This question is related to my previous question.
After knowing that $X \cong S^3$ and $A \cong S^1$, with $X/A \cong S^2$, I attempt to construct the long exact sequence of a pair. I need to use the exact sequence of a pair to find the relative homology groups $H_n(X,A)$.
$... \rightarrow \tilde{H}_n(S^1)\rightarrow \tilde{H}_n(S^3) \rightarrow \tilde{H}_n(S^2)\rightarrow...\rightarrow \tilde{H}_3(S^1)\rightarrow\tilde{H}_3(S^3)\rightarrow\tilde{H}_3(S^2)\rightarrow\tilde{H}_2(S^1)\rightarrow...$
There seems to be something really fishy here. I know that $\tilde{H}_3(S^3) \cong\mathbb{Z}$, but $\tilde{H}_3(S^2)=0$! How can this be? Where have I gone wrong?
There is no such sequence with reduced homology $\widetilde H$. But there is a long exact sequence of homotopy groups $\pi$. And, indeed, $\pi_3(S^2)=\mathbb{Z}$. It sounds like you are confusing the long exact homology sequence of a pair of spaces $A \subset X$ with the long exact homotopy sequence of a fibration.