Identifying Antipodal Points
A quotient space is formed by treating specified points as equivalent and giving the resulting classes a topology determined by the original space. For real projective space, the points to be identified come in antipodal pairs on a sphere. The quotient map records which pair a point belongs to; its topology is defined by asking whether the full preimage of a set of classes is open on the sphere.
For \(n\geq 0\), let \(S^n=\{x\in\mathbb{R}^{n+1}:\|x\|=1\}\), with the subspace topology inherited from Euclidean space. The antipode of \(x\in S^n\) is \(-x\), which also belongs to \(S^n\), since \(\|-x\|=\|x\|=1\).
First, the relation really is an equivalence relation. It is reflexive because \(x=x\). It is symmetric because if \(y=x\), then \(x=y\), and if \(y=-x\), then \(x=-y\). For transitivity, suppose \(x\sim y\) and \(y\sim z\). There are signs \(\varepsilon,\delta\in\{1,-1\}\) such that \(y=\varepsilon x\) and \(z=\delta y\). Therefore \(z=(\delta\varepsilon)x\), and \(\delta\varepsilon\in\{1,-1\}\), so \(x\sim z\). Also, no point of \(S^n\) equals its own antipode: \(x=-x\) would imply \(x=0\), contrary to \(\|x\|=1\). Thus every equivalence class has exactly two points, including when \(n=0\).
The Quotient Topology and the Quotient Map
For a subset \(A\subseteq S^n\), write \(-A=\{-x:x\in A\}\). The preimage of the classes represented by points of \(A\) is
Indeed, a point belongs to \(q^{-1}(q(A))\) precisely when its class agrees with the class of some point of \(A\). That means it is either that point or its antipode. In particular, a subset is a union of whole equivalence classes (also called saturated) exactly when \(q^{-1}(q(A))=A\).
With this topology, \(q\) is continuous and surjective. Surjectivity follows because every class \([x]\) is the image \(q(x)\). For continuity, if \(O\) is open in \(\mathbb{RP}^n\), the definition says exactly that \(q^{-1}(O)\) is open in \(S^n\). The same definition gives the quotient-map test, so \(q\) is a quotient map.
Proof. Let \(U\) be open in \(S^n\). Define \(a:S^n\to S^n\) by \(a(x)=-x\). This map is continuous, since it is the restriction of the continuous linear map \(x\mapsto -x\) on \(\mathbb{R}^{n+1}\); it is its own inverse, so it is a homeomorphism. Hence \(a(U)=-U\) is open in \(S^n\). The preimage formula gives \(q^{-1}(q(U))=U\cup(-U)\), which is open as a union of two open sets. By the definition of the quotient topology, \(q(U)\) is open in \(\mathbb{RP}^n\). This holds for every open \(U\), proving that \(q\) is open. \(\square\)
Openness of \(q\) is useful, but it does not mean that \(q^{-1}(q(U))=U\) for every open \(U\). An image under \(q\) forgets which of the two representatives was chosen. Its full preimage generally adds the antipodes, as the formula shows.
Worked Example: The Zero-Dimensional Case
The sphere \(S^0\) consists of the two real numbers of absolute value \(1\), so \(S^0=\{-1,1\}\), with its subspace topology from \(\mathbb{R}\). The equivalence class of \(1\) is \([1]=\{1,-1\}\), and the class of \(-1\) is the same set: \([-1]=\{-1,1\}=[1]\). Thus there is exactly one class, and \(\mathbb{RP}^0=\{[1]\}\) is a one-point space.
The quotient map has the values \(q(1)=[1]\) and \(q(-1)=[1]\). To check the quotient topology, the only subsets of the one-point target are \(\varnothing\) and \(\{[1]\}\). Their preimages are \(\varnothing\) and \(S^0\), respectively, both open. So both target subsets are open, as expected for the one-point topology.
Check: the two points of \(S^0\) are distinct, but they have the same image. Also, the class has two representatives, not one; the quotient has one point because those representatives have been identified.
Worked Example: A Class in \(\mathbb{RP}^1\)
Here \(S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\), with the subspace topology. Consider \(x=(3/5,4/5)\). Its norm is verified by \((3/5)^2+(4/5)^2=9/25+16/25=1\), so \(x\in S^1\). Its antipode is \(-x=(-3/5,-4/5)\), and therefore
There are no other representatives in this class: by the definition of \(\sim\), a point equivalent to \(x\) must equal \(x\) or \(-x\). The norm check also holds for the antipode, since \((-3/5)^2+(-4/5)^2=9/25+16/25=1\). The two displayed points are distinct, since their first coordinates, \(3/5\) and \(-3/5\), differ.
Check: applying \(q\) to either representative gives the same class. In particular, \(q(3/5,4/5)=q(-3/5,-4/5)\), in accordance with the equivalence relation.
Worked Example: An Open Set and Its Full Preimage in \(\mathbb{RP}^2\)
Take \(S^2=\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_1^2+x_2^2+x_3^2=1\}\), with the subspace topology, and let \(U=\{x\in S^2:x_3>0\}\). This set is open in \(S^2\), since it is the intersection of \(S^2\) with the open half-space \(\{(x_1,x_2,x_3)\in\mathbb{R}^3:x_3>0\}\).
The image \(q(U)\) consists of the classes with a representative whose third coordinate is positive. If \(x_3>0\), then \(x\in U\). If \(x_3<0\), then \((-x)_3=-x_3>0\), so \(-x\in U\) and \([x]\in q(U)\). If \(x_3=0\), both \(x\) and \(-x\) have third coordinate zero, so neither is in \(U\), and \([x]\notin q(U)\). Therefore
This preimage is open in \(S^2\), because it is the intersection of \(S^2\) with the open subset \(\{x\in\mathbb{R}^3:x_3\ne0\}\) of \(\mathbb{R}^3\). The quotient-topology test confirms that \(q(U)\) is open in \(\mathbb{RP}^2\). For instance, \((0,0,1)\) and \((0,0,-1)\) both have norm \(1\), and they represent the same class in \(q(U)\), even though only the first is in \(U\).
Check: the class of \((1,0,0)\) is not in \(q(U)\), because both representatives \((1,0,0)\) and \((-1,0,0)\) have third coordinate zero. This agrees with the preimage calculation.
A Common Mistake: Forgetting the Other Representative
A tempting but incorrect calculation is \(q^{-1}(q(U))=U\). In the preceding example, this would say that the full preimage of \(q(U)\) contains points with positive third coordinate only. But \((0,0,-1)\notin U\), while \(q(0,0,-1)=q(0,0,1)\in q(U)\). Thus \((0,0,-1)\in q^{-1}(q(U))\), contradicting that calculation. The correction is to include the antipodes: \(q^{-1}(q(U))=U\cup(-U)\). Only when \(U\) is already a union of whole antipodal pairs does \(q^{-1}(q(U))=U\).
This distinction matters whenever one tests a set in the quotient using its preimage on the sphere. The quotient topology tests openness of a set of classes by checking its full preimage, not by checking one chosen representative from each class. In practice, first identify which points are equivalent, then make sure the set on the sphere includes both members of each class that it represents.
| Object | Meaning | Useful test |
|---|---|---|
| \([x]\) | The equivalence class \(\{x,-x\}\) | \(q(x)=q(y)\) exactly when \(y=x\) or \(y=-x\) |
| \(\mathbb{RP}^n\) | The set of antipodal classes in \(S^n\) | Its topology is defined using preimages under \(q\) |
| \(q^{-1}(q(A))\) | The full preimage of the classes represented by \(A\) | It equals \(A\cup(-A)\) |
| \(q(U)\) for open \(U\subseteq S^n\) | The classes with at least one representative in \(U\) | It is open because its preimage is \(U\cup(-U)\) |
Check Your Understanding
For each question, use the antipodal equivalence relation and, where relevant, the quotient-topology test.
- Write the equivalence class of \((0,1)\in S^1\), and verify that both representatives lie on \(S^1\).
- For \(A=\{(1,0),(-1,0)\}\subseteq S^1\), compute \(q(A)\) and \(q^{-1}(q(A))\).
- In \(S^2\), determine whether the class of \((0,1,0)\) belongs to the image of \(U=\{x\in S^2:x_3>0\}\). Justify your answer using both representatives.
- Explain why \(q^{-1}(q(A))=A\) holds exactly when \(A\) contains either both members or neither member of each antipodal pair.
- Use the quotient-topology test to decide whether the empty set and the whole space \(\mathbb{RP}^n\) are open.