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What quotients can I take of $\mathbb{R}^{3}$ that give me $\mathbb{R}^{3}$ back?

Thankyou

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If you take $\mathbb{R}^{3}/\mathbb{D}^{3}$, then you get $\mathbb{R}^{3}$ back as a topological space. The same applies to an arbitary disjoint union of $\mathbb{D}^{3}$s, since you can deform all of them in the same time to a point.

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    What is $\Bbb D^3$?2012-12-09
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    @anon: The closed unit ball centred at the origin.2012-12-09
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    @anon:corrected, thanks.2012-12-09
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    @Brian Right, that condition works out, I should have thought that this was being used.2012-12-09
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    And also closed disks and finite non-intersecting paths?2012-12-09
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    You can choose any subset of $\mathbb{R}^{3}$ which can be deformed to a point. But note this is not the same thing as contractible. If you assume $A$ is a CW-complex, then $A$ can be deformed to a point if and only if $\pi_{i}(A)=0,i=1,2,3$.2012-12-09