<s>[INST] Proving the injectivity of function I have a subset $B = \{(x,y) : x^2 + y^2 \leq 1 \} \subseteq \mathbb{R}^2$.
I am trying to prove that function 
$g(x,y) = (\frac{x^2 + y^2}{x^2 + y^2 + 1}x, \frac{x^2 + y^2}{x^2 + y^2 + 1}y)$
is an injective function $g: \mathbb{R}^2 \rightarrow B$.
I attempted to prove this by directly applying the definition of injection.
Suppose we have $(a,b),(c,d) \in \mathbb{R}^2$ such that 
$g(a,b) = g(c,d)$. 
If I can prove $g(a,b) = g(c,d) \implies (a,b) = (c,d)$,
then $g$ is an injective function. 
$g(a,b) = (\frac{a^2 + b^2}{a^2 + b^2 + 1}a, \frac{a^2 + b^2}{a^2 + b^2 + 1}b)$, $g(c,d) = (\frac{c^2 + d^2}{c^2 + d^2 + 1}c, \frac{c^2 + d^2}{c^2 + d^2 + 1}d)$
Since $g(a,b) = g(c,d)$, $\frac{a^2 + b^2}{a^2 + b^2 + 1}a = \frac{c^2 + d^2}{c^2 + d^2 + 1}c$.
Furthermore, $0 \leq \frac{a^2 + b^2}{a^2 + b^2 + 1} < 1$ and $0 \leq \frac{c^2 + d^2}{c^2 + d^2 + 1} < 1$.
I want to have $\frac{a^2 + b^2}{a^2 + b^2 + 1} = \frac{c^2 + d^2}{c^2 + d^2 + 1}$,
and for that I need
$\frac{a^2 + b^2}{a^2 + b^2 + 1} \leq \frac{c^2 + d^2}{c^2 + d^2 + 1}$ and $\frac{c^2 + d^2}{c^2 + d^2 + 1} \leq \frac{a^2 + b^2}{a^2 + b^2 + 1}$.
which I am not so sure how to get.
Is this a valid approach to reaching the conclusion $a = c$?
Thank you for reading.
 [/INST] We have $g(10,0)=\frac{1000}{101}(1,0)\not\in B$. So, $g$ does not map the plane to the disc $B$. Nevertheless, $g$ is an injective self-map of the plane.
To prove injectivity, try to pass to the polar coordinates $x=r\cos\theta$, $y=r\sin\theta$. Then $$g(x,y)=\frac{r^3}{r^2+1}(\cos\theta,\sin\theta).$$
Assume that $g(x_1,y_1)=g(x_2,y_2)$, i.e.
$$\frac{r_1^3}{r_1^2+1}(\cos\theta_1,\sin\theta_1)=\frac{r_2^3}{r_2^2+1}(\cos\theta_2,\sin\theta_2).$$
First, if $r_1\ne r_2$, then the above equality fails, because $r\mapsto\dfrac{r^3}{r^2+1}$ is an increasing function and then the euclidean norms of both sides are not equal. Then $r_1=r_2$. If $r_1=r_2=0$, then obviously $(x_1,y_1)=(x_2,y_2)=(0,0)$. So, we can now assume that $r_1=r_2>0$ and we arrive at
$$(\cos\theta_1,\sin\theta_1)=(\cos\theta_2,\sin\theta_2),$$
hence $\cos\theta_1=\cos\theta_2$ and $\sin\theta_1=\sin\theta_2$. Together with $r_1=r_2$ this means that $(x_1,y_1)=(x_2,y_2)$. This proves the injectivity of $g$.
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