IRL_ABooklet

Clearly, $f$ is real valued, non-negative, and, by Cauchy-Schwarz, $$f(x,y)=x\cdot \sqrt{1-y^2}+\sqrt{1-x^2}\cdot y\le \sqrt{x^2+(1-x^2)}\cdot \sqrt{(1-y^2)+y^2}=1,$$ so that $f$ maps $S$ into $[0,1]$. But if $0\le z\le 1$, then $(z,0)\in S$ and $f(z,0)=z$, and so $f$ assumes every value in $[0,1]$. Hence $f$ is also onto (surjective).

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Alternative solution.

For $x,y\in [0,1]$ there are unique $\alpha ,\beta \in [0,\frac{\pi }{2}]$ such that $x=\sin \alpha$ and $y=\sin \beta$. Because $\cos \alpha$ and $\cos \beta$ are non-negative for such $\alpha ,\beta$, we have $\sqrt{1-x^2}=\cos \alpha$ and $\sqrt{1-y^2}=\cos \beta$. Therefore, $$f(x,y)=x\sqrt{1-y^2}+y\sqrt{1-x^2}=\sin \alpha \cos \beta +\sin \beta \cos \alpha =\sin (\alpha +\beta ).$$ For $\alpha ,\beta \in [0,\frac{\pi }{2}]$ we have $\alpha +\beta \in [0,\pi ]$ and so $\sin (\alpha +\beta )\in [0,1]$ so that $f$ maps $S$ into $[0,1]$. Surjectivity follows from $f(z,0)=z$. Clearly, $f$ is not one-to-one (injective) since $f(x,y)=f(y,x)$, for example $f(1,0)=f(0,1)=1$.