Homotopically trivial $2$-sphere on $3$-manifold 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 $\partial D^3=S^2$?
 A: Let $M$ be a closed $n$-manifold, $n>2$. Suppose $S = S^{n-1} \hookrightarrow M$ is a (locally flat) embedded, null-homotopic sphere. Then because it is null-homologous it must separate $M$ into two pieces, so that $M = M_1 \cup_S M_2$. To prove this, if $S$ is smoothly embedded, one can note that if $S$ did not separate $M$, picking points that locally look to be on "opposite sides" of $S$ and picking a path from one to the other in its complement, then joining this with the path that intersects $S$ once transversely, gives us a curve that intersects $S$ precisely once, hence has nonzero mod 2 intersection number with $S$, hence $S$ is homologically nontrivial. In general, still assuming that $S$ is locally flat, we have a tubular neighborhood and can apply the general Alexander duality to its complement. 
Claim: Either $M_1$ or $M_2$ has trivial fundamental group.
Suppose not. Then $\pi_1(M) = \pi_1(M_1) * \pi_1(M_2)$ is infinite. Pass to the universal cover $\tilde M$ and lift $S$ to it; again $S$ disconnects $\tilde M$, say into two pieces $N_1$ and $N_2$. Both $N_i$ are noncompact. 
Lemma: let $N$ be a non-compact $N$-manifold with boundary $S^{n-1}$. Then the inclusion $S^{n-1} \hookrightarrow N$ induces an injection on homology.
The long exact sequence in relative homology gives exactness of $H_n(N,S) \to H_{n-1}(S) \to H_{n-1}(N)$. Because $N$ is noncompact, $H_n(N,S) = 0$, and hence the lemma.
Corollary: The inclusion $S \hookrightarrow \tilde M$ induces an injection on homology.
First, note that $H_n(\tilde M) = 0$ (noncompactness). The Mayer-Vietoris sequence implies that the kernel of the map $H_{n-1}(N_1) \oplus H_{n-1}(N_2) \to H_{n-1}(\tilde M)$ is generated by $([i_1(S)], [i_2(S)])$, and the previous lemma implies that both of these homology classes are nonzero. So $[S] \in H_{n-1}(\tilde M)$, which is the image of $([i_1(S)],0)$ in the Mayer-Vietoris sequence, is nonzero.
If $S$ was null-homotopic in $M$, we could lift the disc realizing this null-homotopy to $\tilde M$, and hence it would be null-homologous. This gives a contradiction, proving the claim.
This proves the desired result in the case of 3-manifolds; for if $M_1$ is a compact manifold with sphere boundary and trivial fundamental group, we can cap off its boundary with a disc to get a 3-ball by the Poincare conjecture. Deleting this again shows us $M_1$ was a 3-ball itself.
A: 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.
