Construct a homemorphism $\phi : T^2/A \rightarrow X/B $

Question

Construct a homemorphism $\phi : T^2/A \rightarrow X/B $

• $T^2=S^1 \times S^1$ and $A \subset T^2$ is given by $A=S^1 \times\{1\}$.
• $X=S^1 \times [-1, 1]$ and $B = S^1 \times\{-1, 1\}$.

$T^2/A$ is a torus with the circle collapsed to a point

I am not sure what the space $X/B$ looks like, some sort of deformation of the sphere?

How could I go about constructing a homeomorphism?

Thank you

Answer 1

Consider a two step construction. First, consider $C=S^1\times [-1,1]\to T^2$ that wraps the interval $[-1,1]$ inwards to obtain the torus. This is injective, except at the top and bottom of the cylinder, which is mapped to the inner circle $S^1$ in $T^2$. Call this map $f$. You can now compose $f$ with the projection to $T^2/S^1$. This passes to the quotient in the domain, to give a continuous bijection $f: C/\partial C \to T^2/S^1$. It suffices you now show these two spaces are compact Hausdorff.

If you're collapsing a circle in the "other" sphere, you can do them same but glue the cylinder as you'd always do to obtain a torus. This then passes to the quotient, and gives an homeomorphism, and proves all "four" spaces are homeomorphic.

(Visually, the homeomorphism between the spaces obtained by collapsing the different spheres turns the torus/cylinder inside out, if you may. You need more room to do this, but you can.)

Answer 2

These spaces are not homeomorphic as you have described them. $T^2/A$ is, as you say, the torus with a circle collapsed to a point (whose fundamental group is $\mathbb{Z}$), which is like a sphere with the two poles identified. However, $X/B$ is homeomorphic to a sphere, whose fundamental group is trivial. There exists a continuous map $X/B$ to $T^2/A$ which identifies the two "collapsed copies of $S^1$", but a continuous map in the opposite direction will be null-homotopic.