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Imagine $\Bbb{R}^3$ with this equivalence relation $(x,y,z) \sim (x',y',z') \iff \exists t \in \Bbb{R}^3 - \{0\}: x'=tx, y'=ty, z'=tz$.

In this case how can I show that $\Bbb{R}^3/\sim$ is compact and hausdorff but it isn't orientable?

Thank you in advance!

Kevin.S
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geo sd
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  • Your definition of $\sim$ does not make sense ($t \in \mathbb R^3$). Moreover I am sure that you consider $\sim$ on $\mathbb R^3 \setminus {0}$. – Paul Frost Jan 17 '21 at 13:16

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This space is the same as $\mathbb R \mathbb P^2 \cup \{ 0 \}$.

But let us try to see it more directly. Let us call your space $X$ (and let us also forget the origin, since it just makes the computations messier). So $X= \mathbb R^3 - \{ 0\} / \sim $.

First, I claim that there is a map $\pi: \mathbb R^3 -\{0\} \to S^2$ and a map $\Pi:S^2 \to X$ such that the following diagram commutes: $$ \begin{array}{cccc} \mathbb R^3 - \{ 0\} & \xrightarrow{\pi} & S^2 \\ &\searrow & \downarrow \\ && X \end{array} $$

Indeed, $\pi$ is defined by $x \mapsto x/|x|$, and $\Pi$ is defined by sending $x$ to its equivalence class. Note that it is a $2:1$ map.

This proves that $X$ is compact, since it is the image of a compact space. It also proves Hausdorffness, because $\Pi$ is locally a diffeomorphism (explain why and why this is enough).

So $X$ can be identified with $S^2/\mathbb Z_2$, where $\mathbb Z/2$ acts by $x \mapsto -x$.

Here's a sketch of the non-orientability: the group action of $\mathbb Z / 2$ is not orientation-preserving on $S^2$, then use this old math.stachexchange answer.

Fredrik Meyer
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  • Would you please explain more about $\pi$? I don't know why it is enough to prove $\pi$ is locally a diffeomorphism so that $X$ is hausdorff. Besides I don't know what a $2:1$ map means. I'll be grateful if you tell me more about it. – geo sd Jan 17 '21 at 10:41
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    If we do not omit the origin, then $P = \mathbb R^3/\sim$ is not Hausdorff. In fact, the only open neigborhood of $ 0 \in P$ is $P$ itself: If $p : \mathbb R^3 \to P$ denotes the quotient map and $U$ is an open neighborhood of $0 \in P$, then $p^{-1}(0)$ is an open neighborhood of $0 \in \mathbb R^3$ which clearly contains points of all $p^{-1}(x)$ with $x \ne 0$. Thus we must have $p^{-1}(0) = \mathbb R^3$. – Paul Frost Jan 17 '21 at 13:20
  • @paul-frost Thank you for your answer. It can’t be compact, am I right? – geo sd Jan 17 '21 at 14:38
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    @geosd My space $P$ is compact because any open cover of $P$ must contain a neighborhood of $0$. – Paul Frost Jan 17 '21 at 15:13
  • Thank you very much – geo sd Jan 17 '21 at 16:18