Subspaces of the Torus homeomorphic to $S^1$

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

• 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. Commented Sep 24, 2019 at 17:54
• I totally thought $(2,1)$ was not a summand of $\mathbb{Z}^2$ which is wrong. Commented Sep 24, 2019 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.