The Projective Plane from a Disk

Pairing the Boundary of a Disk

One model of the projective plane starts with a closed disk and glues each point of its boundary to the point directly opposite it. The interior is not identified with any other point. This distinction matters: the construction pairs boundary points, rather than collapsing the entire boundary to one point. We will define the quotient precisely and show that it agrees with the model obtained from the sphere by identifying every point with its antipode.

Give \(D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}\) the subspace topology from the Euclidean plane, and give its boundary \(\partial D^2=\{(x,y):x^2+y^2=1\}\) the subspace topology. The sphere \(S^2=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\}\) also has its Euclidean subspace topology. For a space \(X\) and an equivalence relation \(\sim\) on it, \(X/{\sim}\) denotes the set of equivalence classes with the quotient topology: a subset \(V\subseteq X/{\sim}\) is open exactly when its preimage under the quotient map \(q:X\to X/{\sim}\) is open in \(X\).

Thus each interior point has a one-point equivalence class. Each boundary point has a two-point class, since no point on the unit circle equals its own negative. The topology on the quotient is part of the construction: it is not enough merely to list which points are paired.

Relating the Disk to a Hemisphere

Let \(H=\{(x,y,z)\in S^2:z\geq 0\}\), the closed upper hemisphere with its subspace topology. The map \((x,y)\mapsto (x,y,\sqrt{1-x^2-y^2})\) takes the disk onto \(H\). It keeps the first two coordinates and chooses the nonnegative third coordinate.

Proof. For every \((x,y)\in D^2\), the square root is defined and nonnegative, and the squared norm of \(h(x,y)\) is \(x^2+y^2+(1-x^2-y^2)=1\). Thus \(h(x,y)\in H\). Conversely, if \((x,y,z)\in H\), then \(x^2+y^2=1-z^2\leq 1\), and \(z=\sqrt{1-x^2-y^2}\) because \(z\geq0\). So \(h\) is onto. If \(h(x,y)=h(x',y')\), equality of the first two coordinates gives \((x,y)=(x',y')\), so \(h\) is one-to-one. The coordinate functions and the square root are continuous on their stated domains, so \(h\) is continuous. Its inverse is the restriction of the coordinate projection \((x,y,z)\mapsto(x,y)\), which is continuous. Hence \(h\) is a homeomorphism. If \(x^2+y^2=1\), then \(h(x,y)=(x,y,0)\) and \(h(-x,-y)=(-x,-y,0)= -h(x,y)\), proving the boundary assertion. \(\square\)

Define the antipodal relation on \(S^2\) by \(u\sim_S v\) if \(v=u\) or \(v=-u\), and give \(S^2/{\sim_S}\) the quotient topology. Every sphere class has a representative in \(H\): for a point with negative third coordinate, take its antipode; a point on the equator is already in \(H\). Among points of \(H\), two distinct points can belong to the same sphere class only when both are on the equator and are antipodal. Indeed, if \(u\in H\) has positive third coordinate, \(-u\) has negative third coordinate and is not in \(H\).

The Disk Quotient Is the Antipodal Sphere Quotient

We now make the agreement between these constructions precise. The map from the disk to the sphere quotient sends a point first to the upper hemisphere and then to its antipodal class. It respects exactly the identifications made in the disk.

Proof. Let \(q_D:D^2\to D^2/{\sim_D}\) and \(q_S:S^2\to S^2/{\sim_S}\) be the quotient maps. The composite \(q_S\circ h\) is continuous. If \(a\sim_D b\), then either \(a=b\), or \(a,b\) are boundary antipodes. In the latter case, \(h(b)=-h(a)\). Thus \(q_S(h(a))=q_S(h(b))\). The universal property of quotient spaces therefore gives a continuous map \(F:D^2/{\sim_D}\to S^2/{\sim_S}\) satisfying \(F\circ q_D=q_S\circ h\).

To prove that \(F\) is onto, take any sphere class. It has a representative \(u\in H\), as shown above, and \(h\) is onto \(H\). Thus \(u=h(a)\) for some \(a\in D^2\), and the class is \(F(q_D(a))\).

To prove that \(F\) is one-to-one, suppose \(F(q_D(a))=F(q_D(b))\). Then \(h(a)\sim_S h(b)\). If \(h(a)=h(b)\), injectivity of \(h\) gives \(a=b\). Otherwise \(h(b)=-h(a)\). Since both points lie in \(H\), their third coordinates must be zero: nonnegative numbers can be negatives of one another only when both are zero. Hence \(a,b\in\partial D^2\), and the first two coordinates give \(b=-a\). Therefore \(a\sim_D b\), so \(q_D(a)=q_D(b)\). This proves injectivity.

For completeness, the target \(S^2/{\sim_S}\) is Hausdorff. Define a distance between classes by \(\delta([u],[v])=\min\{\|u-v\|,\|u+v\|\}\), where \([u]\) denotes \(\{u,-u\}\). This is independent of the representatives, is symmetric, and is zero exactly when the classes agree. For the triangle inequality, choose a representative of the middle class that realizes the minimum distance from a chosen representative of the first class; then choose a representative of the third class that realizes the minimum from that middle representative. The Euclidean triangle inequality gives \(\delta([u],[w])\leq\delta([u],[v])+\delta([v],[w])\). Thus \(\delta\) is a metric. Its open balls have open preimages under \(q_S\), since the preimage of the radius-\(r\) ball about \([u]\) is the union of the radius-\(r\) balls about \(u\) and \(-u\), intersected with \(S^2\). Conversely, if a set of classes has open preimage, then for each representative \(u\) in that preimage a sufficiently small ball around \(u\) lies in the preimage; the corresponding \(\delta\)-ball lies in the set of classes. So the quotient topology is the metric topology, which is Hausdorff.

The disk \(D^2\) is compact by the Heine–Borel theorem, and its quotient is compact: pull back any open cover along \(q_D\), take a finite subcover of \(D^2\), and then use the corresponding finite collection of quotient sets. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism. Indeed, every closed subset of a compact space is compact, its continuous image is compact, and compact subsets of a Hausdorff space are closed; hence the bijection is closed and its inverse is continuous. Applying this result to \(F\) proves the theorem. \(\square\)

What the Identification Does—and Does Not Do

The boundary of the disk is paired in opposite points, while all interior points remain separate. The circle calculation shows that the set of boundary classes is itself a circle. In the full quotient this boundary circle is the image of the disk boundary; the quotient description specifies its topology as part of the gluing, rather than as a drawn outline.

Common pitfall. A frequent incorrect description says that all boundary points are identified with one another. That would identify \(p=(1,0)\) with \(s=(0,1)\), since both lie on the boundary. But \(s\neq p\) and \(s\neq -p=(-1,0)\), so the stated disk relation does not identify them. Only \(p\) and \(-p\) are paired. Collapsing the entire boundary is a different quotient construction.

Another tempting mistake is to use the upper hemisphere without accounting for its equator. The open upper hemisphere has no antipodal pairs within it, but the closed hemisphere does: its equator contains both \(u\) and \(-u\). Those equatorial pairs are precisely why the disk boundary must be identified in antipodal pairs. Omitting that boundary identification would give a different space.