A 3-manifold $M$ is *irreducible* if every embedded 2-sphere bounds a 3-ball. Thanks to Papakyriakopoulos's sphere theorem, irreducibility is the same as having $\pi\_2(M)=0$. Does irreduciblity imply that the manifold is in fact aspherical, i.e. that $\pi\_k(M)=0$ for all $k \geq 2$?


(Or maybe I should say that the universal cover $\tilde M$ is aspherical, but the question is the same in terms of homotopy groups.)


As pointed out by @Matt Zaremsky in the comments, there is an obvious counterexample in $S^3$. But perhaps this is the only counterexample, or the counterexamples are easy to classify?


Given all of the tools we have about geometric classification of 3-manifolds, I expect someone would have a quick answer. I'm just not enough of an expert to make those arguments myself.



Higher homotopy groups of irreducible 3-manifolds
An irreducible 3-manifold $M$ is aspherical if and only if it's not a finite quotient of $S^3$, which in turn is equivalent to having infinite fundamental group. Essentially you've already outlined the proof: the universal cover $\tilde M$ is a simply-connected 3-manifold with trivial $\pi\_2$, and so also $H\_2(\tilde M) = 0$; if $\tilde M$ is not compact, then $H\_3(\tilde M) = 0$ (because of non-compactness) and $H\_k(\tilde M) = 0$ for higher $k$ (because it's a 3-manifold), so by the Hurewicz theorem $\tilde M$ is aspherical.


On the other hand, $\tilde M$ is compact if and only if $\pi\_1(M)$ is finite. In this case, Perelman proved that $\tilde M$ is $S^3$. 3-manifolds covered by $S^3$ are classified, and they correspond to finite subgroups of $SO(4)$. I think that Scott's *The geometries of 3-manifolds* has a precise statement.


