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:
$A$ is not a retract of the Torus
$A$ 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?
Best Answer
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.