Prove that: 

(i)     $f$ is an injection,

(ii)     $g$ is a surjection.

Given that $X = {\mathbb{R}^ + } \cup \{ 0\}$ and $Y = \mathbb{R}$, choose a suitable pair of functions $f$ and $g$ to show that $g$ is not necessarily a bijection.

(i)     to test injectivity, suppose $f({x_1}) = f({x_2})$     M1

apply $g$ to both sides $g\left( {f({x_1})} \right) = g\left( {f({x_2})} \right)$     M1

$\Rightarrow {x_1} = {x_2}$     A1

so $f$ is injective     AG

Note:     Do not accept arguments based on “$f$ has an inverse”.

(ii)     to test surjectivity, suppose $x \in X$     M1

define $y = f(x)$     M1

then $g(y) = g\left( {f(x)} \right) = x$     A1

so $g$ is surjective     AG

[6 marks]

choose, for example, $f(x) = \sqrt x$ and $g(y) = {y^2}$     A1

then $g \circ f(x) = {\left( {\sqrt x } \right)^2} = x$     A1

the function $g$ is not injective as $g(x) = g( - x)$     R1

[3 marks]

Total [9 marks]

Those candidates who formulated the questions in terms of the basic definitions of injectivity and surjectivity were usually sucessful. Otherwise, verbal attempts such as '$f{\text{ is one - to - one }} \Rightarrow f{\text{ is injective}}$' or '$g$ is surjective because its range equals its codomain', received no credit. Some candidates made the false assumption that $f$ and $g$ were mutual inverses.

Few candidates gave completely satisfactory answers. Some gave functions satisfying the mutual identity but either not defined on the given sets or for which $g$ was actually a bijection.