# 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!

• deflating the tire gets me $A_1 = \{1\}\times S^1$ (strong deformation retract and thus homotopy equivalent) but i dont see how to get my tire to become $A_2 = S^1 \times \{1\}$ – Zest Mar 11 '19 at 16:18
• @NajibIdrissi you are correct. – Randall Mar 11 '19 at 17:38
• Apologies to @Zest for not reading closely enough. It's a good question. – Randall Mar 11 '19 at 19:56
• i still appreciate your help @Randall. Thank you very much. – Zest Mar 12 '19 at 1:29

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.

• Hello Najib. Thank you very much for your help. How exactly do i need to imagine the space $S^1 \times \{1\}$ in the first place? I always imagined it to be simply $S^1$ but with the confusion about the current topic i started questioning my entire imagination. – Zest Mar 11 '19 at 16:32
• @Zest Yes, $S^1 \times \{1\}$ is just a circle $S^1$. But there are several circles embedded in $S^1 \times D^2$, and the way they are embedded matters a lot. – Najib Idrissi Mar 11 '19 at 17:22
• Hello Najib. I just got back to this topic. What's the usual way too see a map such as the inclusion $A_2 \to T_s^2$ is not a homotopy equivalence? (i.e. has no homotopy inverse) - does it immediately follow from the induced homomorphism of the fundamental groups? Or how did you know? And why is the inclusion homotopic to a constant map? – Zest May 1 '19 at 19:47

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, $$D^2 \times \mathbb{S}^1 \simeq \mathbb{S}^1 \times \{ 1 \}$$.

• that's a very helpful insight! thank you. your first paragraph simply follows from $\{1\} \subset D^2$ being a deformation retract, doesn't it? The rest made sense immediately. Thanks again. – Zest Mar 11 '19 at 16:35
• Yes. As Nijab says in his answer, the point the book is trying to make is that the inclusion $D^2 \times \mathbb{S}^1 \rightarrow \mathbb{S}^1 \times \{ 1 \}$ is not a retract, but the two spaces are homotopy equivalent. – Joseph Martin Mar 11 '19 at 16:40