$f\left(x,y\right)=\left(\frac{x}{x^{2}+y^{2} } , \frac{y}{x^{2}+y^{2} } \right) $
$R^{2}-(0,0)$
Can someone help me with shwoing that this function is surjective?
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How is the function defined at $(0,0)$? – carmichael561 Jun 18 '16 at 17:59
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Hint: how do you get a large value of $\frac{x}{x^2+y^2}$? – almagest Jun 18 '16 at 18:00
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It is not defined. It is $R{^2 }-(0,0)$ – Ana Matijanovic Jun 18 '16 at 18:00
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@almagest when $x^{2}+y^{2}$ is smaller as it can be? – Ana Matijanovic Jun 18 '16 at 18:02
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1@AnaMatijanovic So it would help if you stated in the question that $f$ was a function $f:\mathbb{R}^2\to\mathbb{R}^2$. – almagest Jun 18 '16 at 18:02
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Yes, you need $y$ small as well as $x$ small to get a large value of $\frac{x}{x^2+y^2}$. Can you see how to develop that train of thought. – almagest Jun 18 '16 at 18:03
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But I still dont understand how to show that is surjective? – Ana Matijanovic Jun 18 '16 at 18:04
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Asked and answered-here you go. Is function invertible? – Mathemagician1234 Jun 18 '16 at 18:05
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Could you please refrain from asking minor variants of the same question? Thank you. – Andrew D. Hwang Jun 18 '16 at 18:12
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But I dont understand previous answers! – Ana Matijanovic Jun 18 '16 at 18:14
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@AnaMatijanovic: If you don't understand an answer someone has left, you're welcome (even encouraged!) to ask for clarification in the comments beneath the respective answer. By contrast, asking the same question anew causes the community to duplicate effort, and tends to create confusion as people reply to comments addressed elsewhere. – Andrew D. Hwang Jun 18 '16 at 18:56
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As in the answer to your previous question check that $f(x,y)=f^{-1}(x,y)$ on $\mathbb{R}^2\backslash \{0\}$. Since $f(x,y)$ is defined on every value of $\mathbb{R}^2\backslash \{0\}\ \therefore$ $f^{-1}(x,y)$ is also defined on every value of $\mathbb{R}^2\backslash \{0\}$. Since this means that to every image there is a preimage $\therefore$ $f$ is surjective
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But I should show that function is surjective, before I found inverse? – Ana Matijanovic Jun 18 '16 at 18:12
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Suggestion :put $x=r cos \theta \y= r sin \theta$ so you will have $f =(\frac{cos \theta}{r},\frac{sin \theta}{r})$ and , this function is surjective – Khosrotash Jun 18 '16 at 18:13
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@AnaMatijanovic You can always find out the inverse first and check on which points it is defined and whether every image has a preimage.. – Qwerty Jun 18 '16 at 18:22
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Before you can ask if the function is surjective, you have to specify what is the codomain of the function (now I read that the domain is $R^2\setminus (\mathbf 0)$). However, interpreting the lack of such specification as an underlying assumption of the largest possible codomain, that is $$ f:\mathbb R^2\setminus (\mathbf 0)\to \mathbb R^2, $$ then $f$ would not be surjective because $\mathbf 0$ is never attained. However $$ f:\mathbb R^2\setminus (\mathbf 0)\to \mathbb R^2\setminus (\mathbf 0), $$ is surjective. In fact, in vector notation, $f(\mathbf r)=\mathbf r/|\mathbf r|^2$, therefore, for $\mathbf y\neq\mathbf 0$ $$ \frac{\mathbf r}{|\mathbf r|^2}=\mathbf y\qquad \iff\quad \frac{\mathbf y}{|\mathbf y|^2}=\mathbf r. $$ Therefore $f$ is actually bijective on that domain and $$ f=f^{-1}. $$
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