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Say I have a pair of links $L_0,L_1\subset S^3$ and an embedding $F:L_0\times I \rightarrow S^n$ such that $F(L_0,0) = L_0$ and $F(L_0,1)=L_1$ ($F$ is a concordance). Intuitively, the complements $S^3\setminus L_i$ should be homotopy equivalent because $F$ is an embedding, so throughout the concordance, the homotopy type of the complement is unchanged. However, a concordance doesn't seem to give me a homotopy equivalence in the usual way one would expect to find one.

Basically, Im trying to prove something seemingly basic (also seemingly true, perhaps it isnt), but aren't sure where to start, do you have any hints?

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up vote 2 down vote accepted

No, this is not true. The fundamental group of the knot complement is not invariant under concordance. For example, the square knot is a slice knot (i.e., concordant to the unknot), yet the fundamental group of its complement is not $\mathbb Z$.

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Is there a canonical counter-example worth working out in detail? – AnonymousCoward Feb 18 '13 at 18:31
@AnonymousCoward I edited an example into my answer. – user53153 Feb 18 '13 at 18:43

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