It is well known that exist knots which are not slice, however I came up with the following argument contradicting this fact and I fail to see where the mistake is, can somebody help me?

Let $K_0$ and $K_1$ be two knots in $ S^3 = \partial D^4$. Since $K_0$ and $K_1$ are homotopic in $D^4$ they are isotopic in $D^4$. By the isotopy extension theorem, we can therefore find an ambient isotopy $$F:D^4\to D^4$$ sending $K_0$ to $K_1$.

Now suppose that $K_0$ is slice with slice disk $\Delta\subset D^4$, then we conclude that $K_1 = \partial F(\Delta)$. This would imply that every knot is slice.


1 Answer 1


Evidently your application of the isotopy extension theorem is flawed: every ambient isotopy $D^4 \to D^4$ restricts to an ambient isotopy $S^3 \to S^3$, so your same proof would show that any two knots in $S^3$ are ambient isotopic, which is false.

You did not include an actual statement of the isotopy extension theorem, nor any explanation of how you verified its hypotheses before applying its conclusion. I suspect that if you attempted to write that out then you would be able to identify more precisely the flaw in your logic.


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