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Am I allowed to use an index of $y \in \mathbb{R}^n$ as constant, when showing onto ($\exists$ s.t. $f(x)=y$)?

Particularly,

I'm trying to show

$\bigg( \frac{x_1}{1-x_{n+1}},..., \frac{x_n}{1-x_{n+1}} \bigg)=y$

And I find an expression

$$x_i=y_i(1-x_{n+1})=y_i-y_ix_{n+1}$$

$\iff$

$$x_i+y_i x_{n+1}=y_i$$

But by previous examples on surjective functions I think I'd need to have all $x_i$ on the other side, while all $y_i$ on the other side.

There are some topological shortcuts for this, but I was just wondering, whether this algebraic way is feasible.

mavavilj
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2 Answers2

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You have the function $f : \mathbb{R}^{n+1} \to \mathbb{R}^n, f(x_1,\ldots,x_{n+1}) = \bigg( \frac{x_1}{1-x_{n+1}},..., \frac{x_n}{1-x_{n+1}} \bigg).$ It is obviously onto because $f(y_1,\ldots,y_n,0) = (y_1,\ldots,y_n)$.

Paul Frost
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If $f(\vec x)=\bigg( \frac{x_1}{1-x_{n+1}},..., \frac{x_n}{1-x_{n+1}} \bigg),\ $ you can check directly that

$g(\vec y)=\left ( \frac{2y_1}{1+\|y\|},\cdots \frac{2y_n}{1+\|y\|},1-\frac{2}{1+\|y\|} \right )$ is inverse to $f$.

In fact, $f$ is a homeomorphism: $S^n\setminus \left \{ (0,0,\cdots, 1) \right \}\to \mathbb R^n.$

Matematleta
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  • Yes I know, but I was unsure as to how does that show bijection. Because I think exists functions that are continuous and have inverse, but are not bijective. – mavavilj Sep 24 '18 at 22:30
  • But maybe this is the theorem about the equivalence of surjective,injective,bijective to use here, since this is in finite spaces: https://math.stackexchange.com/a/63073/248602. It's easy to show injectivity, thus it's enough. – mavavilj Sep 24 '18 at 22:39
  • $f$ is a homeomorphism onto $\mathbb R^n$,which means, among other things, that it is surjective. It is the stereographic projection. Surjectivity is easy, as the other answer shows. I thought I would just add some more information about $f$,since it is one of those "common" functions. I will be happy to delete my contribution if you wish-- – Matematleta Sep 24 '18 at 22:41
  • Yeah, but the question is about proving homeomorphism. Your answer seems to claim it without proof. – mavavilj Sep 24 '18 at 22:41
  • Do you want a proof of the fact that $f\circ g=id$ and $g\circ f=id$? Because $f$ and $g$ are clearly open maps, so that's enough to show that $f$ is a homeomorphism. – Matematleta Sep 24 '18 at 22:44
  • By what theorem? – mavavilj Sep 24 '18 at 22:45
  • hint: $f$ maps open disks to open disks. – Matematleta Sep 24 '18 at 22:51