Real Projective Space Homeomorphism to Quotient of Sphere (Proof) I need to construct a function $f : (\mathbb{R}^{n+1}-\{0\})/{\sim} \to S^n/{\sim}$,  by 
$$f ([x]_{\mathbb{RP}^n}) = \left[\frac{x}{\|x\|}\right]_{S^n/{\sim}},$$ 
where  $S^n = \{ x \in \mathbb{R}^{n+1} : \|x\| = 1\} $ , $\sim$ on $S^n$ by $x \sim y \iff x = -y \lor x = y $.
${\mathbb{RP}^n}$ for real project space on $\mathbb{R}^{n+1}$.
I have shown that this function is bijective and continuous. However, in order to prove it's homeomorphism, I need its inverse to be continuous, and I find it's very hard to prove this part. Any hint is appreciated.
 A: 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).
A: 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. $g \circ i = \text{id}_{S^{n}}$

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
