# Pre-images of Seifert surfaces are incompressible?

Consider a knot $$K \subset S^3$$ and let $$M_K$$ be the associated double branched cover. The pre-image $$S$$ of a Seifert surface is a surface without boundary inside $$M_K$$.

Can $$S$$ be incompressible? If yes, how is this related to the topology of $$K$$?

I'm particularly interested in the case where $$M_K$$ is hyperbolic.

The incompressible Seifert surface $$\Sigma$$ will have incompressible preimage $$S$$ in the double branched cover if and only if the complement of a tubular neighborhood of $$\Sigma$$ in $$S^3$$ has incompressible boundary. This condition was termed "partially unknotted" by Jaco.
Now, draw a knot on this handlebody which bounds a genus 1 Seifert surface $$\Sigma$$, using the fact that for a genus one surface $$\Sigma$$, $$\Sigma\times [0,1]$$ is a genus 2 handlebody. Any such knot will have a Seifert surface which is not partially unknotted.