Take a self-intersection one $S^2$ in $\mathbb CP^2$ -- it appears to me that several people have made the observation in this thread that the complement is simply-connected. If you don't see this right away then a nice way to see this is that the unit normal bundle is $S^3$ and the Hopf fibration $S^3 \to S^2$ then gives a relator that kills the $S^2$-linking elements of $\pi_1$ but these generate $\pi_1 (\mathbb CP^2 \setminus S^2)$ (generalized Wirthinger presentation). So Poincare/Alexander duality tells you $\mathbb CP^2 \setminus S^2$ is contractible. So "blowing-down" on this embedded $S^2$ basically amounts to a twisted sphere construction -- gluing two discs $D^4$ together along their common boundary. There's the issue of whether or not one of the discs has an exotic smooth structure or not, but that seems to be that's the only issue.
So I guess there's a question lurking in this -- is there a "natural" way to unknot a self-intersection one embedded $S^2$ in $\mathbb CP^2$? If there were, you couldn't produce any interesting $S^4$'s via this construction.
Take a self-intersection one $S^2$ in $\mathbb CP^2$ -- it appears to me that several people have made the observation in this thread that the complement is simply-connected. If you don't see this right away then a nice way to see this is that the unit normal bundle is $S^3$ and the Hopf fibration $S^3 \to S^2$ then gives a relator that kills the $S^2$-linking elements of $\pi_1$ but these generate $\pi_1 (\mathbb CP^2 \setminus S^2)$ (generalized Wirthinger presentation). So Poincare/Alexander duality tells you $\mathbb CP^2 \setminus S^2$ is contractible. So "blowing-down" on this embedded $S^2$ basically amounts to a twisted sphere construction -- gluing two discs $D^4$ together along their common boundary. There's the issue of whether or not one of the discs has an exotic smooth structure or not, but that seems to be the only issue?