I'm wondering if we can describe the effect of quotienting by some subgroup of a homotopy group by actually replacing the space with the mapping cone of the generators of the subgroups.
To be more precise, take $X$ say a $CW$-complex and $\phi$ some map $\mathbb{S}^k\to X$ representing a non zero element in $\pi_k(X)$.
If you look at $cone(\phi)$ then certainly you obtain a surjective map $j_\star:\pi_k(X)\to \pi_k(cone(\phi))$ given by the fact that $(cone(\phi),X)$ is $k$-connected, and of course almost by definition $j_\star[\phi]=0$. However this is not sufficient to conclude that $\pi_k(cone(\phi))=\pi_k(X)/<\phi>$ and i have the feeling that it won't be true in general.
Am I missing something here? Is there an example when $\pi_k$ of the cone will actually be smaller than the quotient by the subgroup generated by the map?
Edit: Here is a proof that it actually works when $X$ is $1$-connected.
If $X$ is one connected, as $(cone(\phi), X)$ is $k$-connected we know, by homotopy excision, that $\pi_j(cone(\phi), X)\to \pi_j(cone(\phi)/ X, \star)$ is an isomorphism for $j=2,...,k+1$. But $cone(\phi)/ X\simeq \mathbb{S}^{k+1}$ and thus $\pi_{k+1}(cone(\phi), X)=\mathbb{Z}$ and is generated by $\phi$. We thus get an exact sequence $$\mathbb{Z}\to \pi_k(cone(\phi))\to \pi_k(X)\to 0$$ where the left map maps $1$ to $[\phi]$ giving us the desire isomorphism.
So now all that is left is finding a counter example when $X$ isn't $1$-connected, but I have the feeling that it shouldn't be too complicated.