There is no retraction of the solid torus $S^{1} \times D^{2}$ onto the torus $S^{1} \times S^{1}$.

Question

I'm trying to use the fact that I know that there is no retraction of $D^{2}$ onto $S^{1}$ (since there can be no injection $\pi_{1}(S^{1}) \rightarrow \pi_{1}(D^{2})$) to show that if there were a retraction $r:S^{1} \times D^{2} \rightarrow S^{1} \times D^{2}$ onto $S^{1} \times S^{1}$ then there would be a retraction of $D^{2}$ onto $S^{1}$ but am stuck.

Answer 1

If there were such a retraction then it would be a left inverse to the inclusion map. But there is no monomorphism $f:\mathbb{Z} \times \mathbb{Z} \cong \pi_1(S^1 \times S^1) \to \pi_1(S^1 \times D^2) \cong \mathbb{Z}$ because such a homomorphism is uniquely determined by $a = f(1,0)$ and $b = f(0,1)$, and $ab \in \mathbb{Z}$ equals both $f(b,0)$ and $f(0,a)$. Or, alternatively, every subgroup of $\mathbb{Z}$ is cyclic, but $\mathbb{Z} \times \mathbb{Z}$ is not.

Answer 2

One can also use the second homology group $H_2(; \mathbb{Z})$, since $S^1 \times D^2$ is homotopically equivalent to a circle. By functoriality, the map $(r \circ i)_*: H_2(S^1 \times S^1; \mathbb{Z}) \to H_2(S^1 \times S^1; \mathbb{Z})$ is the identity, but it factors through $H_2(S^1 \times D^2; \mathbb{Z})$ by $r_* \circ i_*$, and $H_2(S^1 \times D^2; \mathbb{Z})=0$, since $S^1 \times D^2$ is homotopically equivalent to a circle. This argument is very similar to one used to prove that the sphere is not a retract of the disk.