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Define on $ \partial \mathbb{D}^2$ the relation: $x \sim y \iff x=y \quad or \quad x=-y$

$S^1$ is homeomorphic to $\partial \mathbb{D}^2/\sim$ ?

The solutions say it is, but I don't understand why.

My thoughts: Since we identify opposite points on $\partial \mathbb{D}^2$, then for example $\partial \mathbb{D}^2/\sim= \{[x]:x=(\cos(\pi t),\sin(\pi t))|0\leq t<1 \}$ which is a half circle. This set looks contractible to me, hence it has a trivial fundamental group which is not the case for $S^1$. Therefore they are not homeomorphic. Maybe even simpler $S^1$ is closed and the set above is not.

amWhy
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The Lion King
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    Working in $\Bbb C$, the map $z\mapsto z^2$ induces the homeomorphism. Note that ${,[x]: \ldots,}$ is not the same as ${,x: \ldots,}$ – Hagen von Eitzen Dec 02 '22 at 19:25
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    And re your final argument: $[0,\infty)$ is a closed subspace of $\Bbb R$, and the homeomorphic $[0,1)$ is not – Hagen von Eitzen Dec 02 '22 at 19:27
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    $\partial \mathbb{D}^2 / \sim$ is not exactly the half circle. Draw $S^1$ and for now, ignore a pair of antipodal points. If you identify all the other antipodal pairs, you do get a half circle. But we still have that initial antipodal pair we ignored. Under the identification, these two points get joined, resulting in $S^1$ (up to homeomorphism). – Dalop Dec 02 '22 at 19:31
  • $\partial \mathbb D^2/\sim$ is compact as the continuous image of a compact space, but the half-open semicircle is not. You have a $1$-$1$-correspondence between both sets, but this is no a homeomorphism. – Paul Frost Dec 02 '22 at 20:05

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It is true that if you "glue" related points in $S^2$ the result seems a half circle (for example the part of circle that have non negative first coordinate in $\mathbb{R}^2$), but also extreme points of this, $(1,0)$ and $(-1,0)$ are related, and you can pass with continuity from the points near $[(1,0)]$ to those near $[(-1,0)]$. So if you "glue" also these two points you obtain the close circle, omeomorphic to $S^1$. Try the map which send $[(\cos(\theta),\sin(\theta))]\in\partial\mathbb{D}/_\sim$ into $(\cos(2\theta),\sin(2\theta))\in S^1$.

Lorenzo Morandi
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