Separating spheres in $3$-manifolds of positive scalar curvature and mean convex boundary Recently, A. Carlotto and C. Li proved a complete topological classification of those compact, connected and orientable $3$-manifolds with boundary which support Riemannian metrics of positive scalar curvature and mean-convex boundary. Namely, if $M^3$ is such a manifold, then there exist integers $A, B, C, D \geq 0$ such that $M$ is diffeomorphic to a connected sum of the form
\begin{align*}
P_{\gamma_1} \# \cdots \# P_{\gamma_A} \# \mathbb{S}^3/ {\Gamma_1} \# \cdots \# \mathbb{S}^3 / {\Gamma_B} \# \left(  \#_{i=1}^C \mathbb{S}^2 \times \mathbb{S}^1 \right) \setminus \left( \sqcup_{i=1}^D B_i^3 \right),
\end{align*}
where $P_{\gamma_i}$, $i \leq A$, are genus $\gamma_i$ handebodies; $\Gamma_i$, $i \leq B$, are finite subgroups of $SO(4)$ acting freely on $\mathbb{S}^3$, $B_i^3$, $i \leq D$, are disjoint $3$-balls in the interior.
My question: Can we classify, in terms of $A,B, C, D$, the $3$-manifolds $M$ of the form above in which any smoothly embedded $2$-sphere in the interior separates $M$?
For instance, if $(A,B,C,D) = (1, 0, 0, 0)$, then this holds. Indeed, if $M = P_{\gamma_1}$, then $H_2(M) = 0$, so the connecting homomorphism $H_2(M, \partial M;\mathbb{Z}) \to H_1(\partial M; \mathbb{Z})$ is injective. Since an embedded $2$-sphere has no boundary, it lies in the kernel of this map, and thus equals to $0$ in $H_2(M, \partial M; \mathbb{Z})$. This means it separates $M$.
 A: Every embedded 2-sphere will separate $M$ if and only if $C=0$.
Proof: Suppose $C=0$, and let $\Sigma$ be 2-sphere embedded in $M$. Let $\{S_j\}$ be a collection of 2-spheres which decompose $M$ into prime summands. Look at the intersection of $\Sigma$ with $\{S_j\}$. Let $\Delta$ be an innermost disk on some $S_k$. We can surger $\Sigma$ along $\Delta$, which will decompose $\Sigma$ into two 2-spheres. Repeating this process until there are no intersections with $\{S_j\}$, $\Sigma$ is decomposed into a collection of embedded $2$-spheres which we call $\Sigma'$. Then $\Sigma$ will be non-separating in $M$ if and only if some component of $\Sigma'$ is non-separating in $M$.
Each component of $\Sigma'$ is contained within a single prime summand of $M$. We then cut $M$ along $\{S_j\}$, and glue $3$-balls onto each $2$-sphere boundary component of the resulting $3$-manifolds. It is well-known that handlebodies and closed spherical $3$-manifolds are irreducible, which means that every embedded $2$-sphere bounds a $3$-ball. Thus each component of $\Sigma'$ is separating in its respective prime summand and hence in $M$. Removing the $D$ $3$-balls from $M$ which are disjoint from $\Sigma$ and $\Sigma'$ does not affect whether $\Sigma$ is separating. Therefore $\Sigma$ is separating in $M$.
Conversely, if $C\neq 0$, then we can find a non-separating 2-sphere $\Sigma''$ in some $S^2\times S^1$ component which is disjoint from each $B_i$ and each $S_j$. Furthermore the dual $1$-sphere to $\Sigma''$ is disjoint from each $B_i$ and each $S_j$. Therefore $\Sigma''$ is non-separating in $M$.
