[proofplan] We send a unit vector $x \in S^n$ to the projective point determined by the line $\mathbb{R}x$. The key algebraic point is that two unit vectors span the same line exactly when they differ by sign, so this map has precisely the antipodal equivalence classes as its fibres. The universal property of the quotient then gives a canonical bijection $S^n/{\sim} \to \mathbb{R}\mathbb{P}^n$. Finally, we compare quotient topologies using the radial normalization map from $\mathbb{R}^{n+1}\setminus\{0\}$ to $S^n$ to prove that this bijection is a homeomorphism. [/proofplan] [step:Define the map from the sphere to projective space] Let \begin{align*} p: \mathbb{R}^{n+1}\setminus\{0\} \to \mathbb{R}\mathbb{P}^n \end{align*} denote the standard quotient map sending a nonzero vector $v$ to the one-dimensional subspace $\mathbb{R}v$. Define \begin{align*} q: S^n \to \mathbb{R}\mathbb{P}^n \end{align*} by \begin{align*} q(x) := \mathbb{R}x. \end{align*} Let \mathbb{R}^{\times} := \mathbb{R}\setminus\{0\}. Equivalently, if \begin{align*} \iota: S^n \to \mathbb{R}^{n+1}\setminus\{0\} \end{align*} is the inclusion map, then $q = p \circ \iota$. Since $\iota$ is continuous and $p$ is continuous by the definition of the [quotient topology](/page/Quotient%20Topology) on $\mathbb{R}\mathbb{P}^n$, the map $q$ is continuous. [/step] [step:Identify the fibres as antipodal pairs] Let $x,y \in S^n$. We prove that $q(x) = q(y)$ if and only if $x \sim y$. If $x \sim y$, then either $y = x$ or $y = -x$. In both cases $\mathbb{R}y = \mathbb{R}x$, hence $q(y) = q(x)$. Conversely, suppose $q(x) = q(y)$. Then $\mathbb{R}x = \mathbb{R}y$, so there exists $\lambda \in \mathbb{R}^{\times}$ such that $y = \lambda x$. Since $x,y \in S^n$, we have $|x| = 1$ and $|y| = 1$. Taking Euclidean norms in $y = \lambda x$ gives \begin{align*} 1 = |y| = |\lambda x| = |\lambda| |x| = |\lambda|. \end{align*} Thus $\lambda = 1$ or $\lambda = -1$, and therefore $y = x$ or $y = -x$. Hence $x \sim y$. [guided] We need to compute exactly when two points of the sphere map to the same projective point. The map $q$ forgets the length and keeps only the line through the origin, so equality $q(x)=q(y)$ means precisely that $x$ and $y$ span the same one-dimensional subspace. First suppose $x \sim y$. By definition of $\sim$, either $y=x$ or $y=-x$. If $y=x$, then $\mathbb{R}y=\mathbb{R}x$. If $y=-x$, then \begin{align*} \mathbb{R}y = \mathbb{R}(-x) = \mathbb{R}x, \end{align*} because multiplication by $-1 \in \mathbb{R}^{\times}$, where $\mathbb{R}^{\times} := \mathbb{R}\setminus\{0\}$, does not change the line. Thus $q(y)=q(x)$. Conversely, suppose $q(x)=q(y)$. Since $q(x)=\mathbb{R}x$ and $q(y)=\mathbb{R}y$, equality of projective points says \begin{align*} \mathbb{R}x = \mathbb{R}y. \end{align*} Because $x$ and $y$ are nonzero vectors, equality of the two one-dimensional subspaces gives a scalar $\lambda \in \mathbb{R}^{\times}$ with \begin{align*} y = \lambda x. \end{align*} Now the sphere condition is the extra information that forces $\lambda$ to be only a sign. Since $x,y \in S^n$, both have Euclidean norm one. Taking norms gives \begin{align*} 1 = |y| = |\lambda x| = |\lambda| |x| = |\lambda|. \end{align*} Therefore $\lambda = 1$ or $\lambda = -1$. In the first case $y=x$, and in the second case $y=-x$. Hence $x \sim y$. [/guided] [/step] [step:Pass to the antipodal quotient and obtain a bijection] Let \begin{align*} \pi: S^n \to S^n/{\sim} \end{align*} be the quotient map. Since the fibres of $q$ are exactly the equivalence classes of $\sim$, the universal property of the quotient gives a unique map \begin{align*} \overline{q}: S^n/{\sim} \to \mathbb{R}\mathbb{P}^n \end{align*} such that $\overline{q}\circ \pi = q$. Explicitly, \begin{align*} \overline{q}([x]) = \mathbb{R}x. \end{align*} This formula is well-defined because $x \sim y$ implies $\mathbb{R}x = \mathbb{R}y$. The map $\overline{q}$ is injective because $\overline{q}([x])=\overline{q}([y])$ implies $q(x)=q(y)$, hence $x \sim y$, so $[x]=[y]$. It is surjective because every projective point in $\mathbb{R}\mathbb{P}^n$ has the form $\mathbb{R}v$ for some $v \in \mathbb{R}^{n+1}\setminus\{0\}$, and the vector \begin{align*} x := \frac{v}{|v|} \end{align*} belongs to $S^n$ and satisfies $\mathbb{R}x=\mathbb{R}v$. Thus $\overline{q}$ is a bijection. [/step] [step:Compare quotient topologies using radial normalization] Define the radial normalization map \begin{align*} r: \mathbb{R}^{n+1}\setminus\{0\} \to S^n \end{align*} by \begin{align*} r(v) := \frac{v}{|v|}. \end{align*} The map $r$ is continuous because the Euclidean norm is continuous and $|v|>0$ on $\mathbb{R}^{n+1}\setminus\{0\}$. For every $v \in \mathbb{R}^{n+1}\setminus\{0\}$, the vectors $v$ and $r(v)$ span the same line, so \begin{align*} p(v) = q(r(v)). \end{align*} Hence $p = q \circ r$. We show that $q$ is a quotient map. Let $A \subset \mathbb{R}\mathbb{P}^n$ be a subset such that $q^{-1}(A)$ is open in $S^n$. Then \begin{align*} p^{-1}(A) = r^{-1}(q^{-1}(A)). \end{align*} Since $r$ is continuous, $p^{-1}(A)$ is open in $\mathbb{R}^{n+1}\setminus\{0\}$. By the quotient topology defining $\mathbb{R}\mathbb{P}^n$, this implies that $A$ is open in $\mathbb{R}\mathbb{P}^n$. Therefore $q$ is a quotient map. Since $\pi$ is also a quotient map and $q = \overline{q}\circ\pi$, the bijection $\overline{q}$ is continuous by the universal property of $\pi$. To prove that $\overline{q}^{-1}$ is continuous, let $B \subset S^n/{\sim}$ be open. Since $\pi$ is a quotient map, $\pi^{-1}(B)$ is open in $S^n$. Because $q$ is a quotient map and \begin{align*} q^{-1}(\overline{q}(B)) = \pi^{-1}(B), \end{align*} the set $\overline{q}(B)$ is open in $\mathbb{R}\mathbb{P}^n$. Thus $\overline{q}$ is an open bijection, so $\overline{q}^{-1}$ is continuous. [/step] [step:Conclude the canonical identification] The map \begin{align*} \overline{q}: S^n/{\sim} \to \mathbb{R}\mathbb{P}^n \end{align*} is a bijection, is continuous, and has continuous inverse. Therefore it is a homeomorphism. Its defining formula is \begin{align*} \overline{q}([x]) = \mathbb{R}x, \end{align*} so it is precisely the canonical identification of real [projective space](/page/Projective%20Space) with the antipodal quotient of the sphere. [/step]