertex $w$, all crossing dual lines represent primal points on the single primal line $w^{\ast }$, which is itself contained in a cubic. Include one further cubic through $p$ if it has not already been covered. Every other primal point is accounted for by where its dual line intersects $p^{\ast }$. ThereforeThe absolute bound on $b$ comes from $t_2<5n$ and does not depend on the number of points. This completes the requested proof sketch.

The exterior derivative is an $\mathbb{R}$-linear map $d:{\Omega }^p(M)\to {\Omega }^{p+1}(M)$, agrees with the differential $df$ of a smooth function, satisfies the graded Leibniz rule $d(\alpha \wedge \beta )=d\alpha \wedge \beta +(-1{)}^p\alpha \wedge d\beta$ for $\alpha \in {\Omega }^p(M)$, and has $d^2=0$. These properties determine it locally and hence globally. To see locality directly, if a differential form $\alpha$ vanishes near $x$, choose a smooth bump function $\chi$ equal to one near $x$ and supported where $\alpha =0$. The identity $d(\chi \alpha )=d\chi \wedge \alpha +\chi d\alpha$ gives $(d\alpha {)}_x=0$. Thus global forms can be computed using local extensions in a manifold chart.

In coordinates write $\alpha ={\sum }_I{a}_{I}d{x}^{i_1}\wedge \cdots \wedge d{x}^{i_p}$. Since $d(d{x}^i)=d^2{x}^i=0$, the graded Leibniz rule forcesThis is the uniqueness of the exterior derivative from its axioms. The coordinate formula also establishes existence: it has the stated properties, and the chain rule shows that coordinate changes give the same operator.

The de Rham cohomology is the real quotient vector spaceThus a cohomology class records a closed differential form modulo an exact differential form. In degree zero there are no exact forms; closed functions are locally constant. The Poincare lemma says that every closed form of positive degree on a star-shaped open subset of ${\mathbb{R}}^n$ is exact, and hence that positive-degree closed forms are locally exact on a smooth manifold.

For the first de Rham cohomology of the two-sphere, let $U=S^2\setminus \{N\}$ and $V=S^2\setminus \{S\}$. Stereographic projection identifies each with ${\mathbb{R}}^2$, so a closed one-form $\alpha$ has primitives $f_U,f_V$. On the connected overlap $U\cap V$, the derivative of $f_U-f_V$ is zero, so this difference is a constant. Subtracting that constant from $f_U$ makes the primitives agree. They glue to a global smooth primitive. Therefore $H_{\mathrm{dR}}^1(S^2)=0$.

Let $q:S^2\to {\mathbb{RP}}^2$ be the double covering map and $a$ the antipodal map. The pullback of a differential form identifies forms downstairs with $a$-invariant forms upstairs. Averaging $(\eta +a^{\ast }\eta )/2$ commutes with $d$. If an invariant form is exact upstairs, averaging its primitive proves it exact downstairs. Conversely an invariant cohomology class has an invariant representative by the same averaging. This proves the de Rham cohomology of a finite quotient identification $H_{\mathrm{dR}}^p({\mathbb{RP}}^2)\cong H_{\mathrm{dR}}^p(S^2{)}^{a^{\ast }}$. In degree zero the sphere is connected and $a^{\ast }$ fixes constants; degree one is zero; in degree two the granted action is multiplication by $-1$, whose invariant real subspace is zero. Forms of degree greater than two vanish. HenceThis is real de Rham cohomology; it does not detect the integral two-torsion of the real projective plane.

The real projective plane has fundamental group $C_2=⟨a\mid a^2=1⟩$, while the circle has fundamental group $\mathbb{Z}=⟨b⟩$. Apply the Seifert-van Kampen theorem to open neighborhoods of the two wedge summands whose overlap is contractible. This givesChanging the basepoint within this path-connected space gives an isomorphic group.

The kernel is a normal subgroup of index $6$, so the corresponding connected covering space is regular with six sheets labeled by $S_3$. Write $a=(12)$, $b=(123)$ and use right multiplication for path lifting. Above the real projective plane, right multiplication by $a$ pairs the labels, giving three copies of its two-sheeted covering space $S^2$. Each sphere has two marked antipodal points corresponding to its pair $\{b^i,b^{i}a\}$, $i=0,1,2$. Above the circle, right multiplication by $b$ gives two cycles of length three. Attach their six marked points to the identically labeled points of these three spheres.

The upper circle follows $e\to b\to b^2\to e$; the lower follows $a\to ab\to a{b}^2\to a$. The point paired with $b^i$ is $b^{i}a=a{b}^{-i}$, explaining the reversed lower labeling in the figure. On each summand these are genuine covering maps, and the attachment matches the six lifts of the wedge point. A small neighborhood of that point lifts to six disjoint copies of its neighborhood in the wedge. This verifies the covering space locally, and the action of $a,b$ is transitive, so it is connected. Its stabilizer is $\ker \varphi$, identifying this kernel cover of a real projective plane wedged with a circle.

A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

For $S_1$, interior points and points in the interiors of paired edges plainly have disc neighbourhoods. The corner identifications form two classes; in each class two quarter-discs are glued to make a half-disc, and the adjacent identified edge neighbourhoods complete a disc. Equivalently, this polygon is a cell decomposition of the real projective plane, hence a quotient of $S^2$ by the antipodal group. Thus $S_1$ is a topological surface.

In $S_2$, three distinct sides labelled $a$ are identified. A point in the interior of their common image has a neighbourhood made from three half-discs meeting along their diameters. Removing the common diameter leaves three local sides rather than two, so this neighbourhood is not a disc or half-disc. Therefore

For general odd $m_{+}$ and $m_{-}$, the same rectangular circuit givesEvery elementary two-dimensional nematic defect has $|q|=1/2$, so the minimum number is

In three dimensions the director takes values in the real projective plane, whose fundamental group is $\mathbb{Z}/2\mathbb{Z}$. The signs of half-charge line defects are no longer distinct topological classes, and two such lines can annihilate by escape into the third dimension. Hence only the parity of $N_{2D}$ remains:When one is required, it is a disclination line extending through the unbounded $z$ direction.

The antipodal action fixes the origin. Away from the origin it acts freely, so the quotient is locally Euclidean there. A small punctured neighborhood of the image of the origin, however, has linkthe real projective plane, whereas the link of a point in a three-dimensional smooth manifold is $S^2$. Equivalently, the neighborhood is a cone on ${\mathbb{RP}}^2$ and cannot be a three-ball. Thus

A topological surface is a Hausdorff, second-countable space in which every point has a neighbourhood homeomorphic to an open subset of ${\mathbb{R}}^2$.

The antipodal action on $S^2$ is free. The surface quotient by a free finite action applies: Around each point choose a small open disc disjoint from its antipodal image; the quotient map restricts to a homeomorphism from that disc onto an open neighbourhood in the quotient. Compactness gives Hausdorffness and second countability descends from the sphere. Thus the quotient is the real projective plane, in particular a topological surface.

For the second quotient write a point away from the poles asThe mapin these cylindrical coordinates extends continuously over the poles and identifies exactly $\varphi$ with $\varphi +\pi$. It therefore induces a continuous bijection $S^2/\approx \to S^2$. The domain is compact and the sphere Hausdorff, so this bijection is a homeomorphism.