Let $X$ be a closed, simplyconnected fourmanifold. Let $X'$ be obtained from $X$ by removing a point. Is $X'$ homotopy equivalent to a wedge of $S^2$s?
2 Answers
Here is a hopefully better answer which is copied from page 104 of MilnorHusemoller's book on symmetric bilinear forms.
$X^\prime$ is also simply connected (Seifertvan Kampen reversed) and thus has torsionfree $H_2={\mathbb Z}^r$ (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 CWcomplex with one 0cell, no 1cell, some 2cells, no 3cell, one 4cell. Removing a point is homotopy equivalent to removing the 4cell, you get a CWcomplex with one point and several 2cells attached to it, that is, a wedge of 2spheres.
EDIT: This actually seems to be known only in dimensions $\ge5$, which is the case handled in Milnor's book on the hcobordism theorem. The 4dimensional case seems to be open, according to Existence of Morse functions on simply connected manifolds

1$\begingroup$ 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 hcobordim theorem handle only the higherdimensional case. I will edit the answer accordingly. $\endgroup$– ThiKuOct 17, 2016 at 16:46