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

Question

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.

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

Answer 1

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 $f(y_1,\ldots,y_n,0) = (y_1,\ldots,y_n)$.

Answer 2

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: $S^n\setminus \left \{ (0,0,\cdots, 1) \right \}\to \mathbb R^n.$