So, I'm being asked the following:

Is there a subspace $A \subset S^1 \times S^1$ such that $A \cong S^1$ and has the stated property:

  1. $A$ is not a retract of the Torus

  2. $A$ is a deformation retract of the Torus

For the first property, I'm tempted to say no, but I'm not sure. I know that if we fix $m \in S^1$, the subspace $S^1 \times \{m\}$ is homeomorphic to $S^1$ and is a retract of the Torus via the mapping $f: S^1 \times S^1 \to S^1 \times \{m\}$ given by $f(x,y) = (x,m)$. So, if a subspace $A$ of the Torus, is homemorphic to $S^1$, we would necessarily need to have $S^1 \times \{m\} \cong A$ via some homemorphism $h: S^1 \times \{m\} \to A$. I was trying to show that $h \circ f: S^1\times S^1 \to A$ is a retraction, but I have thus far been unable to show that the restriction of $h \circ f$ to $A$ is the identity on $A$. Now, I'm thinking that perhaps there may be subspace $A$ satisfying property 1, but I have no idea how to go about constructing it.

Intuitively, for property 2, I'm also thinking no. My intuition relies on the idea that a deformation retraction $g_t$ of the torus onto a subspace $A$ would inevitably allow us to obtain a deformation retraction $k \circ g_t$ (where $k: A \to S^1 \times \{m\}$ is a homeomorphism) of the torus onto $S^1 \times \{m\}$. But, then, we have essentially deformation retracted one of the circles onto a point, which is a contradiction. Is this intuition correct?


Regarding your attempts at property 1, your explorations regarding $S^1 \times \{m\}$ look interesting, using the retraction $f : S^1 \times S^1 \to S^1 \times \{m\}$. But instead of postcomposing $f$ by a homeomorphism between $S^1 \times m \mapsto A$, you should have considered conjugating $f$ by a homeomorphism of $S^1 \times S^1$: if $g : S^1 \times S^1 \to S^1 \times S^1$ is any homeomorphism, then $g^{-1} \circ f \circ g$ is a retraction from $S^1 \times S^1$ onto the circle $g^{-1}(S^1 \times \{m\})$.

From that, perhaps you can leap to the following guess: if an embedded circle $A \subset S^1 \times S^1$ is a retract then $(S^1 \times S^1) - A$ is connected.

This gives a hint to a counterexample: look for a circle $A \subset S^1 \times S^1$ such that $(S^1 \times S^1) - A$ is disconnected, and prove that $A$ is not a retract.

Regarding property 2, here are a couple of things you may know: every deformation retraction is a homotopy equivalence; and every homotopy equivalence induces an isomorphism on the fundamental group. If you can compute the fundamental groups of the torus and of the circle, I think you'll be able to address property 2.


Let $X = S^1 \times S^1$.

Taking $A = S^1 \times \{m\}$, as you say, $A$ is homeomorphic to $S^1$ via the map $(x,m) \mapsto x$. The map $$r\colon X \to A \colon (x,y) \mapsto (x,m)$$ is a retract as, if we take the map $\iota \colon A \hookrightarrow X$ to be given by $(x,m) \mapsto (x,m)$, then we clearly have $r \circ \iota \colon A \to A$ is the map $(x,m) \mapsto (x,m) \mapsto (x,m)$ which is the identity on $A$.

To see that no subspace $S^1 \cong A \subset X$ is a deformation retract of $X$, note that a deformation retraction is necessarily a homotopy equivalence. As such, if such a deformation retraction existed, then we would have $A \simeq X $ and so $\pi_1(A) \cong \pi_1(X)$. However, $X$ has fundamental group $\mathbb{Z}^2$ and $A$ (being homoemorphic to a circle $S^1$) has fundamental group $\mathbb{Z}$, which is not isomorphic to $\mathbb{Z}^2$.

Oops I misread part 1 as saying $A$ is a retract of $X$. Take $A$ to be a circle which is contained in a small open disk in $X$. That is, $A$ represents a null-homotopic loop. In particular, $\iota^* \colon \pi_1(A) \to \pi_1(X)$ is the trivial map. However, the inclusion map for any retract must induce an inejctive homomorphism on fundamental groups, and so $A$ cannot be a retract.


For part 1:

If you can find a circle such that its inclusion induces the map $\mathbb{Z} \xrightarrow{(2,1)} \mathbb{Z}\oplus \mathbb{Z}$ on fundamental groups, then you are done since if this circle were a retract it would imply that there is a retraction of $\mathbb{Z}\oplus \mathbb{Z}$ onto $\mathbb{2Z}\oplus \mathbb{Z}$ which is not possible.

Such a circle is given by $\theta \rightarrow (2\theta,\theta)$.

  • $\begingroup$ I don't believe your prescribed circle works, because there is a toral automorphism taking that circle to a meridinal circle. What you really need is a null-homotopic subspace which is homeomorphic to the circle. $\endgroup$ – Dan Rust Sep 24 '19 at 17:54
  • $\begingroup$ I totally thought $(2,1)$ was not a summand of $\mathbb{Z}^2$ which is wrong. $\endgroup$ – Connor Malin Sep 24 '19 at 18:11

I include this just because it does not fit in a comment. Maybe it scapes the scope of the question, but I believe it is interesting information. One can characterize de Jordan curves which are retracts of the torus: they are those that do not disconnect the torus, and all can be taken onto a meridian $S^1\times\{m\}$ by a homeomorphism of the torus.

This comes from some results in the beautiful book on knots by Rolfsen. From page 8 to 26 the autor presents a minute analysis of Jordan curves in the torus. It is a bit terse but a fantastic read if guided by someone who helps a bit. I stress what is relevant here:

(i) Two Jordan curves in the torus are taken one onto another by a homeomorphism of the torus if and only if either both disconnect or both do not disconnect the torus, and

(ii) A Jordan curve disconnects the torus if and only if it is nulhomotopic.

Now, if a Jordan curve $A\subset S^1\times S^1$ is a retract, then the inclusion induced in fundamental groups is injective, hence not zero, and so $A$ is not nulhomotopic. Consequently (ii) says $A$ does not disconnect, and since meridians do not disconnect either, (i) says that $A$ can be taken onto a meridian $S^1\times\{m\}$ by some homeomorphism of the torus. The converse is clear, as homeomorphisms preserve all properties involved.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.