Prove homotopy equivalence of two spaces

Question

How can I prove that $ [S^{1} \times D^{2}]/S^{1} \times S^{1}$ is homotopy equivalent to $S^{2} \vee S^{3} $? So far I have proved that $S^{2} \vee S^{3} \cong S^{3}/S^{1} $ Additionaly applying that $D^{3}/S^{2} $ and $ S^{3} $ are homeomorphic may lead to a proper conclusion but I just can't see what's happening in $ S^{1} \times D^{2}$ after we squish $ S^{1} \times S^{1} $

Answer 1

Consider a solid torus $T$ standardly embedded inside $S^3$. If you contract in $S^3$ the boundary of $T$ and its exterior to a point, you get a space which is the same thing as the result of contracting the boundary of $T$ to a point, which is a way of describing $(S^1\times D^2)/(S^1\times S^1)$.

Since the closure of the complement of $T$ in $S^3$ is just another standardly embedded torus, this space is homeomorphic to $S^3/T$ and this has the same homotopy type as $S^3/S^1$ (just thicken the $S^1$ so that it becomes $T$) and you say that you know why $S^3/S^1$ has the homotopy type of $S^2\vee S^3$, so you are done.