Let $X$ be a closed, simply-connected four-manifold. Let $X'$ be obtained from $X$ by removing a point. Is $X'$ homotopy equivalent to a wedge of $S^2$s?

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    $\begingroup$ Yes. This is the first step in the proof of the Whitehead-Milnor theorem classifying such manifolds up to homotopy equivalence. $\endgroup$ Oct 17, 2016 at 6:57

2 Answers 2


Here is a hopefully better answer which is copied from page 104 of Milnor-Husemoller's book on symmetric bilinear forms.

$X^\prime$ is also simply connected (Seifert-van Kampen reversed) and thus has torsion-free $H_2={\mathbb Z}^r$ (otherwise the torsion would contribute nontrivially to $H^3$ via the universal coefficient theorem) and $\pi_2=H_2$ (Hurewicz).

So there is a map $S^2\vee\ldots\vee S^2\to X^\prime$ that induces a homology isomorphism, hence (by Hurewicz) a weak homotopy equivalence, hence (because $X^\prime$ is an ANR) a homotopy equivalence.


A closed, simply connected manifold has a Morse function with one critical point of index 0 and no critical point of index 1. Thus it is homotopy equivalent to a CW-complex with one 0-cell, no 1-cell, some 2-cells, no 3-cell, one 4-cell. Removing a point is homotopy equivalent to removing the 4-cell, you get a CW-complex with one point and several 2-cells attached to it, that is, a wedge of 2-spheres.

EDIT: This actually seems to be known only in dimensions $\ge5$, which is the case handled in Milnor's book on the h-cobordism theorem. The 4-dimensional case seems to be open, according to Existence of Morse functions on simply connected manifolds

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    $\begingroup$ My guess is that the OP would be interested in references (showing the absence of critical points of index $1$ and $3$); otherwise the question is trivial. $\endgroup$ Oct 17, 2016 at 15:38
  • $\begingroup$ I just noticed that this seems to be an open problem in dimension 4, as Milnor's Lectures on the h-cobordim theorem handle only the higher-dimensional case. I will edit the answer accordingly. $\endgroup$
    – ThiKu
    Oct 17, 2016 at 16:46

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