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Let $D^n\subset \mathbb{R}^n$ be the subset consisting of those points $(x_1,\dots,x_n)\in \mathbb{R}^n$ such that $x_1^2+\dots+x_n^2\leq 1$ and let $S^{n-1}\subset D^n$ be the subset of those points $(x_1,\dots,x_n)\in D^n$ such that $x_1^2+\dots+x_n^2=1$ (i.e. the boundary of $D^n$). Define an equivalence relation $\sim$ on $S^{n-1}\subset D^n$ by $$(x_1,x_2,\dots,x_n)\sim (-x_1,x_2,\dots,x_n).$$

Is the quotient $D^n/\sim$ homeomorphic to $S^n$? I think this is true for $n=2$, and am curious about higher dimensions.

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  • $\begingroup$ Have you tried drawing/imagining what the quotient is for $n=3$? $\endgroup$ – Santana Afton Oct 9 '18 at 4:11
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This is true for all $n$: $(D^n/\sim)\cong S^n$.

To see this, first note that $\sum ( D^n/\sim) \cong D^{n+1}/\sim$ where $\sum$ denotes the suspension functor: $\sum X$ is $X\times [-1,1]$ with $X\times \{1\}$ and $X\times \{-1\}$ collapsed to points. One such homeomorphism is $f([x_1,..., x_n],t)\mapsto [(\sqrt{1-t^2}\,x_1,...,\sqrt{1-t^2}\,x_n,t)].$

Now, ($D^1/\sim) = S^1$ because in this case $\sim$ is the usual identification $S^1 \cong D^1/\partial D^1$. Now, the fact that $(D^n/\sim)\cong S^n$ follows from induction: $(D^{n+1}/\sim) \cong \sum(D^n/\sim)\cong \sum(S^n) \cong S^{n+1}$.

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    $\begingroup$ Nice proof! Perhaps it is easier to see if you define $h_n: D^{n+1} = \{ (x,t) \in \mathbb{R}^n \times \mathbb{R} \mid \lVert x \rVert^2 + t^2 \le 1 \} \to \Sigma D^n$ by $h_n(x,t) = [x/\sqrt{1-t^2},t]$ for $t \ne -1,1$ and $h_n(0,\pm1) = [0,\pm1]$. Then $q_n = (\Sigma p_n) h_n$ with $p_n : D^n \to D^n/\sim$ has the property $q_n(x,t) = q_n(x',t')$ iff $(x,t) \sim (x',t')$. $\endgroup$ – Paul Frost Oct 9 '18 at 15:52

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