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.


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 $f$.)


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


$\require{AMScd}$ \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 $S^{n}/\sim$


  • 2
    $\begingroup$ Why is the function g continuous? $\endgroup$
    – Darkmaster
    May 17 '19 at 1:02
  • 2
    $\begingroup$ Because $g$ is a rational function that has no null denominator. $\endgroup$
    – Ilovemath
    Dec 19 '20 at 11:27

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).

  • $\begingroup$ but we haven't talked about compact space yet. BTW, we construct the continuous function by invariant function under $\sim$ $\endgroup$
    – ElleryL
    Jun 27 '15 at 3:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.