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Suppose we have a closed 3-manifold $M$, not necessarily simply connected.

What can I say about the homotopy groups of $M \setminus \text{pt}$? ($M$ punctured by one point)

In particular, what assumptions do I need to ensure that $\pi_2 (M \setminus \text{pt}) = 0$?

Thanks!

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1 Answer 1

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If $\pi_2(M\setminus\{p\})=0$ then $M$ is simply connected (and hence $S^3$ by the Poincare conjecture). To see this, consider the universal cover $q:U\to M$ and let $V=q^{-1}(M\setminus \{p\})$. Then $V$ is a cover of $M\setminus\{p\}$ (in fact, the universal cover) and $U\setminus V$ is a discrete set of cardinality $|\pi_1(M)|$. But if you take any simply connected 3-manifold and puncture it more than once, it will have nontrivial $H_2$ and hence nontrivial $\pi_2$ (the reason you have to puncture more than once is that if your manifold is closed then the first puncture will kill $H_3$ instead of giving you something in $H_2$). Thus unless $\pi_1(M)$ is trivial, $\pi_2(V)=\pi_2(M\setminus\{p\})$ will be nontrivial.

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