Third de Rham cohomology group for the simply connected 4-manifolds

This is a question that has emerged from a discussion about some topological properties of spacetimes.

Suppose that $M$ is a (Hausdorff, paracompact) smooth, orientable, connected, simply connected 4-dimensional manifold, which might be compact or noncompact, with or without the boundary. Is it true that the third de Rham cohomology group $H^3_{\mathrm{dR}}(M)$ is necessarily trivial?

It seems that usual tools of algebraic topology suffice for this conclusion, but I'm not sure about the following details:

(1) Does the Hurewitz theorem (e.g. the theorem 4.32 in Hatcher: "Algebraic Topology") and the de Rham duality (e.g. the theorem 18.14 in Lee: "Introdution to Smooth Manifolds", 2nd ed.) hold for any type of manifold $M$ described above?

(2) I am aware of the Poincaré duality for the closed manifolds (e.g. the theorem 3.30 in Hatcher), the Poincaré-Lefschetz duality for the compact manifolds with boundary (e.g. the theorem 3.43 in Hatcher or the theorem 6.25 in Vick: "Homology Theory"), and the duality with the compactly supported cohomology (e.g. theorem 3.35 and the problem 35 on page 260 of Hatcher) for the noncompact manifolds. But, does the triviality of a cohomology group with the compact support, $H^k_c(M) = 0$, implies the triviality of the corresponding "ordinary" cohomology group, $H^k(M) = 0$, or we need some additional assumption/tool to conclude this?

• Consider $\mathbf{C}^2 \setminus 0$, which is topologically a $3$-sphere. – W. Cadegan-Schlieper Sep 2 '17 at 19:17

No. $\mathbb{R}^4 - \{ 0 \}$ obeys all your conditions, but it retracts onto the unit $3$-sphere, so $H^3(\mathbb{R}^4 - \{ 0 \}) \cong H^3(S^3) \cong \mathbb{R}$.
• @ David Speyer: Thanks for the very simple counterexample! But, if $H_1(\mathbb{R}^4 - \{0\}) \cong 0$, does this imply that at least $H^3_c(\mathbb{R}^4 - \{0\}) \cong 0$? – Ivica Smolić Sep 2 '17 at 19:30
• @abx: Yes, true... I was just hoping to find some additional assumption on the manifold that would force $H^3$ and $H^3_c$ to be isomorphic, or at least simultaneously trivial. – Ivica Smolić Sep 2 '17 at 19:41