Why is $S^1\times \{1\}$ homotopy equivalent to the solid torus $T^2 = D^2 \times S^1$? (see attached picture) I am currently self-studying the basics of algebraic topology and i just learned the definitions of retract, deformationretract and homotopy equivalence. 
Now in my book there is an example of a subspace $S^1 \times \{1\}$ of a topological space $T^2 = D^2 \times S^1$ that is supposed to be homotopy equivalent to the solid torus $D^2\times S^1$ but not even a retract.

However, i fail to see why $A_2 = S^1 \times \{1\}$ is homotopy equivalent to the solid torus $D^2 \times S^1$
Would someone be so kind and help me visualize it? What am i not seeing?
I'd really appreciate a geometrical hint. Thank you very much for any of your help!
 A: The space $A_2 = S^1 \times \{1\}$, considered as a space on its own, is just a circle. A circle is indeed homotopy equivalent to $T^2_s = S^1 \times D^2$; as you said yourself, the solid torus deformation retracts onto its core $A_1 = \{1\} \times S^1$. But the key here is that the homotopy equivalence $A_2 \to T^2_s$ is not the inclusion; rather, you should take for example the composition of the homeomorphism $A_2 \cong A_1$ together with the inclusion $A_1 \subset T^2_s$.
The inclusion $A_2 \to T^2_s$ is not a homotopy equivalence, and in fact it is not even a retract, as your book says. Indeed, the inclusion is homotopic to a constant map, so it induces the trivial map on fundamental groups, for example.
If there is a moral to this story: spaces are important, but maps are even more important.
A: Deflating the tyre $D^2 \times \mathbb{S}^1 \rightarrow \{ 1 \} \times \mathbb{S}^1$ is a homotopy equivalence, so $D^2 \times \mathbb{S}^1 \simeq \{ 1 \} \times \mathbb{S}^1$. I'll leave you to check this.
Then there is an obvious homeomorphism $\{ 1 \} \times \mathbb{S}^1 \cong \mathbb{S}^1 \times \{ 1 \}$. In particular this means $\{ 1 \} \times \mathbb{S}^1 \simeq \mathbb{S}^1 \times \{ 1 \}$
Therefore, $D^2 \times \mathbb{S}^1 \simeq \mathbb{S}^1 \times \{ 1 \}$.
