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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.

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    $\begingroup$ A similar open question (as far as I know) is if the orientation-preserving diffeomorphism classes of compact smooth $4$-manifolds, that is a monoid with the connect-sum operation. Are there any invertible elements other than $S^4$? Are there irreducible 4-manifolds? Are there "small" homotopy 4-spheres, i.e. the Schoenflies problem? $\endgroup$
    – Ryan Budney
    Commented Jun 21, 2021 at 20:23
  • $\begingroup$ Sorry, I meant a slightly different question. I changed it. Here is the old question: "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?". $\endgroup$
    – Victor
    Commented Jun 22, 2021 at 0:25

1 Answer 1

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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.)

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  • $\begingroup$ Do you mean that the fundamental group being $\mathbb Z$ implies that the complement is homotopy equivalent to $S^1$? I know it is not true in higher dimensions. $\endgroup$
    – Victor
    Commented Jun 21, 2021 at 20:42
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    $\begingroup$ Yes, that is true. It follows pretty readily from duality in the infinite cyclic cover (= the universal cover). $\endgroup$
    – Danny Ruberman
    Commented Jun 21, 2021 at 21:59
  • $\begingroup$ Sorry, I realized I meant a slightly different question. I changed it. It is quite obvious that $\tilde C_{K_1\cdot K_2}\simeq\tilde C_{K_1}\vee\tilde C_{K_2}$, where $\tilde C_K$ is the infinite cyclic cover of a knot complement. That's the standard argument that works in any dimension. $\endgroup$
    – Victor
    Commented Jun 22, 2021 at 0:22
  • $\begingroup$ Regarding the "I think" bit, as long as $G_1 *_\mathbb{Z} G_2$ means an actual amalgamated free product (i.e. the edge group $\mathbb{Z}$ injects into the vertex groups $G_i$), then this is true by Britton's lemma, which asserts that the vertex groups also embed into the total group. (See this MO question for further discussion of Britton's lemma: mathoverflow.net/q/343958/1463 .) $\endgroup$
    – HJRW
    Commented Jun 22, 2021 at 13:00
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    $\begingroup$ @HJRW Yes this is the case; the edge group injects into the $G_i$ since the inclusions are isomorphisms on $H_1$. This is implicit in Victor's comment above about the infinite cyclic cover. $\endgroup$
    – Danny Ruberman
    Commented Jun 22, 2021 at 13:17

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