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 latter property is obvious from the picture, however I would like to have a strict proof of nontriviality. What bothers me is that *nowhere* in the literature there is an example of proving nontriviality of any Brunnian $n$-link.

Let us denote this link by $L=K_1\cup K_2 \cup \dotsb \cup K_{n-1}\cup K_n$. Also, it might be useful to examine $L'=L\setminus K_n=K_1\cup K_2 \cup \dotsb \cup K_{n-1}$. I tried to look at the link group $\pi_1(L').$ Since $L'$ is equivalent to $n-1$ disjoint circles, I conclude that $\pi_1(L')=F_{n-1}$. Then I ask what $[K_n]\in \pi_1(L')$ would be in the case if $L$ were trivial. In that case I would be able to move $K_n$ a little bit and get a homotopic loop that is a clean circle, thus $[K_n]\in \pi_1(L')$ is the identity. Now, for contradiction I would like to calculate explicit form of $[K_n]\in \pi_1(L')$ (or to somehow just prove that it is not the identity). I tried doing it with Wirtinger presentation, but it was messy and, even if I did find explicit form for $[K_n]\in \pi_1(L')$ in terms of group’s generators correctly, determining whether $[K_n]$ is equal to the identity in this finitely generated group is a well-known undecidable problem.

Is there a way to improve my approach in order to prove nontriviality? If not, do you see another way to prove that? Actually, just an example of proof of nontriviality for any $n$-component Brunnian link would be enough for me, chances are the idea would work for many Brunnian links if not for all.

arbitraryfinitely presented group, but is easily decidable in afreegroup, which is your situation. You do this by writing your conjugacy class as a cyclically reduced word. $\endgroup$2more comments