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, $0 \leq \frac{a^2 + b^2}{a^2 + b^2 + 1} &lt; 1$ and $0 \leq \frac{c^2 + d^2}{c^2 + d^2 + 1} &lt; 1$.  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.
Consider the subset $B=\{(x,y):x^2+y^2\leq 1\}\ \subseteq \mathbb{R}^2$. Show that $\vert B \vert = \vert \mathbb{R}^2 \vert$.  It is easy to see tht $f:B\rightarrow\mathbb{R}^2$ defined as $f(x,y)=(x,y)$ is injective.   However, in the book I'm working out of (Book of Proof by Richard Hammack), the other injection noted is  $g:\mathbb{R}^2\rightarrow B$ defined as  $$g(x,y)=\left(\frac{x^2+y^2}{x^2+y^2+1}x,\frac{x^2+y^2}{x^2+y^2+1}y\right).$$  In the book I'm asked to verify this is indeed an injective function.  My approach was to assume $g(x,y)=g(m,n)$ and through basic algebra I was able to deduce  $x(m^2+n^2+1)(x^2+y^2)=m(x^2+y^2+1)(m^2+n^2)$  and  $y(m^2+n^2+1)(x^2+y^2)=n(x^2+y^2+1)(m^2+n^2)$  but how does this show $(x,y)=(m,n)$? Or is this the wrong approach?
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