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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?

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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.

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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.

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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 $\theta \rightarrow (2\theta,\theta)$.

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  • $\begingroup$ 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
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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.

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