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Ways to prove that $n$-component Brunnian link is nontrivial

The attached image shows a way to construct an $n$-component Brunnian link for any $n\geq 3$. That is, this link is not trivial, but deleting any of its components makes the new link trivial. The ...
Haldot's user avatar
  • 214
9 votes
1 answer
235 views

Links and non-orientable surfaces

Let $\Sigma \subset \mathbb{R}^3$ be a compact embedded surface with boundary $\partial \Sigma$ and $i:\Sigma\setminus \partial\Sigma \to \mathbb{R}^3 \setminus \partial\Sigma$ the inclusion. Is the ...
mmen's user avatar
  • 443
7 votes
2 answers
566 views

Fundamental group of the space of smooth embeddings of $S^1$ into $\mathbb R^3$

Has the fundamental group of the space of smooth embeddings of $S^1$ into $\mathbb R^3$ been completely computed? Say the basepoint is an unknot. Maybe something is known for other components? If yes,...
Ivan Sergeev's user avatar