TITLE: Homotopically trivial $2$-sphere on $3$-manifold
QUESTION [9 upvotes]: Let $S^2$ be an embedded sphere in a $3$-manifold $M^3$ such that $[S^2]$ is trivial in $\pi_2(M)$. Can we find an embedded disk $D^3$ in $M$ such that $\partial D^3=S^2$?

REPLY [6 votes]: EDIT: This answer appears to reprove about half of this note by Danny Ruberman with a bit more detail. After proving the results below, Ruberman provides manifolds that do satisfy these requirements, such that the connected sum sphere is null-homotopic - but neither side is a ball. The construction, approximately, is to take maps $S^n \to M_1$ and $S^n \to M_2$ of positive degree - which he demonstrates are possible - and "connect sum them" in an appropriate sense to get a map $B^n \to M_1 \# M_2$ with $\partial B^n$ mapping to the connected sum sphere. (It's a fun construction to read.) An explicit example is $(SU(3)/SO(3)) \# (SU(3)/SO(3))$.

$\require{AMScd}$One can say more than in the first answer. Again $M$ is an $n$-manifold with an embedded null-homotopic sphere of dimension $n-1$. First, let $S$ be the embedded sphere, realizing $M$ as a connected sum $M_1 \# M_2$. (I have changed the notation a bit from what I was doing above.) Suppose $S$ is null-homotopic. Then we get a map $f: D^n \to M$ realizing this null-homotopy; by projecting onto $M_1$ and $M_2$ we get maps $f_i: S^n \to M_i$. We want to get some control on the degrees of these maps. From the map $(M_1 \# M_2, S) \to (M_1 \vee M_2, *)$ given by collapsing the sphere we get a commutative diagram 
$$\begin{CD}
H_n(M) @>>> H_n(M,S) @>>> H_{n-1}(S) \\
@VVV @VVV @VVV\\
H_n(M_1) \oplus H_n(M_2) @>\cong>> H_n(M_1) \oplus H_n(M_2) @>>> 0\\
\end{CD}$$
The second vertical map is an isomorphism; the first map sends $[M]$ to $([M_1],[M_2])$. We can thus identify that in $H_{n-1}(S)$, $\partial[M_i] \neq 0$ and $\partial[M_1] + \partial [M_2] = 0.$ Indeed $\partial [M_1] = 1$ because the map $H_{n-1}(S) \to H_{n-1}(M)$ is zero. Now looking at the induced map $H_n(D^n,S) \to H_n(M,S)$, and collapsing $S$ (denoting the generator of $H_n(S^n)$ by $x$) we can identify the image of the generator of $H_n(D^n,S)$ in $H_n(M,S)$ with $(f_1)_*(x) + (f_2)_*(x)$; that is, with $(\deg f_1)[M_1] + (\deg f_2)[M_2]$. Following the two different directions of 
$$\begin{CD}
H_n(D^n,S) @>>> H_{n-1}(S) \\
@VVV @VVV\\
H_n(M,S) @>>> H_{n-1}(S)\end{CD}$$
we see that $\deg f_1 - \deg f_2 = 1$.
Now using the pairing given by Poincare duality and the induced map $H^*(M_i) \to H^*(S^n)$ we see that if $f_i: S^n \to M_i$ has nonzero degree, $H^*(M_i)$ must be a rational homology sphere. If one of the maps has degree zero, the other has degree $\pm 1$, and thus by the same argument with $\Bbb Z/p$ coefficients we realize that $H^*(M_i) \cong H^*(S^n)$. Because the map factors through the universal cover of $M_i$, but the degree is one, $\pi_1(M_i) = 0$, and hence by Whitehead $M_i$ is homotopy equivalent to $S^n$, and by the topological Poincare conjecture $M_i$ is homeomorphic to $S^n$. Thus $S$ bounded a ball on that side.
Therefore if $S$ bounds a ball on neither side, the degree of both $f_i$s are nonzero, and hence both $M_i$s are rational homology spheres. One of them has to be simply connected by my first answer.
Because simply connected rational homology 4-spheres are 4-spheres, this plus the original argument implies the desired result for 4-manifolds. I don't know yet if there is a counterexample for 5-manifolds.