3

Let $I^2$ be the square $\{(x, y) ∈ \Bbb R^2: 0 \leq x, y \leq 1\}$

$C$ be the circle $\{(x, y) \in \Bbb R^2: 1 \leq x^2 + y^2 \leq 4\}$, regarded as subspaces of $\Bbb R^2$ in the usual topology.

Let us have equivalence relations $∼$ and $≈$ on $I^2$ and $C$ respectively by

$(x, y) ∼ (x, y)\ \forall (x, y) \in I^2$, $(0, y) ∼ (1, y)$ and $(x, 0) ∼ (x, 1)$ if $0 \leq x, y \leq 1$

$(x, y) ≈ (x, y)\ \forall(x, y) \in C,\ (x, y) ≈ (2x, 2y)$ if $x^2 + y^2 = 1$.

How to show that $[S^2]_∼$ and $[C]_≈$ are homeomorphic in their respective quotient topologies?

I am trying to visualise a picture proof i.e. view quotient spaces as homeomorphic to the torus; but not sure if that helps

Andrews
  • 4,293
user82479
  • 71
  • 2
    nitpick: $C$ is not a "circle", more like a solid ring. Or "annulus" if you like fancy words. – Henno Brandsma Feb 24 '19 at 10:23
  • Annulus would be absolutely right haha :) May I ask what is the difference in the conditions between my question and https://math.stackexchange.com/questions/365920/prove-a-square-is-homeomorphic-to-a-circle that would change the homeomorphism proof? – user82479 Feb 24 '19 at 14:08
  • that linked question has no quotients. It just shows a circle and square (filled) are homeomorphic – Henno Brandsma Feb 24 '19 at 14:28
  • Ah yes but it suffices to show the original spaces are homeomorphic since that implies the quotient spaces are homeomorphic as well right~ Is the hom proof the similar tho for the case of circle $x^2+y^2\le1$ vs annulus $1 \leq x^2 + y^2 \leq 4$? – user82479 Feb 24 '19 at 14:43
  • the annulus is not homeomorphic to the square. And different quotients of the same space need not be homeomorphic even. – Henno Brandsma Feb 24 '19 at 14:46
  • Ahh I see thanks for clarifying, hmm according to suggestion below, "it is sufficient to show that the Disk and the Solid square are homeomorphic." So does this still hold or the way to prove? – user82479 Feb 24 '19 at 14:55
  • I don’t think that suggestion works, no. – Henno Brandsma Feb 24 '19 at 15:28
  • Your last paragraph is the most important one : both quotients are a torus so they're "obviously" homeomorphic (intuitively). Now the point is to see why they're both tori. I think you should try to see why $I^2$ with only $(0,y)\sim (1,y)$ is homeomorphic to $C$, and why the relation $(x,0)\sim (x,1)$ becomes $(x,y) ≈ (2x,2y)$ – Maxime Ramzi Feb 24 '19 at 15:46

1 Answers1

0

I am not writing an answer but the comment section is not appropriate for this, you may take it as an answer.

Using the fact in the answer here, it is sufficient to show that the Disk and the Solid square are homeomorphic. See here for proof. For visualization of how square is homeomorphic to circle here is an image. Hope this works.

Sujit Bhattacharyya
  • 3,058
  • Ah thanks! Would the extra equivalence relations specified in my case compared to https://math.stackexchange.com/questions/365920/prove-a-square-is-homeomorphic-to-a-circle change the homeomorphism proof? – user82479 Feb 24 '19 at 14:08