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Is there a "nice" embedding of $\mathbb{C}\mathbb{P}^2/\,\mathbb{C}\mathbb{P}^1$ into $\mathbb{R}^8$?

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    $\begingroup$ What does $/$ mean here? $\endgroup$ – Igor Rivin Nov 16 '18 at 6:23
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    $\begingroup$ @IgorRivin: I think $X/Y$ means that we collapse $Y$ to a point, and take the quotient topology from the obvious map $X \to X/Y$. $\endgroup$ – Ben McKay Nov 16 '18 at 9:26
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    $\begingroup$ Of course, quotients depend on the way a subspace is embedded. But if I take the standard embetting, then I obtain $\Bbbk P^n$ from $\Bbbk P^{n-1}$ by attaching one cell. So should the quotient not just be a sphere $S^{n\dim\Bbbk}$? Or do you mean something different? $\endgroup$ – Sebastian Goette Nov 16 '18 at 11:17
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    $\begingroup$ @User371 The attaching map may be twisted, but if you contract its target space to a point, you don't see that any longer. Also, from the long exact reduced homology sequence you get $H_1(\mathbb R P^2/\mathbb R P^1;\mathbb Z/2)=0$. $\endgroup$ – Sebastian Goette Nov 18 '18 at 11:54
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    $\begingroup$ The comments remain: if $/$ means what Ben McKay says, then $\Bbb{CP}^2/\Bbb{CP}^1 \cong S^4$, and surely this is not what you're asking about. So what do you mean? $\endgroup$ – mme Nov 19 '18 at 0:05