Let $F_1, F_2 \subset S^4$ be two smoothly embedded disjoint closed orientable surfaces. Assume that $H_1(F_1; \mathbb{Z}) \to H_1(S^4 - F_2; \mathbb{Z})$ is the 0 map - does it follow that $F_2$ bounds a Seifert solid that is disjoint from $F_1$.
For context, this result holds down a dimension for pairs of knots in $S^3$. If this is true, I would be interested to hear if it also holds in $n$-dimensions for disjoint orientable compact $F_1^{n-2},F_2^{n-2}$ in $S^n$.
Edit: I figured this out in the case of surfaces in $S^4$ - although I am still interested in the corresponding result for say pairs of 3-manifolds in $S^5$ - where say $H_1(M_1; \mathbb{Z}) \to H_1(S^5 - M_2; \mathbb{Z})$ is trivial.
For the case of surfaces, we can pick any Seifert solid $M^3$ that $F_2$ bounds and we can look at how it intersects the surface $F_1$. Think of the usual cell decomposition of $F_1$ with one 2-cell. By assumption on the map on $H_1$ we know that each of the 2-cells intersects $M^3$ in a bunch of signed points and that the sum of these signs is 0. Then by changing $M^3$ by tubing it to itself along arc in this circle on $F_1$ between + and - points, we can change $M$ so that it does not intersect the 1-skeleton (note: this is exactly what we do to get the result a dimension down). Now $F_1$ intersects this new 3-manifold in a bunch of circles in the 2-cell, but we can surger the new $M$ along these (starting with the innermost) to produce a Seifert surface for $F_2$ that is disjoint from $F_1$.