In this question (and also this one) we have a result that $S^n / \langle x\sim -x\rangle$ is homeomorphic to the space $D^n$ after we identify antipodal boundary points. The answer seems correct to me, but I find it really confusing that when we look at $D^1=[-1,1]$ and identify boundary points, i.e. we consider $[-1, 1] / \langle 1 \sim -1 \rangle$, we are supposed to end up with a space homeomorphic to $S^1 / \langle x\sim -x\rangle$. But the space $[-1, 1] / \langle 1 \sim -1 \rangle$ is known to be homeomorphic to $S^1$. The proof seems to be applicable to the case $n=1$ as well, but I just don't see how $S^n / \langle x\sim -x\rangle$ can be homeomorphic to $S^1$. Is there anything wrong? If not, could you help me with getting some intuition and constructing an explicit homeomorphism?
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1A homeomorphism from $S^1 / \langle z \sim -z \rangle$ to $S^1$ is $z \mapsto z^2$, where $S^1$ is seen as the unit circle in the complex plane. – Zerox Feb 07 '22 at 14:55
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1https://math.stackexchange.com/questions/433912/the-space-obtained-by-identifying-the-antipodal-points-of-a-circle – wormram Feb 07 '22 at 14:55
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Take the map$$\begin{array}{rccc}f\colon&S^1&\longrightarrow&S^1\\&z&\mapsto&z^2.\end{array}$$Then $f$ is continuous and surjective. Furthermore,$$f(z)=f(w)\iff z^2=w^2\iff z=\pm w.$$So, $f$ induces a homeomorphism from $S^1/\sim$, with $z\sim w\iff z=\pm w$, onto $S^1$.
José Carlos Santos
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