# Another representation of $S^n$ as a quotient of disk

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.

• Have you tried drawing/imagining what the quotient is for $n=3$? – Santana Afton Oct 9 '18 at 4:11

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: $$(D^{n+1}/\sim) \cong \sum(D^n/\sim)\cong \sum(S^n) \cong S^{n+1}$$.
• 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 $(x,t) \sim (x',t')$. – Paul Frost Oct 9 '18 at 15:52