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I am wondering if the following is true:

Let $U, V \subset \mathbb{R}^n$ be open sets. Suppose $\pi_i(U) =\pi_i(V) = {1}$ for all $i = 0,1,2,3, ...$ Then $U$ and $V$ are homeomorphic.

I came about this question after reading Hatcher's proof that the "house with two rooms" is contractible.

wellfedgremlin
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2 Answers2

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Open subsets of $\Bbb R^n$ are homotopy equivalent to CW-complexes, so by Whitehead's theorem, you're asking whether contractible open subsets of $\Bbb R^n$ are homeomorphic to the open $n$-ball $D^n$.

This is not true: the Whitehead manifold is a contractible open subset of $\Bbb R^3$, not homeomorphic to $D^3$. (Whitehead asked this very same question while trying to prove the Poincaré conjecture, which is why he ran into this particular manifold.)

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    I don't see the relevance of Whitehead's theorem to this argument. The Whitehead manifold is already a counterexample before you mention Whitehead's theorem. – Qiaochu Yuan Feb 23 '15 at 06:17
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    It's not relevant to the argument. I just prefer to replace weak contractibility with contractibility whenever possible, as I find the latter more conceptually reasonable. I mention it only so the OP has a better grasp on the space he's considering. –  Feb 23 '15 at 06:22
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Here is a positive result in this direction: any open simply connected subset of $\mathbb{R}^2$ is homeomorphic to the open disk $D^2$. This is a consequence of the uniformization theorem, and as Mike Miller's answer shows, the corresponding statement with $2$ replaced by $3$ is false.

Qiaochu Yuan
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