TITLE: Invertible 2-knots in $S^4$
QUESTION [6 upvotes]: Q1: Is it true that a knot $S^2\hookrightarrow S^4$ has an inverse iff it is trivial? Or it is also an open question?
See relatedly Unknotted $S^{n-2}$ in $S^n$.
Q2: It is easy to see that if a knot $f\colon S^2\hookrightarrow S^4$ has an inverse than its complement $C_f\simeq S^1$. Has the converse been proved?
Both questions are answered below by Daniel Ruberman.

REPLY [7 votes]: Q1: This is true in the topological category and unknown in the smooth setting. In the topological setting, the fundamental group of $S^4 - K_1 \# K_2$ is $G_1 *_\mathbb{Z} G_2$ where $G_i$ are the fundamental groups of $S^4 -K_i$. If this is $\mathbb{Z}$ then I think the $G_i$ are both $\mathbb{Z}$. By the arguments in your earlier question this means that both $K_i$ are unknotted.
Q2: It is also true in the topological category and unknown in the smooth category that the complement being a homotopy circle implies that the knot in question has an inverse. The argument in the topological case is that the knot is trivial (since the group is $\mathbb{Z}$ as noted previously.)