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 $S^n$](https://mathoverflow.net/questions/395745/unknotted-sn-2-in-sn).


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.*



Invertible 2-knots in $S^4$
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](https://mathoverflow.net/questions/395745/unknotted-sn-2-in-sn) 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.)


