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?



Four manifold without point homotopy equivalent to wedge of two-spheres?
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.


