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Is this function injective and surjective?

Let $f(x)=x^2$. In each of the following cases, is this function injective and/or surjective?

  1. $f: \mathbb{R} \longrightarrow [0,\infty)$, I know this is surjective but not injective.

  2. $f: \mathbb{C} \longrightarrow \mathbb{C}$. This one, I dont know how to see. I mean the function evaluated in the imaginary unity is -1, so I don't know how to deal with this.

  3. $f: \mathbb{R} \longrightarrow \mathbb{R}$. This is neither surjective nor injective.

  4. $f: \mathbb{R} \cup \{x \in \mathbb{C} : \mathrm{Re}(x) = 0\} \longrightarrow \mathbb{R}$. I dont really know how to deal with complex function definitions.

  5. $f: \{z=x+iy: i^2=-1, y>0\} \cup \{z=x+iy: i^2=-1, y=0 \text{ and } x \ge 0\} \longrightarrow \mathbb{C}$..

Thanks a lot. I do know what surjective and bijective means, but I don't know how to prove it over complex subsets.

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marked as duplicate by Gerry Myerson, t.b., Jennifer Dylan, LVK, J. M. Aug 20 '12 at 19:31

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Better to edit your original question than to repost it. Better yet to accept some answers to your other questions. Please see meta.math.stackexchange.com/questions/3399/… –  Gerry Myerson Aug 18 '12 at 10:48

1 Answer 1

$\Bbb R\subseteq\Bbb C$, so if $f$ is not injective on $\Bbb R$, it can’t be injective on $\Bbb C$, either. That is, if there are two different real numbers $x$ and $y$ such that $f(x)=f(y)$, those two real numbers are also complex numbers showing that $f$ is not injective on $\Bbb C$. Thus, for (2) all that’s left is to decide whether $f:\Bbb C\to\Bbb C$ is surjective. In other words, if $z\in\Bbb C$, is there always a complex number $w$ such that $w^2=z$? This is most easily answered if you know the exponential representation of complex numbers: every complex number can be written in the form $re^{i\theta}$, where $r,\theta\in\Bbb R$ and $r\ge 0$. Given such a number $re^{i\theta}$, can you find real numbers $s$ and $\varphi$ such that $\left(se^{i\varphi}\right)^2=re^{i\theta}$?

In (4) the domain of $f$ is the two coordinate axes in the complex plane. Alternatively, it’s the set of all $x+iy$ such that at least one of $x$ and $y$ is $0$, so it’s the set of all real numbers together with all purely imaginary numbers, i.e., numbers of the form $yi$. You already know that $f$ is not injective on the domain $\Bbb R$, which is included in this set, so you can conclude right away that $f$ is not injective on this set either. Thus, once again it comes down to deciding whether $f$ is surjective. Suppose that $r\in\Bbb R$. If $r\ge 0$ there is certainly an $x$ in the domain of $f$ such that $x^2=r$. What if $r<0$? For instance, is there a number of the form $yi$ whose square is $-4$?

Try (2) and (4), and if you get them, see whether the experience helps you with (5); if not, let me know, and I’ll add some hints for (5).

Added: For (5) let $D=\{x+iy:y>0\}\cup\{x:x\ge 0\}$, and consider the function $$f:D\to\Bbb C:z\mapsto z^2\;.$$ Pictorially speaking, $D$ is everything above the real axis in the complex plane together with all of the non-negative real numbers. In terms of the exponential representation of complex numbers, $$D=\left\{re^{i\theta}:r\ge 0\text{ and }0\le\theta<\pi\right\}\;.$$

Suppose that $se^{i\varphi}$ is any complex number. Show that you can always choose $s$ and $\varphi$ so that $s\ge0$ and $0\le\varphi<2\pi$, just as you can when you choose the polar coordinates of a point in the plane. Then find an $re^{i\theta}\in D$ such that $\left(re^{i\theta}\right)^2=se^{i\varphi}$; this will show that $f$ is surjective.

To decide whether $f$ is injective, suppose that $re^{i\theta},se^{i\varphi}\in D$ and $\left(re^{i\theta}\right)^2=\left(se^{i\varphi}\right)^2$, i.e., that $r^2e^{2i\theta}=s^2e^{2i\varphi}$. Knowing that $r,s\ge 0$ and $0\le\theta,\varphi<\pi$, can you show from this that $r=s$ and $\theta=\varphi$? If so, you’ll have proved that $f$ is injective.

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Hi Brian, i really appreciate your help, what i can look now is that the third one is neither injetive nor surjective, the fourth is not injective but indeed is surjective, the last one i really dont have any idea. so am i ok? can you help me with the last one? –  Sebastian Griotberg Aug 17 '12 at 15:20
    
@Sebastian: You’re right about (2) and (4). Give me a few minutes to add something about (5) to my answer. –  Brian M. Scott Aug 17 '12 at 15:25
    
Thanks a lot. i really appreciate your help. –  Sebastian Griotberg Aug 17 '12 at 15:28

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