# Check the injectivity and surjectivity of the following functions: $f : Z\to Z$ given by $f(x)\; = x^3$

A) $f$ is injective and surjective
B) $f$ is injective only
C) $f$ is surjective only
D) $f$ is neither injective nor surjective

Note: This is the 5th part of a 5 part question, which is split as 5 separate questions here.

Toolbox:
• A function $f: X \rightarrow Y$ where for every $x1, x2 \in X, f(x1) = f(x2) \Rightarrow x1 = x2$ is called a one-one or injective function.
• A function$f : X \rightarrow Y$ is said to be onto or surjective, if every element of Y is the image of some element of X under f, i.e., for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
Given $f:Z \rightarrow Z$ defined by $f(x)=x^3$.
Let $x$ and $y$ be two elements in $Z$.
Step1: Injective or One-One function:
For an injective or one-one function, $f(x) = f(y)$
$\Rightarrow x^3 = y^3$$\Rightarrow x = y.$
Therefore $f:Z \rightarrow Z$ defined by $f(x)=x^3$ is one-one or injective.
Step 2: Surjective or On-to function:
For an on-to function, for every $y \in Y$, there exists an element x in X such that $f(x) = y$.
$\Rightarrow$ For every $y \in Z$ there must exist $x$ such that $f(x)=x^3 = y$.
However, we see that for $y=2$, there is no $x \in Z$ such that $f(x) = x^3 = 2$.
Therefore $f:Z \rightarrow Z$ defined by $f(x)=x^3$ is not onto or surjective.
Solution: $f:Z \rightarrow Z$ defined by $f(x)=x^3$ is one-one or injective but not onto or surjective.

edited Mar 19, 2013