I stumbled upon the attached image. It 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. I have not managed to come up with such a proof, although I do have some kind of a plan.
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 look at the link group (i.e., fundamental group of the link's complement) $\pi_1(L').$ 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, 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?
Thank you.
One thing we can take advantage of is the $n$-fold rotational symmetry about an axis that passes through the center of the link. Let's say $L\subseteq S^3$ is our Brunnian link (or $L\subseteq \mathbb{R}^3$ if you wish). The quotient of $S^3$ by this rotational symmetry is still $S^3$, and the image of $L$ through this quotient is the following knot $K\subseteq S^3$:
By putting the basepoint along the axis of rotation, one can see that the induced map $\pi_1(S^3\setminus L)\to \pi_1(S^3\setminus K)$ is surjective: given a loop in $S^3\setminus K$, lift the path up to one of the $n$ lifts in $S^3\setminus L$, then notice you get a loop. (The quotient map $S^3\setminus L\to S^3\setminus K$, by the way, is an example of a branched cover, where the image of the axis of rotation is a branch/singular locus of order $n$.)
Hence, $L$ is nontrivial if $K$ is nontrivial. One can identify $K$ as a connect sum of two trefoil knots, and hence $L$ is nontrivial.
You might also want to see that $L$ is non-split -- I think this follows from the fact that it's Brunnian.
This is true for the general word problem, but the word problem for 3-manifolds is decidable.