I'm an undergraduate student currently studying Algebraic topology. I've been struggling to find all relative homotopy groups $\pi_k (S^n, S^1)$ for $n\geq 3$, $k\leq n$.
Here are my thoughts: If either $S^n$ or $S^1$ were contractible, this would be easy to solve. However, they are not. The only other thing that comes to mind is to check if $S^1$ is a strong deformation retract for $S^n$ because then $\pi_n(S^n, S^1)=1, n\geq1$. But I suppose $S^1$ is not a strong deformation retract for $S^n$? (At least I wouldn't know how to prove that it is.)
Also, since $(S^n, S^1)$ is a CW pair maybe there is a way to approach this issue using the long exact sequence, but I'm not sure how to go about it. Any ideas?
You may consider the homotopy exact sequence $\cdots\to \pi_k(A,x_0)\xrightarrow{i_*}\pi_k(X,x_0)\to \pi_k(X,A,x_0)\xrightarrow{\partial_n}\pi_{k-1}(A,x_0)\to\cdots\to \pi_0(X,x_0)$.
And $\pi_k(\Bbb S^1)=0$ for $k\geq 2$ due to universal covering of $\Bbb S^1$ is contractible.
So the problem reduces to compute $\pi_k(\Bbb S^n)$. One can show, using simplicial approximation theorem that, $\pi_n(\Bbb S^n)=\Bbb Z$ and $\pi_k(\Bbb S^n)=0$ for $k<n$. But, $\pi_m(\Bbb S^n)\not=0$ for $m>n$. For example, there is a map $\Bbb S^3\to \Bbb S^2$ which not homotopic to constant, called Hopf map. For further calculation of higher homotopy group a useful tool is Spectral Sequence, but it depends on homology theory.