# Is this function injective and surjective?

Let $f(x)=x^2$. Is this function injective and surjective if the function is defined as:

1. $f: \mathbb{R} \longrightarrow [0,\infty)$.

2. $f: \mathbb{C} \longrightarrow \mathbb{C}$.

3. $f: \mathbb{R} \longrightarrow \mathbb{R}$.

4. $f: \mathbb{R} \cup \{x \in \mathbb{C} : \mathrm{Re}(x) = 0\} \longrightarrow \mathbb{R}$.

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!

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Please, consider learning to write in TeX. –  Sigur Aug 17 '12 at 1:06
$f:\mathbb{R}\to[0,\infty)$, $x\mapsto x^2$ is not 1-1 since $f(-1)=f(1)=1$. –  Sigur Aug 17 '12 at 1:07
Can you not do any part of this? Do you know what injective means? What surjective means? It's best if you show us what you can do and where you get stuck, so we know where to start giving help. –  Gerry Myerson Aug 17 '12 at 1:46

First you should understand the definition of injective and surjective.

A function $f:A\rightarrow B$ is a map on sets that assigns every element of $A$ to an element of $B$. We actually call $B$ the codomain and $A$ the domain. $B$ is usually mistaken as the range, but the range, which we can denote as $f(A)$ is the set of elements of $B$ that actually have a preimage in $A$, (ie. For every $b\in f(A)$, there exists an element $a\in A$ such that $f(a)=b$).

$f$ is called onto/surjective, if $B$ equals the range of $f$. It is called one-to-one/injective, if $f(x)=f(y)$ implies $x=y$. When $f$ is injective and surjective, it is called a bijection. In this case we can invert $f$, and define $f^{-1}:B\rightarrow A$, by sending $b\in B$ to its preimage under $f$. It then follows that $(f^{-1}\circ f)(a)=a$ and $(f\circ f^{-1})(b)=b$.

1) $f$ is not injective (Sigur explained why) but is surjective (every non-negative real number has a real square root).
2) $f$ is not injective (for the same reason as 1), but is surjective (every complex number has a square root).
3) $f$ is neither injective nor surjective (The image is $[0,\infty)$).
4) $f$ is again not injective, but is surjective (since the square root of the negative reals lies on the imaginary axis).
5) $f$ is injective and surjective (and hence a bijection). It is easiest to see this by writing an element of the domain in polar coordinates and observing that applying $f$ has the effect of doubling it's angle from the positive real axis.