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Let $\sim$ denote the equivalence relation on the $n$-sphere $S^n$ defined via $$ (x_1,\dots,x_n,x_{n+1})\sim(x_1,\dots,x_n,−x_{n+1})\:\text{ for all }\: (x_1,\dots,x_{n+1})\in S^n. $$

Show that the quotient space $S^n/\sim$ is homeomorphic to the $n$-disc $D^n$.

I know that Ii have to show that there exists a bijective function $f: (S^n/\sim) \to D^n$ that is continuous and that the pre-image $f^{-1}$ is continuous as well, but I can't advance from here, any tips?

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Hint: Define$$\begin{array}{rccc}f\colon&S^n/\sim&\longrightarrow&D^n\\&\bigl[(x_1,\ldots,x_n,x_{n+1})\bigr]&\mapsto&(x_1,\ldots,x_n).\end{array}$$

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  • $\begingroup$ Okay, i see, but my struggle is to advance from here, how do i show that the function is continuous and furthermore that the pre image is? $\endgroup$ – Math Matich Dec 11 '18 at 13:40
  • $\begingroup$ Two things. First of all: it is not pre-image, it's the inverse function of $f$. Secondly: you asked for a tip and I provided one. $\endgroup$ – José Carlos Santos Dec 11 '18 at 13:44
  • $\begingroup$ Okay, thanks. I'll try. $\endgroup$ – Math Matich Dec 11 '18 at 14:10
  • $\begingroup$ @MathMatich the inverse function is clear: map $(x_1, x_2,\ldots, x_n)$ to the class of $(x_1,x_2,\ldots,x_n, 1-\sqrt{\sum_{i=1}^n x^2_i})$. $f$ being continuous is clear (it's a projection in essence) and compactness does the rest as soon as you know it's a bijection. $\endgroup$ – Henno Brandsma Dec 11 '18 at 15:42
  • $\begingroup$ I suppose that it's a typo and that it should be$$\left(x_1,x_2,\ldots,x_n,\sqrt{1-\sum_{n=1}^n{x_n}^2}\right).$$ $\endgroup$ – José Carlos Santos Dec 12 '18 at 18:04

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