Consider the disk $D^2$ and its boundary $ \partial D^2 = S^1$. Then $(D^2, S^1 )$ is a good pair. This means there is a boundary map induced
$ \tilde{H}_2(D^2 / S^1 ) \xrightarrow{\ \partial \ } \tilde{H}_1 (S^1)$
But we have $D^2 / S^1 \cong S^2$. We also have that $\tilde H_2 (S^2) \cong \mathbb Z$ and $\tilde H _1 (S^1) \cong \mathbb Z $. So the above boundary map corresponds to a map $ \mathbb Z \to \mathbb Z$. I would very much like this map to be multiplication by $2$. Is this true?
In my current course we do not give a proof of what the boundary map $\partial$ is or how to construct it; we just acknowledge that it exists. Is it possible to show that it corresponds to multiplication by $2$ in a simple way?