Real Projective Space Homeomorphism to Quotient of Sphere (Proof)

A standard trick at this point would be to use the fact that a continuous bijection from a compact space to a Hausdorff space must be a homeomorphism (see here for example).

As for how to prove that $\mathbb{R}\mathrm{P}^n$ is compact, usually you do that by showing that $S^n$ is compact and that there is a continuous surjection $S^n\to \mathbb{R}\mathrm{P}^n$ (which, it turns out, induces the inverse to the function you're working with).

I'm curious to know how you proved that $f$ is continuous, since the proof (or at least the one that I'm aware of) for proving continuity is almost as hard as proving the continuity of its inverse.

Here, I'll prove $f^{-1}$ being continuous. The proof for $f$ being continuous uses similar ideas. (Let me know in the comments what you did for $f$.)

Claim:

$U$ is open in $\mathbb{RP}^{n}$ $\Longrightarrow$ $f(U)$ is open in $S^{n} / \sim $

Proof:

$\require{AMScd}$ \begin{CD} \mathbb{R}^{n+1}\backslash\{0\} @>g>> S^{n} \\ @V\pi_{1}VV @VV\pi_{2}V \\ \mathbb{RP}^{n} @>>f = \pi_{2}\circ g\circ\pi_{1}^{-1}> S^{n}/\sim \end{CD}

• Let $\pi_{1}: \mathbb{R}^{n+1}\backslash\left\{0\right\} \longrightarrow \mathbb{RP}^{n}$ and $\pi_{2}: S^{n} \longrightarrow S^{n}/\sim $ be the standard quotient maps.

• Let $g: \mathbb{R}^{n+1}\backslash\left\{0\right\} \longrightarrow S^{n}$ be s.t. $g\left(x\right) = \frac{x}{\left\|x\right\|}$

• Let $i: S^{n} \longrightarrow \mathbb{R}^{n+1}\backslash\left\{0\right\}$ be the standard inclusion map, $i\left(x\right) = x$, s.t. $g \circ i = \text{id}_{S^{n}}$

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By property of quotient maps, we have $\pi_{1}^{-1}\left(U\right)$ open in $\mathbb{R}^{n+1}\backslash\left\{0\right\}$.

Next, we claim that $g\circ\pi_{1}^{-1}\left(U\right) = \pi_{1}^{-1}\left(U\right) \cap S^{n}$:

• Let $x \in \pi_{1}^{-1}\left(U\right) \cap S^{n}$.

  Then, $g\left(x\right) = \frac{x}{\left\|x\right\|} = x \in g\circ\pi_{1}^{-1}\left(U\right)$

  $\Longrightarrow \pi_{1}^{-1}\left(U\right) \cap S^{n} \subseteq g\circ\pi_{1}^{-1}\left(U\right)$

• Let $x \in g\circ\pi_{1}^{-1}\left(U\right) \subseteq S^{n}$.

  Then, $i\left(x\right) = x \in \pi_{1}^{-1}\left(U\right)$. Thus, $x \in \pi_{1}^{-1}\left(U\right) \cap S^{n}$.

  $\Longrightarrow g\circ\pi_{1}^{-1}\left(U\right) \subseteq \pi_{1}^{-1}\left(U\right) \cap S^{n}$.

Therefore, $g\circ\pi_{1}^{-1}\left(U\right) = \pi_{1}^{-1}\left(U\right) \cap S^{n}$ is open in $S^{n}$.

Next, we claim that if $x \in g\circ\pi_{1}^{-1}\left(U\right)$, then $\forall$ $y \in S^{n}$ s.t. $y \sim x$, $y \in g\circ\pi_{1}^{-1}\left(U\right)$.

• $y \sim x$ in $S^{n}$ $\Longrightarrow y = \pm x \Longrightarrow i(y) \sim i(x)$ in $\mathbb{R}^{n+1}\backslash\{0\}$

  $\Longrightarrow i(y) \in \pi_{1}^{-1}\{x\} \subseteq \pi_{1}^{-1}\left(U\right) \Longrightarrow g(i(y)) = y \in g\circ\pi_{1}^{-1}\left(U\right)$

In other words, $x \in g\circ\pi_{1}^{-1}\left(U\right) \Longrightarrow \{y \in S^n: y\sim x\} \subseteq g\circ\pi_{1}^{-1}\left(U\right)$. Thus, $g\circ\pi_{1}^{-1}\left(U\right) = \bigcup_{x \in g\circ\pi_{1}^{-1}\left(U\right)}{\{y \in S^n: y\sim x\}}$. Therefore, $\pi_{2}\circ g\circ\pi_{1}^{-1}\left(U\right) = f\left(U\right)$ is open in $S^{n}/\sim$

QED