Is $S^1$ a retract of the torus $T$?

Question

I'm having trouble deciding whether the circumference $S^1$ is a retract of the torus $T$ or not.

At first, I tried to arrive to a contradiction by working with their fundamental groups, but that's not possible (I can't see how, at least).

In the second try, I looked for an explicit retraction. In order to do so, I saw $T$ as a cylinder $C$ with its border properly identified, and retracted $C$ to its middle $S^1$. The problem is, is that map continuous at all? Since I cut $T$ in order to construct it, I'm not sure at all. The thing is both parts of the border of $C$ end up going to the same points, so I think it should be. Would any confirm or disprove my thoughts?

Thanks in advance.

(PS: This is my first post in the page. I'm happy to be now part of this community, which has helped me to pass my math tests over the past 4 years.)

Answer 1

You can represent $T$ as $S^1\times S^1$. Pick $a_0\in S_1$ and then the map $F$ taking $(x,y)\in S^1\times S^1$ to $(a_0,y)$ is a retraction of $S^1\times S^1$ to $\{a_0\}\times S^1$, which of course is a copy of $S^1$ within the torus.

OK, this retraction is not a deformation retraction....